Thanks for checking out this movie review video. So this is for the 2015 release, The Invitation, which when I watch this and I'm now reviewing it and putting the video up, is available for streaming on Netflix. I know I usually do stuff streaming on Shutter, but I got a stray every now and then because I want to watch what I want to watch. And here's the review. So The Invitation is one that I had had in my Netflix DVD queue to check out and then found out it was on Netflix streaming. So awesome on that. But it's one that kind of popped closer to the top of my queue just because I had seen some people out there put it on their top 10 or top 20 or top 50 films of the 2010s related to the horror genre. And I was like, oh, I've been meaning to watch The Invitation, so maybe I should bump that up since some people are saying it's really, really good. So I will say up front, I did like it. I am going to do spoilers for this review just because it's almost like a five-year-old film at this point.  So if you're watching this and you haven't seen it yet, I would say I definitely recommend it. It's definitely a good film. So stop this here. Go watch it. Then come back because going forward, it's spoilers because it kind of begs to be talked about in a spoilery way because you really want to break things down. And that's one of the things I really like about this film is that it made me think a lot. It brought up a bunch of themes, and that's what I like when I watch horror films.  Like, I'm going to be doing a review for Godzilla, King of the Monsters, and I'm fully expecting when I watch that that it's just going to be straightforward, dumb horror because it's supposed to be set up kind of like the Toho Godzilla film. So, you know, two different modes. But this I particularly like, the thinker's horror films. So let's get into the backstory on this. It was directed by Karen Kusama, who directed the film's Girl Fight, which I have not seen, Eon Flux, which I have also not seen and not heard very good things about, and a very underrated horror film that I should review at some point, Jennifer's Body. That is, that movie got panned so hard when it came out, but it's actually starting to kind of gain more steam now since it's been out for a while. It's got a bit of a cult following, and rightfully so. It is a good film. So the writers for this were Phil Hay and Matt Manfredi. These two individuals have worked on a lot of scripts together. They worked on Eon Flux scripts, so they knew Karen Kusama. They did the newer Clash of the Titans, which I have not seen, and I don't know if I really want to. I might at some point. I really love the original, the really old one from the 70s, I think. That one's a lot of fun, but the newer one looked dumb. They did RIPD, I think is Ryan Reynolds in it. That always looked kind of interesting to me, so I will check that out at some point, but it's not like high up on my list. And then they did the movies Ride Along and Ride Along 2, which I just didn't have interest in. I won't see those. So kind of not a success story with this film. It had a $1 million budget, and at the box office it made $354,835. I wonder why that is, because, well, I have a theory. I feel like maybe they didn't market it really, because I didn't know about this film back in 2015. I didn't even think going into watching this, and before I did my research on it, I was under the impression it was just like a straight to video type deal or straight to streaming. I had no clue actually had a box office release, so that kind of tells you that probably they didn't market it enough to get people into the theaters to make money in the box office, so kind of mishandled in my opinion. Or maybe they just didn't market it the right way, because there's an issue with like when you do trailers or when you're really trying to push a film, you kind of give up too much about the actual film, and that's what gets butts in seats for the theater, but the problem then is you don't really want to ruin it, especially not with a film like this, where there's a lot of uncertainty. You're kind of uncertain like what is really going on. You have theories and you're kind of like, well, it could go this way or it could go this way, and until a certain point in the film towards the end, you're not 100% sure. So you really don't want to give that kind of stuff up, so it gets tricky. Also, this could have been one of the issues is when it was originally announced, they were saying that it was going to have bigger names in the film, because almost all of it is like kind of no-name actors, which for me, I'm totally fine with it. I have no problem with that, but for your common moviegoer, they want to see bigger names. That makes them feel a little more comfortable like, oh, there's a big name in it, or a few big names must be a good film. I don't think that way, but I know people who do. So originally, they said it was supposed to have like Luke Wilson, Johnny Galecki, and Tofer Grace. So I searched online, I couldn't find what happened with that situation. My theory is that maybe they had all of them tentatively signed up to do it, and someone canceled on it, which made another one cancel on it, which then the third one was like, well, if they're not doing it, then I'm not. That's just speculation on my part, because I've heard plenty of instances of those types of things happening with films. So maybe that's part of their problem as well. So I put in here, by the way, I really like John Carroll Lynch, the bald dude in this who played the character of Pruitt. He is good in everything he does, and I know everyone's seen him in at least something. I think he's an underappreciated actor, not in the sense that he doesn't get enough work, because if you look at his IMDb credits, he is continually working, it's just people don't really talk about him much. Whenever I see him, I am totally down. I think he does an outstanding job. He does an outstanding job in this film, once again. And I do want to say that one of my favorite roles of his that I've seen was his role in the Channel Zero season of Channel Zero called The Dream Door. He did an outstanding job with that. He had a cool character to work with, and he just did it really well. So if you haven't, first of all, watched Channel Zero. I haven't seen all the seasons, but it's a good time. Specifically the Dream Door one for John Carroll Lynch. He does an awesome job. So the beginning of the film kind of sets the tone with it being kind of like messed up and uncomfortable. And I say that from the standpoint of, you know, when Will and Kira are in their car and they hit the coyote, it's very uncomfortable, especially because he's like, well, gotta put it out of its misery. And he like beats it to death with a tire iron. So this scene does a few things. Like I was saying, it starts the film uncomfortable. It gives you a foreshadowing that the film is not afraid to go to very uncomfortable places, and it will, and it will also go to violent places. But it also sets up the idea that Will has the ability. He has the mental ability to kill. And that's important for later on, but they also kind of switch it later on where you assume that Will will be even more of a badass and take out Pruitt, but it ends up being Kira who does that. So in essence at the end, Kira is the one who beats the coyote to death with the tire iron and Will doesn't. And so it's an interesting juxtaposition because in the beginning when Will is killing the coyote, Kira is very disturbed by it. But then you have that switch towards the end, which I thought was a really cool, interesting touch. The other thing that kind of ratchets up the tension and disturbing nature of that opening scene is there's no music. And that's one of the wonderful things about this film in general is the use of no music. I talk about this all the time in reviews. You can overdo it with music. And I feel like a lot of people need to take a more minimalistic approach to the use of music in film because too often you're telling people how to feel. And one of the issues with that is it makes things feel predictable. And with this one, they use a lot of silence which creates audience confusion in a sense. Because you don't have that auditory cue to tell you this is how we're telling you you're supposed to be feeling right now.   Like how should I be feeling right now? And that plays particularly well with this film because there's a lot of uncertainty in is Will just losing it or is there really something going on with this kind of cult the invitation cult thing? And are they planning on killing people or are they planning on just killing themselves or are they planning something else? Who knows? And so I feel like instead of telling you through music how you should be feeling at any moment, they just let you sit with your uncertainty and that works really well. I love it. I really like that aspect of it. So well done. So when they finally do play music in the opening, it sounds a bit off which is a really interesting for shadowing technique. You know having the music kind of go along and it feels very melodic and like it makes sense musically and then there will be like a note or two like here or there that are just kind of off. So it's enough to kind of lay out the landscape that things seem normal and then there's this really weird thing that kind of like breaks your mind out of the song and you're like oh something's not right here which is the whole thing with the actual story and that's where the foreshadowing piece comes into. I really like that touch. That's very cool. So the host talking about enjoying what you have and pouring crazy expensive wine gives the impression in the beginning that people won't be alive much longer to live it up now. Yeah it comes off as more of like a either they're planning on killing themselves which I mean they really were or they were planning on killing everyone else or they were going to plan on killing themselves and everyone else which was kind of the ultimate thing. It just kind of seemed like they were like giving away this great greatly expensive wine and they were like pushing it on people and it was this big emphasis on well I mean might as well live it up and you're just like and I feel like that was almost pushed a little too strongly in the beginning. Maybe you should have pulled that back just a little bit. It also made you feel like because they were pushing the wine so hard that and this is what I thought immediately when they were pushing the wine hard I was like there are drugs in there that are either meant to sedate people or kill people and I ended up being totally right on that one. So they should have not pushed it as hard. Eden slapping Ben is the moment that you understand she doesn't fully have it together and is capable of violence. She seems very together when you first meet her like oh maybe she's gotten things together in her life and that moment when she slaps Ben is that little bit of an inkling that they're giving the audience to say she's not okay and she's capable of snapping at any moment so it makes you feel more comfortable and that helps with the tension going forward because you then know as an audience member she has that ability to get over the top to go violent. The door locking incident is another small thing that feels off and helps increase the intention. The tension because you know something's coming where they kind of make a big point of oh we like to keep the door locked and then they're like oh why? Well because there was a home invasion in the canyon and someone was like I didn't hear about that so they make a very pointed effort to once again make things feel off to drive the tension up and that makes a lot of sense. I like that because they just continually do that with things throughout the film which just gradually steps up the tension and steps it up and steps it up until you get to the blow up that you have at the dinner table which is satisfying. Watching the weird cult video in a group like that is uncomfortable and they did a good job of translating that to the audience. Partially by showing the characters looking uncomfortable and partially by not using music. Once again the not using music thing to make you feel uncomfortable to make you feel uncertain about what's going on and it makes you focus even more on the facial expressions of the party attendees who are very very uncomfortable with this very weird cult video. So excellent job on the filmmaking. The people at the door thing just looking for a party was super suspicious. They don't really come back to that though. I think that was just kind of in there as one of those extra things to like move the tension up make you feel even more suspicious and in a lot of the things it was kind of like it could go either way and at some point it feels that way because you do start to kind of question like is will the one who's losing it? Because especially with the moment where Choi actually shows up and I'll talk more about that later. That's when you start to question things and then you're just like oh so then you look back on some of these things like the people coming to the door and you don't view it as being as suspicious as you initially did. It's all about the context of what happened before it. The identification of the phenobarbital pills bolsters the suspicion about the wine situation that I was talking about about drugs being in there. That made it very clear to me that that's what was going on. I had a strong suspicion like I said initially with them really pushing the wine and then when they identified the pill I'm like they're making a very big point about this so obviously it's in the wine. One of the best tension increasing moments is when Will finally gets the message from his buddy Choi and he isn't actually there and he said oh I got here early but I need you to pick something up for me and then he's just like well where's this dude?  This is nuts and I feel like a lot of movies like this don't go to that point. They don't go to the moment where a character starts calling things out and saying look this is suspicious. I think there's some crazy stuff going on. Usually they go the route of the person's like no I'm gonna maintain my composure. I'm gonna be polite and we'll just see where things go. Which isn't all that realistic. You know if you were dealing with as many things as Will was at that point you probably would just kind of come out with it and be like what is going on? And this is a wonderful moment not just because it feels very realistic but because it leads to the moment where you start to question Will's sanity because leading up to that point he kept having these flashbacks showing that he's stuck in the past and he can't get past the death of his child whereas he sees his ex-wife and he feels like at least the way she's acting that she's kind of over it and that's kind of making him feel like how is this even possible? So then you start to especially question Will's sanity when Choi actually shows up at the party and then that kind of casts so much doubt in your mind as an audience member of oh wow well then maybe the angle they're going with this is he's kind of losing it and has been losing it and then that makes you rethink about all those moments like the people coming to the door and the door being locked and all of those things and the pushing of the line and you're just like maybe they just framed it in that particular way to make it feel suspicious but it's just suspicious to Will because the film is from his perspective so it makes you think well maybe we're just seeing it through his mind and he's a bit unhinged and that's what we're gonna find out. So they did a good job of kind of creating that uncertainty and I really enjoy that.    If they needed this excuse so badly they should have set it in the early 90s or before and I think this is a good point that if you wanna have people not complain about the no reception situation which I feel like a bunch of people would because they want it to feel realistic and that doesn't feel realistic they should have set it in an earlier decade because then you could say it was only landlines at that point and they didn't have a phone there or they cut the lines or whatever. It just makes it a little more neat. So I do think though in the very end maybe there's an explanation for the no cell reception and that is the fact that it kind of appears that all the houses in that area were involved in the same cult activity.  So it could have been a concerted effort and they made sure that there was no cell reception there so you know I mean they could have cut wires somewhere and whatever. So maybe that explains it but at least at first the no cell reception thing just rubs you in the wrong way. At least it did for me. You would think, oh like I said you would think that Will would be the one to end up killing Pruitt when that ends up going down but it ends up being Kira and I really like that switch because like I was saying from earlier where she's very appalled at what's going on with the Coyote and he's just like mindlessly like I gotta take care of this and he kills and it switches and she's like I gotta take care of this and she puts Pruitt down like the Coyote. Just beats him in, beats his head in. I dig the twist at the end with all the red lights indicating that there are a lot of other households involved and like I said that could account for the cell reception situation. So at that moment it made me feel like maybe it turned into a little bit more of a movie like The Purge where it's like these people will potentially not just get done within their house but leave the house and maybe kill some more people. I don't know. So it kind of sets it up where there could be a sequel which I would be kind of interested in to see what happens because I think they did a good job of handling this although not all stories lend themselves to having sequels really and I don't know what you would do but I'd be open to see what they would do with a sequel. I don't know. So let's talk about some themes and some overall stuff about the film. Directing and cinematography looked really good and I think it visually kept people engaged. The camera movements were really nice the way they framed a lot of the shots were good. The only thing that I really didn't like visually is a lot of the lighting was super low. So if it was in high definition it didn't really look like high definition. This is one of the problems with having things shot in high def in low lighting is it then doesn't look high def at all. You get a lot of like blurring that goes on and it just looks like digital. So that was kind of something that bothered me about the film. This film uses music sparingly like I talked about and I love that. Like I said it increases your uncertainty about a lot of stuff and that's important for the audience member to sit with. Big kudos for doing that. It was a great, great choice. So thematically this shows the tensions of how it can become so toxic when someone involved with the trauma deals with it and someone else doesn't. They feel alienated and resentful that the other person has worked past it because they're still stuck and suffering. And you see this especially and this theme comes through especially when you don't fully know what's going on yet. You feel like well maybe it's Eden and her new spouse that are the problem or maybe it's Will that's the problem. You see that they're butting heads because he even makes some comments to her about, oh you're just like over it, like how can you be over it? Because obviously she's acting like she's moved on and she's in a better place and she's gonna be better which I don't, that isn't actually the case in the end because she does admit that she really misses a kid and she keeps thinking about it before she dies but she's giving the appearance that she's moved on and that kind of sets Will off even more because he's obviously stuck. Like he cannot get past it. He cannot, it's like he hasn't even fully grieved. He doesn't know how to deal with it really. He doesn't know how to move on and that's clearly seen too in all the flashbacks that he sees when he goes around the house. I mean that's a very good visualization of showing you that he is stuck. He is 100% stuck in those memories and stuck in the pain and the trauma of that time. So this speaks to a social issue actually where people judge how others grieve and deal with trauma especially deaths within the family and that scene in the, you know, the budding of heads between Will and Eden when he's kind of like, how do you, how are you past this? Like our kid died, like how can you be okay with this? And it gets the heart of a social issue that I've thought about a lot which is people judging how other people grieve because people grieve differently and then the issue of people judging when someone moves past a death or they do they're grieving and they get through the process and they find a way to move on. And I see it most when there are situations where it's a spouse that passed away and I hear it from people. I've heard it many, many times where someone's like, oh, they're seeing someone else now. That feels kind of soon, doesn't it? But it's like, how are you gonna pass that kind of judgment? Like you're not that individual person. You don't know how they deal with that type of thing. And maybe they didn't even deal with that type of thing. Maybe they jumped into a relationship because that's their way of distracting because they can't deal with it at that point. Doesn't mean that the relationship's necessarily a good one, but that's how they're choosing to deal with it, choosing to grieve, choosing to try and move past it. And some people don't wanna move past it. Some people like being reminded of what was or what the trauma is. It's not necessarily healthy, I don't think, but everyone's got their own way of doing it. And I think that people need to give everyone some space and be understanding that you are not anyone else. You are one unique individual. Everyone is a unique individual. And everyone thinks about things differently and deals with things differently. We have similarities as human beings, but everyone's unique. And the last thing I had a quote that I thought was really great from this film that kind of exemplifies what I was just talking about, is said, letting go is not a betrayal. Just kind of signaling that you don't have to judge. It's fine for you to move past something. And I just thought that that's a very important quote from the film that people should take away from it. So that said, this is a good film. I highly recommend it to people that should definitely check it out. Is it a perfect film? No, but I quite liked it. I don't, having seen it now, I would not change the list that I have for my top 20 horror films from the 2010s, which, by the way, that video is on this channel. Go check that out, because I think I did a pretty good job with my list, to be honest. But having seen the invitation, it wouldn't crack my top 20, but top 50, yeah, potentially top 50, maybe even top 30, I don't know. Yeah, potentially top 30, actually. It's good. So that said, rating-wise, five stars with half stars in play, I'm gonna give it a four-star rating. I think it is a solid four stars. Quite good film. So, I would love to hear what other people have to say about this film, because it's one of those thinker films. So I probably missed something, or people are seeing it from a different perspective. So put some comments down there, if you've seen it, really, really want to hear from you on this one, because I love kind of deep diving on these types of movies. So comment about it, let's talk. Do me a quick favor, hit that subscribe button. If you like any of the videos that I do, that's your best way to motivate me and say thank you, is just hit that subscribe. And if you've already subscribed, just hit the like button, that'd be nice. But if you haven't subscribed, don't hit the like, hit the subscribe, or if you wanna go the extra mile, you can do both. And I would appreciate that. But thanks everyone for checking this out, and until next time, keep it brutal.


When dealing with stress components, we have to apply a sine convention. We will be using the following sine convention throughout this course. Now consider shear stress component tau xy. We know from the subscripts that this stress acts on a surface that has a normal vector in the x direction, and the resultant force from that shear stress is acting in the y direction. To apply the sine to this stress, we will first take the magnitude of that stress. We will then multiply it by the direction of the surface normal relative to our coordinate system. So if the surface normal is in the positive x direction, we would multiply it by positive 1. We then multiply that by the sine for the resultant force direction. So if it's in the positive y direction, this would be a positive 1 value. And remember, if we're dealing with normal stresses, the single subscript can be expanded to a double subscript in order to determine the sine. Let's look at this for an example. Here we see two stress states that are the same in direction but opposite in sine. If we look at the stress state on the left, our outward surface normal is in the positive z direction, and our normal stress sigma z has a resultant in the positive z direction as well. So our stress becomes sigma z magnitude of sigma z multiplied by positive 1 multiplied by positive 1, which gives us a positive normal stress. Remember, the first positive number is the sine of the outward normal, and the second positive number is the sine of the resultant force. On the right, we have a positive outward normal direction, but our resultant is in the negative direction. So in this case, sigma z would be negative. If we apply that to the shear stress components on the left example, we see that they are all in the positive direction, while on the right they are all in the negative direction.


The main results of the presentation I gave in the last lecture on the direct-algebraic method for solving the hydrogen atom in minus ulcer this morning. The direct-algebraic method for solving the harmonic oscillator I think you're probably familiar with most of this. The essential ingredients of the creation and annihilation operators, there are matrix elements with respect to the energy in the state. The expression of the Hamiltonian in terms of the creation and annihilation operators also solved with the position of momentum in terms of the A's and A's daggers, just the inversion of the definitions, which is also sometimes useful. All right. Now, I'd like to begin today with the subject of wave functions. In particular, let's talk about the ground state wave function, zero is the ground state. And if I form this camera product with the position I can ask, we definitely cannot consider it to be the ground state wave function called side zero of S. The question is, what is side zero of S? Well, this follows easily from the fact that the annihilation operator interacting with the ground state annihilates in just gives you zero. The annihilation operator, however, is X plus I t over the square root of 2. And if we put that in the wave functional language, X plus I t goes over into the X, of course, as both of them have to emphasize these are operators. X goes into the multiplication by X, T goes into minus I d dx. So excuse me. So as an operator, acting on the wave function, this goes into X plus d dx. And therefore, the ground state wave function satisfies the differential equation that X plus d dx acting on the side zero of X is equal to zero. It's a simple differential equation. And it's easy to solve. And if you solve it, and if you normalize the result, you find that side zero of X is 1 over pi in the one-quarter power times e to the minus X squared over 2, pi stuff is the normalization. And this is the ground state wave function, which is, I guess you know, as a Gaussian function, a function of X, it's a Gaussian function centered before it connects to the ground state of the chromatic oscillator, side zero of X, here like this. Now, one of the things we're going to be interested in, many of our attention to today, is expectation values of position momentum, as well as versions, which are expectation values of first moments, both of the interesting and second moments, which are related to the dispersions. And in particular, the Q piece for the ground state is easy to show that the expectation value of X and the expectation value of momentum here, both equal to zero, which is before the ground state, for the ground state. And one can also show quite easily that the dispersion delta X is equal to dispersion also in momentum of P, which is 1 over the square root of 2. This means, obviously, the dispersions are equal, and it also means that the product of delta X times delta P is equal to 1 half. This is all the units in where h bar is equal to 1. So this would be h bar over 2 at ordinary units. And this is the minimum value of delta X and delta P, which are allowed by the uncertainty principle. So the result of this is the ground state of the harmonic oscillator is actually a minimum uncertainty wave packet. It's a Gaussian wave packet. All minimum uncertainty wave packets in terms of delta X and delta P have to be Gaussian, but this is a particular example of it. And so that's an interesting conclusion also. While I'm talking about momentum, by the way, it's interesting to take the Fourier transform of a size 0 in order to obtain a momentum space wave function, phi 0 of P. That's straightforward. It leads to the same 1 over the pi over the 1 quarter normalization, and then it becomes p to the minus p squared over 2. The result is, as you see, that for the ground state anyway, the momentum space wave function has the same functional form as the configuration space wave function. This is an example of a wave function which goes into itself in a Fourier transform. All right. Now, so those are some basic facts about the ground state. Now, let's take a look at the excited states. Let's see, maybe before I do that, let me just provide some motivation for where we're going here. The harmonic oscillator is a particularly useful system for exploring and understanding the correspondence in classical mechanics, to understand how far it goes and what needs to be changed by going to quantum mechanics. Makes the relationship between the two particularly clear. And that's really going to be the main theme that we'll be following in the rest of the lecture today, is to understand the relation in classical and quantum mechanics. Here's just one thing to say, is that because this ground state of the harmonic oscillator has these dispersions, this is obviously related to the statistical nature of a quantum measurement. We can visualize the ground state of the harmonic oscillator in quasi-classical terms in the following way. If I plot the classical phase space x and p, then they can think of the ground state of the harmonic oscillator as a kind of blob or a distribution, which is centered in the origin in base space. It's centered at the origin because the expectation, the average value is x and p are 0. And, but on the other hand, it has some spread around it because of non-zero dispersions. Spread of course is necessary by quantum mechanics because x and p don't commute. And so they can't be measured simultaneously with infinite precision. This blob that I've drawn here is not intended to be a rigorous representation of ground state. It's really intended just for the purposes of intuition to give you an idea, represents in some way the statistics of measurement that come out of the ground state rate function. One could make this more precise by wanting the Wigner function, which was actually a part of last week's homework problem. But I won't go into that. I'll just use this for the illustrative purposes. All right. Anyway, that's what kind of the ground state looks like. Now, if we're thinking about the classical limit of the harmonic oscillator, or if that matter ain't one of the problem hydrogen atom, one is struck by the fact that in classical mechanics you have particles moving on orbits. And in quantum mechanics you've got these energy eigenfunctions. At least that's what people talk about most of the time. And an energy eigenfunction doesn't look at all like a particle moving in an orbit. And the question is, why? What's the difference in how we understand this? Well, just for the case of the harmonic oscillator, let's shift the tension down to the excited states in, say in is one of the excited states. And let's consider the expectation values of position and momentum for the excited states, just to see, because expectation values of position and momentum are ordinary numbers, and they could sort of look like classical quantities. So let's consider the expectation value of position. Let's say in an energy eigenstate. And I'll put a hat on the x to emphasize that that's an operator. Similarly, we can do this for the momentum. Look at the average value of momentum in an energy eigenstate. By the way, my use of the hats here is that I tend to put hats on something to indicate that it's an operator when there's some danger of confusion between the operator and something else like an eigenvalue or a classical quantity. And I tend not to do it when there's no danger of confusion. That's the rule for the hats for the time being. Well, in any case, these expectation values are easy to work out. Because the easiest way to do it is to write in terms of a's and a's and a's and a's and a's and a's and if you do that, you'll find quite easily that both these expectation values are zero. So as far as expectation values are concerned, an energy eigenstate doesn't look at all like a classical particle in an orbit. If you have a classical harmonic oscillator and you give it some definite energy, it's going to swing back and forth within amplitude. So the question is, where is that classical time dependence in the quantum problem? It's not appearing here in the eigenstates at all. In fact, the energy eigenstates of quantum mechanics are what they call stationary states. That means the probability density in configuration space is independent of time. And so any expectation value you take of any operator is going to be independent of time in an energy eigenstate. Therefore, there is no time evolution at all. In particular, it stays zero for all time. So it's disappointing if you're trying to make a connection with a classical orbit. Now, how can we understand this from an intuitive standpoint? Here's one way to do this. Let's go back to the classical problem. Here's a classical phase space like this. The orbit of the classical harmonic oscillator in phase space is a circle, something like so. And let's say we've got some initial condition called x0p0, which lies on this circle. Then the motion is a clockwise motion around the circle, this classical motion. So a clockwise motion around the circle. Let's suppose in a later time, we've got x of t and p of t. Another point in the circle, when there's an angle in between, and that angle is actually equal to the time. Or if we want to restore ordinary units, then the time that's replaced by omega t is basically the phase angle of the harmonic oscillator. Simply speaking, this looks like the hands of a clock even go through the right direction clockwise. All right. Now, so what this suggests is, OK, so the question is, why do we have expectation values at zero for the energy eigenstate? Here's an intuitive way of understanding it. It's related to the time, energy, and uncertainty relation. It says that the product of delta e is equal to, or greater than 8 to the h bar over 2. This looks just like the delta x, delta p uncertainty relation for position of momentum. We haven't really talked about time energy uncertainty relations yet, but we did talk about delta x and delta p. And in fact, the mathematics is deriving this is quite similar to what she used for the delta x, delta p uncertainty relations. But the physical interpretation is different, because t is not an operator. Whereas x and p are, at least in the non-oldivistic theory, that's how it is. And so the interpretation of this uncertainty relation actually depends on the context, and it's very various from one application to another. But in the present application, it means the following. Is that if we have an energy eigenstate, then delta e is equal to 0, because in an energy eigenstate, the energy is exactly low. And in some sense, that means that delta t must be less than infinity. The sense in which that's true is that the idea is that instead of having something that's localized on a classical orbit, which you could tell the time if you followed that at a point in terms of 2 o'clock, it's like in order to make a measurement of energy, precise measurement of energy so the energy is known exactly, the measurement process must extend over an infinite amount of time. That's another way of saying it. So if we look at the classical orbit and we try to measure over an infinite amount of time and blur our time clock, what happens is this particle gets smeared out over the entire orbit in a uniform manner. And if we do that, then even the classical average value is a maximum theory of 0, because it averages to 0 over this oscillation. And that, in some sense, is the reason why we have these expectation values. What it shows is that if we want to get some quantum object that looks like the classical motion, we can't use energy eigenstates. We've got to use a linear combination of energy eigenstates, then those different terms, and the expansion will have different time evolutions, and we'll get a non-trivial time evolution for the quantum state. So in particular, let's think about putting a minimum uncertainty in the white packet, let's say in some different position in the phase space, some different expectation value besides 0 and 0, different from what we have in the ground state of the harmonic oscillator. All right, that's kind of an introduction to where I want to go for the rest of the lecture. I'm sorry I have to cover this up now. But that's the location. So now this takes me to the next topic that I want to tell you about, which is coherent states. First of all, the definition of a coherent state in the context of mechanical oscillators, like we're talking about now, a coherent state is a wave function, which is, first of all, satisfies the fact that this version in both position and momentum are equal, and they're equal to 1 over the square root of 2. And this means in particular, these are minimum uncertainty, these are minimum uncertainty wave packets. This would be h bar over 2 higher stored ordinary units. But for simplicity, I'm going to get rid of the physical constants by choice of units in this convention earlier, so it's called 1-hash. All right, so coherent states are all minimum uncertainty wave packets. And also, as far as the expectation value position in momentum are concerned, these are allowed to be anything. But in fact, the expectation values of X and P are parameters of coherent state. And you can see in particular that the Brown state and the harmonic oscillators, an example of coherent states, what are these expectation values are 0 and 0. But what we like to do then is to take this, in fact, take this Brown state and the harmonic oscillator, and think of them in a space like this, and we like to move it to other places in the base space to change these expectation values. Now, the nice way of doing that is to use translation operators. The translation operators we introduced before over the top of the board, excuse me for covering things again, let me draw your attention to them. These are the translation operators that, as we defined previously, the first slide here is the definition of the translation operator. This is in one dimension, and the A here is the displacement for the translation goes into the violation operator. I'm sorry, it's the same symbol. The second line gives the effect of the translation operative wave functions, and the third line gives the expression of the translation operator in terms of its generator, which is the momentum. Putting a hat here to indicate that's the momentum operator, the A here is an ordinary number or a C number, as the rat would say, multiplying p as well as displacement. Anyway, these are the things we've covered before. All right. Now, the first question I'd like to address is how does the translation operator affect expectation values? So let's suppose we've got a state psi, and this has an expectation value called Lx0, which is just psi sandwiched from x hat. Let's define a new state called psi prime, where it's just a translation operator, T of A, acting on psi. Well, let's consider the expectation value position operator with respect to the prime state. How does that change compared to the expectation value of the O prime state? Well, you can easily see by just substituting in here the definition of psi prime, this is the same thing as psi times T of A dagger times x hat times T of A times psi like this. And so working on the answer, it's going to involve working out this conjugation relation of the position operator by means of the translation operator. So let's make a place over here where we're going to write the answer, T of A dagger x hat T of A is equal to, and we'll fill this in in just a moment. The easiest way to work out this conjugation relation is just to use this second line here for what translation operator suits the wave functions. We'll sketch it for you. The wave function is psi of x, so we want to apply T, translation operator, first in multiplication by x and then the initial translation. So first of all, if I apply T of A, to this, this comes up with psi of x minus A. And if I then apply x hat, that's multiplication by x, so this goes there when the x times psi of x minus A. If I then apply T of A, well, it's really T of A dagger. But these are unitary operators. So T of A dagger is T of A inverse. And that's the same with T of minus A. So I'm going to write it this way. T of minus A, which is T of A dagger. So here I use this middle rule here except I change the psi of A, so it's x psi of x plus A. So x here has to be replaced by x plus A. It goes in the x plus A, the multiplying of psi of x. And this is the same thing as an operator x hat plus A acting the way function psi, which is then evaluated at x. And so this is a sequence here that shows you the effect of these three operators. And it shows you that the product of these three operators is the same as x hat plus A right here. So there you are. There's the conjugation relation. And now if I take expectation values of both sides, which is speculated by the initial state psi here, you can see this turns into the original expectation value of x hat 0 plus A. So expectation value is shifted by forward by A under a translation. Now this is pretty obvious because the translation I've already done is that the way function is just pushes it down to the x-axis. And so expectation value is having shifted in the same way. Anyway, while we're at it, let's also look at momentum. What does T of A do to the momentum operator when we conjugate? And the answer is it doesn't do anything because T of A is a function of momentum. That's so it can use with it. So I can bring the T of A dagger past the p or it cancels itself out and just take p. This is a position space translation operator and it doesn't do anything to the momentum or only to position variables. What does it do to dispersions? Delta x and delta p, what happens to them? It's delta x going to under translation. Well, the dispersion, we just learned the statistics. It's a measure of the dispersion about the mean. It's really a difference between the mean and the actual value of x as an average of the square root of the difference square root is the RMS difference. And the translation affects the x, but it also affects expectation values in the same way. And so the difference between the two cancels is out when you do a translation. Translations don't affect dispersions. So delta x goes in the delta x and delta p goes in the delta p. Now, this is important because it means that if you have a minimum uncertainty wave pack, that the translation operator maintains the minimum uncertainty condition that doesn't change at all. The only thing that changes is the expectation values. All right. And so, in particular, we took the ground state of the harmonic oscillator in this space-based picture or image. I'm just going to blog here at the board and here's the ground state zero right there. And we apply a translation operator to it to get the state of t of a acting on the ground state zero. We can think of this as a blog here that's centered in a position that positions x equals a on the x-axis. In fact, it will just be that Gaussian wave bunch which is just that Gaussian root value of x-axis. That's all it is. But let's look at it in a face-face and think of it this way. So by doing this and using t of a, you can change the expectation value of x. That's not quite what we wanted, though. We wanted to change the expectation value of both. See, do I have it somewhere here? Both position x and y now because we're interested in starting some kind of wave packet out of the orbit here in an arbitrary initial. I sort of have a p zero. So to take care of the momentum translation, allow me now to introduce momentum translation operators.  Allow me to define. This is just a definition. I'll call it s of v. I'll use s just to make it a different letter from t. But this is a translation operator momentum. And this is defined as a definition of it. It's e to the plus i v times x hat. v here is just a number and it's going to be the momentum displacement and the x hat is the operator. It sort of looks like this except x and v have been swapped and the minus sign has been answered by the number used. So let's see what s of v does to wave functions. So first of all, if I take s of v and let it act on a wave function psi and evaluate it at x. Well, s of v is a function of the operator x hat which is just a multiplication by x on an x space wave function. So this is just the same thing as e to the i v x times psi of x. Both by psi of x by a position dependent phase factor. What does this do to momentum space wave functions? Let's call this psi prime of x. What does it do to momentum space wave functions? Let's write phi prime of p to be the momentum space wave function corresponding to psi prime of x. Well, that's the Fourier transform and it will be x over square root of 2 pi h bar and then it's e to the minus i p x h bar is 1 versus just 2 pi square root of 2 pi h bar. e to the minus i p x and then we need to multiply by psi prime of x which is in the line above here and I'll write e to the i v x times psi of x when we place these psi prime of x like this. Now you can see that this is the same as the Fourier transform of psi of x except the parameter p has been replaced by q minus b. So this is the same thing as phi of q minus b where phi is the Fourier transform of the original psi. So the net effectiveness is that I'll write this this way. So s and b hacking on a momentum space wave function as a function of p is equal to phi of q minus b. Notice how it's analogous to the position space translation operator. Notice that this is an effect that moves the momentum space wave function down the momentum axis by displacement of p. Likewise, one can easily show that s and b can act a lot in a momentum of the state p, and that's the key plus b. This way we have a parallel now between parallel set of coordinates in position and momentum displacements. So if we took our Brown state wave function and we're displacement in x, we can then follow that by displacement in p that's called as s of b times t of a acting on zero. We can filter it as a blob in face space which is centered on position coordinates a, b like this. Before I say more about this, let me go back to the question of expectation values which I've covered for the translation operators t of a over this panel. What does this new momentum space displacement operator, what does that do to expectation values? Well it's an obvious analog of what t of a does. It works like this. I kind of like put it over near this so I'm not going to erase this. It works like this. Because that s of b, dagger s of b, s of b dagger x times s of b so we conjugate the position operator by the momentum displacement. That just does nothing to it. The reason is that s of b is a function of x hat, so it commutes from there. That's just like this formula over here with t of a of p. If we take s of b dagger using the particle momentum operator and it goes into itself plus the displacement which just indicates how momentum space expectation values are shifted forward. And then finally we have the effect on dispersions, deltas and deltaki, both of them going into the cell just to say they did not change for the same reason that they did in the case of the X-paces displacement. That's a yes question. No, the momentum position translation operators do not commute and that's the effect going to be by next point. The just one remark before I enter that is that by using these products, these operators, we can in effect take this ground state with a harmonic oscillator which is an example in the current state and move its expectation values ground without changing and thus maintaining the minimum uncertainty wave packet property of these sort of equations in these states. So by doing this we can create what we call coherent states with any expectation values of the position momentum in which there's a two parameter camera that can effect cover all the base space. However, because of this point that was raised here a minute ago there is an issue we need to deal with which is that getting with this point a d in base space direction and then p direction what if we went the other direction applied momentum translation operators first and then position operators second would we get the same answer? The answer is no because these operators t of a and s and p don't commute with each other. That's no surprise because they're exponentials of operators namely p hat and x hat that don't commute. So there's a difference between these two paths and getting there. Turns out the difference between the two paths is only a phase factor and some people would say that's not very important but nevertheless I'm going to devote some attention to giving a more symmetrical way of getting the final location that hits around the dual to the space factor and the manner which is symmetrical under exchanges of x and p. This is the idea of what we're doing now about creating coherent states and the basic idea of coherent states is that there are minimum and certain wave packets based based parameterized by the expectation values of position of momentum. By the way before I go further with coherent states let me say that in current physics I think there's two main uses for coherent states one of them is really conceptual or theoretical if they're used for exploring the classical limit. The reason is that a coherent state has a minimum dispersion to an x and p we know because the uncertainty principle is that you can't make measurements of position of momentum with infinite precision simultaneously as you can in classical mechanics. However a coherent state is as close as you can come to that and so in some sense it's the quantum object and some people think of it this way it's the quantum object that this is as close as possible to a single point of a definite classical state. And so the use of all kinds of studies, very common in literature nowadays to use them for this purpose is used for studies trying to understand the classical limit of quantum theories. There's a second purpose of coherent states which is in the case of quantum objects they are they are quantum states that can actually be produced with lasers and they're different than these energy eigenstates which in the case of lasers are states of a definite number of photons coherent states have a variable number of photons that's not an eigenstate and a number operator but in the case they're interesting and experimentally for things that one can do with them I'm mainly here going to concentrate on the use of them for understanding classical correspondence alright in any case the main problem I want to address now is the non-computativity of these operators S of A and S of B excuse me, S of B and T of A in order to work around this I'm going to find a new operator which is called a Heisenberg operator and it's just kind of a combination of translations and positions of momentum taken at the same time we'll call it W it's parameterized by both A and B and this is a defined and behavior with I times B Bx hat minus AP hat so it looks like the T of A and S of B are the exponents of mind that's all it is however this is not equal to E to the I Bx hat times E to the minus I AP which is the same thing as S of B times T of A it's not equal to it because X and P don't commute if X and P were ordinary numbers which do commute then the usual rule of explanations would give you an equality but because it ordering matters this is actually not true there's some other stuff that has to be taken into account however the Heisenberg operators can be expressed in terms of these S and T operators it's not too complicated and the way to get from one to the other is to use what is what's called Glauber's theorem which I'll now tell you about Glauber's theorem has rather restricted conditions on it so it's not applicable in many circumstances but this happens to be one of them so it's useful for this longer by the way I see physicists in Harvard won a Nobel Prize for this and worked on quantum optics which they did in the 1960s alright it is all related to this kind of stuff alright Glauber's theorem says this is that if you have two operators A and B which commute with their commutator then there's a nice identity that says that e to the a times e to the b is equal to e to the a plus b that's all you'd have if they commuted but then there's an extra term which is one half the commutator of A and B so in particular if they do commute the commutator is zero you're just going to be usual rules of exponentials alright let me go over here and prove this give me the proof let's define, this proof will follow what we did in the first Homer problem one of the first Homer problems first Homer set let's define an operator which depends on a parameter lambda which is equal to e to the lambda a times e to the lambda b and we'll get a differential equation for it I think for edge a df to the lambda is equal to a df to the lambda b plus e to the lambda a times b times b times e to the lambda b like this by the way let's note that f of zero is equal to the identity operator it's kind of an additional condition for this differential equation we're going to get now in this line here for df to the lambda I brought these operators a and b down to the left when I did the differentiation in this expression for this derivative allow me to insert here between the b and e to the lambda b we put in e to the minus lambda a times e to the plus lambda a which of course cancels out and the reason for doing that is that we get on the right hand side here in this term e to the lambda a e to the lambda b and likewise in this term e to the lambda a e to the lambda b in the same term and that's just the same thing as f itself so effectively we've factored out after the right of this and so this becomes this expression of a plus e to the lambda a times b times e to the minus lambda a the whole thing multiplied on to f so this is our differential equation now at this point we haven't made any assumptions about a and b all a and b but again going back to that first homework exercise this product here which is exponential is conjugated around b this can be developed into a power series of iterated commutators in particular it's equal to b plus the commutator of here's a multiply by lambda so it's lambda the commutator of a with b plus lambda squared between factorial times the iterated commutator of a is the commutator of a with b plus dot dot dot infinite series like this and that's also almost true however while our theorem says we've been assumed that a and b can use with their commutator in particular that means that this lambda squared term goes to zero in fact all the higher order terms are all the iterated commutators for zero and so we're just left with two terms of this series and so this gives us then to rewrite the differential equation as the f of lambda is equal to a plus b plus the lambda times the commutator of a with b times f of lambda now if this differential equation were numbers instead of operators it would be easy to solve it would be that f of lambda would be equal to lambda times a plus b plus lambda squared number two times the commutator of a with b this is assuming f of zero is equal to one but we have to put a question mark on this until we can check to see that this is true even for the case of non-commuting operators well in fact it will be true for non-commuting operators if the derivative of the exponent commutes the exponent itself the derivative of the exponent is a plus b plus the lambda times the commutator of a with b and you can see that you commute a plus b in operator commutes with itself so a plus b commutes with a plus b a plus b commutes with a commutator because that's what we're assuming commutator commutes with a plus b and the commutator commutes with itself so the derivative of the exponent commutes with the exponent and so in fact the solution is correct this works and this is the solution for this function that the lambda now finally all we need to do is just set lambda equals to one and it turns into the derivative which is stated over here that's the proof of the derivative and it appears so much better proof anymore let's erase that let's apply the derivative so let's do this let's let a be equal to i v x hat and let's let b be equal to minus i a key hat using these products of these displacements in the momentum and position then the commutator of a with b is equal to i times minus i is one more case a times lower case p all right down what's left is the commutator of x hat with p hat but that's equal to i as far as one here so this just becomes i a b commutator at least to operators is constant it's a c number c numbers can mean with everything so in particular they can mean with a and b and so the conditions of lovers pair are matched and the result of this is that this operator e to the i v x hat times e to the minus i a key hat is equal to e to the i v x hat minus a p hat combined together in the x component plus i a b over 2 that's the result of the product today momentum times a position displacement you see it's a Heisenberg operator in the first term we find up here and then a correction term which is a face factor in fact since this is a face factor it commutes with everything and I can just put it as e to the i as a separate face factor here and bring it back over to the other side and if we do this then we get this result is that the Heisenberg operator w a b is equal to s of b times t of a or even the face factor first e to the minus i a b over 2 times s of b times t of a and so this then it allows us to express the Heisenberg operator in terms of position momentum splices alright now having done that let's now define a general coherent state which is going to be parameterized by its expectation values a and b these are expectation values in position momentum this will be defined as the Heisenberg operator w of a b acting on the ground state and if we work out the wave function of this let's call it psi a b of x which is equal to the scalar product of x with the state a b like this it's straightforward to work this out by just applying the position momentum of translations we'll find that this is the same normalization whatever pi is in one quarter e to the minus x minus a squared over 2 that's the x shift in that when you get plus i b x that's the momentum shift and then minus i a b over 2 which is this extra face factor here and so this then is the wave function they compare and state minimum uncertainty wave packet with given expectation values of x and p think of it as a blob in base space and by using the Heisenberg operators it's symmetrically treated in position momentum that's where this extra face goes from now go back to our original problem then which is we're back to this board what we'd like to do and understand the classical rhythm is to take the initial condition for the quantum harmonic oscillator which is the coherent state centered in some position x zero and p zero I've been calling this a and b but now I'll call it x zero and p zero because we'll think of it as an initial point on a classical orbit like this in the classical base space and we'd like to do the time evolution of classical mechanics maybe before I do that let me say something about the time evolution of classical mechanics while I already showed you what it is just a rotation by an angle but there's a convenient way of expressing this that I'd like to tell you about let us to introduce complex coordinates in the classical base space let's introduce a coordinate I'll call z which is defined to be x plus i p with a square root of two and except for the square root of two this is really identifying the base plane x p plane with a complex plane so it's the real and imaginary parts of z square root of two is just convenient to see here you might notice however that this is a class of definition now so the x is a piece of the square root of zero which you may notice that this is a classical formula which is exactly the same as the definition of the unilation operator in quantum mechanics so the z is the classical analog of the unilation operator likewise let's introduce the complex conjugate of z which is equal to x minus i p over the square root of two I'll note the complex conjugate with a bar here so that's like the a dagger alright now if we do this then this x zero p zero can be translated into a complex initial condition called z zero and this x of t and p of t can be translated into a complex final position called z of t and it's easy to see what the solution of the classical equations is it's in z of t it's equal to it's just the phase factor which is e to the minus i t multiplying on the z to the zero in complex variables the time of evolution can become really simple by the way I might mention that this is a classical analog the solution of the Heisenberg equation is the motion of quantum mechanics for the unilation operator if you solve those which you find this is an a of t is equal to e to the minus i t times a is zero this is in the Heisenberg picture exactly the same formulas no wonder the equations are exactly the same too except they're reinterpretation as soon as alright anyway this is a useful way of talking about the classical solution in terms of these complex in terms of these complex coordinates now if I may go back to the where do I have it these Heisenberg operators are they here allow me to make some further slight changes in notation this is really pretty simple let's first of all let's take a and replace it by x zero and b and replace it by p zero because we're going to be thinking about these displacements as moving us to some additional condition and then we've got a Heisenberg operator which is parameterized by x zero and p zero and just as by the way in notation we can also write this as w of z zero so z zero is the complex coordinate x zero plus i of b zero over the square root of two just a different way of parameterizing the same operator but similarly as far as the coherent state itself is concerned which I call a b over here let's call x zero p zero is w of x zero p zero applied to the ground state zero and let's also just define this to be z zero for short again parameterizing the coherent state also by the same complex number complex coordinate in the phase space so let's do that now then here then is the problem that we'd like to solve as we like to say suppose the psi initial time zero is equal to this initial coherent state the problem is what is the psi a later time t equal to this is an initial value problem for the Schrodinger equation by the way according to the error test relations which we went through before the expectation values follow the classical motion they do because the Hamiltonian is a quadratic function the potential is a quadratic function and so in particular we know that whatever this solution is the expectation values of x and p will follow this is the same classical orbit here however, that still leads up in the question about whether the initial coherent state remains a coherent state does it remain a minimum and certainly weight packet actually the dispersion only involves the second moment what we'd really like to do is to solve Schrodinger equation and get the whole solution and the quantum problem so it's actually more than just looking at expectation values at least we know something already about the solution excuse me, from the error test relations alright, so anyway to go back and this is going to be the problem we'd like to address so the answer of course is psi of t is the unitary time evolution of u of t acting on this initial coherent state z zero and this of course is the Hamiltonian given the minus i t h what if h had an operator acting on z zero the strategy I'm going to use for solving this problem is to decompose the initial coherent state that location z zero is the linear combination of harmonic oscillator eigenstates and then we'll evolve that forward in time by the usual method of basically normal load at the time so the first task is to take the initial coherent state and to decompose it into an energy eigenstates so here's what we've got we've got z zero then is equal to the same thing as x zero p zero which is the same thing as w of x zero p zero acting on the ground state which is the same thing as e of i p zero x hat minus x zero p hat applied to the ground state that's the Heisenberg operator now allow me to do this let's go from the x zero p zero or z zero as ordinary numbers and the x hat and p hat are operators let's express these in terms of their corresponding complexified versions so the x hat is equal to a plus a dagger over the square root of two p hat is equal to a minus a dagger over i the square root of two x zero is equal to z plus z bar over the square root of two this is the classical correspondence and then we've got p zero is equal to c minus z bar over i the square root of two this is just using defining relations to go over from x's and p's to a's and a's daggers and one of the operators or z's and z bars in the classical picture I should put knots on the z's here because it referred to the initial conditions at zero and p zero and if you do this and you transform this exponent in terms of this in terms of e to the power of z zero z zero times a dagger minus z zero bar times a applied to the ground state it's just the algebra to transform that exponent for these equations if you do it in terms of this so this is what we need to evaluate now for this purpose I'm going to use Flouder's theorem again let's consider the product z zero a dagger times even the minus z zero bar times a and let's let our first operator a be z zero a dagger capital A z zero times annihilation operator a dagger capital D is equal to minus z zero bar times annihilation operator a


It goes and sticks in your column, so the waste material which should have easily flown out will not flow out of the system. It's a hotbed for illness. That means your respiratory action and your thought process and your sense of smell will get lowered over a period of time. It definitely renders to your dullness. If you eat too much of tuberous material like potatoes and other tuberous substances, you will generate gas, which is below the navel. Once you generate gas there, your prana between the navel and the pit of the throat will not function as it should and the dynamism levels in your system will go down. Among the vegetables, things like radish, these are one of the few things which grow underground and still positive, carrot, radish. All the other tuberous material like potatoes and potato family are zero-pronic, but they cause what is called the stomach. Stomus means it brings a certain inertia into you. We generally advise students and meditators not to eat potatoes and its family because they increase the sleep quota. Both for the student who is studying and for the meditator, the biggest enemy is the sleep, you know. If you close your eyes and sit, if you open the textbook, textbook is a great tranquilizer, isn't it?  Dairy products have certain enzymes which are used today to manufacture certain textile adhesives. These adhesives are not water soluble. So when you consume lot of dairy product, it goes and sticks in your colon. So the waste material which should have easily flown out will not flow out of the system. The general alertness in the body and in the mind will go down by consumption of dairy products.  Yes or no? If somebody is sleepy, it's a common thing in South India, naa, thair saadam saapto vandeya. Everybody knows it makes you dull, but the teacher also eats, that's a problem. It is an established fact medically that in your present system, if you eat raw meat, it would take anywhere between seventy to seventy-two hours to pass through the system. If you eat cooked meat, it would take fifty to fifty-four hours to pass through the system. If you eat cooked vegetarian meals, it will take twenty-four to thirty hours to pass through the system. If you eat an uncooked vegetable, it will take twelve to fifteen hours to pass through the system. If you eat a fruit, it will take one-and-a-half to three hours to pass through the system. Why is it taking so much time? Because essentially the alimentary canal in a human being is about twenty-four to thirty-six feet long. This whole pipe that's going from lip to the anal outlet. In all the herbivorous animals, this is a feature that their alimentary canal is approximately five to six times the length of their body. In all the carnivores, it's only two-and-a-half to three times the length of their body. Or in other words, all the meat-eating animals have a very short intestinal length. All the herbivores have a long intestinal length. So because vegetarian food passes through the system quickly, so the design factor of length has been designed like this. So when you keep… eat meat and keep in the system for a long time, definitely it's a hotbed for illness. In many ways you will create it. Eating meat has become a part of our, you know, in the culture in the world because in many societies in the past, survival was the mode. When it's a question of survival, eat anything that is necessary to be eaten, that's a different thing. Once survival is taken care of, that's when you choose.  This is not about a taboo about eating meat. This is about eating more consciously. This is eating by choice. In the yogic tradition, if you cook anything, maximum within one-and-a-half hours you must consume it. If there are certain types of items, up to four hours is okay. Beyond that you don't touch. It doesn't matter how tasty it is, how nice it is, you don't touch anything because then you will develop gases in the system. Gas is not only in the stomach that you will feel obviously, but you will develop vayus in the system which will work against your prana vayu. That means your respiratory action and your thought process and your sense of smell will get lowered over a period of time. These three are very vital for you to be a vibrant human being.


Hello people of the internet, my name is Johnny and welcome back to another FNAF News video. But this time we're kind of focusing on just one topic with like two bonus news at the start of this video, which I'm going to talk about right now. But first if you are not subscribed to the channel boys, please subscribe. I cover so many, probably too much FNAF news, now that I think about it probably too much, but I cover basically everything there is when it comes to this series. So if you want to stay up to date, subscribing is the best way. And also, since you're here, smash like. It takes like two seconds. Anyways, getting that out of the way, let's talk about some bonus quick news before we hop into FNAF AR, which is our main topic we're going to talk about today. So first news first, Help Wanted has officially released physically on PlayStation 4 and Nintendo Switch. If you pre-order them, they should be shipping out very very soon and you can actually go into stores and buy them right now. This is the first time FNAF has ever been released physically where you can actually hold the game in your hands. So this is very exciting. The core collection, which includes FNAF 1 all the way through Sister Location, will be released January 12th. Kind of a ways away, but since we're already talking about physical copies, I thought I'd mention it here. Also Sister Location received an update today for the Nintendo Switch. They got a new pause menu with cursor speed options, additional door shortcuts with a D-pad, Lullbit analog stick fix, bitty bab fix, and many other fixes and tweaks. They are still working on the PS4 release. This is very exciting news because Custom Night on Sister Location on consoles was basically kind of unbeatable, especially when it came to Lullbit because you had to click on the keypad, which is very small on the screen. So I'm very very excited that this is finally out and I'll probably return to the game just to see if we can beat Custom Night. So anyways, now let's finally move on to FNAF AR, which is the big discussion point for the last couple of days. If you don't know, right now Olimax is hosting a community quest where if we can collect enough event currencies, we can actually fill up a bar and we can unlock a secret character early. Last time I checked, we were only like three bars away from getting the character. So if you're not collecting your event currencies, do so right now. We are so, so, so close. So I'm going to quickly read Olimax's post before the interview with Qnhappen, where Darko interviewed her, asking a couple of questions. There's a lot to go over there. So I'm going to quickly read this, then we'll hop over to the interview. We hope you're having fun with all the Winter Wonderland festivities. And yes, Arctic Ballora will be making her debut at 4 p.m. today. Obviously, this was five days ago. She's not returning today. Now about that quest. Some of you have noticed the new community quest pop up in game and wondered what it was all about. Well, here you go. How does it work? You just have to collect as many Winter Wonderland event tokens as possible over the next week and the amount that you collect will be reflected in the community token thermometer, which will be updating in the game. If the thermometer is filled up, a very special delivery will be sent out next week. Y'all are already on your way there. How do we get tokens? There's a few different ways you can do so. You'll get most of the tokens through daily challenges, followed by the event rewards, defeating animatronics and map balloons. I'm game. What's the very special delivery? Drumroll, please. The early release of a new animatronic character, which we do know who that is and we'll talk about very shortly. Well, I mean, he's in the title and the thumbnail, so you know who I'm talking about already. We'll still be releasing a new animatronic skin next week. That is this week now as part of the Winter Wonderland event, but wanted to give you all the opportunity to come together and unlock this character as a community. We're grateful for all, as always, and hope you have fun with this quest. We'll be keeping you updated so keep those eyes peeled.  If this was posted five days ago, we only got like one or two more days. Like if we don't fill up the progress bar, we would still get the character, but it would not be next week. It would be the final week of December. So I'm just curious if we complete it this week, do we get two characters tomorrow like the skin and then the actual character? Is that how it would work? I don't know. I'm a little bit confused about how this is worked because clearly we're going to get the character no matter what. Elimix has clearly taken the time to make the character. So it's not like they're just not going to release the character. So I guess getting it early is the prize here. But if we get it no matter what, it's like we're only waiting an extra week. I don't know. It's a little bit weird, but I still respect it. When they did a event like when the game first came out, that was great. I love when they do stuff like this and I hope they keep doing it in the future. And so now let's move on to Darko's interview with Kieran. We got a lot of information. The stream was like an hour and a half. It was pretty long. But I actually have a Reddit post here made by the GG Code 2014. So I'm getting my information from a six-year-old. That's great. They have a whole list here of basically all the news that we're going to talk about right now. So let's talk about that. So like I've said, it kind of already the community character's hint for the next character being added into FNAF VR was it's me. Now I'm not even going to try and sugarcoat it. We're just like, oh guys, it's funny. If Kieran said it, then she's being added to FNAF VR. What's her mechanic going to be? Clearly it's Golden Freddy. How he's going to walk. I'll speculate a bit here. So this is the way. I think he might play out. I don't think he'll be like Plus Strap. And that's based off of reasons I'll get into shortly. From information from the interview. I feel like he probably, it would be weird if he stood up. Because in basically everything we've seen about Golden Freddy is always like slumped over. So I feel like what they're going to do is they are probably going to use the FNAF 2 Golden Freddy. I feel like that is kind of the one that's being used a lot more than the FNAF 1 Golden Freddy. You know, even Scott prefers FNAF 2 withered Golden Freddy over FNAF 1. He was in Custom Night. So Ultimate Custom Night, I mean. So I definitely do think they'll use FNAF 2 Golden Freddy. And in FNAF 2, again, talking about the interview here, Illumix says they try and take influence from past appearances of the characters in previous games. So a big thing about Golden Freddy in FNAF 2 was that he was kind of a ghost. You know, much like FNAF 1, he had a huge head that appeared in the hallway that would like fly at you if you didn't put the mask on on time. So I hope they do something like that. That would be amazing to see. Like for his charge attack, it's just his floating head flying at you. That's going to be awesome. I hope that's what they do. And then I hope he has a feature where it's like hallucinations.  And then if you hover over them, they like fade away. I feel like I would want them to focus more on that. Because that's a mechanic that doesn't really get used a whole lot in the main game.  They're distractions. So I feel like that could be a good new mechanic for Golden Freddy because he is a hallucination. So seeing him in different places at the same time would fake out a lot of people. Or you can see him over here, but the static is over here.  I hope they do something like that. That's probably going to make me upset in the future because I'll be just like, oh, I don't understand what he's doing. But I think it'll be cool. So yeah, that's what I hope they do with Golden Freddy. Moving on, more characters that take you to areas kind of like Plush Trap will be arriving. However, since Plush Trap took quite a long time to get into the game, I feel like they probably won't be as regular as just normal characters and skins. But then again, a lot of that time was just the making of the engine. That made that whole thing possible. So who knows? Maybe it takes a lot less time than we think it does. There will not be any holiday specific winter skins clarifying nothing Christmas related. So sadly, alas, no Santa Spring Trap. But it does make sense. You know, Illumix, I will say, they have been very, very good at bringing in different cultures, different religions into the game. And that's something I really, really love about this game. You know, you got what's it called? Katrina Torchica. That was amazing. You know, I doubt there's that many Mexican FNAF fans, but the fact that they did that was amazing. And there's not me saying, oh, there's Mexicans. They hate FNAF. That's me saying the large portion of the FNAF community is not Mexican. Hopefully I'm not offending anyone there. Anyways, moving on. A new gameplay type will arrive as early as next week, something with the coming soon curtain. Yeah, so if you don't know, in the game with the balloon image used to be, there's a red curtain with the Fazbear Entertainment logo on it saying, coming soon. And that apparently is a brand new gameplay update involving the camera. There's a few hopes that I have with the gameplay update, but I don't really want to say them because then I'll get upset when it's something really bad. I hope it's something good. In early 2021, a huge gameplay update will come out. And when we say huge, we mean huge. Right? I believe Kirin said that it was going to like change the entire game, which I really, really hope it does. Because this game really needs some shaking up. It's been mostly the same ever since it came out over a year ago now. A lot of new skins are coming. Of course, there was no doubt in my mind that they were going to stop doing that. The reason why we have so many skins will connect to the 2021 update. Yeah, so I'm very curious, because when Kirin revealed this, I was like, okay, that's a bit interesting. The reason why they've been releasing so many skins will be revealed in this big update. And she said, if you're not collecting the skins, I highly recommend doing so. So I guess I gotta get working on skins. Boy, send me your skins placed. Something big next week. Kinda ominous. I don't remember what this is referring to. So I guess we're just gonna have to wait and find out. That's so vague. Anyways, moving on. Sometime in the new year, Illumix will be taking a break once the 2021 update releases. And this is something I'm very happy about. If you know me, I'm not a huge fan of the skins. Even when they say in this big update, we're gonna find out why we've released so many skins. Even then, I'm still not the biggest fan of them just pumping out skins after skins. Even if it has a purpose that we just now learned about, months and months later, I'm still not a huge fan of it. Especially when the skins cost more than the price of like one singular FNAF game. So yeah, I'm happy that they're taking a break. They've worked their butts off this year. Illumix, you do deserve it. And at the same time, I'm just happy we're not gonna have to do skins after skins after skins. Because they can get pretty repetitive. The 2021 update more than likely won't release in January yet. I believe they said, uh, actually no, I'm not even gonna spit out a release date because I don't fully remember and I don't want to say something because if it's wrong, people will hold it against me. So yeah, not January. Probably. Oh god, what did she say? I can't remember what she said. Go watch the interview, it'll be linked down below. More FNAF-era exclusive characters like Frostbear will be coming in the future. So you know how Frostbear never appeared in the past game? Nor is he a, you know, a skin. Like Ape It Baby technically issues from the sister location Cupcake minigame. So she is not a exclusive character to FNAF-era, but Frostbear is. Hopefully you understand where I'm coming from with that. That is what they're talking about. Characters that were not seen in the past that are exclusive to FNAF-era that are not skins. This is another hyped one. Security bridge characters will be coming to FNAF-era eventually after SB is released.  Like, oh my god, it's like the second most knowing thing after them constantly asking for plus trap. And I realized I was one of those people. But like, at least we had evidence to back us up when we were like give us plus trap, give us plus trap. Here people were just saying give us SB characters because I don't know, it was just knowing. So I'm kind of glad that they cleared this up. I'm hyped for it, though they're not going to appear until after FNAF security breach. So keep that in mind. So many skin concepts are created or currently backlog, meaning we'll be getting a lot of skins in the future. I don't remember them talking about this, but it's something that again, I'm not too surprised about. They're going to keep pumping out skins, you know, so much longer. More lore will be coming in 2021. I was very interested about the lore questions because normally you can't ask, you can't ask questions about the lore. Even Kyuren said in the interview that she knows lore that we don't know yet. So she kind of tends to stay away from questions about the lore. And I was even more interested about them talking about the FNAF lore because we had those emails leaked. The in-game emails leaked not too long ago. So I wonder, I was wondering, I was wondering if she was going to bring up those. She didn't, but she did say that more lore is coming, which is kind of surprising. I don't know if they're going to release the emails that already got leaked if they're going to create new ones. I'm not sure, we're just going to have to wait and find out. Well, FNAF, the armor which has come in the future, yep, totally saw that coming. Characters who have had voices in the past might have their voices, if possible, return to apprise their role. Meaning if the actor-actress is available, they will try to get them to apprise their role. This was something that we already knew. Right, Jack O Chika, her voice actress came back, same thing with. So baby, Afton, all those boys. Interestingly enough, and it talks about this later on, Ballora, she is an example of a character who did not have her voice actress come back to apprise the role. And we'll talk about that a bit more later on. A way to fight your own animatronics might come in the future. The keyword here is might. This is something that I personally have wanted for a long time. So the possibility of it coming to the game is very exciting. But again, it does say might. That's not confirmed. They have said that they've looked into that possibility, but they don't know 100% if it's happening. More new voices will be seen in the future. So what it means by this is, you know how Toy Bonnie didn't have a voice before this game? That's what they're talking about here. So characters who have not had voices in the past, again, Toy Bonnie will sometimes. I say sometimes because Jack O Bonnie didn't have a voice, but he was in the game and he still didn't have a voice. So sometimes characters will get new voices. Skins for Plus Trap will come eventually, but it is more work to do them. So don't expect them soon. Yeah, because Plus Trap is basically like the God among all these other like puny, mechanic characters. You know, like he has a full 3D environment that you have to take into account. His CPU doesn't work with any other character besides his own skin. So that's why it's going to take a bit for Plus Trap Skins. But I'm hyped for them. More shadow related characters might come in 2021 with some things having to do with Remnant. Again, this is just a possibility. She said not 100% confirmed. Yeah, this is something I remember them talking about. It's kind of warded a bit weird here. I don't know if they're talking about Shadow Freddy. I don't know if they're talking about like Agony and all that stuff. But more shadowy content is maybe coming in the future. And that is exciting because that is a aspect of the game. Let's kind of meditate. You know, it hasn't really been touched upon basically ever since the whole AR thing, where Shadow Bonnie could like appear behind certain items. Which by the way, I don't know what happened to that. It feels like that aspect of the game just kind of is forgotten. Did they ever release that? I don't know. Anyways, last thing. It is currently unknown if Blora will get her voice in FNAF AR. So just keep in mind that there is a possibility that some actors, actresses, might not come back to reprise their role. Also for new voices, Scott must approve of them as well as a Lumix. And some characters will not get voices due to lore and etc. And etc. You gave one example and then you said and etc. You can't like make a list out of anyways. Yeah, so this is again what I was talking about. Blora, Toy Bonnie and Jack O' Bonnie are very interesting characters to talk about when it comes to voice actors. Because Toy Bonnie, he didn't have a voice and he got one. Jack O' Bonnie, he didn't have a voice and he didn't get one. Blora, she had a voice but she didn't get one. I'm not entirely sure why Michelle Moss, the voice actress for Blora, did not come back for this game. So we're just gonna have to wait and find out what happened. And it's so strange too because like Blora, she's been in so many of the games. Sister Location, Ultimate Custom Night. She's been in two games. Two? I don't remember. In both those times, Michelle came back to do a voice line. So not sure what happened there. So that was kind of a long video. But there's a lot of stuff to talk about. Very exciting stuff. Kevin said that they are going to keep walking on this game for a very long time. So I'm very excited to see where the game goes. And from this information here, we know just kind of where it's gonna go in the future. At least for next year. So thanks so much for watching boys. And I'll see you all on the flip side. Goodbye. Gregory, be still. I think she's found us.