So, as I discussed yesterday, the topic for today is going to be state discrimination. So the question is, I have two or more states and I would like to distinguish between them. And so, what is the optimal way of doing this and what are the different regimes of interest for this question, this is what we'll look at today. So we'll start with something that I'm guessing also most of you are familiar with, is maybe the most famous distance measure between states, it's the trace distance. Okay, and we'll see how this relates to state discrimination. Okay, so let's do that. So the trace distance between two quantum states, rho and sigma, is defined as the following. So I'll denote it by delta between the two states, rho and sigma, as one half of the one norm between rho and sigma. So what I mean by this is I take the sum of the absolute value of the eigenvalues of rho minus sigma. Okay, you can also write it like this. So let's see a few properties of this trace distance. So it's between zero and one, and it's zero if and only if rho is equal to sigma. So it's a valid distance. It's invariant under unitary transformation. So if I apply the same unitary to the two states, the distinguishability, the distance between them doesn't change. If I restrict myself to the classical case, so in other words, if rho and sigma are commuting and there is a common eigenbasis, so I'm denoting a here. Okay, and so the eigenvalues are probability vector P of A and Q of A for sigma. Then this trace distance just becomes the following. So one half of the sum of P of A minus Q of A, absolute value. Okay, and this is again very well used in, used a lot in probability theory, and it's called sometimes the total variation distance or the statistical distance between the two distributions P and Q. Okay, what other important properties are satisfied by this distance measure is what I would call the data processing inequality. Okay, and this is an inequality which is really fundamental for any distance measure in general, and this may be the most fundamental inequality of a distance measure, and we'll see it, we'll see this inequality several times for the different measures we'll introduce during this lecture. Okay, so what do I mean by data processing? It just means that if I process further my two states with the same map, then the distance between them can only decrease, okay, or their distinguishability can only decrease. Okay, so what is a valid physical process here? It's a quantum channel as we saw yesterday, so it's modeled by quantum channel that I denote E, and the trace distance, when I apply a channel E, the same channel E to the two states can only decrease. Okay, good, so why is this total variation distance or the trace distance playing such an important role, and why do we use the one norm and not, for example, some other, the two norm or some other notions of distance? Okay, so the main reason is that it has a very natural operational interpretation, in terms of distinguishing between the two states, rho and sigma. Okay, so what is the setting for state discrimination or distinguishing between states? The following is that I have a quantum system A, and I know it's in one of two states, right? Or I have two hypotheses about the state of the system. Okay, so I know that the hypothesis which I'll call zero is that the system is in the state rho zero, okay? Or the hypothesis one is that it's in the state rho one, okay? And the question is that I would like to know in which setting I am. So is it hypothesis zero or hypothesis one that is true? Okay, and I have access to the system A. Okay, so what is my task here is to try to achieve this distinguishing with a minimum error probability. Okay, so what is a strategy? What is the thing that I can do in order to achieve this task is that because I have access to my system A, I have to do some sort of valid quantum operation which allows me to say, am I in hypothesis zero or in hypothesis one? Okay, and so it turns out as we saw yesterday, the most general way of modeling that is a POVM was just two outcomes, zero and one, right? Because my objective is just to output either zero or one, hypothesis zero or hypothesis one. Yeah, so this model by POVM is zero and E1. So as you saw, as you remember, so E0 and E1 should both be positive operators that sum to the identity. So okay, so there is various different error probabilities you can look at. And so for now as a start, I'll look at a simple measure which you can call the average error probability, which is that I'll put a prior on my hypothesis that I assume that with probability a half it's H0, which is correct, and with probability a half it's H1. As we'll see later, this is not necessarily the case. It depends on the applications. But let's think of this natural setting where we'll look at the average error probability. Okay, so in this case, if I put this prior, then let's see what is my total error probability. Okay, so my error probability is one half. So this one half corresponds to H0 being true. To the fact that I'm in hypothesis H0, which means that my state, the correct state, is rho zero. Okay, so what is the error probability in this case? The error probability is that I output one. Okay, so that's why I say it's the trace of E1 times rho zero. Okay, so this is the probability, again I wrote it here, that the state is rho zero, but my procedure wrongly says that it's H1. Okay, this is sometimes we call it the type one error. Okay, this is one of the two types of errors. And then there is the other type of error, that the true state is rho one, but I say zero. Okay, so this again is the type two error. Okay, so I should say that often, so here H0 you see in the expression, even in the expression that H0 and H1 play a symmetric role, but often the case that H0 and H1 do not play a symmetric role. And this is actually what we'll see in Stein's lemma in a bit, that H0 will treat H0 and H1 differently. Okay, so let's go back to the trace distance. So what's the relation between the trace distance and this question of discriminating between two states rho zero and rho one? So it turns out that the minimum error probability that I can achieve, so if I optimize over all possible E0s and E1, over all possible POVMs, then the minimum error probability is given exactly by this trace distance. Okay, and the quantity is exactly this quantity. It's one half minus one half, the trace distance between rho zero and rho one. Okay, so I won't prove it, this is one of the problems in the problem session. Okay, good. So that was for the trace distance, so let's now get to the, let's say the more general setting, where now I want to understand the fundamental trade-off between the two types of errors. Okay, so now I want to not just understand what is the minimum of the average of the two errors, I want to understand really the trade-off. Okay, so if I fix, let's say, the type one error to be at most epsilon, what can I say on the type two error? And so in order to understand this, it will be useful for this lecture to introduce a quantity that quantifies this, and it exactly quantifies this thing, is given a fixed type one error, what can I say on the type two error? Okay, so it's parameterized in some way, but to make it look like a divergence, but it's only saying that. Okay, so let's go over this definition. So I will call it, the hypothesis testing relative entropy. Okay, or also divergence. I will use divergence and relative entropy interchangeably. Okay, so it has some parameter epsilon. Okay, you should think of this parameter epsilon as exactly the type one error probability I aim at achieving. Okay, and it's defined by the following expression. Okay, so for fixed row and sigma, it depends on row and sigma, and it has some parameter epsilon, and I just maximize or minimize the type two error probability, but here the way I'll parameterize it is that I'll maximize minus log of the type two error probability. Okay, so yeah, so this condition corresponds to the fact that the type one error probability is at most epsilon. Okay, my POVM, remember, E0 and E1, it's given by just E and identity minus E, and yeah, E corresponds to E0. Okay, and so this exactly corresponds to the minimum type two error I can achieve. Okay, so a few remarks on this. So notice that this quantity is between zero and plus infinity. Okay, because this is between zero and one. Yeah, so if this is zero, then I just, this is counted as plus infinity. And as I previously said, multiple times even is that two to the minus this hypothesis testing divergence is the minimum type two error if I fix the type one error to at most epsilon. So let's look at some like border cases. Okay, so if I fix the type one error to at most, if I fix epsilon to one, then this is not putting any constraint and so of course I can get plus infinity in this definition. If I pick epsilon equal to zero, right, then I have to have here the trace of E times rho is equal to one. Right, so what I have to do for E, I have to pick E to be to be the whole support of rho. Right, at least the whole support of rho. And so this gives me this quantity. Right, so minus log of trace of the projector onto the support of rho times sigma. And yeah, so something maybe the support because this will come back again. So when I say the support of rho or the projector onto the support of rho I just mean the sum of the projector onto EI where EI is an eigenvector with a nonzero eigenvalue. Okay, so yeah, another setting of interest is when rho is equal to sigma. Okay, so this is supposed to be a distance measure, right? When rho is close to sigma we expect this quantity to be small. Okay, so for rho is equal to sigma it gives you exactly zero but it gives you something which is close to zero as epsilon goes to, as for epsilon small. Okay, so, okay, another remark is that if rho and sigma have orthogonal supports then it's very easy to distinguish them. You just take the projector onto the support of rho, let's say, for E, then you get plus infinity. Okay, so this gives you an idea of this quantity. So let's make a few further remarks about this. So yeah, in general, so I wrote the way I define this quantity is via an optimization problem, right? And this, in general, this optimization problem does not have a closed form expression, right? So you have to optimize over all these different E's but you may notice that this is a nice problem in the sense that it's a convex optimization problem. And more precisely, it's even a semi-definite program. Right, so the constraints are some operators being positive semi-definite. Okay, so this can be computed efficiently in general as a function of the dimension of the systems that are present. Okay, good. So to get maybe a little bit more intuition, maybe it's useful to consider the classical case, right? So if I take rho and sigma to be diagonal in the same basis, okay, I call it x here, okay? So and p and q are the corresponding probability vectors. Then what would be a natural test, right? So I want to test whether I have p or q, okay? What will be in this case a natural test? We didn't talk about tests so far, but so maybe this is a good point to talk about how I would test whether I get, I have distribution p or distribution q. Okay, so I claim here that the natural test, and this is used a lot, is to, if you get the sample x, okay, because what is a test doing, right? It's getting a sample x and should output, do I have p or q? Okay, so a natural test, and this is what actually we'll use in a minute, is that we compute this ratio, p of x over q of x, okay? And so the natural thing to say is that if this ratio is bigger than one or bigger than some threshold value, okay, which we'll determine depending on the problem, I will output p, and if it's below some threshold, this same threshold value, I'll output q. Okay, so I hope this gives a general idea of this quantity. So, okay, so let's prove, so remember I told you that an important property of a given distance measure is the data processing inequality, right? So saying, for example, that if I forget some part of my system, then I shouldn't lose in terms of distinguishability. Okay, and so here for this quantity, it does satisfy the data processing inequality, and it's relatively simple to show. So, yeah, the statement here is that this satisfies the data processing inequality, and again, I remind you that this just says that for any quantum channel, if I apply this channel to rho and to sigma, then the distance measure can only decrease. I said this several times, so I'll repeat it. Okay, so how do we prove this? This is very simple. So I will take an optimal E, remember our divergence is defined in terms of an optimization problem. So if I define E to be the problem corresponding to E of rho and E of sigma, right? Then by construction, so the hypothesis testing entropy between E of rho and E of sigma is minus log trace of E times E of sigma, right? And I have, of course, as a constraint, it should satisfy the trace of E times the channel E applied to rho as at least one minus epsilon. I just use the definition, okay? And so, okay, so now what I will do is I will construct an E prime, if you want, for the optimization problem corresponding to rho and sigma. Okay, so how will I do that? There is a very natural choice, and so this uses this idea of the adjoint of a quantum channel E, which you saw in the problem session yesterday. For those who didn't reach this, so I can see the set of linear operator as a Hilbert space itself, so I can define an inner product, and so I can use exactly the same definition, remember of an adjoint that I defined for a regular operator. I can also do it for a super operator, okay? And so, so by using this property of the adjoint of a map E, I can take this, the trace of E times the map E applies to sigma, remember this is the quantity which appears here, okay? Sigma, so okay, what's the natural thing to do is that I just put this operator, apply its adjoint on the operator E, okay? So by using also the fact that E is her mission, you get that the trace of E times the channel E applied to sigma is equal to the trace, the adjoint of E applied to the operator E times the operator sigma to rho as well, right? And so if you remember from the problem session from yesterday, so if E is completely positive, then the adjoint is also completely positive, and this is if and only if, and for the trace preserving property, this is equivalent to the unital property if I go to the adjoint, okay? And remember, unital just means maps identity to identity, okay? So what do I have to show? So now I have a natural candidate for the optimal or for one feasible choice for the operator E for the problem between rho and sigma, it will be just the adjoint E star applied to the operator E. I should have chosen a different name for her, but okay. So, yeah, so I choose this E star applied to E and it satisfies the property that it's between zero and identity. Remember, this was the constraint in my problem. So if I go back, I should now verify that I satisfy these constraints. I satisfy this and I satisfy this, okay? That E star of E satisfies this. And so I just check these two conditions, right? So E star of E is positive, right? Because my map is completely positive and it's at most E star of identity, which is a self-identity because E star is unital, okay? And so, yeah, also the condition of being bigger than when I run this epsilon follows immediately from here. And so I get that E star of E is a feasible solution for the program for the program that defines this quantity, okay? And so I have that this quantity is at least the value which is given by this fixed feasible solution that I gave you, okay? Okay, so this proves the data processing for this hypothesis-testing relative entropy. Okay, I hope this was clear. I tried to go relatively slow so that you follow this calculation. Okay, good. So now I will look at a special case of interest. So here I just gave you a quantity which tells me what is the type two with respect to the type one. We just saw a way of writing it and a few of its properties. So now we will look at a specific setting which comes of interest and whether we can say more about this quantity, whether we can say more about what is the tradeoff between type one and type two. Okay, and so the specific setting of interest that I will look at is imagine that I have multiple copies of rho and sigma. Okay, so imagine that I have a box, right? Which when I press the button, it either gives me rho or it gives me sigma. Okay, and I want to know if I can use this box several times, right? So I can use it n times and think of n as being very large. Okay, and I want to know whether I'm in the setting where the box outputs rho or it outputs sigma. Okay? No, no, no, it's not half-half. No, no, here. So from the start when I defined this hypothesis testing relative entropy, I look really at the two quantities. I don't put any prior on rho zero or rho one on the two hypothesis. I just say that I have rho zero or I have here I call it rho and sigma and for any strategy, I have a type one error and a type two error. Okay, so I have these two numbers and of course there's a tension between these two numbers. I cannot get both of them to be zero in general. Okay? And my, okay, so again, my trace distance was saying that if I want to minimize half of this number plus half of this number, okay, what is the minimum that I can achieve? Now I will ask a different question and this is what we'll do in Stein's Lemma. I assume the first number is at most epsilon. What can I say on the second number, on the second error probability? Okay, so as I told you, so this is a setting which is not symmetric between the two types of errors, right? So, and this does happen a lot, because the two types of errors, like the false positives and the false negatives, they're not always symmetric, sometimes on this to be much smaller. And so, yeah, so here I will fix one of the two errors to be a constant epsilon, constant at most epsilon, and I look at the other error and I will look more specifically how it behaves as a function of n, right? So remember that I look at the setting where I have rho tensor n and sigma tensor n and I fix one of the two errors to be at most epsilon, which is just a fixed parameter, which I see as a constant, and I look at the other one, naturally it will go down with n, right? So as I have more copies, as I can use my box more times, I can distinguish more easily, right, between the two situations. And so the other kind of error probability will go down as n goes to infinity and the question that we'll ask exactly is how quickly okay, as n goes to infinity. Okay, and so this is exactly the statement of quantum Stein's lemma, right? So it's... So again, I fix one... I fix my epsilon, I fix constant and I take rho and sigma, okay, and I look at rho tensor n and sigma tensor n, right? And so remember this quantity is... So dh of epsilon, I take it as the log of the error probability, right? Or minus the log of the error probability, right? So here, what I'm doing is that... what we'll show is that the second error probability goes down to zero exponentially and I want to know exactly what is the rate of it going down exponentially, right? So again, recall, so the first type of error is that most epsilon, the second type of error, type 2 error goes to zero exponentially in n and I want to understand what is the exponent, exactly, okay? And what Stein's lemma is saying is it exactly characterizes what is this exponent, okay? And it's given exactly in terms of the quantum relative entropy, which I'll define in a minute, okay? So again, the 1 over n is to say that I look at the constant in front of the... like the error goes down as 2 to the minus some constant times n and so I look at this constant and it will be given by this quantum relative entropy. Okay, yes. The rate of convergence so... Yeah, of course, as epsilon goes to... as epsilon goes to zero, the convergence is slower but we actually understand quite well the convergence in the sense that so we even understand like the second order term, right? So we know that here it's completely independent of epsilon. The first order term is completely independent of epsilon and then if you want to be more precise here and you just don't look at the limit but say 1 over n, this is equal plus 1 over square root n times some quantity which depends on epsilon here. Okay, so in some sense the first order term does not depend on epsilon and this is what appears here because I said it works for any epsilon but the second order term depends on epsilon so the dependence on the dimension I mean so the second order term has so I mean but it depends on actually the states themselves here so so but this quantity can be infinite, right? So or so yeah the okay so here usually in these expressions the dimension does not appear explicitly, right? So we look really at the fixed row and sigma and we want a quantity which really depends on just row and sigma and so in the first order term what appears is this quantum relative entropy and in the second order term there is a term which depends on epsilon and a term which depends on some variance between row and sigma but the dimension does not explicitly appear, okay? Any further questions about the statement? Because okay so what we'll do in the remainder of the lecture is to try to prove the statement okay so I'll go a little bit quickly probably over the proof but yes so yeah so the classical one yeah good question so the classical one is exactly the case where row and sigma are probability distributions or if you want row and sigma commute okay so they are they can be diagonalized in the same basis and so yeah you take to it's exactly the same question right so you have the box and now it outputs a sample okay from P or from Q and you want to distinguish between them and yeah further questions okay good so okay so now I have to tell you what the quantum relative entropy is I didn't say that yet so okay so what is the quantum relative entropy so I take two states row and sigma actually it's useful to have sigma not necessarily be a state operator okay the definition works equally well for the case row it's quite kind of important for it to be normalized otherwise there's many definitions you can come up with if row is not normalized but for sigma is not normalized this is fine okay so I take a finite dimensional Hilbert space and yeah the quantum relative entropy is defined as follows so trace of row log row minus log sigma okay and so if you see here if you look a little bit at this quantity you will see that if the support of row is included in the support of sigma then this is well defined and there's no problem but if it's not the case then you see there is a problem right between this log sigma and the row and so that's why we set it to plus infinity okay so just yeah recall that what do I mean by the log of an operator several times but let me repeat it in case so I take the log of an operator is just I take the log of its eigenvalues and I neglect or the zero eigenvalues I completely remove them okay good so the yeah so the classical case it's useful to look at the classical case so yeah when row and sigma have this form so they written as the sum of px and sum of qx and projector on x then the relative entropy is just the sum of px log of px over qx and this is as you might expect it's very well studied in classical say probability theory and statistics so it's called relative entropy or sometimes it's called cool back library divergence and you can see this quantum relative entropy as a quantum generalization of the cool back library divergence okay yeah I can say here that this is one one generalization one non commutative generalization of the relative entropy but there are actually others but this one is the most most famous or most well the most used one reason I would say it's the most used is exactly this shine lemma that I that I will show right so because the shine lemma does not have any divergence it has this specific one okay then this is why or one of the reasons let's say it's the most used even okay so the paper that proves this it's called the proper the proper generalization of the of the relative entropy or something like this it's a paper by pets and here I believe okay okay good so yeah before getting into the proof let's see a few properties of this quantum relative entropy so first so if I take two states right so if row and sigma are both states then the relative entropy is is non negative it's almost non negative so it's a valid measure and we have equality if and only if row is equal to sigma particularly if row is equal to sigma then it gives zero and zero only in this case okay and so and again as I told you what is the most important inequality is what the data processing and this also holds for this for this for this relative entropy and so if I apply a channel on the two sides then the quantity can only decrease okay so as I said so I won't prove this even though we'll use it actually we'll use this in the proof of the Stein's lemma I will only prove this tomorrow this will actually be the objective of tomorrow's lecture will be to prove this inequality okay so this quantum divergence plays an important role also because you can derive from properties of this quantum relative entropy properties of the von Neumann entropy which you might be more familiar with so yeah this you can see this relative entropy as a sort of parent quantity from which you can you can get the von Neumann entropy okay so how do you do that so for a state for bipartite state row AB okay you define the von Neumann entropy this is the way I define it you can check that it corresponds to the usual definition you define it as minus you put a minus sign between the relative entropy of row A and identity so you might be wondering why this minus sign what this minus sign is doing we just saw that it's positive but remember it's positive only for if the states are normalized if the operators are normalized so it's an identity so in general this quantity actually will be negative so when I take minus sign it gives me the the usual von Neumann entropy okay so I can also define the conditional entropy as minus the relative entropy between identity on A tensor the marginal on B and I can also define the mutual information which is the relative entropy between row AB and the product of the marginals okay I mean in the I won't discuss this more in this lecture but in the tutorial you'll play a bit with these entropies this will be useful for the following lectures okay good so I hope yes of these no I there are but this is it will come I mean so I mean we'll see one maybe not very direct one but in the fourth lecture on Thursday but now I won't discuss okay so the mutual information at least it has a natural one in terms of right so in terms of the Stein's lemma if you say I want to distinguish a state row AB and the product of its marginals and you can ask about the type 2 error for a fixed type 1 error then this gives an operational interpretation if you want but here these also have other interpretations in terms of compression and things like this like as in the classical setting but we won't cover this in this course okay good so I hope this is clear so now what I will try to do is to try to do a proof of Stein's lemma so now I will go a little bit more quickly maybe if you don't follow the all the details of the calculations it doesn't matter so much but maybe if you get the overall idea of what the structure of the proof is and then you can look back at the notes if you want to see the details okay so again recall the statement the statement is as follows so it can be written in the following way so it's a limit okay good so I have to show the two inequalities right so I have to show this inequality which I'll call an achievability result which is to say that I can make the type 2 error sufficiently small I can make the type 2 error of order 2 to the minus the relative entropy between row and sigma and then I so this is what I will start with okay and this is what we call achievability is that I have to give you a strategy okay and so what we'll do is we'll start with the case the classical case where row and sigma commute okay so this is then a statement about just probabilities so okay so yeah let's take the setting where I have row and sigma which have eigenvectors eigenvalues p and q respectively and I take now tensor copies so I take row tensor n and sigma tensor n okay then this naturally corresponds to the product distribution where I take yeah so the eigenvalues are just the product of the corresponding eigenvalues and then so then I will have to define a test right so what is my test for this problem okay so I can see it classically so it's useful here to think classically it's maybe simpler to think of a test in this case I see n samples x1 to xn again either they come all from p or all from q okay so I will compute as I discussed before I will compute this likelihood ratio so I will compute r and compute what is the probability that I have of seeing this particular sample I saw right so what is the probability it's p of x1 times p of x2 etc times p of xn okay and I divide it by what would be the probability if the distribution was q okay and I just multiply them okay and now I will have to just set the parameters so okay so I will have to put my threshold right so remember I told you that we will put a threshold if r is bigger than something I will say p if r smaller than something I will say q okay so here it turns out that the right threshold to put so that we have the right errors is this okay so I will just say 1 over n times the log of r is bigger than this quantity okay and delta is a small parameter that will let go to 0 at the end of the proof okay so if this r is bigger than the threshold I say the samples are from p otherwise I say that samples are from q so if you prefer to okay so this is a strategy that is explained in classical terms if you prefer to use the quantum notation and the definition of the optimization problem that I gave you for this d h epsilon the corresponding e has the following form right it's the sum over all the x1 xn the project around to x1 xn of the particular x1 xn that have the right such that the ratio of the probabilities is bigger than this threshold that I chose okay good okay so of course yeah I wrote here that d depends on n I didn't put it explicitly but of course it depends on n okay so now I have to analyze this test and to say okay so what does it mean to analyze this test is I have so given a test it defines type 1 air probability and it defines type 2 air probability so I just have to compute these two numbers now and to check that it satisfy that the type 1 air probability is more than epsilon and the type 2 air probability is smaller than 2 to the minus n times the relative entropy okay and so yeah so I have to compute these two things so let's compute it so it's simpler here to compute the probability of success rather than the probability of error okay so if the samples are from p so again I mean hypothesis 0 then I just it's useful here to use the just the probability notation okay if the threshold is bigger than d and now I just use the fact that the p and the q are is a product distribution so and I write this as the sum of the logs of the probabilities okay and now you see that now these random variables are independent and identically distributed and they have an expectation right I mean remember that I'm in the case where the samples are from p here they have an expectation which is given by the relative entropy between p and q okay and so again by the now by the law of large numbers so this is a sum of independent random variables which have an expectation which is given by the relative entropy the probability that it's bigger than the relative entropy minus a little bit goes to 1 as n goes to infinity okay so yeah so even the type this or here I should say the type 1 success probability goes to 1 as n goes to infinity so in particular for large enough n this will be bigger than 1 minus epsilon okay good okay so this was the first type of type 1 air probability now I have to analyze the second type of air probability and now this I have to show that it it goes down very quickly to 0 as a function of n okay so right so yeah I have to analyze this quantity and so I just write it like this so it's the sum over x1 xn of the probability that these specific x1 xn satisfy this condition okay so I wrote it in a slightly different notation here where I put the condition here okay and if you just rewrite this condition it just tells you that the product of the q's is at most sum number which is directly related to the threshold which is exactly the threshold that I fixed or 1 over the threshold I fixed times the product of the p's okay so and then what I do is I replace this product of q's with this bound that I have okay and so this bound that I have stays and then the sum over the p of p of x's is at most 1 here even because I forgot the constraint that I had okay and so I have that this type 2 error probability is bounded by 2 to the minus n times the relative entropy okay so here I just manipulation I just take minus log of that it gives at least n times the relative entropy and so I get basically the desired statement okay so I have for large enough n 1 over n the hypothesis testing relative entropy is at least the relative entropy between p and q okay and then I let delta go to 0 okay good so that was we did the classical case achievability okay classical achievability case okay so now to finish the achievability we have to I only give you the classical case okay so this is how do I go to the quantum case now okay so for that we'll use a technique which is quite useful sometimes I mean useful in quantum information theory it's this technique of called pinching okay and it's a general way of reducing the non-commuting case to the commuting case okay as we'll see it's I mean we get some losses but at least for this particular problem these losses we can handle them okay so what does pinching mean okay so what does pinching mean it means so pinching map is depending on some her mission over let's say positive operator here sigma and I just do a spectral decomposition of sigma okay so let's assume its eigenvalues are lambda okay so and pi lambda is just the projector onto the eigenspace corresponding to lambda okay and what I do here what this pinching map is so it's a quantum channel it's a bad quantum channel what it does is it takes an operator s and it maps it to the sum over lambda of p lambda s times p lambda of course here lambda is just in the spectrum of sigma I didn't rewrite it again but yeah so p lambda is this one okay so in some sense I'm so if I take these eigenspaces what I'm doing is I'm removing these two diagonal terms that are corresponding to this to these eigenspaces okay so for example if we imagine that pi lambdas are rank one projectors so okay they correspond to to some one dimensional projectors then this would mean what what p sigma of s is doing is that it's removing all the off diagonal terms okay but notice that and this is important they are not necessarily rank one they can be of much bigger rank and in this case I'm removing the off diagonals corresponding to so I have blocks I have diagonal blocks and I'm removing the off diagonal blocks okay good so why is this useful why is this map useful so the thing which is useful here is that notice that p sigma of s commutes with sigma okay this is easy to see you will discuss this in more detail in the problem session and so of course it's easy to construct an operator that commutes with sigma alone but we want another property that it retains some properties of the operator that I apply the map to right and so this is quantified by this operator inequality right it says that if I apply the pinching map relative to what was to the state or to the positive operator row then I cannot lose too much in the sense that this operator is at least some quantity times row and this quantity what does it depend on it depends on the number of distinct eigenvalues of sigma okay or the size of the spectrum of sigma yes yes I include it yes but yeah so just if an eigenvalue appears several times I don't include it multiple times this is the key fact yes yes I mean her mission so that you have to diagonalize but yeah yes okay good so this is what I'll use okay I see I'm a bit late on time so let me go quickly the important maybe is to see just the overview of this part so okay so I start with row and sigma which in general don't commute so what I'll do is I'll use the data processing inequality for dh epsilon and I'll apply pinching to the two sides okay so I apply pinching relative to sigma tensor n to the two sides okay so if I of course if I pinch a state relative if I pinch relative to a state and apply to the same state it gives me the same state right so it doesn't do anything okay but of course on row it does do something okay but now the advantage is that here these two states they commute okay so now I'm in a classical case right so I can pick an eigen basis where both are diagonal and so now everything is diagonal and I can apply the the classical theorem the classical result that we saw before okay so so yeah so this is what I wrote here so this because these two commute then this corresponds to the hypothesis testing between two probability vectors p and q okay so now if I use the classical result at the classical achievability result that we have just seen now if I take the limit as m m is different from n okay so I have two of these parameters so I take the limit as m goes to infinity and I take multiple copies of p I will get that this goes to the classical relative entropy between p and q okay good so now how do I but okay now this is not exactly what I wanted right I don't I don't care about this this row after which I apply pinching I want to get back row itself right so okay so what do I do what we will do is this so I will take I will apply the pinching to n copies of of row and then I take that m times okay and then I use this result to get or okay here I just I just rewrote the statement I just replace this quantity by p and this quantity by q and now I use this result here to say that this is at least I mean here it's just this inequality at least this is what we're interested in this goes to like the one over n stays and the relative entropy comes from here this inequality okay and now I can again go back from the the probability vector to the to the corresponding quantum states commuting quantum states okay and then I want to get back now I have a relative entropy so it's good I don't have a hypothesis testing right of entropy but but the usual relative entropy but what I want to do is I want to get rid of this pinching operator and so now I here is where I will use this pinching property the second pinching property is that if I apply this pinching to rho tensor n then this is at least one over the size of the spectrum of sigma tensor n times rho tensor n okay and now the important thing to observe is that what is what can I say on the size of the spectrum of sigma tensor n right so notice that that the eigenvalues of sigma tensor n are of the form I take a product of eigenvalues of sigma right but how many different eigenvalues are there right so I can count for each eigenvalue how many times it appears right and so how many times it appears is between some number between 0 and n right so I have n plus 1 possibilities and this is for each one of the eigenvalues okay so that's why the number of different eigenvalues is at most n plus 1 it's because it's 0, 1, 2 up to n and I have at most d of them d where d is the underlying dimension so it's to the power d or d minus 1 because the last one you can infer it from the others okay good so that's good because this quantity now is that I have one over some polynomial in n times rho tensor n and this is what is the crucial here is that it's polynomial in n not exponential okay and so now so if I look back at now my relative entropy between the pinched rho tensor n and the sigma tensor n okay I can do some basic manipulations with this pinching and the operator monotonicity of the log right to use now okay so yeah let's go directly here okay so I have log rho tensor n times log of p sigma rho tensor n and I want to replace this by rho tensor n but remember yesterday in the problem session you saw that log is operator monotone right so the I have so this this operator inequality if I apply the log to it it still holds right and so this is where I can replace this p sigma of rho with just rho tensor n but I will lose some factor this I mean here it should be divided by the spectrum of sigma tensor n but of course yeah log of the product is the sum of the logs and so I can get this minus log outside okay so globally what did we do so if we put this together we have 1 over n the relative entropy after apply the pinching is at least 1 over n the relative entropy before apply the pinching minus sum lost term here but this lost term is very reasonable because this thing is polynomial in n and I take the log and then I divide by 1 over n so this thing goes to 0 as n goes to infinity okay so notice that one thing I used here that I didn't that you might wonder about is here I use the fact that this quantity is additive so the the quantum relative entropy on tensor powers it's the sum of the relative entropy it's relatively easy to check so I included the idea here of how to show this okay so okay so I hope this was clear so I have two minutes remaining so maybe it's not worth doing the the converse or maybe I do it very quickly the converse let's say in two minutes okay so for the converse actually we'll prove something slightly weaker we'll not prove the full result I only prove that this was sometimes it's called a weak converse I'll just prove that if I take the limit as epsilon goes to 0 then this gives me the relative entropy but it is true that it works for any epsilon this is what is called a strong converse but here it's I mean we need to introduce some new tools in order to do that so I thought it's simpler to just focus on this okay and so the main idea if you want to prove that is just applying data processing so it's a very simple proof the main thing that you do is just apply data processing in equality to the right map so what is this map so that we'll apply the data processing in equality 2 it's the following map so right okay so yeah maybe I should so what do I mean by converse right so I want to prove the opposite inequality right so and what is a converse it's a sort of no-go result it's saying that if I take any valid strategy then the type 2 error cannot be too small it cannot be too small so what I have to do is I take an arbitrary strategy E okay which has to satisfy this this is by construction by definition of what the strategy is okay and I will apply now the data processing inequality for a well-chosen quantum channel which depends of course on the strategy and so how what is this channel doing is it's mapping an operator T into the following so it will of course depend on the strategy that I look at so I will just output 0 with probability trace of E times T and I will output 1 with like 1 minus this probability okay so you can see that if I apply E to this channel then I will observe easily all the types of error probability and the success probabilities right and okay so and what will we do now is we'll say that on one side I'll just compute this quantity right the relative entropy between rho tensor n and sigma tensor n okay so it's just by additivity is equal to n times the relative entropy between rho and sigma and on the other hand I can find a lower bound on this quantity by using data processing right so I apply this previous map E and what this allows me to do is it allows me to give a lower bound on this relative entropy as a function of the type 1 and type 2 error probabilities right because remember this is the case and so I just do that and yeah by just then doing some simple computations we get the desired result okay let me just finish with a very quick remark on what this quantity is so the reason I wanted to define a specific quantity for this task and call it a divergence this d eps on h because this is done a lot in information theory and such quantities are sometimes called one-shot entropies or one-shot relative entropies and they're usually measures that have an operational interpretation for any states for any choice of states so for example here it has an operational interpretation directly in terms of this hypothesis testing problem okay and there are many others that are particularly relevant in settings where you want measures that work for any states not necessarily for specific classes of states but a particular one very important one is sometimes called the mean entropy or the smooth mean entropy which is particularly relevant in cryptography okay and these these entropy measures that you might be more used to like the von Neumann entropy and the corresponding relative entropy they usually have an operational interpretation when we're in a setting when we're in an IID setting like exactly what we saw in this lecture remember we have n and sigma tensor n and we want to characterize the type one type two error probabilities in this case okay in this IID setting or in the average setting okay and I'll stop here for today sorry for being a bit long


Hi, everyone, Joe for Jaspy's case breaks.com coming at you with four boxes of 2021 hip parade autograph triple-play baseball Series number five. This is random team break number one big. Thanks to this group right here for getting in on the action There are the teams right here. Let's roll it. Let's randomize it five and a one six times for names and teams one two three four five And oh one six and final time We got Adam St. Anish down to Adam Covernment five and a one six times for the teams one two three four five And oh one six and final time. We got the Cubs down to the Phillies right after six times great Now trades are allowed but remember to trade at your own risk. We discussed this We also discussed something that we started before the video We discussed That we're gonna be using our college rules Right, especially for the autographed baseballs or a non no checklist rules in this case for the autographed baseballs Unless there's like a year or a team clearly stamped on the baseball, right? Not not the inscription But like on the actual official baseball itself It'll go to the team that they're currently on if they're active All right, this is especially for the baseballs the autographed baseballs or the team they played for the longest if they're retired All right Adam with the Cubs PJ with the Rangers Mike with the Rockies Alan with the Red Sox Victor with the Angels Michael with the White Sox TJ with the Reds Adam K with the Mets Alan with the Marlins PJ with the Stroze Travis with the Mariners Matthew with the Rays PJ with the Padres and the Royals Eastwood with the Blue Jays Cody You got my Dodgers PJ with the Yankees Michael P with the Indians Oliver with the A's Tyler with the Giants Cody with the Brew Crew TJ with The O's Michael P last but Mojo twins Tyler with the Cardinals PJ with the Braves Jake with the Nationals David The big boys 007 with the Pirates Matt hedges with the Tigers Alan with the Diamondbacks Adam with the Philadelphia Phillies All right, let's sort by column B again trade at your own risk But feel free to trade we're gonna pause the video when we come back We're gonna see if there's any trades and then the break stick around be right back All right, welcome back everybody as you can see on your screen. There were two deals done So Adam is out of the Cubs spot and into the Red Sox spot put a tee there for trade trading with Alan Who now has the Cubs? There was another trade Michael P is out of the White Sox spot and into the Orioles spot We'll put a lowercase tee there and TJ is out of the Orioles spot and now into the White Sox spot TJ normally wouldn't trade the Orioles. He says but for the White Sox We'll make that exception. All right. Now the trade window is closed. Let's print. Let's rip Let's see what happens in this four box break. So the four boxes are right back here all hand numbered Triple play Series five We're doing so that I think this was the only series we had in the back, but yeah series five four box random team one. All right Here's the official printout hot off the presses Thanks very much for spending your Thursday with me ladies and gentlemen. Let's see what we have in here, right? Let's do the eight by ten photo first. Oh, they have stickers on the back that tells me who this is. Okay Let's play a quick round of who am I I think we can play it with that with the photos as well Make the photo a little bit more to more of a Easier one here. All right. Who am I? retired Left fielder first baseman third baseman and DH three time MLB all-star a career 442 home runs and 1210 runs that batted in in my 16 year career Who am I a righty? Ended up with a 236 batting average as well He would have fit right in in modern-day baseball played with the well I don't know what team he's on on this picture. It'll go to the team. He's on with the picture, but Giants play with the Giants Mets Padres Angels Yankees Cubs Mets eight And a two-time home run leader as well two-time NL home run leader righty from Pendleton, Oregon Some of you remember that Tommy La Sorda had some choice words about this guy Back in the day in an infamous press conference Michael P. Yep, TJ's right We do have repack products like hit parade. We've done them a lot our repack products, right a big hit express Mystery crates all that sort of stuff very very similar to this Who am I? The Tommy La Sorda not not gonna give it away for anybody He played from I played from 1971 to 1986 Only one person in the chats got it so far. No, not Kevin Mitchell Eric Jennings is the only person who got it. It's Dave Kingman Cubs edition so that'll go to Alan Murdoch who got the Cubs. He's drawing first blood in the trade There's all the Dave and Adams authentication right there. You can see Dave Kingman right there Nice Dave Kingman. Yeah played with the Cubs from 78 to 1980 this will be the next up is I don't know if some some of you may saw may have seen the name think this All right. I'm a baseball player This team signed me at 16 years old as an international free agent in July 2017 for signing bonus of 1.75 million dollars Made my professional debut with the Dominican summer league of this team in 2018 primarily a right fielder Supposed to be pretty good Let's see in 2019 went to class A in West Virginia Promoted to high A To Modesto if you know your minor league teams, you know what team this will be In high A became one of the only three 18 year olds in all of minor league baseball to play high a ball in 2019 I think I kind of showed you the name right there I was in the futures game this year too. Yeah, Adams got it. It's Julio Rodriguez There's the Beckett certificate in there nice clean baseball. There's the Beckett sticker there And I don't know which way Think Julio Rodriguez's autograph goes that way Adam. You saw the name. Yeah, I think his autograph good does go this way That's kind of the J right there Nice Julio Rodriguez. That's for the Mariners and that'll be for Travis Yeah, he's supposed to be He's supposed to be real good. All right the autograph Jersey See if I could spy who this is. Oh Okay, all right, who am I another retired baseball player a Right-handed Righty third baseman lifetime 279 average 223 home runs 952 runs batted in played for three teams my entire career Ended up with four all-star appearances and two world series Championships including a world series MVP a one-time gold glover and a silver slugger as well Who am I I do a little work on the on on the MLB network as well? All right, there you go Michael use that as a clue Not Jim Laird. It's not Longoria not Beltran. I don't think Longoria is a two-time world series champion Beltran is not a two-time world series champion Getting closer with Kevin Mellar Eric Jennings got it yet Mike Lowell. I see a TJ got it too. There's Mike Lowell silver ink autograph right on his number There's the David Adams authenticity card. There's Lowell back here. It looks like a an official Jersey as well So it must have all the officially licensed stuff It's worth taking a look some additional Tags right there as well. Nice. It's got the majestic stuff right here Nice Boston on the front nice majestic tags on the sleeves There you go. Oh nice. Yeah for TJ the MLB network clue gave it away. All right, very cool-looking Jersey. Let's Kind of fold this up nicely Let's keep it in the box here and that's gonna go to Adam with the Red Sox So now that that trade is evened out So both the trader and the tradie have gotten Something and Adam's a fun fact. He went to his retirement game. He's a big Mike Lowell fan. Yeah, wait, there's more We've got three more boxes to go. Oh, yeah, it's even Alan. You have a hit He has a hit.  But once I just gets absolutely nothing. All right 8x10 first. All right 8x10 active baseball player Who am I active baseball player? I've a bats left throws right first baseman from Canada Lifetime as of now 302 batting average. I have 1,962 hits 306 home runs and a thousand five runs batted in play for one team my entire career Was a six-time all-star An NL MVP An NL Hank Aaron award winner and a gold Glover Who am I yep Eric Jennings and Oliver's got it. It's Joey Votto and that's for TJ There he is you got the PSA right here. There's his autograph doing what he does at first base Nice and Adam was saying hey, let's get TJ something that is for TJ Adam some sort of sorcery. There's some wizardry There's all the PSA stuff right there, too. I like Joey Votto the baseball I want to reveal the name like I did last time. I think I I'm a retired baseball player. I think I've played For multiple teams my entire but I played the most for that team. Okay. All right. Who am I? I've got a lifetime retired baseball player 83 years old. I'm still with us today righty What position did I play first baseman and a member of the Hall of Fame voted in by the Veterans Committee in 1999 Played for one two three four five six teams in my entire career an 11-time all-star a world series champion an NL MVP an NL rookie of the year an NL home run leader a two-time NL RBI leader Lifetime 297 batting average 2351 hits 379 total home runs 1365 runs batted in who am I and Eric Jennings is really good. Are you looking these up Eric Jennings? Am I making this too easy for you? All right It is indeed Orlando Sapella who played for the Giants the longest played for a lot of teams Atlanta for four seasons St. Louis for a few seasons, but played with the Giants mostly nine seasons with the San Francisco Baseball Giants Got the Dave and Adams card in here. It's a Dave and Adams sticker right here And there is he's got a Hall of Fame inscription and Zodagrath All right, so that's for Tyler and the San Francisco Giants and the Jersey That edition of this player it is All right, who am I? Let's see retired catcher 73 years old a righty lifetime 269 batting average 376 home runs 356 hits and 1330 ours be I Played for two teams my entire career an 11-time all-star AL rookie of the year Gold glove award winner three-time silver slugger the numbers retired by both teams that I played for and a Hall of Famer in 2000 with 79.6 percent of the vote in my second ballot Yep, PJ and Oliver. I think got it pretty quickly It's Carlton Fisk and it's White Sox edition of Carlton Fisk There you go Thank you number 27 with the red socks and then with the white socks flipped it 72 Got a Hall of Fame inscription down here as well With the fanatic sticker and everything. Yeah, TJ's gonna like this product even more He's got a couple hits right here and TJ. This is another nice. This is a Now they're nice officially licensed Jersey got all the officially licensed stuff right here as well And of course Carlton Fisk autograph Nice And I'll I'll refold that Once I'm done with this break So that goes to TJ who got the red socks or the white socks. I'm sorry in a trade right two boxes to go The PJ's saying his mom ran into Carlton Fisk at a pharmacy in Massachusetts back in the late 70s Nearly swooned over him seeing him live And it's talked about his thighs for decades Let's catch her thighs Next box See who this is I like this guy All right um 43 year old Inactive he's not active anymore at 43, but I'm 43 now Lefty batted left through left lifetime 295 Average 2,217 hits 18 home only only ended 18 home runs in my career 517 runs batted in but 614 stolen bases Played for a lot of teams one two three four five six seven teams in my entire or six teams in my entire career Two teams twice A world series champion and a three times stolen base leader Sean green had a little more than 18 home runs all over And I don't think he was ever world series champion I was a world series champion in the early 2000s I was named my team's most valuable player after that that uh that world series Received the I've got a lot of accomplishments. Who's working on my wikipedia page Like one pierce kids or something like that received the james cool pop of bell legacy award From the negro leagues baseball museum in the early 2000s led the nl Installed in bases with 46 and 65 one year The only player in baseball to play every inning of all my team's games in one year and only the third player to do it Since 1971 That's crazy um In 2006 I robbed berry bonds Of a career 400 714th home run, which would have tied berry bonds with bay roof for the second on the all-time list Catching the ball right before a top defense I was the first uh player for this club To record 200 hits in a season Wow, he stole his 100th base with my dodgers becoming only one of the four players mlb history to steal At least 100 bases with three different teams stole 100 with the rockies 167 while I was with this team Tommy harper brett bowler and otis nixon are the only others who have accomplished this feat. Yeah, two lz's got it one pierre This guy was he was on the dodgers for a little bit. Yeah, and it's marlin's edition. There he is um A pretty fun player to watch whenever he gets on base So he was on the dodgers for a little bit so I I enjoyed watching one pierre play Oh, did I say one? I said his name I was just reading and I was just like this guy There's a whole bullet pointed accomplishment section So I was a little taken aback there All right, allen murdoch has the fish Yeah, he had Yeah, he had no arm though He would he he threw like like Peyton Manning in his last last year of his career But fun to watch on the base paths. All right the autograph baseball Who am I an active baseball player lifetime 288 60 home runs 257 runs batted in played with uh Played with the mariners started my career with the mariners But currently on this teams which is where the baseball is going to go because I'm an active player a one time all star um And I am the nephew of wilson valdez and married to the cousin of lad guerrero jr He's on my fantasy team and just can't Stay healthy this season, which is annoying Because if he was healthy Uh, he'd be pretty good Like he was in 2019 when he hit 329 hit 32 home runs and had 92 rbi's Uh, a handful of almost a dozen stolen bases and finished fourth in mvp voting That was also my all star season Who am I If any of you care, I uh I traded him with a couple other players to get sandy alcantara Yep kettle marté erie jennings Getting it.  That's pretty cool It's like the autograph has smeared a little bit, but still really nice Eric what what gave it away? Eric's really good at this game And that's alan murdoch again with the arizona diamondbacks The big season and mariners nice Yeah, I what happened in that On november 23rd 2016 the mariners traded marté and taiwan walker to the diamondbacks for gene suger and mitch hannagher and zack hurtus taiwan walker right now having a good season All right, who are you? Oh, is that a number? I guess it was All right, who am I? I'm a retired switch hitting shortstop currently 54 years old lifetime 272 uh batting average 2,877 hits 80 home runs and 951 runs batted in played for one two three four five six teams my entire career And uh after I retired I did a little coaching with the angels and the tigers three time all star 11 time gold glove shortstop And the current man. I think i'm the current manager of the doros de tijuana in the mexican league Apparently I had a feud with jose mesa Yeah, they had a he had a feud with jose mesa After is after the publication of his autobiography That eric jennings again Omar viskell So there things autograph goes this way There's the authentication sticker right there. This does not look this looks like a custom jersey. So i'm gonna leave it folded in here Yeah custom jersey But you got the diskel right here because number 13 And once again his autograph Cleveland this is for you michael p The last one indians pj. Those are the indians colors It wouldn't be mariners color. I don't think there's any dispute, right? Not mariners not giants not rangers not white socks not bridges All right, this is box 34 out of 100 our fourth and final box 8 by 10 first All right, who am I an active baseball player 37 years old or righty? um My mlb stats Uh, I a shortstop second baseman third baseman mostly first base lifetime uh mlb stats 291 81 home runs 354 runs batted in has played for only one team my entire career Won a world series and a world baseball classic Representing cuba Played for one team my entire career since 2016 I think you're actually on the right track pj. I think they might have been teammates um, and I've got a uh I think I don't have a brother in In baseball as well Yeah, I've got a brother Uh in the mlb who plays for the blue jays Yep, oliver's got it See you jesse. You have a good one, man. All right, there's his Tiny autograph right here julie guriel Which goes to the astros that'll be pj. My brother lourdes guriel plays for uh lourdes plays for the blue jays Well, the only guriel that won a world series According to some pj. That's the one all right Next one Uh, oh, let's go baseball. All right. Who am I? I'm an active starting pitcher in the mlb Uh played for three teams my entire career I've posted a uh win-loss record of 134 to 95 losses lifetime 344 e r a in a lifetime 1677 strikeouts. I'm a righty. Let's see. I don't want to give this away to All right, how about I give you uh, I'll reach the pitching style section I throw a variety of pitches although my main ones are a four seam fastball a two seam fastball slider Which I supplement with a cutter a changeup and a curve ball I don't know why he has such a big pitching style section Um, I only throw my change to lefties rarely use my curve ball likes to use my slider with two strikes I have a distinctive wind-up which on some pitches begin to by spinning back towards second base So my back faces the batter and I've been compared to that of louise tiant and hideo nomo My unorthodox delivery Where I would uh turn towards second base then shimmy before the ball was thrown Led tigers manager brad osmos to question its legality during and after my shutout of the tigers in 2015 I argued umpire, uh, joe west during the game and to report after the game that my Sometimes stops in the wind-up There by thereby making my pitches illegal Through most of my career, I've used pitches from four different wind-ups The traditional wind-up the tiant the quick pitch and the rocking chair I have names for these which have contributed majorly to his big league's success I'm also able to trick batters by using a variety of wind-ups using an arsenal that has a depth of almost 20 different types of pitches Which has helped me make me one of the top 10 pitchers in baseball since 2010 Yeah, oliver and uh, eric jennings got it Johnny kweto. There you go. You got the the steiner authenticity card here And the baseball There you go. There's the autograph There's the sticker right here Well, he's the active player. So he'll go to the giants. Is he active? He's still active. He's still playing, isn't he? Yeah, he's still with the giants So that will go to tyler and The giants all right and the final hit coming up the jersey. All right, and the final jersey Is gonna be Is it oh, okay Another active baseball player 28 years old A switch hitting third baseman and I throw righty Lifetime batting average 279 145 home runs 488 runs batted in and 135 stolen bases played for one team my entire career And a three-time all-star three-time silver slugger And part of the 30 30 club As well Yeah, i'm pretty good a doubles leader Two-time extra base hits leader A run scored leader runs created winner Sacrifice hits leader 30 30 club two-time al player of the month three-time al player of the week bob feller man of the year All-star and three-time silver slugger at third base Not matt chapman. No I think both matt chapman chris bryan eric janes would love to have the speed that I have All right, who has this player by the way? Okay back play for the al From the dominican republic Yeah in 2018 hit 39 home runs and sold 34 bases to enter the 30 30 club And i'm under contract with This team through 2021 He might he might get a decent paycheck At the uh at the at the conclusion of that season nautland door What am I hitting this season? Yeah, i'm hitting pretty well this season. I'm hitting this season. I'm hitting 260 with 19 home runs 52 rbi's and eight stolen bases I have a two 2.9 war already at this stage of the season. Yep, oliver's got it. Jose Ramirez There's the autograph right there. There's the davin adam sticker. There's Ramirez is number 11 There's the r is there. I assure you this looks like a custom jersey Yeah, it's a jose Ramirez custom jersey. So it doesn't have all the officially licensed stuff So we'll just leave it nicely folded in the bag And cleave lad. This is for you. That's for michael and the tribe And there you have it ladies and gentlemen that was a really fun break. That was uh The uh hit parade autograph triple play baseball edition four boxes of some really fun stuff Thanks for watching. Thanks for breaking with us. I'm joe. I'll see you next time for the next hit parade break. Bye. Bye


I have a new spider. I found her the other day walking across the pavement outside my apartment. They live underground and they're only active during the night. I will need a camera that works at night. If I can build myself an infrared lamp and if I can remove the infrared filter from this camera, which will allow me to record what the spider does during the night. Like that. I believe that I now have a working infrared camera, which is nice. The next step is to light the spider's enclosures somehow and for that I will need an infrared floodlight. So let's build ourselves an infrared light. Now I was originally going to sacrifice this, which is a lighthouse unit for my Roomba robot vacuum cleaner. I have three of these and they've never been much use, so I thought that I could remove the infrared LEDs from this. However, I then had a better idea and instead visited my favorite junk shop, which sells things that other people have thrown away and got these. These are remote controls. In each one of these should contain an infrared LED, which I should be able to harvest. And there may also be other useful and or interesting things in here. So let's start ripping these apart and see what's inside. These will all be the same. There's a nasty rubber mat with carbon pads that push onto tracks in the PCB. A closing of contact and then a single chip. Wow, there's actually quite a lot of this. This must be an old one. There's quite a lot of circuitry. Normally there's a single chip, which is frequently a blob that does the work. What it does is when you press a button, it then outputs a series of pulses to the infrared LED here, which transmits to the CD player by the look of it. So let's just remove this LED. See if I can get it off intact. Excuse me, my soldering iron is acting up. The last time I used the soldering iron was to poke holes in the plastic lid of an ice cream container to produce air holes for an enclosure for crickets, which I feed to my spiders. The spider, by the way, did some more excavation and has eaten a cricket. Well, I assume it's eaten a cricket. The cricket is no longer in the enclosure. It is possible the cricket escaped. There we go. One infrared AD. What have we got here? Some diode transistor, a mysterious thing. The plastic box might be handy for a battery holder, but mostly that is now e-waste. What about this one? We've managed to get all our LEDs. We have a pretty decent hole, actually. So let me do some clear up and then we'll see if these actually do anything. I now have a nice handful of infrared LEDs and one vintage red LED. I took this one off in order to demonstrate that it is really useful for testing. If I plug it into my tester, let me just put this in a place where it doesn't get too much camera glare. Here, between pins 1 and 2, and press the button, you can see the light flash as the tester figures out that it's a diode. This also has the advantage of pinpointing the positive and negative side, which is not always terribly obvious. Let's try that with the infrared ones. I have the infrared camera here set up so hopefully we can see the flashes. There we go. Blinking nicely. And that shows the positive side is on pin 1. And I'm not sure the camera will focus adequately, but the LED does have a flat side, which in this one is pin 1. And we'll go through all the others just to make sure that it's that way around and that they all actually do infrared.  All of them in parallel so I can drive them simultaneously. So I believe I finished the bulk of the soldering. It did in fact change the layout to make it all fit. It now uses a single track for the cathode, the flat side of each LED down the middle, and then staggers them in the opposite direction. I did have some problems where one of the LEDs was the wrong way around, but my trusty component tester here detected it. So now if I stick these two pins in and push the button, you will notice that it shows up as a single diode, which is good. All right. So the next thing is to actually drive it properly. So I will just remove these two pins I soldered on for testing. I want to run this thing of five volts. And so since LEDs need series resistors, I did the calculations or at least found a website that did them for me. And it seems I need a 38 ohm resistor. So this is indeed a approximately 38 ohm resistor. So I will just double check that. I might have mislabeled them. There we go. 38 ohms. So let us solder that in. So push through the ground and solder. So with luck, I also need a wire for the other pin. Okay, we've got some wire. So let us tin the ends. I had to guess at what sort of current these LEDs like. And it seems that infrared LEDs typically run at what, 100 milliamps? So a 105 volts at 100 milliamps requires a, it doesn't actually work, requires a 38 ohm resistor. Now it's possible that since I have LEDs in parallel, now I think of it, the current may need to be higher. I've got 10 of them. Do I actually need a 3.8 ohm resistor? I should probably check up on that. I went and read up how LEDs handle being in parallel and how to calculate the resistor. The answer is you can't put LEDs in parallel. I wish I'd thought to do that before doing all this work. So what is going on is, I'll just turn the power supply off. LEDs are not ohmic devices. So the higher the current through them, let me rephrase. With an ohmic device, the current is proportional to the voltage being dropped across the device. A classic ohmic device is a resistor. A resistor will draw twice the current at twice the voltage. However, LEDs don't do this. LEDs either conduct with low resistance or they don't conduct with high resistance. And the actual current through them goes up very sharply as the voltage goes above a particular critical threshold known as the forward voltage. When you connect multiple LEDs in parallel, then they will all have very slightly different forward voltages. And you end up with the case where one LED will turn on, its resistance will drop sharply, and then all the current will go through that one LED and not any of the others because they haven't turned on. With the result that the LED goes pop. It's a situation known as thermal runaway. And as this thing is going to be running through the night, I wish to avoid it. This is safe because this resistor here is sized to run any one LED. So even if one turns on and all the others stay turned off, it will limit the current to the appropriate value. If I put the 3.8 ohm resistor in and one LED turned on, then that would be 10 times too much current and the LED would die. What you have to do instead is use one resistor per LED. So I'm going to need a completely different set of wiring. Fabulous. So now I'm going to have to desolder this, and of course desoldering is my favourite task. How hard can wiring up 10 LEDs be, I thought? This will be a nice simple job suitable for a short video, I thought. Bah. Well. So I'm going to have to rethink why I'm doing things. Each of these LEDs, and I hope they're still going to work after I've desoldered them, needs its own separate resistor. Luckily I have 10. Just the right number. So each one will need a resistor. Each one will need one pin connected to a common ground or power. And the other pin connected to the resistor. And the other end of the resistor will be connected to the other power rail. So I'm wondering whether it would be best to do this with floating resistors. Or I've got this prototyping board. Prototyping board is similar to VRaboard, except each pad is on its own and disconnected from the others, rather than strips like VRaboard. This means that if you want to connect pads together, you have to bridge them with solder. Okay. I'm going to take a break and then come back to this fresh, I think. So it's the next day. I'm reasonably refreshed and more coherent. I think it is time for a time lapse. So now let's hook this up to our good old tester. Now this won't work quite as well because there are resistors in line with the LEDs. So who knows what it will show up as. And let's get the infrared camera set up, which should be here. Press the button. No blinking lights. It's showing up in the tester as a resistor. There are 4.4 ohms, which given that I have 10, 39 ohm resistors in parallel is about right. So let's try it with the power supply. That's on. 5 volts. Nothing. Well, it's during an amp, which is what I expect. That's 100 milliamps each. But I'm not getting any results. So I've actually cut a couple of the bus bars, isolating all except this LED here and this LED is also still not lighting. So I think the only course of action now is to desolder it and see whether it still works. Okay, I did a bit more reading and I am somewhat confused. Let me show you something. I've taken one of the LEDs off and patched on this 220 ohm resistor. And when I power it up, it lights, as you can see in the camera footage. Now, I found a datasheet for the infrared LEDs, which say that they will max out at 50 milliamps, which is way lower than the 100 I have calculated these resistors for. So really, I should have exploded the LEDs on this board. But I haven't. They just didn't light at all. But this one is fine. When it connects it up to 5 volt supply with a high resistor with a lower current, it lights fine. So what is going on with this board? Well, the answer is I don't know. So I'm going to try putting this LED back onto the board. But I'm not going to bother with that for now. Instead, I'm going to mend these broken connections. So I just need to straighten these wires I cut out so they touch and then add a solder blob like so. So now we stick some... The end seems to fall on off my soldering iron. I think this soldering iron might not be too long for this world. I might need a new actual iron part of it. So we put lots of solder on here. This is turning into a mess. I think the plastic screw thread is not in great shape. That explains some problem I've been having with soldering because the tip may not be getting hot enough. There we go. That's better than this one. I think that should do it. So now all the other LEDs should be connected up. So I am going to crank the voltage way down as far as it will go. Hook it up and then wind it up a little and see what happens. You see, it's not lit. It's still not lit. That is just not lighting. So why was this lighting? So let me just double check which side the flat is. So this is ground. I'm sure that's not shorting out and this is... And you can see lots of infrared light coming out.  The focus is all dreadful because I haven't adjusted it. Adjusting the focus and the camera is pretty dreadful. So let's just... I need to hold this button down and do one of these. That's the wrong way. Anyway, you can see that light is coming out. So I am deeply confused about what's going on here. The only thing I can think of is that I've somehow managed to screw up the connections in such a way that the electricity is not actually passing through the LEDs at all. That way they are not being subjected to the high current and therefore aren't failing. I wonder if one of the LEDs are shorted through, but the LED itself, once it's lit, once it starts conducting, it's got negligible resistance. So yeah, I just don't know. So I think what I need to do now is to take all these LEDs off again and check the board. So I got the hot air gun out and laboriously desoldered this. Not all the holes are clear, but most of them are. And I have rapidly figured out what the problem was. Do you see these white marks in between the pads on each side? Well, that indicates that there is a conductive track between them. So because the LEDs were attached across these, the LEDs were being shorted across. That was why it wasn't working. So after all that, I went through and I retested all the LEDs and surprisingly enough, nearly all of them work if I attempt to get this in. Unfortunately, I ended up having to cut the leads short. But if I hold that in front of the outer focus camera and run the tester, it flashes. And in fact, they all work except probably this one, assuming I didn't manage to... Let me just get these a bit more on camera. Unless I managed to get them mixed up. Yep, that one is dead. I probably ended up dumping 100 milliamps across this and fried it. So this one goes in the bin and we go back to the drawing board. I have 10 40 ohm resistors and nine LEDs. What can we do with this? Well, it turns out that one of the recommended ways to run groups of LEDs is to put them in series. I want to run this a five volt so I can use a USB power supply. So because each one has a voltage drop of... Well, the tester will tell me about 1.1 volts. That's actually quite low. There's supposed to be about 1.5 to 2 volts for an infrared LED. But this is saying 1.1. So we can put them in parallel until you reach five volts and then we need a single current limiting resistor. So at 1.1 volts, four of them gets to 4.4 volts. So we need to drop 50 odd milliamps over the remaining 0.6 volts. So that should be fairly easy to calculate. So luckily it turns out that with a string of three LEDs then in order to get a current of about 30 milliamps which seems reasonable for one of these LEDs you need a 40 ohm resistor so I can make use of these although at least only three of them. So I think I'd better solder this up now. And you know what this means. So let's plug it into the tester and see what the tester thinks it is. Okay, the infrared camera is on so if anything comes out we'll see it. Of course only three lit. But that does believe it's a diode with a forward voltage of 2.4. 2.4, interesting. Why 2.4? It should be 1.1 times 3, 3.3. So I suppose it could be two diodes but three of them are lit. Okay let's do some continuity testing. There's probably a dodgy joint to the bus bar or possibly two dodgy joints to the bus bar. Well those all make connection. Although it might be that the voltage is sufficiently low it's only causing, well clearly I have two different types of LED here the blue ones and the white ones. It may be that the blue ones have a lower forward voltage which means that the meter which is not producing very much current at all is not sufficient to light three at once. So let's get out of the power supply and wind this all the way down to try and prop it up so the camera can see it. Let's get some of these wires out of the way. This is all turning into a horrible mess. Okay what do we get? Okay two of the white ones have lit. We will just wind this up a bit. Okay there knows the light. We're at 3.2 volts, 4 volts, 8 milliamps, 5 volts. Okay that one has not lit and none of these three have lit. So the current must be going through this one because they're all in series and if that was not lighting there's a piece of gunk under the LED if I can get that out. I know why these haven't lit. I failed to bridge a joint there. Okay so that's light done. That looks clean. Let's stick it back on the vice, line the voltage back down. That should be okay with 5. I just want to see them light one at a time. Okay we've got those. And that is drawing a 150 milliamps total which I think is alright. Now for that LED to not be working maybe I've shorted through it. Is there a solder bridge? There isn't one here. It would be between these two pads. Actually I can detect that using the meter. So if we do this, that's the connection. That's the right way round. So black to black, red to red. So if we do it the other way round there's no connection. I wonder if that LED has failed through. It's actually much more likely because I can see something under it. It's much more likely that there's a solder bridge on the other side. The holes in the board are kind of prone to using capillary action to pull solder through the board. That's happened several times when I was soldering it up. Okay well, I'm just going to have to take it off again. Okay, so let's stick this thing in the tester and see what happens. The LED is fine. That means the problem is with the board. What do we have here? We've got a big blob of solder which is my fault from the work I just did. Let's get rid of this then. Get rid of some of the whiskers. Okay, it looks good. Let's use the meter to check that there aren't any unexpected conducting pathways. There are not. So look at the underside of the LED. It looks fine. Whatever that black blob was is no longer there. So we just need to arrange it. The right orientation. Come on. And slot it in. And solder it up. And just see if that's flush. It's not what I wanted to do. All right. So that should now be soldered back into place. Let's stick it in the vice again. What do we get in the camera? Great. Now none of these are working, which is not what I wanted. That's because I forgot to solder up the bridges that I had to remove in order to get the LED out. So it's fact not connected to anything. Okay, that's one. And that's two. Okay. It's getting quite late here.  So let's give this one last try. Fantastic. All nine LEDs work. And yes, I'm aware they're showing up in purple. That is just, that's the color that the infrared shows up on the CCD.  And as you can see, can you see anything coming out of that? My real camera has a decent infrared filter. So very little should be coming through. But that is working. Finally, 120 milliamps, nine LEDs. Shame about the dead one.  So next time I actually get to try and take photos with this. And I'm actually going to have to make another modification to this camera that means taking it apart again. But until then, I hope you've enjoyed this video. As always, please let me know what you think in the comments.