So what's the plan for today? So as I announced on the first day, my plan for today is to do the proof of the data processing inequality for the quantum relative entropy. So remember, this was like the fundamental property of distance measures in general. So it's relatively simple to prove for the trace distance, for example. For the quantum relative entropy, it's a little bit more complicated. And this is what we'll see today. And this is really, I would say, it's a central result in quantum information theory. It's used everywhere. And it's, I think, almost every proof of a converse theorem, like no-go kind of result, uses a form of data processing. So remember yesterday when we proved the Stein's lemma. So we had two parts. We had the part which shows an achievability, which gives a strategy, which achieves a given type 2 error. And then we had to show that you cannot do better than this. And if you remember, the place where we use the data processing inequality is when we prove that you cannot do better than the quantum relative entropy. And for that, we used the data processing. And this is basically in almost all information processing tasks that you try to analyze and try to compute the optimal rates for, you use the data processing inequality or a form of data processing inequality. OK, so yeah, OK, so let's get to it. So again, the main objective today is to show this inequality. So I take two states, rho and sigma, or two even positive operators in general. And I apply quantum channel, the same quantum channel, to both sides. And this says that the quantum relative entropy can only decrease. OK, and so the plan for this is as follows, is that first we'll show a convexity result. So this is a function that takes as input two positive operators and outputs a number. And it has a nice property of being convex. We often say also jointly convex because it has two inputs. So it's jointly convex in the two inputs. I'll say in a minute explicitly what this means, but it's the natural thing you would expect. So and then we'll see that convexity is very related to the data processing inequality. So we'll see how we can get from convexity to the data processing inequality, to the fact that when I apply quantum channel, it can only decrease. And I should say that this is a very general statement. And so in general, for a general distance measure, when you have convexity, you can get data processing and the other way as well. I wouldn't say that this is under some conditions. It depends on some particular properties of the divergence we consider. But in general, it's almost always the case that these two statements are equivalent. But you can go from one to the other and I'll present a proof that is relatively generic that allows you to go from convexity to data processing. The other side you can also do again under some conditions. So once we do that, so this is the second point. The third point will be to explore implications of it. And namely, this will be, namely, we'll discuss strong subadditivity. So this is maybe one of the main implications. So it's fundamental inequality on the von Neumann entropy. So good. So now let me, OK, so we'll start with the first point. And so let me state it as a theorem. So I consider this function. So as I said, so that maps rho sigma, so recall the expression for the quantum relative entropy is trace rho log rho minus rho log sigma. And my claim is that this is convex. What the theorem says is that this is a convex function. So notice that, yeah, so it takes two inputs. So what do I mean by convex here? Is that if I take a convex combination of rho 0 and rho 1 in the first input of the relative entropy and the same convex combination in the second input, then this is upper bounded by the convex combination of the corresponding quantum relative entropy. OK, so how will I show this? So I will even show you this in a way that makes it very explicitly convex. And what do I mean by this? I mean that I will write an equality. So I will write the quantum relative entropy using a different expression than its definition that makes it very clear that it's convex. OK, and what do I mean by this? So I mean that I will write it as a supremum of functions that are linear in rho and sigma. OK, so let's go a bit over this expression. The exact details of the expression don't matter for now. But the important thing is that, OK, so it's a supremum over some operators, over an infinite family of operators, Zt1 for every t from 0 to infinity. And then I take an integral from 0 to infinity of an expression. And this is the important part, an expression which is linear in rho and sigma. So the crucial thing that I want to emphasize here, at least at this stage, is that this expression is linear in rho and sigma. So of course, a function that is linear is in particular convex, right? And it's a very simple kind of convex function. And the supremum of linear functions and even the supremum of convex functions is convex as well. So if you accept this equality, then convexity becomes obvious. And this is even a certificate if you want a certificate of the convexity of this function. So yeah, I'm reiterating this point here, is that this expression that I'm giving here in star makes it manifestly convex as a supremum of convex functions. And OK, so I even included a brief proof of this. So if you have a set f of linear functions, right, and you take a supremum over all these functions f in this family script f and of this function applied to the convex combination, then just by linearity here, you get that this is 1 minus pf of rho 0 sigma 0 plus pf of rho 1 sigma 1. And then of course, you can separate the two supremum. So we have a supremum of a sum. This is smaller than the sum of the two supremum. So this is obvious.  This is something you're probably used to. So in 1D, if you take a function that is convex, you can write it as the supremum of these linear functions. OK, see what I mean. So all these linear functions, OK, my function is not very nice. If you consider, for example, the tangents at every point, so these are linear functions. And the function of interest is a supremum of all of these tangents. So in some sense, the expression I'm giving is an analog of this, right? OK, so the other remark I wanted to make is that, yes, for this expression, I mean, I never said what the base of the log is. Usually it doesn't matter because we were only comparing relative entropies. But here, for this expression, it matters. So I take the log base E. If you want to get the log base 2, you just divide by log 2. But yeah, this expression is for this log base E. OK. So yeah, another thing is that this expression is quite nice is in that it allows you also to prove data processing directly without going through this standard proof of going from convexity to data processing. I mean, both proofs are relatively easy, but I'll present both of them just so that you see that this expression, that you see the potential uses of this way of writing the quantum relative entropy. Yeah, so this way of writing the quantum relative entropy is not very well known, usually, but I think it's useful. And I have been using it recently for computational purposes. So I wanted to present it to you. OK. Good. So yeah, so given this, so the only thing I have to prove now is this expression, right? So yes. I thought it was that expression? Yes. Yes, I would actually prove it. I would prove it. Yes, yes, yes. So what I was saying is that the whole proof, actually, of strong sensitivity will be to prove this expression. Yes. Ah yeah, there's a T missing. Yes, you're right, thanks. Let me fix that. Thanks. Other questions? OK, so yes, so again, this is my objective. I want to write this expression. And so yeah, the first thing to note, maybe, I guess you all know this, but just to see the difficulty here, the main difficulty here is that we have logs of operators that are not commuting. So rho and sigma, in general, don't commute. So I don't have that the log of rho minus log of sigma is log of rho sigma inverse or something of this form. OK, so this even doesn't make sense because rho sigma inverse is not a positive operator. But yeah, I just wanted to make sure that you're all aware that this is the main difficulty. OK, so OK, so what are we going to do in order to get this expression? So we would like to have just a single log, right? So having a difference of log is a bit difficult to handle. And so you'd like to have an expression which involves just a single log. And so one way of doing this is by using this simple trick, where you write the trace of a product of two operators, in terms of its bilinear in A and B. So you write it in terms of the tensor product of A and B. This is a simple trick. I'm sure you've seen variants of it. Yeah, it's very simple. You just check that this works. So you write A tensor B. And here I chose the version where you sandwich it with a fixed maximally entangled state or an unnormalized version. So some of our I of Ii were I some fixed basis of my Hilbert space. And here the transpose is in the same basis. OK, so now using this, I will write the recall the expression for the quantum relative entropy is trace of products. So I will just use this expression. And so I write it as my operator A will be just row log row and tensor identity for the first part. And for the second part, I would take A to be row and B transpose to be the log of sigma transpose, which is log of sigma transpose. OK, so now notice that I did it slowly. So I just rewrite it. So as row tensor identity times log row tensor identity plus identity tensor log sigma transpose. And here notice that because the row and the sigma are commuting, I can combine them into a single log. And what I'm using here is this expression. So the fact that log of A tensor B is equal to log A tensor identity plus identity tensor log B. And so now I have written the quantum relative entropy as this row tensor identity times log of row tensor sigma inverse transpose. OK, so now starting from this, OK, we have a single log. What we'll do is that we'll use this integral representation for the log. This is often useful in this area. And you've seen this in the first problem session. This was quite useful in order to prove, for example, the operator convexity and the operator monotonicity of the log. Because you prove the operator monotonicity of the inverse function. And then you use this integral representation to directly get the operator convexity and monotonicity of the log function or minus log function. OK, so here we'll do the same. We'll use even the same expression. So remember, you used exactly this expression for in the exercise session of the first day. So let's use this. And so here I will apply to, so OK, this is valid for scalars, but by the functional calculus that we've seen, right? So for matrices, you can apply function for matrices as well by just applying the corresponding function to the eigenvalues. OK, so the A tensor identity, I just leave it on the side. And now I applied the integral representation to this log, OK? So yes, so OK, so I get, so A tensor identity, the integral of 1 over, I don't know why I put the identities here. OK, so this is just a constant here, right? It doesn't depend on x, OK? So it's a constant function. I can put identity here if I want, OK? Yeah, I say as identity, because remember, we're working on the tensor product Hilbert space, H tensor H, OK? OK, so now what about the second term? So this 1 over t plus x, OK? So the way we can see it is that this is, so x, I replace x by A tensor B inverse, OK? And this is identity if you want. OK, so now I will just rewrite this expression. So I put the A tensor identity inside, OK? And then the A tensor identity here cancels with the A which is here, and so I get an A inverse in the first term, OK? So here it's just I put the A tensor identity inside the integral. OK, so good. So now we're starting to be in good shape. Why? Because, OK, so this part is simple because it's linear in A. So OK, so notice that I, so A will be rho, and B will be sigma transpose. I just wrote it in general because I didn't want to write the transposes all the time. OK, so let's look at this expression. So this part is linear in A, so that's good. This part is not linear in A and B, OK? But this is a very well-studied quantity in matrix analysis, right? It's sometimes called the parallel sum, OK? So and it's sometimes written in this way. So A parallel sum B is just 1 over the sum of the inverses of A and B, OK? And so this is directly related up to a factor of 2 to the harmonic mean, OK? So if A and B are scalars, right? So you might be aware of the harmonic mean of two scalars, OK? So this is up to a factor of 2, an operator version of this harmonic mean. And so this is very well-studied in matrix analysis. So we know very well properties in general of these B. There is a big theory of operator means in general. So you know the geometric mean, for example. There is a generalization of that to operators, which has also very nice properties. And there is a whole theory related to this, OK? So good. So you might be worried because I'm taking inverses here about how about if some operators are not invertible and things like this. So I don't want to discuss this during this lecture. This is one can handle it. But yeah, these are technicalities. I don't want to get into it. And basically all the properties are satisfied, even in the general case, where they're not invertible, OK? But you should see basically the inverse as the inverse on the support, OK? So one of the many nice properties we know of this parallel sum or of this operator harmonic mean is a nice variational expression, OK? And we will use that to get our expression then. So let's go over it. How does it work? So it says the following. So I take two positive operators, E and B. I take the parallel sum, OK? So A parallel sum B. And I apply it to some vector X in my Hilbert space, OK? So then it turns out that this has a very simple expression or a very simple variational expression which is linear in A and B, OK? And so what is this expression? So yeah, I get something linear in A and B. And on what I sandwiched them with which vectors was vectors Y and Z, OK? And the only constraint is that Y and Z should sum to X, OK? So this is the expression. So I even included a proof of it for completeness. I mean, so I say, OK, what is hard in this is coming up with the expression. But once you come up with the expression, it's relatively simple to check that it's the case, right? So it was just a simple manipulation. OK, so yeah, let's go over this quickly. So the parallel sum, you can write it in this way. So the inverse of the sum of the inverses and by some standard manipulation, you can rewrite it as this, OK? So B minus B times A plus B inverse times B, OK?  For an arbitrary Y and Z, that sum to X, and just observe that it's always non-negative, OK? OK, so yeah, this is what I'm doing. So I'm writing this expression, so it was Y. And now Z, I take it to be X minus Y, OK? And I'm subtracting the parallel sum between A and B sandwiched with X, OK? So I'll just write it down. So where does the sandwiching with X appear with B, OK? So with Y, I get one from A and one from B. And then I get the cross terms, which are here, OK? So then the parallel sum term, I just replace the parallel sum with this expression, OK? So yes, so this is what I get, OK? So this cancels with this part, and this is what remains, OK? So now I observe that this is, I can see it as a two norm of some operator by taking half of this operator, basically. So everything is a mission here, right? It's positive, even, right? So A and B are positive operators. So I can take square roots, OK? Here, too, I can do the same thing. And here as well, OK? I'm just starting to write it in a way that will make it obvious that it's a square, OK? By just multiplying and dividing by A plus B inverse, A plus B to the power of half, OK? And then, OK, by combining these two, you see that it's just a two norm, right? So it's a two norm of a difference of some vectors, OK? So X to which I apply some operator, and Y to which I apply some other operator, OK? So this is obviously positive, right? Non-negative. And I also see with this expression that it's obvious how to make it an equality, right? To make it equal to 0. I just have to pick Y so that this thing is equal to 0, OK? And this is just Y is equal to A plus B inverse B times X. No, unfortunately, not here. I don't have much. Oh, you mean to prove it? You could try to prove it in this way, like say that B is identity and then doing the general case, or? Yeah, so sorry, I don't have much insight on it. OK, so now let's use this expression back in our, OK, so we just get back to our expression. Remember, it was this. So the relative entropy to norm sigma. Yes, OK, so I just rewrote the expression I had in the previous slide. And then I just put in the phi inside the integral, OK? And so, OK, so let's look at each one of these terms. So this is rho tensor identity sandwiched with phi. And this is the parallel sum between some operator on rho and some operator related to sigma. And I sandwiched it with phi. Remember, phi is the maximum integral state. OK, so let's compute each one of these expressions. OK, so the first one is easy, right? This is just rho tensor identity sandwiched with phi. This is just trace of rho, OK? Up to this scalar factor 1 over t plus 1. OK, so this is just trace of rho. This is nice. Now what about the other expression, right? So this is a parallel sum sandwiched by a vector. So this is exactly the setting of the proposition that I showed here. And so I will apply to just x being this vector phi. And I will optimize over the y's and the z's. OK, so here I just did that. So again, the parallel sum between rho tensor identity and identity tensor sigma sandwiched with phi is the infimum over all operator z. Now I'm in the tensor product Hilbert space. And so, yeah, my first term is rho tensor identity over t sandwiched with z. And the second term is identity tensor sigma transpose sandwiched with phi minus z. OK, so what I will do now to get to the original expression is I will just write z in a fixed basis. So I'll choose the same fixed basis of h that I took before. And I write z as z ij i tensor j. And I just express each one of these terms. So the rho tensor identity sandwiched with z. So yeah, rho tensor identity I sandwiched with z. You see this identity, I get the delta on jj prime. And so by just reshuffling the vector z into a matrix, I get this expression, trace of rho times z z star, where z is now an operator, which is obtained by just flipping from ket j to a broad j. And you observe that z star is, of course, the adjoint. And so I get this expression. So already, hopefully, you start seeing where this is going. This is, so we already have seen this expression, the trace of rho. This was the first term. The second term is exactly this one, the z z star. And with the other expressions, we'll get exactly the other terms here. OK, so let's do this quickly. So if I take phi itself, then phi with identity tensor sigma transpose, this is also a trace of sigma. Then if I sandwiched i tensor sigma transpose with z, it's exactly similar to the one with rho. I just get trace of sigma times z star z. Now for the cross terms, where I have one phi and one z. So these will be terms that are linear in z, not quadratic in z. And because this phi has no z parameters. And so in particular, with this choice of z that we took here, this will just be trace of sigma times z. And the other cross term will actually be trace of sigma times z star. OK, and so now if I get back to the whole expression, to this expression, and putting everything together, I get exactly what I want. And note that, OK, here I didn't put t in the dependence on z. But notice that the operators for which I'm taking the mean depend on t. So there is a t appearing here. So the optimal z can depend on t. So I have to pick a different operator for the different parallel sums, OK? So yeah, so just putting everything together, we get what is inside the integral has this form. Yeah, OK, here the t was not needed. But OK, it's better to let me put it here. OK, and I conclude by just taking the integral. I'm going to get back this expression. Good. So I hope this was clear. Yeah, I agree that the formula for the parallel sum is a little bit magical. But so the main idea is to reduce this relative entropy by an integral representation to an expression which is related to this parallel sum and then use the known results about the parallel sum. So in particular, the parallel sum is known to be operator concave. And this is one way of showing it was this variable. An explicit way of showing it that it's operator concave is via this variation expression. OK, good. So we have now proved the joint convexity of the relative entropy. And see, this was the proof was not very long, right? So and I think I did all the calculations. I didn't skip any details. So it's relatively short. And again, it's in a way that makes it explicit that it's jointly convex. OK, so now remember our objective was not only the joint convexity, it was the data processing inequality, or sometimes called monotonicity under quantum channels. So if I apply quantum channels, the relative entropy can only decrease. Again, as I said before, also, this is a rather generic argument. If you look at this argument, it doesn't, in any way, depend on the quantum relative entropy. This is very generic. OK, so for example, and OK, so I should say that there are many other useful divergences, as I mentioned. So for example, one of the most known is there is rainy divergences, right? Where it was a parameter where you have a parameter alpha. And so for all of these, this kind of a relation between joint convexity and data processing holds as well. OK, good. So let's go over this standard argument. So OK, so for this argument, it's useful to use the Stein-Spring dilation of the quantum channel E. So remember, I can always write a quantum channel E in terms of applying first an isometry, which I call V here, and then doing the partial trace. OK, so I wrote it here in a compact form that E can be seen as applying an isometry. So this quantum channel V is just applying an isometry followed by partial trace. OK, so the isometry is easy to handle, right? Because it just keeps the relative quantum relative entropy exactly the same. It doesn't change it. OK? So this doesn't change anything. So the only thing I have to analyze is, and I put the calculation here, it's just using the definition. And so the only thing I need to analyze is the partial trace. OK, so I need to show that when I apply the partial trace to some joint state, the relative entropy can only decrease. OK, so yeah, this is what I'm saying. So here I'm saying I take two states on B and E, rho and sigma. And what I would like to do is to transform this rho BE into the marginals, rho B and then sigma B. Of course, in such a way that I use the convexity result that I proved before. Right? OK, so the question now is, is there a simple way of mapping rho BE to just rho B? OK, and more specifically, we'll do it to rho B tensor the identity on the E system. OK, so if I had that, then the relative entropy between rho BE and sigma BE is just this. And the relative entropy between rho BE tensor identity and sigma BE tensor identity is just because of the additivity of the relative entropy under tensor product, this is just equal to the relative entropy between rho BE and sigma BE. OK, and so the question is, what kind of map can we use in order to do this transformation? And so you might be familiar with this. This is sometimes called a quantum one-time pad, if you want, because you erase the E system, you make it uniform. So there's a simple way of doing this by taking Pauli operators, or if you're not in a set of qubits, there's generalized Pauli operators, which is the x just shifts my fixed basis, my computational basis by 1, so k maps to k plus 1, and the z adds a phase. And so I can define this quantum channel, which is by applying a z followed by an x to some power m. And I guess many of you would have done this calculation so far, would have done this calculation at some point, is that it's easy to see that this map, this quantum channel, what it does is that on k, k prime, you get 0. So on the half diagonal terms, you get 0. And on the diagonal, OK, so why is this? It's just because the sum over z of these phases, if k is different from k prime, the sum of these phases is equal to 0, except if k is equal to k prime. So if k is equal to k prime, now you let x act, like the Pauli x act. And then if you apply a random shift to a fixed operator on the diagonal, you get the identity. Good. OK, so again, what have we seen here is that if I take my operator rho b e and I apply this depolarizing map d on e and do nothing on b, what I get is rho b tensor identity on e. And same thing for sigma, of course. OK, and now remember, I want to use the joint convexity. And so how will I use it? So this channel has a very simple form. It's a mixture of unitary channels. So I pick l and m at random. And I apply a unitary, which is the Pauli z fold by the Pauli x to the power m and l. So I can use the convexity here. OK, so let's see it explicitly. So now I will use convexity to establish the following inequality. So here I have the relative entropy of some mixture and the same mixture here. And I bound this by the mixture of the relative entropy. And so this mixture is nothing but what I introduced before is I apply to the state rho b e. I apply the randomly chosen Pauli on the e-system. And as we've seen by the previous calculation, this operator as a whole, if I take the average, it's equal to nothing but rho b tensor identity. OK, and same thing for here, it's sigma b tensor identity. OK, so now let's look at the other side of this inequality. So I have now an average. The average is now outside of the relative entropy. And now I have a relative entropy between rho b and sigma b, but to which I apply this fixed Pauli l and m. But now this is not really what I was interested in. I was interested in the relative entropy between rho and sigma itself. But then these two are equal because x and z are unitary. So and we saw before that applying a unitary to the same unitary to rho and sigma keeps the relative entropy the same. OK, and that's it, and that concludes the proof. So yeah, we've seen that. So this side is just the relative entropy between rho b tensor identity and sigma b tensor identity, which is nothing but the relative entropy between rho b and sigma b. And the right-hand side is the relative entropy between rho b e and sigma b e. So we've shown that if you take the partial trace, the relative entropy can only decrease. I hope this was clear. OK, good. So I told you also that we'll see somehow more direct proof of. So you see, this is very generic. So there is nothing I used about d except a few properties, like invariance under isometry and joint convexity. You see, if I replace d here by any divergence which satisfies these two things, this proof goes through. So if I have convexity, I get to data processing. OK, so now I'll look at another alternate proof of the same thing, except that it will be specific to this expression. So I will try to tell you how that just by looking at this expression and doing some very minor manipulations, you get directly the data processing. OK, so remember, this was the expression. So the relative entropy looks like this. So it's the supremum of an integral of things which are linear in rho and sigma. So now I apply e to rho and sigma. So I apply this operator e to rho and sigma. And so the natural thing to do here is to apply the adjoint on the other side of the trace. So the adjoint of e, I apply to this other operator. So I will apply the adjoint to identity and the adjoint to zt star, et cetera. OK, here I again forgot the t here. OK, so now we only need a small claim. Is that so e star, adjoint of e, apply to identity is identity, right? Because this is e star is unital. OK, so the only thing which is a bit tricky to handle is this thing. So it is the adjoint of e applied to a product of two operators, zt, zt star. And of course, what I want to do, right, what do I want to do is I want to relate this to the expression for rho and sigma itself, not for e of rho and e and sigma. OK, so what I would like to do is to relate this thing to a product of two operators. And here also this thing, these are the difficult parts. But this is a well-known property, a simple property of quantum channels. Sometimes channels that satisfy this are sometimes called Schwarz maps. It's the following. It says that if I apply some unital completely positive map to zt star, this is lower bounded in the positive semi-definite ordering by f of z star times f of z. And then, to see this, it's relatively simple. You just use the complete positivity for some positive operator which is constructed from z, z star, and z star, z. OK, if you're familiar with Schwarz complements, this is exactly what is going on. OK, so now what do I do? So I just use this inequality, apply to e star. And I write yt to be just e star applies to zt.  And so because I have a minus sign, then this becomes an upper bound. And so by just using this inequality and plugging it in, I get this expression. So here I get yt, yt star. And I do exactly the same on the other side. So the identity is just by unitality. Here it's just linear. So I don't have to use this Schwarz inequality. But here I use it again, but just inverting z and z star. OK, but this is nothing but the original expression for the relative entropy. OK, I just called yt equal to zt. OK, so again, I wanted to show this to show you how useful it is, let's say, to have such expressions which are linear in row and sigma. Good. OK, so the last thing I wanted to discuss today is an implication of this data processing inequality to just entropies. So forget relative entropies. So many of you might be dealing with just von Neumann entropies, but not really relative entropies. So let's see what are the consequences. And yeah, so this is sometimes called strong subadditivity of the von Neumann entropy. So what is the way of writing it, which justifies this name of strong subadditivity? So subadditivity would be the version without conditioning on C. So if I don't condition on C, it's just saying that H of A plus H of B is at most H of AB. And the strong version is when you condition on C. OK, so sometimes you can write it in this way. This makes it obvious why it's called strong subadditivity. Just by basic rewriting, this is equivalent to saying that if I condition on an additional system, so imagine I look at the entropy of A condition on C, and now I condition on an additional system that I call B here, then the von Neumann entropy can only decrease, which is natural. And this is again equivalent to some other quantity, which is the conditional mutual information being non-negative. So these are all rewritings of the same thing. OK, I'm not trying to define the conditional mutual information in general setting. It's just equal to the difference between these two quantities, just by definition, the difference of these two. So just briefly, why is it the case that the data processing inequality implies these three inequalities? It's immediate, but let's go through it. So if I compute the von Neumann entropy of A condition on BC, this is, remember the definition, it's minus the relative entropy between the joint state on ABC and identity on A tensor, row BC. And what is the version where I condition? Ah, I don't know. OK, it's a symmetry, but it doesn't correspond to here. Maybe let me. A plays exactly the same role consistent with what I wrote here. Let me just correct this. And this is fine. I think all is good. OK, so, ah, yeah, thanks. OK, so, yeah, now written in this way, it's obvious how to go from here to here. I just need to partial trace B. OK, so if I partial trace B, I get from here to here. And this quantity is smaller than this quantity. And so there was a minus sign. It's the opposite. OK, good. OK, so how to see that two implies one. This is, again, very simple von Neumann entropy manipulation. And I think you've seen a bit of that in the exercise session yesterday. So, yes, I can write, remember, by definition, if you want, or by the simple properties of the log, H of A conditioned on BC is equal to the joint von Neumann entropy minus the entropy on BC. And I can add and subtract H of C. And so this gives me H of AB conditioned on C minus H of B conditioned on C. And so by combining, yeah, so this is exactly the same as, is exactly equivalent to this inequality. I should say that this equality is often quite useful. It's called the chain rule for the von Neumann entropy. And, yeah, so it allows you to decompose the entropy, why is it useful? And even though it's sort of an obvious identity, is because it allows you to decompose the entropy of a joint system, AB, as a sum of the entropies of the parts, right? So the entropy of B conditioned on C plus the entropy of A provided you condition on both B and C. This is often a useful tool. OK, and, yeah, concerning the third one, as I said before, the conditional mutual information is anyway defined as the difference of these entropies. So in this case, obvious that these are all equivalent. OK, good, so that ends what I wanted to say for today. So let me just briefly recap is that I hope you have seen that the data processing inequality, it's not too hard to prove it. And it's related to its joint convexity, and that it is a sort of a general inequality that can be used to get the useful inequalities we know about for the von Neumann entropy. And so I'll stop here for today. Thank you.


Okay, so the class is going to be taped for the rest of the quarter, I think. Shant back there is in charge of the video. If you do not want the back of your head videoed for any reason, then move to the back of the class. So, you know, as I mentioned in discussion section yesterday, I totally honor you for going up in front of the class and presenting problems. So you can imagine how terrified I am to have myself on video that will probably eventually be on YouTube. Okay, so I want to start off by continuing where we left off. We're starting on our next lecture on molecular orbitals. And remember lecture one, that was more than a one-hour lecture. We talked about the fundamental idea behind arrow pushing, that it represents the interaction of filled orbitals with unfilled orbitals. And I gave you seven rules for mechanistic arrow pushing that we are going to agree to follow as a class when we do problem sets, when we do exams, those are the times when you really have to pay attention to those rules. And so, because arrow pushing represents the interaction of filled orbitals with unfilled orbitals, we have to make sure we all are on the same page with orbitals. So today we're going to talk about orbitals. And I want to start off with a simple problem for you to solve. Let's just imagine that we have two particles that are both positively charged on the atomic scale. And to make things meaningful, let's make these two protons that are ready to form a bond. And we all understand what electrostatics does. We understand Coulomb's law. If you were an electron with a negative charge, what's the ideal position for you to occupy relative to these two-point positive charges? An electron has a negative charge. I think we have this idea. I have this notion that I want to put that negatively charged electron right there. With all of my being, I want to put that electron there. But if you do that, you have violated the Heisenberg uncertainty principle. You can't specify the location of an electron. You have to put that electron all over the place. And that's what results in bonding. So we have two simple outcomes to that. One is we need orbitals. We need some sort of mathematical description of the probability of finding an electron smeared all over the place around those two hydrogen atom nuclei. So there's a second problem here. Whenever we put electrons into orbitals, we put them in two at a time. We put in electron pairs. And because we're putting in electron pairs, we need some way to show that those electrons will avoid each other. And so as a result of that, we need phasing. We need to phase those orbitals to show that there's a way for the electrons to get out of the way. And so let's go ahead and draw the canonical types of orbitals that we will use in this class. And there's really only two types.  We'll also use P orbitals. And we'll represent them with this kind of lame representation like this. They don't look exactly like that, but this is good enough. And they could have various orientations with the X or Y or Z axes. So these are P orbitals. And so note the importance of phasing. Here I've separated the two different S orbitals. You have to imagine them on top of each other with that different phasing. Okay, so phasing allows us to think about the idea of having two electrons around the same place in space but avoiding each other. And so we need that. Okay, so we're going to talk about three types of orbitals today. And it's supposed to be reviewed for you. Supposedly these are ideas that you've already seen before. And so the intention is to review. So first we're going to review atomic orbitals. This is the stuff that you know is SPD and F orbitals relative to the structure of atoms. Then we're going to talk about combining those things into hybrid atomic orbitals. That's SP2, SP3, SP, stuff like that. So you should be familiar with that. And then finally we're going to talk about combining those things to make molecular orbitals. Okay, so let's go ahead and start off with atomic orbitals. Well hopefully we can add some new insight. Hopefully you can take these ideas that I'm sure you heard at some point in time. But when you take those ideas and you reapply them to reactions you already know, then hopefully you'll be able to do some more stuff with that. Okay, so let's go ahead and start off with atomic orbitals. So there's four types of canonical atomic orbitals that we use in chemistry commonly. There are potentially other shapes and other formulations but these are the ones that you should be familiar with from freshman chemistry and from sophomore organic chemistry. From general chemistry and organic chemistry. And that's SPD and F. Now fortunately we can rationalize everything in this class with just S and P orbitals. Now there will be examples of organometallic chemistry, transition metal chemistry where we will need to put in DNF orbitals but we will be able to explain everything in this class just by focusing on S and P orbitals. In a way that makes our life incredibly simple but what you'll learn is just handling these two types of orbitals and the way they interact is going to be immensely complex and yet immensely powerful. Okay, so we're just going to focus on S and P orbitals in this class. And I have one super important statement to make. And that is that P orbitals are higher in energy than S orbitals. I think every one of you knows this already. But what you don't know is just how powerful that is. And so let me go ahead and start off by graphically representing the fact that P orbitals are higher in energy than S orbitals. And I'm going to start by using a construct that we're going to use many times in this class called a molecular orbital or an orbital energy diagram. I'm going to have some mystical orbital axis here and I'll call that E. Sometimes I'll write E with a subscript like MO for molecular orbitals.  So let's graphically represent the fact that P orbitals are higher in energy than S orbitals. And I want to focus on P orbitals on a carbon atom because this is organic chemistry. We care most about carbon. Usually everything that I talk about is going to be carbon or else nitrogen, oxygen. And then sometimes things from slightly lower levels. So this is a P orbital energy. There's three types of P orbitals. They're orthogonal to each other. If I draw the shapes out for those orbitals.  You'll see why that's important that I specify that in just a moment. So let me go ahead and try to represent those P orbitals. This would be a P orbital aligned with the Y axis. And there's two other orthogonal types of P orbitals that can coexist in space on a carbon atom. These are the other types of atomic orbitals. There would be one on the X axis. And I'm showing one of two arbitrary phasing combinations here. And then finally there's one sticking straight out at you that would look something like this. And I'll try to draw the hashed part going in the back of the plane of the board. Okay, so those are P orbitals. The important point is that these are way higher in energy than an S orbital on carbon. And I'll put a subscript there. Vastly higher in energy. If you were an electron in a P orbital, you'd be much more energetic and likely to attack things than an electron in an S orbital. That's the importance of energy. And that's what you care about. You care about which electrons are likely to attack other things. It's the most energetic ones. Conversely, if I were a pair of electrons and wanted to put myself somewhere, I would want to go into the S orbital because it's lower in energy. Electrons always want to go into orbitals that are lower in energy. That makes sense. They're more stable. That's a more stable place for them. So, let's draw a little comparison there here just to let me make some room here. And I'll draw two different cations. Let's think about these two cations momentarily and try to imagine which of these two cations is more reactive. I just gave you enough information to answer this question. You probably have an idea for which of these types of cations is more reactive or maybe you don't. But I just told you in essence how to answer this question. So, I've got two different cations and the way to answer this question is we finish this up by drawing the fact that there's an empty p orbital on carbon centered on carbon here. And here there's an empty p orbital located on nitrogen. Let me draw it in. It's empty. It would love to have a pair of electrons there. And what we need to do is think about the effect of electronegativity on the energy of atomic orbitals. And all orbitals associated with that nitrogen atom. So, the effect of electronegativity is that all of these energy levels will drop. If I were an electron, I would much rather be on one of the p orbitals on nitrogen than on one of the p orbitals on carbon. The effect of electronegativity is that these p orbitals will be lower in energy. The effect of electronegativity on the s orbitals that they will be lower in energy on nitrogen. And that makes sense. Let me sketch out the periodic table for you just in miniature form. As we march across the periodic table from boron to carbon to nitrogen to oxygen to fluorine, things become more electronegative because there are more protons in the nucleus. There is one more proton in the nucleus of carbon. There's yet one more proton in the nucleus of nitrogen. There's one more proton in the nucleus of oxygen. Those protons have a positive charge. And if you were an electron, you would enjoy those positive charges more. So this makes total sense that as we go to more and more electronegative atoms, that we should expect all the orbital energies to decrease if they are involved with that nitrogen atom. So that's a powerful idea. So you can guess from this that this empty p orbital is more reactive, more electrophilic than the empty p orbital that I drew on that carbocation. Just based on electronegativity and we represent that with that orbital energy diagram. So a very powerful idea to extend. Okay, so let's talk about hybrid atomic orbitals. We're done talking about just plain simple atomic orbitals, SP, DNF. And let's talk about what happens as we mix together different atomic orbitals. Okay, so I'm going to give you a rule for mixing orbitals, for thinking about the way the orbitals mix together. It's best to think of them as mixing two at a time. And the rule is pretty simple. If you mix two orbitals, you get two orbitals out. I take two orbitals and I mix them together. And there's two different ways that they can combine together. And this is a result of that phasing idea. There's two different phasing combinations for every orbital. When you mix them together, there's two possible ways that they can interact based on phasing. So let me start off by drawing out just an atomic orbital energy diagram. Once again, I'm basically repeating what I showed you before and I've got this sort of arbitrary axis. It used to really bother me when I was a graduate student. I was like, why don't they draw units on there? The units are actually in electron volts. If you'd like, you can convert those into K-cal's per mole but it won't do you any good. The only thing energies in K-cal's per mole are good for are converting into number ratios, 10 to 1, 30 to 1, 0 to something. That's the only thing K-cal energies are good for. And we're never going to do that with orbitals. We're never going to divide into ratios. This orbital is 99% more populated than the other one. So don't worry about the fact that there's no energies on this axis. Again, they're measured in electron volts typically and you don't need to worry about that. So I'm going to start off by drawing out the orbitals. This could be on a second row atom and it could be carbon, nitrogen, oxygen. It doesn't matter and those are the orbitals that matter are the 2s orbitals and the 2p orbitals. There's a 1s orbital that's way down below the level of the floor. That's never going to be involved in bonding. Don't worry about that. So I'm going to start to soon abbreviate like I did over here. It was just writing p and s. But you just have to remember that's the 2s and the 2p orbitals that are important for second row atoms like carbon, boron, nitrogen, oxygen. So if I represent these orbitals, let me once again draw out these canonical orbital shapes and the phasing is arbitrary here. So one is aligned with the x-axis. We would call that the p sub x orbital, sorry, y-axis. One is aligned with the z-axis coming straight out of the board. That would be the p sub z orbital, et cetera. And the s orbital, we would draw sort of this sphere. Actually, the 2s orbital has this egg yolk inner core that has the opposite phasing. But I never represent that inner core. I just draw a circle and that's good enough for us. There's two arbitrary phasing combinations to that. What I want to do is I want to think about the fact that in order to get ready for bonding, we can think about this in a different way. We can take exactly that same carbon atom if this is carbon and we can think about a different set of four orbitals by mixing orbitals together. And we're going to call this hybridization. When we mix orbitals on just a single atom, we're not forming bonds. I'm just talking about a single carbon atom. I have a different way of thinking about this carbon atom, for example. And what I can do is I can think about just invent little games like what would happen if I mixed this 2s orbital with this 2p orbital. If I take a low energy orbital like the 2s orbital and I mix it with a high energy orbital like the 2p orbital, on average, I ought to get two orbitals out because I mixed two orbitals and they ought to be somewhere right in between. And that's exactly the picture that we get. So let's imagine that I take this orbital here, the py orbital and I mix that together with the, let me call it the px orbital, that looks like an x-axis. If I take the px orbital and I mix that with the 2s orbital, what I should expect is these other two orbitals will stay at the same energy. I'm not talking about mixing those. But I should get two new orbitals out and they should be halfway in between the high energy and the low energy. So I'll try to draw my new orbital energy lines here. And here's my two hybrid atomic orbitals. It's just a different way to think about this carbon atom. I'm not changing the carbon atom in any way. And so what I get out of this are two non-bonding orbitals. I mean, if I draw them that way, two new orbitals that are ready to form bonds. And the hybridization of those orbitals is not s, it's not p, it's 50% s and 50% p. And we represent that with this sp sort of symbolism here. So a carbon atom normally has four electrons ready for bonding. So now I would have one electron in each of these orbitals and one electron in each of these p orbitals ready for bonding. And so you know this to be the hybridization, I hope, for something like an alkyne. Those two bonds sticking out at the sides are the result of these hybridized orbitals, these sp hybridized orbitals. And then of course you have room for pi bonds. Let me try to draw these in sort of a bluish color. And then we have room for pi bonds if we wanted to form an alkyne like this, that's what the p orbitals would do. We'll talk about that deconstruction in just a second. The point is that we can take the atomic orbitals and we can think about hybridizing them with each other. So let's go ahead and look at another hybridization combination. There's other ways that we can hybridize these and I hope you're familiar with these. So let's suppose I take one of these orbitals and I mix it with those others, I'll be left with one of the p orbitals but I still have three other orbitals left. And now these orbitals, if I mix two of these p orbitals with one of the s orbitals, I'll have three more orbitals. And now with that extra p orbital mixed in the energy, the average energy ought to go up. And these are sp2 hybridized orbitals. I can draw these with a little non-bonding symbolism there. We'll talk more about that in just a bit. And hopefully you know that whenever you have trigonal carbon, these are the orbitals that are forming those bonds. They have this hybridization associated with them that's sp2. Finally, I'm going to try to move to the other board here. I'll draw the last possible hybridization combination. And I shouldn't say last possible. These are kind of extremes but if we understand these extremes then we can understand things that are in between. And then finally, the last possible combination is to take these three orbitals and mix them with this last p orbital to add even more p character. And so this is supposed to be on the same axis. And hopefully you know that this is sp3 hybridized. Three parts p character, one part s character. That's the recipe. And I'll use a little non-bonding orbital combination. So those would look something like this. And this is supposed to be a tetrahedral carbon atom with four sp3 hybridized orbitals ready for bonding. What's the really important point here? The really important point is that whenever you see orbitals that are higher in energy, they're more nucleophilic. If I were an electron in a higher energy orbital, I would be more nucleophilic. If you're adding to an empty orbital and it's lower in energy, you're more electrophilic. Hybridization tells you about chemical reactivity. It tells you about where to start your arrows. When you do arrow pushing, it tells you where to end your arrows. So last important point, let me keep this on board so it's at the same level, has to do with the strict correlation between hybridization and geometry. Atom geometry correlates with hybridization. And it's, maybe correlates isn't strong enough for it equates with hybridization. So if I were to take a nitrogen atom, if I were super small, and I could reduce myself to the atomic level, and I could grab onto these bonds on ammonia or on amine and stretch them upward so that this were a planar molecule, what would happen is the hybridization would have to change in response to that geometry. If I placed this amine functional group into some molecule that constrained it so it had to be planar, that lone pair would become screaming hot in terms of nucleophilicity because by changing the geometry, I would be changing from an SP3 hybridized orbital to one that's purely P hybridized. And I just told you what the difference is between something that's SP3 versus something that's P. SP3 orbitals are lower in energy than P orbitals. Let me try to draw all these on a single diagram so that you can see what's going on. So if I have an S orbital down here and a P orbital up here, SP3 orbitals are lower in energy and P orbitals are higher in energy.  The energy difference between a pair of SP3 hybridized electrons over there and a pair of P hybridized electrons is massive. The difference is massive in terms of nucleophilicity. And I'll put some quantitative numbers on that in just a little bit, but you need to appreciate this difference in reactivity just by changing the geometry. It's not sufficient to say that that's a lone pair on nitrogen. We have to go beyond that to hybridization. Okay, so hybridization is super important. Understanding the hybridization of every atom and every molecule is essential. Okay, this is a command for you. It's a command that you must do this. Every time you see a molecule, I want you to deconstruct that into atomic orbitals. And I'm going to start with a very simple example so we can see how that works. I'll take hydrocyanic acid. And let's try to think about the atomic orbitals about a picture of bonding that's based on hybrid atomic orbitals. How do I think about what kinds of hybrid atomic orbitals would make a bond between nitrogen and carbon, a sigma bond or a pi bond? So the way I think about that is let me just draw the three atoms here without the bonds. And let's try to draw some hybrid atomic orbitals associated with that. Okay, what's the hybridization of the carbon atom here? It's sp hybridized. And you know that because this is a linear molecule. I told you that geometry correlates with hybridization. If it's linear, that means there's two p orbitals that never mixed in. So I'll try to draw one p orbital that's on the y-axis and another one coming out at us. This is supposed to be aligned with the z-axis coming out of the board, but I'm not very good at drawing that. What's the hybridization of the nitrogen atom? It's sp also. And so again, I'll try to draw one of the p orbitals coming out of the board. And I'm going to label this p orbital on carbon, p orbital on carbon. I'll put a little subscript c there just so that I don't forget that that's a p orbital on a carbon atom. And as we just said, a p orbital on a carbon atom is not the same as a p orbital on a nitrogen atom. P orbitals on nitrogen atoms are lower in energy. And that's how you get those pi bonds. Two of the three bonds here are pi bonds that result from the mixing of those p orbitals. Okay, last, we're going to draw the hybrid atomic orbitals. And I'll change this line here to blue just so I can more easily see it. And I'll remember that there's an sp t orbital, sp 2 orbitals sticking out this side. And I'll hybridize it in this way. And there's a tiny lobe on the back end. And then for the nitrogen atom, there's another sp t orbital that's sticking out. And these two are combining together to make that carbon-nitrogen sigma bond. And there's a little orbital on the back end sticking out. Okay, I've got three orbitals now. Let me go ahead and label these. This is my sp 2 hybrid atomic orbital. I'll call this sp 2. And I'll say that it's on carbon. And then this one here, I'll put a little label line sp 2 on nitrogen. And those two hybrid atomic orbitals get together to make a sigma bond, the blue sigma bond in my Lewis structure. Okay, there's a lone pair in an orbital that's sticking out here. Let me draw that. And that's sticking in another sp 2 orbital with its lobe on the back end. And I can phase it this way if I want. Let me not phase that part. And I'll call that another sp 2 orbital on nitrogen. And I need to do the same sp. Thank you, everybody who got upset by that. These are sp hybridized. That's what we said before. Sorry, I'm getting p orbital happy here. So these are sp hybridized. We said carbon is sp hybridized. Nitrogen is sp hybridized. So these are sp orbitals on nitrogen. I need to do the same thing for that carbon atom. And draw another bond going back here. So let me draw another sp orbital on carbon sticking out on the back end. And then finally, that makes a bond with hydrogen. Hydrogen only has a 1s orbital to contribute. We don't need to worry at all. There is no way to do hybridization with a hydrogen atom. It's only got one orbital. It can't hybridize with its own atomic orbitals. And it's the combination of this 1s orbital and sp orbital that makes the bond. Now, why am I so intensely interested in the hybridization of these individual atoms? Why am I so intensely interested in the hybridization of this bond? Why am I so intensely interested in the hybridization of this lone pair? What I'm trying to do is I'm trying to gauge how much p character there is in that bond and in that lone pair. And you should be asking yourself this every single time. This is the fundamental question that you need to ask for every single structure you draw, for every single bond, for every single lone pair. How much p character? Let me just remind you of why I'm always asking myself how much p character there is. It's because p orbitals are high in energy and the more p character you have, the more nucleophilic you are. I am always asking myself how much p character there is in every single bond and every single lone pair because I care how nucleophilic this bond is. I care how nucleophilic that lone pair is relative to other molecules. Now, let's put some numbers on this. I keep telling you it's important to think about p character. Let's put some numbers on how important it is to think about p character. So, the fundamental idea is that the more p character, the more reactive. I'm going to start off by looking at nitrogen lone pairs because it's easy to rank the reactivity based on pKa or acidity. So, I'm going to draw a little table here. And in my table, I want to compare lone pairs on nitrogen. So, if I had some alkylamine with a carbon bonded to an NH2 group, there's a lone pair on there. If I have some sort of an emine with a lone pair on nitrogen that has two other bonds to it, that's not the same. And if finally I compare that to that hydrocyanic acid or to an organic nitrile, I would end up with something that looks like this. So, here's three different nitrogen lone pairs. They're all lone pairs. They're all on nitrogen, but they're different. So, I want to start off by ranking the hybridization. What's the hybridization of this lone pair on the top nitrogen? It's sp3. How about the second one? The emine or pyridine or emidazole. And then finally I get down to the nitrile and we already said that that's sp hybridized. Now, what's the percentage of p-character in this top case, sp3? 75% p-character. Sp2, what's the percentage of p-character about? It's about two-thirds or 67% and then finally the sp, 50%. These are paltry differences, 75 cents, 66 cents. What's the difference there, right? How big of a difference could that make? Maybe you already know something about the relative basicity of these. I can go to a pKa table and look up the converse of basicity, acidity. What you find out is, I'll assign this a relative value of one, that an sp3 hybridized nitrogen lone pair is 100,000 times more basic than an sp2 hybridized nitrogen lone pair. You know, if you put $1 in front of me and $100,000, I would have no problem seeing the difference between the two piles of money. It would be totally trivial and yet, why is it so hard to see the difference between these two? The difference is just as great, 100,000 times in reactivity. So, when I come over to this other board here, let me raise this up and pick a new board, draw out these two molecules. Which of these is more nucleophilic? Answer should be instantaneous. This one is more nucleophilic. It is about 100,000 times more nucleophilic than pyridine. Lone pair on nitrogen, but they're not all the same. And that's because this is sp3 hybridized and this is sp2 hybridized. Those lone pairs are totally different. Now, if I come down to an sp hybridized, okay, this is like a difference of what? I'm not good at math. 8% versus 17%. How big of a difference could that make? How many zeros did I, does it matter how many zeros I drew really? It's 15 zeros or 10 to the minus 15. There's a huge difference in reactivity. This is the effect of p-character. What looks to you like small differences in p-character translate into phenomenal differences in orbital energies. And this is why, if you from the very beginning of sophomore organic chemistry were not paying attention to p-character, I don't know how you made sense of the subject. They probably told you this at the beginning of the book and at the beginning of the first quarter and then they forgot to emphasize how important this is. So I'm going to ask you to pay attention to hybridization from now until the end of your career as organic chemists. Let's make sure we understand these differences. Which of these oxygens is more, I guess what I'm asking is basic. Which of these oxygen atoms would you expect to be more basic? I've got a ketone and an ether. Yeah, the ether oxygen. They both have two lone pairs but these lone pairs are sp2 hybridized and the lone pair sticking off of this oxygen are sp3 hybridized. Now the important point is we look at a pKa table and the difference is not 10 to the fifth this time. It's not always 10 to the fifth. Here it's only a thousand times more basic. But again, when you're pushing arrows, if you know that one lone pair is a thousand times more reactive than another, that kind of settles the issue. It doesn't have to be 100,000 times. Even a thousand times more reactive is useful to you in terms of thinking about reactivity. This is incredibly powerful stuff. But let me put one sort of fly in the ointment here, one word of caution in applying this very simple idea that hybridization is important and that p character correlates with basicity and nucleophilicity. More p character means more nucleophilic and more basic. I've got two oxygen atoms here. I've got a carbonyl oxygen on top and a carboxyl oxygen on the bottom. Top and bottom. Which oxygen atom is more basic? We mentioned this yesterday in discussion section. The top atom is more basic. I don't know of a single example ever in the field of recorded organic chemistry where the bottom oxygen atom acts as a nucleophile or as a proton acceptor. Now why is that? The only way to understand that is by truly understanding the structure and geometry of this molecule. In order to understand the geometry and the hybridization, we need to remember that there's resonance. And resonance is going to influence our picture because it tells us about the hybridization. The true structure of this molecule really has partial double bond character between this carboxyl oxygen and the carbonyl oxygen. Once we draw a resonance structure for this, we realize that these lone pairs up here are not really sp2. And this lone pair that I've got drawn at the bottom is not sp3. You can't think about the hybridization of these atoms until you've drawn the resonance structure. You have to incorporate resonance before you assign the hybridization to something. So really, this is more like sp3 in this extreme resonance structure and this is more like sp2. And if you draw the charges, it's even less likely you would ever, ever in your life make the mistake of trying to protonate this oxygen atom on the bottom. Who would do that? Nobody. Nobody would make that mistake. So you have to think about resonance before you try to assign hybridization. And as soon as you draw the resonance structure, it's absolutely clear that this is the more reactive set of electrons on the top oxygen. There's no cases where the carboxyl oxygen on the bottom is more reactive. Okay, so we've talked about atomic orbitals. We've talked about hybrid atomic orbitals. We've talked about the supreme importance of p-character. p-character, p-orbitals are higher in energy. More p-character means more nucleophilic. Let's go ahead and talk about mixing hybrid orbitals in order to generate molecular orbitals. Tent was that this is supposed to be review. I'm stunned. I went back and I looked at the sophomore organic chemistry texts that I teach out of and all this stuff is in there. But somehow it just gets lost and dropped. All of these important concepts get lost once you get to the chapters beyond chapter one. Okay, so let's go ahead and build up this picture of starting with a carbon atom that has atomic orbitals on it. So let me go ahead and depict the atomic orbitals on carbon. There's a 2s orbital that's shaped something like this. And I'm not going to worry about the phasing. There's this egg yolk-shaped part of the orbital that's in the center. I'm not even going to draw that because it will make the picture too confusing. Inside here I'm going to draw the picture of a 2x orbital, or sorry, a px orbital. There's a p orbital. I'm kind of overlaying it on top. It looks a little bit confusing. The actual overall shape of a p orbital is, it fits inside a sphere. So ChemDraw draws these orbital shapes that are so badly distorted you'd never know what a real p orbital looks like. So we take two atomic orbitals. Let's suppose we had a carbon atom and we wanted to get that ready for bonding. What we think about is the mixing of the p orbital here with the s orbital here, 2s orbital and a 2p orbital. And we think about the idea that that can hybridize. So we've just been through this, but now I'm not drawing out the energy diagrams. I'm just drawing pictorially the way I think about this. Yeah, an s orbital and a p orbital can get together to make two different sp type orbitals. So here's one of the two sp orbitals.  And I can imagine this interacting with another atom to make a bond. So this is my hybrid orbital here. And I'll draw that interacting with a hydrogen atom. And I'll make sure the phasing combination matters. If I want to represent the formation of a bond, I have to make sure that the phasing, the arbitrary phasing, either hashed or unhashed matches the phasing of this orbital. If they don't match, then it's an anti-bond. It's the opposite of a bond. And when I mix these two orbitals together, when I mix this 1s orbital on hydrogen with this hybridized sp orbital on carbon, sorry, sp orbital on carbon, I get a new bond. And that's a sigma bond. And it has a shape that looks something like this. Let me try to get this right. And if I wanted to give you extra information, I often put subscripts on my orbitals to give you extra information. That's not just a sigma bond. It's a sigma bond between a carbon and a hydrogen atom. And that's different from a sigma bond on a nitrogen and a hydrogen atom. Orbitals on nitrogen are lower in energy and less nucleophilic. CH bonds are higher in energy and more nucleophilic. So this is a molecular orbital. And you should already be familiar with this nomenclature. Double and triple bonds, we call pi bonds, single bonds, we call sigma bonds. You should be familiar with that kind of idea. So I want to return to this the way we started our class. We started our class by talking about arrow pushing. And the idea that we're going to use arrow pushing to help describe one part of a powerful equation for thinking about reactivity. So remember I started off by telling you that any time two things get closer and closer together, if we wanted to think about the interaction between those, the energy we get, either forming a bond or sterically bumping into each other, we have this equation that we use to think about the energetic consequences. We said that this equation obligates us, whenever we have two things interacting with each other, obligates us to add together all the charges. So if I had some sort of an amine interacting with some sort of an electrophilic organic compound like bromomethane, as these get closer and closer together, if I want to calculate how good or bad it is for these things to get closer and closer, I have to think about the charges between the proton and the bromine, the proton and the carbon, the proton and the H. I have to think about the charges on the nitrogen interacting with the proton. You have to add that up for every single atom in your system. If you want to get a full accounting for the overall charge effects as these things come closer together, we have to think about the non-bonding interactions, the steric interactions that don't form bonds, the bumping of this H into that H, the bumping of this nitrogen into that H. In addition to thinking about the formation of a new bond, we have to think about all those bumping interactions that aren't good. And we add all of them together as these get closer and closer together. And then finally, this is our arrow pushing. We agreed that lastly arrow pushing wasn't going to represent sterics, it wasn't going to represent charge, it was going to represent the interaction of this filled orbital, this sp3 hybridized orbital on nitrogen with some sort of an anti-bonding orbital that we usually don't draw with Lewis structures. Now I want to get down to the nitty gritty of this equation because I didn't tell you the form of this equation and now I'm going to have to give you an equation that may look sort of complex but it's not too complex. That's the equation. This is the equation that underlies arrow pushing. It says, what does this equation tell you? That if you want to think about how good your arrow pushing is, is that good arrow pushing or better, is that likely or unlikely? Was that a good idea for me to push arrows like this? You have to think about the interaction spatially of the orbitals. Are they getting close or they're overlapping in space? How good is that degree of overlap? And then secondly, you have to think about the difference in energy between the orbitals. Is this high in energy and is this empty orbital low in energy and it is the difference between the filled orbital energy and the empty orbital energy that determines how much energy you get as these get closer and closer. So let's take a look at that and I want to start off by using something called a molecular orbital interaction diagram. It's a device that you should have been introduced to back in freshman chemistry and then you never used it again. It was like why did we learn that? I have no idea why we learned these molecular orbital interaction diagrams. So let me just start off by reviewing pictorially the molecular orbitals in a molecule. If I were to draw an energy diagram showing the molecular orbitals of ammonia, for example, what would it look like? At the level of resolution that I think. And again, don't worry about the energy on the energy axis. The big point I want to make is you got all these molecular orbitals associated with any molecule. And when you look at the electron occupation, it looks something like this. There's filled orbitals that have two electrons in each and then above that there's empty orbitals. And they're not all mixed together. And always these empty orbitals and I'll call them molecular orbitals, MOs, the empty orbitals are always higher in energy than the filled orbitals. It's just a fundamental fact. If I measure the energy difference between filled and empty, it's there's always some sort of an energy difference associated with them. Okay, so we're going to learn about how to think about the interaction of one molecule here that has energy levels with another molecule that has its own molecular orbitals that are different and how to think about the differences in energy in those orbitals. And instead of thinking about an actual chemical reaction, what I want to do is I want to draw a picture of some sort of a species and I want to ask you what do you think is the geometry of this species? It's a carbocation. Or is it a carbocation? What does this look like? I think that's stabilized and I think you can tell that it's stabilized. But how do you think about the stabilization? Is this like an allyl cation that has this sort of a lillic resonance? It's so that would mean the positive charge is distributed on both of these carbon atoms. But the alternative way of thinking about this is maybe the lone pairs donate into that. And maybe it really looks like this linear molecule that might be best represented like this. And what kinds of tools do you have at your disposal to answer a question like that? They are exactly the same tools that you use to think about this reaction between two molecules. And the difference is here we're talking about two kinds of nucleophiles, a pi bond versus a lone pair that are attached to the same nitrogen atom. It's just the same story. Here I'm talking about a nucleophile on two different molecules. And here I'm talking about two different nucleophiles that are on the same atom. So I'm going to start off by drawing a molecular orbital interaction diagram that will immediately answer that question. And here's what a molecular orbital interaction diagram says. I've got some nucleophile with a filled orbital. I've got some electrophile with an empty orbital. So I'm going to start off by representing the case where the lone pair on nitrogen is interacting or sorry the pi bond on nitrogen, the double bond is interacting so that bond between carbon and nitrogen, the pi bond, if it's donating into the carbocation I need to draw that as my nucleophilic filled orbital. Then on the other atom, the carbon atom, I've got some empty orbital, this empty p orbital, the carbocation orbital on the carbon atom. And I have to think about the interaction of these two orbitals. When I mix two orbitals together, when I donate the electrons from that carbon-nitrogen pi bond into the carbocation, when this donates into that I have to get two new orbitals out. Here's the orbitals in my product. When I mix two orbitals, I get two orbitals out. When I mix a filled orbital with an empty orbital, when I donate these electrons in, I'm going to get two orbitals back out. So here's what the orbitals would look like in my product. One is going to be lower in energy and one is going to be higher in energy. And if I try to diagram, I'm going to try to draw this sort of edge on so you can see what I'm talking about. I'm going to have some empty orbital here on the carbon atom. That's an empty orbital on p carbon. And then there's this pi bond and I'm, I don't know how good of a job I can do to represent this pi bond here, but I'll try to smear some electron density and make that my pi bond. This is the pi bond between the other carbon-nitrogen. So what happens when those two orbitals overlap? What happens when this pi bond between carbon and nitrogen donates into the carbocation there? I'll get two new orbitals out and this one is going to have pi symmetry, the new orbital that I get out. And this is if they're in the correct phasing combination where the top is hashed and the top is hashed, but there's also an anti-phasing combination to the way those two can combine. And when I mix the other phasing combination, I'll get an anti-bonding orbital. We represent that with a star that's just as destabilized as the new bonding combination was stabilized. So I'll generate some new pi bond if I think about that interaction. Now let me contrast that and I'm going to try to quickly draw this diagram because we're running out of time here. Let me contrast that with an alternative scenario and that's the lone pair interacting. And I'm going to try to draw my diagram at the same axis, I didn't label it over here. And I'm trying to try to make sure I draw my p orbital at exactly the same energy because that's not different. The carbocation is the same. But instead of thinking about the pi bond interacting with that orbital, I'm going to think about the lone pair on nitrogen interacting. And what I immediately need to do is think about the idea that non-bonding lone pairs on nitrogen, sp2 hybridized non-bonding lone pairs of nitrogen are higher in energy than pi orbitals. And this is something that we will talk about shortly. I'll tell you how we know that shortly in our next lecture. And if this is higher in energy, I'm going to try to draw a line across here just to show you that I'm trying to draw this non-bonding lone pair on nitrogen being higher in energy. If this is higher in energy, then the difference between these two is smaller. And if the difference between those two is smaller, I've lost my equation here, that equation that I showed you, if the difference in energy is smaller, the overall interaction energy is bigger. In other words, you want more nucleophilic things that are higher in energy. And if these are closer in energy and the energy difference is smaller, the overall interaction energy drop is bigger. You get more interaction energy out when these two are smaller. So this is delta E between our MOs. And I get more interaction energy out, E sub interaction. I get more interaction energy out than when these are far apart. And you kind of knew that. What I'm telling you here is, well, this is less nucleophilic. Of course, it's harder to donate into that empty orbital. This is more nucleophilic. Of course, it's better to donate into that empty orbital. You get more overall bonding energy out. What that tells you is the answer is it's the nitrogen lone pair that you'd want to donate into that cation. This should be a better representation of that structure because nitrogen lone pairs are more nucleophilic. Okay, sorry I went over on my time. We'll get back together on Friday. And I'll tell you about how did I know that this nitrogen non-bonding lone pair would be more nucleophilic than a pi bond? Yeah.