L6.2 Is probability conserved? Hermiticity of the Hamiltonian.

published on July 17, 2020

PROFESSOR: Let's do a work check So main check If integral psi star x t0, psi x t0 dx is equal to 1 at t equal to t0, as we say there,

Then it must hold for later times, t greater than t0 This is what we want to check, or verify, or prove Now, to do it, we're going to take our time So it's not going to happen in five minutes, not 10 minutes, maybe not even half an hour Not because it's so difficult It's

Because there's so many things that one can say in between that teach you a lot about quantum mechanics So we're going to take our time here So we're going to first rewrite it with better notation So we'll define rho of x and t, which

Is going to be called the probability density And it's nothing else than what you would expect, psi star of x and t, psi of x and t It's a probability density

You know that has the right interpretation, it's psi squared And that's the kind of thing that integrated over space gives you the total probability So this is a positive number given by this quantity is called the probability density

Fine What do we know about this probability density that we're trying to find about its integral? So define next N of t to be the integral of rho of x and t dx Integrate this probability density

Throughout space, and that's going to give you N of t Now, what do we know? We know that N of t, or let's assume that N of t0 is equal to 1 N is that normalization It's that total integral of the probability

What had to be equal to 1 Well, let's assume N at t0 is equal to 1 That's good The question is, will the Schrodinger equation guarantee that– and here's the claim–

DN dt is equal to 0? Will the Schrodinger equation guarantee this? If the Schrodinger equation guarantees that this derivative is, indeed, zero,

Then we're in good business Because the derivative is zero, the value's 1, will remain 1 forever Yes? AUDIENCE: May I ask why you specified for t greater than t0?

Well, I don't have to specify for t greater than t naught I could do it for all t different than t naught But if I say this way, as imagining that somebody prepares the system at some time, t naught, and maybe the system didn't exist for other times

Below Now, if a system existed for long time and you look at it at t naught, then certainly the Schrodinger equation should imply that it works later and it works before So it's not really necessary, but no loss of generality

OK, so that's it Will it guarantee that? Well, that's our thing to do So let's begin the work by doing a little bit of a calculation And so what do we need to do?

We need to find the derivative of this quantity So what is this derivative of N dN dt will be the integral d dt of rho of x and t dx So I went here and brought in the d dt, which became a partial derivative Because this is just a function of t, but inside here,

There's a function of t and a function of x So I must make clear that I'm just differentiating t So is d dt of rho And now we can write it as integral dx What this rho? Psi star psi

So we would have d dt of psi star times psi plus psi star d dt of psi OK And here you see, if you were waiting for that,

That the Schrodinger equation has to be necessary Because we have the psi dt And that information is there with Schrodinger's equation So let's do that So what do we have? ih bar d psi dt equal h psi

We'll write it like that for the time being without copying all what h is That would take a lot of time And from this equation, you can find immediately that d psi dt is minus i over h bar h hat psi

Now we need to complex conjugate this equation, and that is always a little more scary Actually, the way to do this in a way that you never get into scary or strange things

So let me take the complex conjugate of this equation Here I would have i goes to minus i h bar, and now I would have– we can go very slow– d psi dt star equals, and then I'll be simple minded here I think it's the best

I'll just start the right hand side I start the left hand side and start the right hand side Now here, the complex conjugate of a derivative, in this case I want to clarify what it is It's just the derivative of the complex conjugate So this is minus ih bar d/dt of psi star

Equals h hat psi star, that's fine And from here, if I multiply again by i divided by h bar, we get d psi star dt is equal to i over h star h hat psi star We obtain this useful formula and this useful formula, and both go into our calculation of dN dt

So what do we have here? dN dt equals integral dx, and I will put an i over h bar, I think, here Yes i over h bar

Look at this term first We have i over h bar, h psi star psi And the second term involves a d psi dt that comes with an opposite sign

Same factor of i over h bar, so minus psi star h psi So the virtue of what we've done so far is that it doesn't look so bad yet And looks relatively clean, and it's very suggestive, actually So what's happening?

We want to show that dN dt is equal to 0 Now, are we going to be able to show that simply that to do a lot of algebra and say, oh, it's 0? Well, it's kind of going to work that way, but we're going to do the work and we're

Going to get to dN dt being an integral of something And it's just not going to look like 0, but it will be manipulated in such a way that you can argue it's 0 using the boundary condition So it's kind of interesting how it's going to work But here structurally, you see what

Must happen for this calculation to succeed So we need for this to be 0 We need the following thing to happen The integral of h hat psi star psi be equal to the integral of psi star h psi

And I should write the dx's They are there So this would guarantee that dN dt is equal to 0 So that's a very nice statement, and it's kind of nice

Is that you have one function starred, one function non-starred The h is where the function needs to be starred, but on the other side of the equation, the h is on the other side So you've kind of moved the h from the complex conjugated

Function to the non-complex conjugated function From the first function to this second function And that's a very nice thing to demand of the Hamiltonian So actually what seems to be happening is that this conservation of probability will work if your Hamiltonian is good enough

To do something like this And this is a nice formula, it's a famous formula This is true if H is a Hermitian operator

It's a very interesting new name that shows up that an operator being Hermitian So this is what I was promising you, that we're going to do this, and we're going to be learning all kinds of funny things as it happens So what is it for a Hermitian operator?

Well, a Hermitian operator, H, would actually satisfy the following That the integral, H psi 1 star psi 2 is equal to the integral of psi 1 star H psi 2 So an operator is said to be Hermitian if you can move it

From the first part to the second part in this sense, and with two different functions So this should be possible to do if an operator is to be called Hermitian Now, of course, if it holds for two arbitrary functions,

It holds when the two functions are the same, in this case So what we need is a particular case of the condition of hermiticity Hermiticity simply means that the operator does this thing Any two functions that you put here, this equality is true

Now if you ask yourself, how do I even understand that? What allows me to move the H from one side to the other? We'll see it very soon But it's the fact that H has second derivatives, and maybe you can integrate them by parts and move the derivatives from the psi 1 to the psi 2,

And do all kinds of things But you should try to think at this moment structurally, what kind of objects you have, what kind of properties you have And the objects are this operator that controls the time evolution, called the Hamiltonian

And if I want probability interpretation to make sense, we need this equality, which is a consequence of hermiticity Now, I'll maybe use a little of this blackboard I haven't used it much before In terms of Hermitian operators, I'm almost there with a definition of a Hermitian operator

I haven't quite given it to you, but let's let state it, given that we're already in this discussion of hermiticity So this is what is called the Hermitian operator, does that But in general, rho, given an operator T, one defines its hermitian conjugate P dagger as follows So you have the integral of psi 1 star T

Psi 2, and that must be rearranged until it looks like T dagger psi 1 star psi 2 Now, these things are the beginning of a whole set of ideas that are terribly important in quantum mechanics Hermitian operators, or eigenvalues and eigenvectors

So it's going to take a little time for you to get accustomed to them But this is the beginning You will explore a little bit of these things in future homework, and start getting familiar For now, it looks very strange and unmotivated

Maybe you will see that that will change soon, even throughout today's lecture So this is the Hermitian conjugate So if you want to calculate the Hermitian conjugate, you must start with this thing, and start doing manipulations to clean up the psi 2, have nothing at the psi 2,

Everything acting on psi 1, and that thing is called the dagger And then finally, T is Hermitian if T dagger is equal to T So its Hermitian conjugate is itself It's almost like people say a real number is a number whose complex conjugate is equal to itself

So a Hermitian operator is one whose Hermitian conjugate is equal to itself, and you see if T is Hermitian, well then it's back to T and T in both places, which is what we've been saying here This is a Hermitian operator

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