Calculate d<p>/dt - 65 Characters

[kq6up]

Homework Statement

Calculate $$\frac { d\langle { p }\rangle }{dt}=\left< -\frac { \partial V }{ \partial x } \right> $$

Homework Equations

$$\hat{H}\Psi=i\hbar\frac{\partial \Psi}{\partial t}$$ and $$\langle { p }\rangle=-i\hbar\int _{ -\infty }^{ \infty }{ { \Psi }^{ * } } \frac { \partial }{ \partial x } \Psi dx$$

The Attempt at a Solution

By the product rule $$\frac{\langle { p }\rangle}{dt} =-i\hbar\int _{ -\infty }^{ \infty }{ { \frac { \partial \Psi ^{ * } }{ \partial t } }\frac { \partial \Psi }{ \partial x } +\Psi ^{ * }\frac { \partial }{ \partial t } } \frac { \partial }{ \partial x } \Psi dx$$

The partial derivatives are commutative, so $$\frac { \partial }{ \partial t } \frac { \partial }{ \partial x } \Psi=\frac { \partial }{ \partial x } \frac { \partial }{ \partial t } \Psi$$ Alrighty, so now we can subsitute
$$\frac { \partial \Psi }{ \partial t }$$ for $$\frac{\hat{H}\Psi}{i\hbar}$$. This yields:
$$\frac{\langle { p }\rangle}{dt}=-\int _{ -\infty }^{ \infty }{ { \frac { \partial \Psi ^{ * } }{ \partial x } }\hat { H } \Psi +\Psi ^{ * }\frac { \partial }{ \partial x } } \hat { H } \Psi dx$$

Subbing in the hamiltonian turns this into a total mess, as does integration by parts. I quickly glanced at the solution, and it didn't look so messy, so I am not sure if I made a mistake or went down a rabbit hole.

If you see any errors, let me know.

Thanks,
Chris Maness

 

[Chopin]

You've made a mistake in your final step. Check the first term of the integral--you've swapped around some derivatives in a way which isn't legal.

 

[kq6up]

So are you saying this bit is false?:

The partial derivatives are commutative, so $$\frac { \partial }{ \partial t } \frac { \partial }{ \partial x } \Psi=\frac { \partial }{ \partial x } \frac { \partial }{ \partial t } \Psi$$

Chris

 

[Chopin]

No, you're correct that $\frac{\partial}{\partial t}\frac{\partial}{\partial x}\Psi = \frac{\partial}{\partial x}\frac{\partial}{\partial t}\Psi$. However, $\frac{\partial}{\partial t}\Psi^*\frac{\partial}{\partial x}\Psi \neq \frac{\partial}{\partial x}\Psi^*\frac{\partial}{\partial t}\Psi$.

 

[kq6up]

No, you're correct that $\frac{\partial}{\partial t}\frac{\partial}{\partial x}\Psi = \frac{\partial}{\partial x}\frac{\partial}{\partial t}\Psi$. However, $\frac{\partial}{\partial t}\Psi^*\frac{\partial}{\partial x}\Psi \neq \frac{\partial}{\partial x}\Psi^*\frac{\partial}{\partial t}\Psi$.

Hmm, that is subtle. I am refreshing my skills, so what math text would you recommend to practice this type of operation? Would this come up in Freshman/Sophomore calculus, or later on (Diff Eq's for example).

Regards,
Chris Maness

 

[Chopin]

Yeah, the interchange of partial derivatives only applies when both derivatives are applied in sequence to the same function. In this case, you have two derivatives applied to different functions, so no interchange is possible. This sort of thing usually gets discussed after partial derivatives have been introduced in detail, which is generally in a multivariate calculus course. For me, that was in third-semester calculus in college.

Backing up a bit, though, I always found doing these sorts of manipulations using the wavefunction approach (where you treat $\Psi$ as a function that has to be integrated over space) to be really cumbersome, because getting the correct answer involves dealing with a lot of really subtle issues like this involving how to order derivative operators correctly. If you follow all the rules correctly, you'll get the right answer in the end, but it gets really tedious. Proving Ehrenfest theorems like this, as well as many other operations in quantum mechanics, gets a lot easier in my opinion if you can learn to deal with the bra-ket notation, where you can work with operators and commutators instead of integrals and partial derivatives. The underlying mathematics are exactly the same, but the notation hides a lot more of the drudgery, and makes the important parts easier to see.

Have you been introduced to this notation yet? The notation I'm referring to is the one which leads to equations such as $\frac{d}{dt}\hat{O} = \frac{i}{\hbar}[\hat{H}, \hat{O}]$ and $[\hat{x}, \hat{p}] = i\hbar$.

 

[kq6up]

Yeah, the interchange of partial derivatives only applies when both derivatives are applied in sequence to the same function. In this case, you have two derivatives applied to different functions, so no interchange is possible. This sort of thing usually gets discussed after partial derivatives have been introduced in detail, which is generally in a multivariate calculus course. For me, that was in third-semester calculus in college.

Backing up a bit, though, doing these sorts of manipulations using the wavefunction approach (where you treat $\Psi$ as a function that has to be integrated over space) can get really cumbersome, because getting the correct answer involves dealing with a lot of really subtle issues like this involving how to order derivative operators correctly. If you follow all the rules correctly, you'll get the right answer in the end, but it gets really tedious. Proving Ehrenfest theorems like this, as well as many other operations in quantum mechanics, get a lot easier if you can learn to deal with the bra-ket notation, where you can work with operators and commutators instead of integrals and partial derivatives. The underlying mathematics are exactly the same, but the notation hides a lot more of the drudgery, and makes the important parts easier to see.

Have you been introduced to this notation yet? The notation I'm referring to is the one which leads to equations such as $\frac{d}{dt}\hat{O} = \frac{i}{\hbar}[\hat{H}, \hat{O}]$ and $[\hat{x}, \hat{p}] = i\hbar$.

Yes, I am familiar. I actually gave up earlier, and tried to do the same thing using braket notation, but got stuck there too. Ok, I don't feel so bad about this problem kicking my butt, but Griffiths gave it one star. I have been at it for several hours.

Thanks,
Chris Maness

 

[Chopin]

Have a look at http://en.wikipedia.org/wiki/Ehrenfest_theorem. It covers a derivation very similar to the one you're trying to do (including the proper way to deal with that first term in your integral), as well as describing how the equation arises naturally from the Heisenberg equations of motion using bra-ket notation.

 

[kq6up]

This is as far as I got with the braket: $$\frac{\langle { p }\rangle}{dt}=\frac{1}{i\hbar}\langle \hat{H} \Psi | \hat{p} \Psi \rangle+\frac{1}{i\hbar}\langle \Psi | \hat{p} \hat{H} | \Psi \rangle$$

Chris

 

[Chopin]

Start with just an operator equation, ignoring the expectation value part. The Heisenberg equations of motion tell us that:
$$\frac{d}{dt}\hat{p} = \frac{i}{\hbar}[\hat{H}, \hat{p}]$$
So now, all of the integral manipulations we were dealing with above reduce to simply evaluating this commutator. For the specific Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x})$, the commutator of $\hat{p}$ with the first term will evaluate to 0, so all we have to deal with is $[V(\hat{x}), \hat{p}]$.

The end result that we find is that $[V(\hat{x}), \hat{p}] = i\hbar\frac{d}{dx}V(\hat{x})$. There may be more elegant ways to prove this, but the one that I'm aware of is to simply use the canonical commutation relations $[x, p]=i\hbar$ to show that $[\hat{x}^n, \hat{p}] = i\hbar \cdot n\hat{x}^{n-1}$. This proves it for all polynomials, and by extension for all functions $V(\hat{x})$ which are equal to their Taylor expansion.

So with that little excursion behind us, we plug everything together, and find that $\frac{d}{dt}\hat{p} = -\frac{d}{dx}V(\hat{x})$. Put all of that inside of an expectation value, and the Ehrenfest Theorem is proven, thus providing a proof at the quantum level of Newton's Second Law in classical physics.

 

[kq6up]

I was able to derive the Ehrenfest theorem like this:

http://en.wikipedia.org/wiki/Ehrenfest_theorem#Derivation_in_the_Schr.C3.B6dinger_picture

However, my math chops are not up to where I can even follow the proof that I was trying to do.

This proof was on that page as well. $$\frac{d\langle p \rangle }{dt}=\left< -\frac{\partial{V}}{\partial{x}} \right>$$

I am debating on whether to put the book down, and wait until I get further into my math methods book (Mary Boas). I just finished chapter three. I felt like I had to do a little physics for a change because just studying the maths can get really dry.

Or maybe it is worth just skipping the bits that take me more than 15 minutes or so, and plugging through Griffiths. Then when my math skills are sharper, go back and look at what I skipped.

So from a pedagogical point of view, what do you think?

Thanks,
Chris Maness

 

[Chopin]

Using the wavefunction approach in which you originally posed this question, you should be able to derive it like so:
$$\frac{\partial}{\partial t} \langle p \rangle = -i\hbar\frac{\partial}{\partial t} \int dx \Psi(x)^*\frac{\partial}{\partial x}\Psi(x) \\ = -i\hbar\int dx \left(\frac{\partial}{\partial t}\Psi(x)^*\right)\left(\frac{\partial}{\partial x}\Psi(x)\right) + \Psi(x)^*\frac{\partial}{\partial t}\frac{\partial}{\partial x}\Psi(x) \\ = -i\hbar\int dx -\Psi(x)^*\frac{\hat{H}}{i\hbar}\frac{\partial}{\partial x}\Psi(x) + \Psi(x)^*\frac{\partial}{\partial x}\left(\frac{\hat{H}}{i\hbar}\Psi(x)\right)$$

where in the last line we use the definition of the conjugate $(\frac{\partial}{\partial t}\Psi)^* = (\frac{\hat{H}}{i\hbar}\Psi)^* = -\frac{1}{i\hbar}\Psi^*\hat{H}^* = -\frac{1}{i\hbar}\Psi^*\hat{H}$

Then substitute in the Hamiltonian:
$$-\int dx -\Psi(x)^*\hat{H}\frac{\partial}{\partial x}\Psi(x) + \Psi(x)^*\frac{\partial}{\partial x}(\hat{H}\Psi(x)) \\ = -\int dx -\Psi(x)^*\left(\frac{1}{2m}\frac{\partial}{\partial x}^2 + V(x)\right)\frac{\partial}{\partial x}\Psi(x) + \Psi(x)^*\frac{\partial}{\partial x}\left(\left(\frac{1}{2m}\frac{\partial}{\partial x}^2 + V(x)\right)\Psi(x)\right)$$

The first terms cancel, leaving:
$$-\int dx -\Psi(x)^*V(x)\frac{\partial}{\partial x}\Psi(x) + \Psi(x)^*\frac{\partial}{\partial x}(V(x)\Psi(x))\\ = -\int dx -\Psi(x)^*V(x)\left(\frac{\partial}{\partial x}\Psi(x)\right) + \Psi(x)^*\left(\frac{\partial}{\partial x}V(x)\right)\Psi(x) + \Psi(x)^*V(x)\left(\frac{\partial}{\partial x}\Psi(x)\right) \\ = -\int dx \Psi(x)^* \left(\frac{\partial}{\partial x}V(x)\right)\Psi(x) \\ = -\left\langle\frac{\partial}{\partial x}V\right\rangle$$

As for how to study this stuff further, I guess I'm not sure what to say. I've only ever self-studied any of this as well, so I don't have any "official" opinion on this, except for what worked for me. In my case, I went back and forth quite a lot on some of these problems--working them out as best I could, but frustrated with some aspects of them, and then moving further along, learning a bit more, and then going back and solving them again with the benefit of some more perspective. That eventually worked out all right, although it certainly took a lot of time to finally get there.

I will say that once you get further along, you often end up learning things in such a different way that you eventually look back on the work you did before, and can blast through in 5 minutes something that used to have you completely stumped. With the benefit of new perspective on what you're actually doing, you can think about it in a different way that completely sidesteps the difficulties you'd had before. So I think there's some value in moving on from these problems before you feel like you have them 100% under your belt, and cycling back later.

In this case, getting some more experience with bra-ket notation and understanding how operator equations work in the Heisenberg Picture (which you'll do more of as you go further into any QM book) will eventually make equations like this fairly simple to solve, where mucking around with all the wavefunction integrals gets pretty unwieldy, and is usually dropped completely in textbooks once the more advanced notation is established. I haven't read Griffiths, so I'm not sure how they do it, but that seems to have been the case with the books I have seen in the past.

 

[kq6up]

Thanks @Chopin. I am not used to having any challenges with my math. I have regained my confidence, and you sharing your experience with this stuff helps. I know E&M will be at least this challenging :D

Thanks,
Chris Maness

 

[Chopin]

No problem--best of luck to you as you venture further down the rabbit hole!

 

[kq6up]

@Chopin After calming down and going through your last proof above line for line. It makes perfect sense. Very cool -- thank you.

Chris

 

[unscientific]

Hmm, that is subtle. I am refreshing my skills, so what math text would you recommend to practice this type of operation? Would this come up in Freshman/Sophomore calculus, or later on (Diff Eq's for example).

Regards,
Chris Maness

I would strongly recommend Mathematical Methods for Physics and Engineering by "Riley, Hobson and Bence". It is widely used as an undergraduate text for physics majors worldwide.

 

[kq6up]

Finding $$\frac{d\left< p \right>}{dt}$$ with Ehrenfest's theorem is pretty straight forward for me now with bra-ket notation. I can also derive Ehrenfest with Schrödinger's equation. However I do have a couple of lingering questions about my result.

1. I am not sure why $$\frac{d\left< p \right>}{dt}$$ would not be equal to zero, but $$\left< \frac{dp}{dt} \right>$$ does equal zero.

Edit, I know the following part is wrong, I just don't understand why.

2. When I take the time derivative of $$\left< \Psi | \hat{p} | \Psi \right>$$ I am not sure why the middle term can't be just $$\frac{1}{i\hbar}\left< \Psi | \hat{H} \hat{p} | \Psi \right>$$ in stead of $$\left< \frac{dp}{dt} \right>$$.

For reference here is my result $$\frac{d\left< p \right>}{dt}=\frac{1}{i\hbar}\left< \Psi | \left[ \hat{p} , \hat{H} \right] | \Psi \right>+\left< \frac{dp}{dt} \right>$$.

Making the above considerations and subbing in the commutator, the theorem yields $$-\left< \frac{dV}{dx} \right>$$ as it should.

Thanks,
Chris

 

[vanhees71]

I haven't followed this thread from the beginning, but #17 is flawed. By the fundamental postulates of quantum theory, if $\hat{A}$ is a self-adjoint operator representing an observable that is not explicitly time dependent but only a function of the fundamental operators $\hat{x}$ and $\hat{p}$, then the operator
$$\mathrm{D}_t \hat{A}:=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]$$
represents the time derivative of the observable $A$.

Now we have
$$\frac{1}{\mathrm{i} \hbar}[\hat{p},\hat{H}]=-\frac{\partial \hat{H}}{\partial \hat{x}},$$
and this implies Ehrenfest's theorem for the momentum operator, which reads
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle p \rangle=-\left \langle \frac{\partial \hat{H}}{\partial \hat{x}} \right \rangle.$$

 

[kq6up]

I haven't followed this thread from the beginning, but #17 is flawed.

What part do you think is wrong? I state in the question that one of the results is incorrect because it does not agree with Ehrenfest, but I am not sure why (thus the reason for the post). However, I don't think the last line before the result is wrong -- because that is the Ehrenfest theorem itself. Just plug in the commutator that gives $$-\frac{\partial V}{\partial x}$$.

EDIT: Take a look at the red font on my post above.

Chris

 

[Oxvillian]

Finding $$\frac{d\left< p \right>}{dt}$$ with Ehrenfest's theorem is pretty straight forward for me now with bra-ket notation. I can also derive Ehrenfest with Schrödinger's equation. However I do have a couple of lingering questions about my result.

1. I am not sure why $$\frac{d\left< p \right>}{dt}$$ would not be equal to zero, but $$\left< \frac{dp}{dt} \right>$$ does equal zero.

Edit, I know the following part is wrong, I just don't understand why.

2. When I take the time derivative of $$\left< \Psi | \hat{p} | \Psi \right>$$ I am not sure why the middle term can't be just $$\frac{1}{i\hbar}\left< \Psi | \hat{H} \hat{p} | \Psi \right>$$ in stead of $$\left< \frac{dp}{dt} \right>$$.

For reference here is my result $$\frac{d\left< p \right>}{dt}=\frac{1}{i\hbar}\left< \Psi | \left[ \hat{p} , \hat{H} \right] | \Psi \right>+\left< \frac{dp}{dt} \right>$$.

Making the above considerations and subbing in the commutator, the theorem yields $$-\left< \frac{dV}{dx} \right>$$ as it should.

Thanks,
Chris

You're working here in what's called the "Schrodinger picture" of quantum mechanics, which means that as time goes by, the state vectors change but operators like $X$ and $P$ stay fixed. So when you evaluate

$$\frac{d}{dt}\langle \psi \vert \ P \vert \psi \rangle$$

you'll only get two terms - one because $\langle \psi \vert$ is changing and one because $\vert \psi \rangle$ is changing. $P$ is a constant operator.

 

[kq6up]

You're working here in what's called the "Schrodinger picture" of quantum mechanics, which means that as time goes by, the state vectors change but operators like $X$ and $P$ stay fixed. So when you evaluate

$$\frac{d}{dt}\langle \psi \vert \ P \vert \psi \rangle$$

you'll only get two terms - one because $\langle \psi \vert$ is changing and one because $\vert \psi \rangle$ is changing. $P$ is a constant operator.

I need a little more unpacking to understand that, can you give a physical example so that I can visualize and satisfy myself that this is indeed what is going (not saying it isn't). I can see that in a closed system P is conserved and therefore time independent. However, if P is conserved and the equations are a complete description of the whole system, then all of the forces should cancel too. I do realize that we are talking about systems that are quantum mechanically fuzzy. Therefore I am referring to the expectation value classically, as Ehrenfest basically states that it is ok to treat expectation values classically -- if I understand correctly :D

Thanks,
Chris Maness

 

[Oxvillian]

kq6up - you shouldn't think of the operator $P$ as "being" the momentum. Think of it instead as a mathematical tool for extracting information about the momentum of the state $\vert\psi\rangle$.

In the Schrodinger picture, $P$ has the representation

$$P = -i\hbar\frac{\partial}{\partial x},$$

a form which is entirely independent of (1) time and (2) whatever system is under consideration, whether it happens to conserve momentum or not.

Note that all of the above pertains to the "Schrodinger picture" where the state vectors change in time and the operators representing observables are fixed. There's another (equivalent) picture called the "Heisenberg picture" (see post #18 above) where the state vectors are fixed and the operators evolve in time. That picture and a hybrid picture called the "interaction picture" are useful in more advanced quantum mechanics, eg. quantum field theory.