Why are <dx/dt> and <dp/dt> 0?

[EquationOfMotion]

Hello,

on the Wikipedia page for a general example of the Ehrenfest theorem, they note that <dp/dt> and <dx/dt> because the operators p and x have no explicit time dependence. But we have

<dp/dt> = <ψ | (d/dt)pψ>
<dx/dt> = <ψ | (d/dt)xψ>

and I can't seem to prove that these inner products (which become integrals) go to 0. I'm definitely misunderstanding something here, so any help is much appreciated.

Thanks!

 

[king vitamin]

Your link doesn't show that they are zero, so I don't follow your question. The link also does not contain the equations you've written out here.

 

[EquationOfMotion]

Your link doesn't show that they are zero, so I don't follow your question. The link also does not contain the equations you've written out here.

"Suppose we wanted to know the instantaneous change in momentum p. Using Ehrenfest's theorem, we have

<[p, H]>/(ih) + <dp/dt> = <[p, V]>/(ih)

since the operator p commutes with itself and has no time dependence." Given H = p^2/(2m) + V and p = -ih(d/dx), [p, H] = [p,V], which implies that <dp/dt> = 0. Ditto for <dx/dt>.

 

[king vitamin]

Given H = p^2/(2m) + V and p = -ih(d/dx), [p, H] = [p,V], which implies that <dp/dt> = 0.

It does not imply that, because V may be (in fact, it usually is) a function of x.

 

[EquationOfMotion]

It does not imply that, because V may be (in fact, it usually is) a function of x.

Do we not have that <[p, H]>/(ih) + <dp/dt> = <[p, V]>/(ih) + <dp/dt>? Why not?

 

[king vitamin]

We do, but I'm not understanding how that leads to <dp/dt> = 0. I feel that you are also not being careful in distinguishing partial and total derivatives... in this post I am using the total derivative; are you asking why the partial derivative vanishes?

 

[EquationOfMotion]

We do, but I'm not understanding how that leads to <dp/dt> = 0. I feel that you are also not being careful in distinguishing partial and total derivatives... in this post I am using the total derivative; are you asking why the partial derivative vanishes?

I am indeed not being careful with partial and total derivatives, apologies for the confusion. More correctly, we have

<[p, H]>/(ih) + <∂p/∂t> = <[p, V]>/(ih) + <∂p/∂t> = <[p, V]>/(ih), which implies <∂p/∂t> = 0.

 

[PeroK]

I am indeed not being careful with partial and total derivatives, apologies for the confusion. More correctly, we have

<[p, H]>/(ih) + <∂p/∂t> = <[p, V]>/(ih) + <∂p/∂t> = <[p, V]>/(ih), which implies <∂p/∂t> = 0.

It's not just that $\langle \frac{\partial p}{\partial t} \rangle = 0$; it's that $\frac{\partial p}{\partial t} = 0$. I.e. the form of the operator does not change over time.

A time-dependent operator would be something like $e^{t}p$, for example.

 

[EquationOfMotion]

It's not just that $\langle \frac{\partial p}{\partial t} \rangle = 0$; it's that $\frac{\partial p}{\partial t} = 0$. I.e. the form of the operator does not change over time.

A time-dependent operator would be something like $e^{t}p$, for example.

Doesn't this mean $\frac{\partial}{\partial t} \hat{p} f(x,t)$ for any function $f(x,t)$? Any intuition on this?

 

[PeroK]

Doesn't this mean $\frac{\partial}{\partial t} \hat{p} f(x,t)$ for any function $f(x,t)$? Any intuition on this?

I don't understand your question. The operator $p$ doesn't change over time. I.e. the operator representing momentum doesn't change, as an operator, over time.

 

[EquationOfMotion]

I don't understand your question. The operator $p$ doesn't change over time. I.e. the operator representing momentum doesn't change, as an operator, over time.

Are you allowed to directly take partial derivatives of operators? I was under the impression you needed a test function.

 

[PeroK]

Are you allowed to directly take partial derivatives of operators? I was under the impression you needed a test function.

Your getting confused. An operator can be made dependent of time, as an operator. You only need test functions to see what an operator does.

If you take my example. Let $L$ be a linear operator. You can define a time dependent linear operator by $T = e^tL$.

Now, $T$ is an operator that changes (as an operator) with time. If you apply it at $t=0$, then $T(0) = L$. But, if you apply it at time $t=1$, then $T(1) = eL$.

The measurements you get over time now depend on two things: the time evolution of the system (which determines the expected value of $L$) and the time dependence of your operator $T$. If the state of the system is $\Psi(x, t)$, then:

$T\Psi = e^tL\Psi$

And, if you take the (total) time derivative of that (exercise for you), the you can see that it depends on the change in the system and the change in your operator $T$.

 

[EquationOfMotion]

Your getting confused. An operator can be made dependent of time, as an operator. You only need test functions to see what an operator does.

If you take my example. Let $L$ be a linear operator. You can define a time dependent linear operator by $T = e^tL$.

Now, $T$ is an operator that changes (as an operator) with time. If you apply it at $t=0$, then $T(0) = L$. But, if you apply it at time $t=1$, then $T(1) = eL$.

So more correctly, we'd have $(\frac{\partial p}{\partial t})Ψ$. But of course, you take the time derivative of $p$ first, which means this becomes $(0)Ψ = 0$? Or is there something else I'm missing?

 

[PeroK]

So more correctly, we'd have $(\frac{\partial p}{\partial t})Ψ$. But of course, you take the time derivative of $p$ first, which means this becomes $(0)Ψ = 0$? Or is there something else I'm missing?

I'm sorry to say that I can't make much sense of that.

 

[PeterDonis]

we have

<[p, H]>/(ih) + <∂p/∂t> = <[p, V]>/(ih) + <∂p/∂t> = <[p, V]>/(ih), which implies <∂p/∂t> = 0.

You have the logic backwards. It is already known that $\partial p / \partial t = 0$; that is not deduced from the equation quoted above. Rather, because it is already known that $\partial p / \partial t = 0$, its expectation value $\langle \partial p / \partial t \rangle$ must be zero as well, so it can be removed from the equation.

How do we know $\partial p / \partial t = 0$? Because the operator $p$ is $- i \hbar \partial / \partial x$, which obviously does not depend on $t$, so its partial derivative with respect to $t$ is zero.