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Consider a system whose hamiltonian isn't explicitly dependent on time. Let A be the operator for the eigenvalue a in the Schrödinger picture and $A_H=U^\dagger A U$ the corresponding operator in the Heisenberg picture, with $U=\exp(-\frac{iHt}{\hbar})$. Let the initial state $\left|\Psi (0)\right >$ be the eigenstate of A.

a) Show that $A_H(-t)\left|\Psi (t)\right >=a\left|\Psi(t)\right>$

b) Consider a one-dimensional (free of forces) motion of a particle of mass m. Solve Heisenberg's equation of motion

$\frac{dA_H(t)}{dt}=-\frac{i}{\hbar}[A_H(t),H]$

for the operators $X_H(t)$ and $P_H(t)$. The constraints are $X_H(0)=X$ and $P_H(0)=P$.

Here's what I got so far:

a): Since A is already the operator to the eigenvalue a in the Schrödinger picture, and I think this applies to $A_H$ in the Heisenberg picture as well (?), all I need to show is that $A_H(t)=A_H(-t)$, right? I thought since $U=\exp(-\frac{iHt}{\hbar})$ the '-' would only change it so that, $A_H=U A U^\dagger$ (the hermitian conjugate of U is now applied from the right instead from the left). But that's the same isn't it? Would that suffice as a solution?

b): First I took out the unitary operator from the commutator

$[A_H(t),H]=U[A,H]U^\dagger$.

Now, I could plug in X for A and H is simply $\frac{p^2}{2m}$ since it's free of forces. Now I just plug in $p=-i\hbar\partial_x$ and solve the commutator

$U(-i\hbar x\partial_x+i\hbar)U^\dagger$ (To be honest I don't know what happens when you apply the position operator on a derivative so I just left it there).

But I don't know how to continue from here.

Someone have any ideas? And is my approach in a) even correct?

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closed as off-topic by yuggib, ACuriousMind, Gert, user36790, John Rennie Jun 24 '16 at 6:14

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    $\begingroup$ Hint for a): $\psi(t)=U(t)\psi(0)$, and $A(-t)=U^*(-t)AU(-t)=U(t)A U^*(t)$ by definition of $U(t)$. Then $A(-t)\psi(t)=U(t) A U^*(t)U(t)\psi(0)$. What does $U^*(t)U(t)$ equals to? If you answer to that correctly, then it is trivial... Concerning b), you need to justify why you can take out the unitary operator $U(t)$ in the commutator (it is true only because this one has a special form); then you have simply to use the usual canonical commutation relations. $\endgroup$ – yuggib Jun 23 '16 at 10:00
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    $\begingroup$ I don't understand the close votes. Sure it's an easy question if you know this stuff, but the OP is trying to learn it and has indeed shown his/her working and effort. $\endgroup$ – WetSavannaAnimal Jun 23 '16 at 13:13
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Use User Yuggib's hint for part (a);

For part b), solve the equation of motion for $p$ first - it's really simple. For the equation of motion for $x$, as you realize, you need to find

$$[x,\,H] \propto [x,\,p^2]=x\,p\,p-p\,p\,x\tag{1}$$

Before you plug anything in, try bringing one of the terms into the same order as the other with the help of the CCR; it sounds as though this isn't something you've either come across or thought of:

$$x\,p\,p = (p\,x + [x,\,p])\,p = (p\,x + i\,\hbar\,\mathrm{id})\,p = p\,x \,p+ i\,\hbar\,p$$

Can you see where this is going? Do this trick again to re-order $p\,x\,p$ and then put the final results into (1). You'll find you'll get a very simple expression, especially in the light of your solution for $p$.

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  • $\begingroup$ Sorry for my late response. I wasn't at home for the last couple of hours. You and yuggib helped me a ton. I wouldn't have come up with the idea to use the CCR in this problem. Nice trick to remember. :). But just to be sure: $[x,p^2]=2i\hbar p$ and $[p,p^2]=0$, right? $\endgroup$ – Rab Jun 23 '16 at 19:05
  • $\begingroup$ @Rab You've got it! That ordering trick is used a great deal in quantum optics and field theory too - one talks of Normal Ordering for raising and lowering operators. I'm really glad we could help! Sorry about the close votes; that kind of thing infuriates me. Enjoy physics and learning about our beautiful World. $\endgroup$ – WetSavannaAnimal Jun 23 '16 at 22:08

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