# Check my working please? Expectation of momentum

1. Jun 8, 2014

### unscientific

1. The problem statement, all variables and given/known data

A state at time t is given by:

$$|\psi\rangle = \frac{1}{\sqrt 2}\left[ e^{-\frac{i\omega t}{2}}|0\rangle + e^{-i\frac{3\omega t}{2}}|1\rangle \right]$$

Where eigenfunctions are $\phi_0 = \left(\frac{1}{a^2 \pi}\right)^{\frac{1}{4}}e^{-\frac{x^2}{2a^2}}$ and $\phi_1 = \left(\frac{4}{a^6 \pi}\right)^{\frac{1}{4}} x \space e^{-\frac{x^2}{2a^2}}$.

I'm supposed to find $\langle \psi|\hat p|\psi \rangle$.

2. Relevant equations

$$\hat p = -i\hbar \frac{\partial}{\partial x}$$

3. The attempt at a solution

By symmetry, we know that $\langle 0|\hat p|0 \rangle = \langle 1 | \hat p | 1 \rangle = 0$.

Starting:

$$\frac{1}{2} \langle 0 | e^{-i\omega t} \hat p |1\rangle$$
$$= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}} \frac{\partial}{\partial x} \left(x e^{-\frac{x}{2a^2}}\right) dx$$
$$= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}}\left(e^{-\frac{x^2}{2a^2}} - \frac{x^2}{a^2} e^{-\frac{x^2}{2a^2}}\right) dx$$
$$= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{a^2}} \left(1 - \frac{x^2}{a^2}\right) dx$$
$$= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \left[ \sqrt \pi a - \frac{1}{a^2} \frac{1}{2} \sqrt {\pi} a^3 \right]$$
$$= -\frac{i\hbar}{2} \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}} \left(\frac{1}{2} \sqrt \pi a \right) e^{-i\omega t}$$
$$= -\frac{i\hbar}{2\sqrt 2 a} e^{-i\omega t}$$

Similarly, for $\frac{1}{2}\langle 1 | e^{i\omega t} \hat p | 0 \rangle$, we take the complex conjugate of above.

Adding them both together, we get:

$$-\frac{\hbar}{\sqrt 2 a} sin (\omega t)$$

Given $\langle \psi |x|\psi\rangle = \frac{a}{\sqrt 2} cos (\omega t)$, then I'm supposed to find:
$$\frac{\partial}{\partial t} \left( \langle \psi|p|\psi\rangle + m\omega^2 \langle \psi |x|\psi\rangle \right)$$

They don't add up to zero, which is strange since the first term is simply the force, and the second term is the restoring force..

Last edited: Jun 8, 2014
2. Jun 9, 2014

### Oxvillian

Surely this would be much easier if you just express $X$ and $P$ in terms of creation & annihilation operators?

3. Jun 9, 2014

### unscientific

They aren't given in the question at all, so I don't think we need to use them.

4. Jun 10, 2014

### strangerep

Yeah, sorry. I did look at this one, but couldn't see anything obviously wrong with your working.

But you make a valid point, so I'm trying again to understand what's going on/wrong here...

Isn't there a dimensional problem here? Momentum has dimensions $MLT^{-1}$ and $\omega$ has dimensions $T^{-1}$.
So $m\omega^2 x$ has dimensions $MLT^{-2}$.
If I got that right, then you can't add $p$ and $m\omega^2 x$.
(In this problem, $a$ must have dimensions $L$ for consistency, so that seems consistent with the expectation values you found.)

Are you perhaps mixing up the partial time derivative with the total time derivative? Cf. Ehrenfest's theorem .

5. Jun 10, 2014

### Oxvillian

Perhaps you mean

$$\frac{d}{dt} \left( \langle \psi|P|\psi\rangle\right) + m\omega^2 \langle \psi |X|\psi\rangle = 0$$

(which would be true)

6. Jun 10, 2014

### unscientific

Yes, I meant that, would that make a difference to the answer? I know it's true, because it commutes with $hat H$, but the problem is how to show it..

7. Jun 10, 2014

### strangerep

From this, I guess you didn't bother to read the link I suggested earlier at the end of post #4 ?

8. Jun 11, 2014

### unscientific

Sorry, I do know Ehrenfest's theorem. What I was trying to say is since all the terms commute with the hamiltonian it doesn't make a difference whether it was a partial or total derivative.

9. Jun 11, 2014

### Oxvillian

I don't know what you mean here, since both $X$ and $P$ fail to commute with $H$.

But the main point is that in the version I posted, the time derivative only acts on the first term. That fixes (among other things) the dimensional inconsistency pointed out by strangerep.

10. Jun 11, 2014

### strangerep

As with Oxvillian, I don't know what you mean by this. Neither x nor p commute with H.

So what is the question?? In your opening post you just wanted to calculate the $p$ expectation. Is there another part to the question that you haven't shown us?

11. Jun 12, 2014

### unscientific

Here is the full question:

I'm trying to find the link between finding $\langle x \rangle$, $\langle p\rangle$ and the last part of the question.

12. Jun 12, 2014

### strangerep

OK, the actual question is simpler than I thought.

So... have you evaluated $$\frac{d}{dt} \langle p \rangle ~=~ ~~?$$If so, what do you get? Also, what classical dynamical quantity does this correspond to?

And what do you get when you add it to $m\omega^2 \langle x \rangle$ ? (And what classical quantity does that correspond to?)

(I'm still having trouble understanding what your problem really is.)

Last edited: Jun 12, 2014
13. Jun 13, 2014

### unscientific

My problem is that they don't sum to zero, as forces should balance.

$$\frac{d\langle p \rangle}{dt} = -\frac{\hbar \omega}{\sqrt 2 a} cos (\omega t)$$

$$m\omega^2 \frac{d \langle x \rangle }{dt} = -\frac{ma\omega^3}{\sqrt 2} sin (\omega t)$$

They are wildly different.

14. Jun 13, 2014

### Oxvillian

unscientific - I think you have us pulling out our hair in frustration here

Why do you still insist on differentiating $\langle x \rangle$?

15. Jun 13, 2014

### strangerep

Yep!

He's seeing parentheses where there aren't any.

16. Jun 14, 2014

### unscientific

But even still, they don't sum to zero:

$$\frac{d\langle p \rangle}{dt} = -\frac{\hbar \omega}{\sqrt 2 a} cos (\omega t)$$

$$m\omega^2 \langle \psi |x|\psi\rangle = \frac{ma\omega^2}{\sqrt 2} cos (\omega t)$$

17. Jun 14, 2014

### Oxvillian

Almost there! Now use the definition of $a$ given in the question.

18. Jun 15, 2014

### unscientific

$a$ is a constant, it's not defined in the question.

19. Jun 15, 2014

### Jilang

Unscientific, your second long equation is not correct. Try the differentiation again. : )

Edit, oh I see the first equation is mis-typed? Belay that!

Last edited: Jun 15, 2014
20. Jun 15, 2014

### Oxvillian

Look about 2/3 of the way through the question, right after the wavefunctions. It's the characteristic length scale of the harmonic oscillator, for what it's worth.