Check my working please? Expectation of momentum

  • #1
1,734
13

Homework Statement



A state at time t is given by:

[tex]|\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] [/tex]

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##.

Homework Equations



[tex]\hat p = -i\hbar \frac{\partial}{\partial x}[/tex]

The Attempt at a Solution



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

Starting:

[tex]\frac{1}{2} \langle 0 | e^{-i\omega t} \hat p |1\rangle [/tex]
[tex] = -\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[/tex]
[tex]= -\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 [/tex]
[tex] = -\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 [/tex]
[tex] = -\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] [/tex]
[tex] = -\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} [/tex]
[tex] = -\frac{i\hbar}{2\sqrt 2 a} e^{-i\omega t} [/tex]

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:

[tex]-\frac{\hbar}{\sqrt 2 a} sin (\omega t) [/tex]

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

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:

Answers and Replies

  • #2
196
22
Surely this would be much easier if you just express [itex]X[/itex] and [itex]P[/itex] in terms of creation & annihilation operators?
 
  • #3
1,734
13
Surely this would be much easier if you just express [itex]X[/itex] and [itex]P[/itex] in terms of creation & annihilation operators?

They aren't given in the question at all, so I don't think we need to use them.
 
  • #4
strangerep
Science Advisor
3,261
1,199
From your other thread...
What's frustrating is that others who put in decent effort get no replies at all...
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...

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

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..
Are you perhaps mixing up the partial time derivative with the total time derivative? Cf. Ehrenfest's theorem .
 
  • #5
196
22
Given ## \langle \psi |x|\psi\rangle = \frac{a}{\sqrt 2} cos (\omega t) ##, then I'm supposed to find:
[tex]\frac{\partial}{\partial t} \left( \langle \psi|p|\psi\rangle + m\omega^2 \langle \psi |x|\psi\rangle \right) [/tex]

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..

Perhaps you mean

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

(which would be true)
 
  • #6
1,734
13
Perhaps you mean

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

(which would be true)

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
strangerep
Science Advisor
3,261
1,199
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..
From this, I guess you didn't bother to read the link I suggested earlier at the end of post #4 ?
 
  • #8
1,734
13
From this, I guess you didn't bother to read the link I suggested earlier at the end of post #4 ?

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
196
22
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..

:confused:

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

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
strangerep
Science Advisor
3,261
1,199
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.
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
1,734
13
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?

Here is the full question:

2psg9as.png


I'm trying to find the link between finding ##\langle x \rangle##, ##\langle p\rangle## and the last part of the question.
 
  • #12
strangerep
Science Advisor
3,261
1,199
I'm trying to find the link between finding ##\langle x \rangle##, ##\langle p\rangle## and the last part of the question.
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:
  • #13
1,734
13
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.)

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

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

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

They are wildly different.
 
  • #14
196
22
unscientific - I think you have us pulling out our hair in frustration here :smile:

Why do you still insist on differentiating [itex]\langle x \rangle[/itex]?
 
  • #15
strangerep
Science Advisor
3,261
1,199
unscientific - I think you have us pulling out our hair in frustration here :smile:
Yep!

Why do you still insist on differentiating [itex]\langle x \rangle[/itex]?
He's seeing parentheses where there aren't any.
 
  • #16
1,734
13
Yep!

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

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

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

[tex]m\omega^2 \langle \psi |x|\psi\rangle = \frac{ma\omega^2}{\sqrt 2} cos (\omega t)[/tex]
 
  • #17
196
22
Almost there! Now use the definition of [itex]a[/itex] given in the question.
 
  • #18
1,734
13
Almost there! Now use the definition of [itex]a[/itex] given in the question.

##a## is a constant, it's not defined in the question.
 
  • #19
1,116
72
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:
  • #20
196
22
##a## is a constant, it's not defined in the question.

:cry:

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.
 
  • #21
1,734
13
:cry:

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.

I can't believe I missed that out! Thanks so much for your help! That looks right now.
 
  • #22
strangerep
Science Advisor
3,261
1,199
I can't believe I missed that out!
Perhaps now you understand why the HW guidelines are so insistent about posting the original problem exactly as it was given to you ?

If you had done that in your opening post, we could have diagnosed the difficulty far earlier and with far less frustration.
 
  • #23
1,734
13
Perhaps now you understand why the HW guidelines are so insistent about posting the original problem exactly as it was given to you ?

If you had done that in your opening post, we could have diagnosed the difficulty far earlier and with far less frustration.

Yeah I get it, but sometimes I don't want to bother you guys with the whole question when I'm only having difficulty with a small part of it. I will post more carefully in the future.
 
  • #24
strangerep
Science Advisor
3,261
1,199
Yeah I get it, but sometimes I don't want to bother you guys with the whole question when I'm only having difficulty with a small part of it.
I can understand that, but I also understand how easy it is to mis-read or mis-interpret the written question... :frown:

It also helps potential helpers if they can see the entire context.

I will post more carefully in the future.
Ok, good.
 

Related Threads on Check my working please? Expectation of momentum

  • Last Post
Replies
5
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
0
Views
2K
Replies
2
Views
911
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
2K
Top