Expectation of the Momentum Operator

[daveyman]

Homework Statement

Here is another True or False question from the same practice test.

Since the expectation of the momentum operator <p>=<n|pn> is zero for an energy eigen state of the harmonic oscillator, a measurement of the momentum will give zero every time (True or False).

Homework Equations

N/A

The Attempt at a Solution

This question sounds very false to me, since an expectation value implies a probability and not an absolute. However, I'm thinking it might be a trick question...

Thanks!

 

[olgranpappy]

Homework Statement

Here is another True or False question from the same practice test.

Since the expectation of the momentum operator <p>=<n|pn> is zero for an energy eigen state of the harmonic oscillator, a measurement of the momentum will give zero every time (True or False).

Homework Equations

N/A

The Attempt at a Solution

This question sounds very false to me, since an expectation value implies a probability and not an absolute. However, I'm thinking it might be a trick question...

Thanks!

Maybe the best way to understand this problem is to rewrite the operator p as the commutator of x with the hamiltonian (times some factor). Can you do this?

 

[daveyman]

Do you mean like this?

p = [x,H]g(x) = xHg(x) - Hxg(x) = 0

 

[olgranpappy]

Do you mean like this?

p = [x,H]g(x) = xHg(x) - Hxg(x) = 0

no. if I follow all those equal signs that equation says that the momentum operator p=0...

Here. Just do this. Tell me, what is this
<br /> \left[<br /> \hat x<br /> ,<br /> \hat H<br /> \right]<br />
equal to? In the above \hat x is the position operator and \hat H is the hamiltonian
<br /> \hat H=\frac{\hat p^2}{2m}+V(\hat x)\;.<br />

 

[daveyman]

\hat{x}\hat{H}f(x)-\hat{H}\hat{x}f(x)

 

[daveyman]

Can you just substitute the Hamiltonian now?

 

[olgranpappy]

presumably "f(x)" is some "test function" that you are letting the operators act on?

Yes. go ahead and substitute in the hamiltonian now and see what you get. what do you get?

 

[daveyman]

Okay I'll omit the test function:

\frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \frac{\hat p^2\hat x}{2m}+V(\hat x) \hat x

This can't be simplified, can it?

(Sorry - I'm very new to QM)

 

[olgranpappy]

Okay I'll omit the test function:

\frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \frac{\hat p^2\hat x}{2m}+V(\hat x) \hat x

This can't be simplified, can it?

(Sorry - I'm very new to QM)

It certainly can.

For one thing you have forgetting to distribute the minus sign into the last term. Your expression should read:
<br /> \frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \left(\frac{\hat p^2\hat x}{2m}+V(\hat x) \hat x\right)<br /> =<br /> \frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \frac{\hat p^2\hat x}{2m}-V(\hat x) \hat x<br /> =<br /> \frac{\hat x\hat p^2}{2m} - \frac{\hat p^2\hat x}{2m}<br />

Better bring back your test function if you want to continue simplifying the rest. Since p=-id/dx,
you now want to determine
<br /> -\frac{x}{2m}\frac{d^2f}{dx^2} + \frac{d^2}{dx^2}\left(\frac{x}{2m}f\right)<br />

 

[borgwal]

You don't need those manipulations at all: What would be the condition for the state |n> to *always* give the result p=0 for a measurement of momentum?

 

[olgranpappy]

You don't need those manipulations at all:

One does not *need* all "those manipulations". But "those manipualtions" are important from a pedagogical standpoint. Do you understand that?

What would be the condition for the state |n> to *always* give the result p=0 for a measurement of momentum?

And besides... I believe that the way you are suggesting to think about this problem (at least in my estimation of what you are getting at) is incorrect.

 

[borgwal]

one does not *need* all "those manipulations". But "those manipualtions" are important from a pedagogical standpoint. Do you understand that?

I don't agree: Poster says he's a beginner at qm. You're using an atomic bomb to kill a musquito.

And besides... I believe that the way you are suggesting to think about this problem (at least in my estimation of what you are getting at) is incorrect.

lol.

 

[olgranpappy]

I don't agree: Poster says he's a beginner at qm. You're using an atomic bomb to kill a musquito.

Perhaps. Perhaps not. But at least I am not misleading the OP.

 

[borgwal]

Perhaps. Perhaps not. But at least I am not misleading the OP.

Misleading?? Soon I'll show my simple elegant pedagogical solution, maybe you'll understand it's not misleading, nor incorrect.

 

[olgranpappy]

Misleading?? Soon I'll show my simple elegant pedagogical solution, maybe you'll understand it's not misleading, nor incorrect.

sure.

 

[borgwal]

Ok, daveyman, here we go:

First, you write the operator p as a linear combination of the operators a^+ and a^-, that I think you will have heard of when discussing the harmonic oscillator.

Second, the given statement that <n|pn>=0 states that p|n> is orthogonal to |n>. You can prove that is true [although the True/False question doesn't ask you to prove this] by using the properties of the afore-mentioned operators a^+ and a^-.

Third, the statement that you'd *always* find zero momentum in the state |n> would be equivalent to stating that p|n>=0|n>. Using the preceding result, you'll find that p|n> is *not* equal to 0|n> for any value of n.

Done.

I think oldgran was letting you prove <n|p|n>=0 in a more complicated way. I also think that perhaps he was thinking that the equation p|\psi>=0 doesn't have a normalizable solution, but that is not relevant here: you only have to prove that |n> is *not* a solution to that equation.

 

[olgranpappy]

Ok, daveyman, here we go:

First, you write the operator p as a linear combination of the operators a^+ and a^-, that I think you will have heard of when discussing the harmonic oscillator.

Second, the given statement that <n|pn>=0 states that p|n> is orthogonal to |n>. You can prove that is true [although the True/False question doesn't ask you to prove this] by using the properties of the afore-mentioned operators a^+ and a^-.

Third, the statement that you'd *always* find zero momentum in the state |n> would be equivalent to stating that p|n>=0|n>. Using the preceding result, you'll find that p|n> is *not* equal to 0|n> for any value of n.

Done.

I think oldgran was letting you prove <n|p|n>=0 in a more complicated way.

sigh. I hate to say this... but you are actually correct; I was trying to help the OP prove that <n|p|n>=0 for eigenstates. But, now that I reread the question, that is not at all what he was asking.

Now I look silly... and it serves me right for trying to answer the question before reading the whole question. I apologize.

Anyways, OP. borgwal is correct to say that you should understand what it means that *every* momentum measurement yields zero. Namely, if this were true the state in question would be an eigenstate of p (with eigenvalue zero). And this is certainly not the case for |n> (which is an eigenstate of H not P).

 

[borgwal]

That's very gracious of you, thanks! Now let's hope daveyman is happy too :-)

 

[cepheid]

I didn't read this whole freakin' thread, but basically eigenstates of the Hamiltonian are stationary states...states that are guaranteed to yield a certain value for the energy (the eigenvalue) when a measurement is taken. So, in order for the value of the observable p to always be the same whenever a measurement is taken, |n> would have to be an eigenstate of p-hat as well as of H-hat. If those operators don't commute, then simultaneous eigenstates of them both cannot be constructed. So maybe all of that fancy math was required to answer the problem, maybe it wasn't (if you already know whether those two operators commute...)