Is there a magnitude of momentum quantum operator?

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Is there a "magnitude of momentum" quantum operator?

Homework Statement



Is the ground state of the infinite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not? What can you say about the magnitude of the momentum?

Homework Equations



Ground state wavefunction of the infinite square well: [itex]\psi(x) = \sqrt(\frac{2}{a})sin(\frac{\pi}{a}x)[/itex] (0 ≤ x ≤ a).

Momentum operator: [itex]\hat{p} = -i\hbar\frac{d}{dx}[/itex].

The Attempt at a Solution



Since the wavefunction is normalizable, if it were an eigenfunction then we would have a state with a definite momentum, violating the uncertainty principle.

We can show this directly by plugging it in:

[itex]\hat{p}\psi(x) = -i \hbar \frac{d}{dx}(\sqrt(\frac{2}{a}) sin(\frac{\pi}{a}x))[/itex]

[itex] = -i \hbar \sqrt(\frac{2}{a}) \frac{\pi}{a} cos(\frac{\pi}{a}x) [/itex]

[itex] = \frac{-i \hbar \pi}{a} cot(\frac{\pi}{a} x) \psi(x)[/itex].

So we can see that it is not an eigenfunction.

The problem I am having is with the magnitude of the momentum. Now I must confess here that I have the solutions manual to the book, so I cheated a bit and looked in there. It says "the magnitude of the momentum [itex]\sqrt(2mE_{1}) = \pi\hbar/a[/itex] is determinate, but the particle is just as likely to be found travelling to the left (negative momentum) as to the right (positive momentum)". I don't understand how we can see this from the above equation. Is there an operator for the magnitude of the momentum, for which ψ is an eigenfunction?
 
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Answers and Replies

  • #2
BruceW
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You can write your wave function as a sum of momentum eigenfunctions. Have you learned about complex numbers and polar representation?
 
  • #3
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Hello, thanks for your tip. I see that you can write

[itex]\psi(x) = \sqrt(\frac{2}{a})sin(\frac{\pi}{a}x)[/itex]

[itex] = \sqrt(\frac{2}{a})[\frac{e^{i \pi x/a}-e^{-i \pi x/a}}{2i}][/itex]

[itex] = \frac{-i}{\sqrt(2a)}[\sqrt(2\pi\hbar)f_{\pi\hbar/a}(x) - \sqrt(2\pi\hbar)f_{-\pi\hbar/a}(x)][/itex]

[itex] = -i\sqrt(\frac{\pi\hbar}{a})[f_{\pi\hbar/a}(x) - f_{-\pi\hbar/a}(x)][/itex]

where [itex]f_{p}(x) = \frac{1}{\sqrt(2\pi\hbar)}e^{ipx/\hbar}[/itex] is the momentum operator eigenfunction with eigenvalue p.

But I think I have made a mistake here because should it not be the case that the sum of the square-magnitudes of the coefficients comes to 1, since each of these tells you the probability of measuring that particular momentum? Was I wrong to use [itex]f_{p}(x) = \frac{1}{\sqrt(2\pi\hbar)}e^{ipx/\hbar}[/itex]? Should it have been [itex]f_{p}(x) = Ae^{ipx/\hbar}[/itex] and then normalize in our specific case to find A?

At least I can see now why the momentum is one of these two values, so the magnitude is known. But can this result be seen from the equations in my original post? The reason I ask if that the author of the book i'm reading just states it likes it's a completely obvious conclusion which doesn't require any extra working out?

Thanks again.
 
  • #4
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Sorry, double post.
 
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  • #5
BruceW
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It's only an obvious conclusion if you've already got some experience with quantum mechanics! I'm surprised the book you're reading didn't go through it in more detail.

And the rule for normalisation is:
[tex]\int_{- \infty}^{\infty} \psi \psi \ast dx = 1 [/tex]
From here, you can work out what the normalisation constant is. (Keeping in mind that psi is zero outside the square well).

Also, you didn't make a mistake, but you need to use the rule above.
 
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  • #6
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Thanks once again for your reply.

I'm aware of the normalization condition, but ψ is already normaliazed in this case. I know that for discrete energies then we have

[itex]\Psi(x,0) = \sum c_{n}\psi_{n}(x)[/itex]

so the [itex]c_{n}[/itex]'s represent the probability of measuring the energy [itex]E_n[/itex]. And so you can show that

[itex]\sum|c_{n}|^{2} = 1[/itex]. (*)

So I was thinking that since we have the same thing but with a sum of momentum eigenfunctions instead of hamiltonian eigenfunctions, then we would again have the sum of the squares of the coefficients equal to 1, and each coefficient representing the probability of measuring that momentum. Is this incorrect?

I think that for (*) to be true then you need that the eigenfunctions are orthonormal, which is true in the case of the hamiltonian for this potential, but untrue in the case of momentum?

Thanks.
 
  • #7
BruceW
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Your momentum eigenfunctions:
[tex]f_{p}(x) = \frac{1}{\sqrt(2\pi\hbar)}e^{ipx/\hbar}[/tex]
Are orthogonal, but they are not normalised. You need to normalise them, then when you re-write psi in terms of the orthonormal momentum eigenfunctions, you'll see that the sum of the squares of the coefficients will equal 1.
 
  • #8
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But you can't normalize [itex]f_p(x) = Ae^{ipx/\hbar}[/itex] can you? It's not square-integrable for any p.
 
  • #9
BruceW
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If that was the function over all space, then you're right that it can't be normalised. But the function is zero outside the square well, so you can normalise it.
 
  • #10
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Ahh okay I forgot about that. So we could have a particle which is in just one of these momentum eigenfunction states, and we would know it's momentum for certain? In that case, would the uncertainty in the position be infinite or something?
 
  • #11
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Oops I see why that wouldn't happen: because the only allowed states are the ones already calculated from the Hamiltonian. Duh!! :blushing:

Thanks again for your help!
 
  • #12
BruceW
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No worries man
 
  • #13
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epsilonjon said:
Oops I see why that wouldn't happen: because the only allowed states are the ones already calculated from the Hamiltonian. Duh!!

Wait, I think what I wrote here was incorrect? Sorry to keep labouring over this, but I just want to understand it properly.

So if the particle is in the ground state of the Hamiltonian a competent measure of energy will definitely get [itex]E_0[/itex]. But at this moment with this energy measurement I only know that the magnitude of momentum is [itex]\pi\hbar/a[/itex], I do not know its direction. However, if I then make a competent measure of the momentum, the wavefunction will collapse to one of either of the two momentum eigenfunctions and I will measure that corresponding eigenvalue (either [itex]+\pi\hbar/a[/itex] or [itex]-\pi\hbar/a[/itex]). So now I have a definite value for the momentum, but the particle is no longer in the ground state of the Hamiltonian so I no longer know the energy. And this all comes from the fact that Hamiltonian and the momentum operator don't share a common set of eigenfunctions, so they don't commute?

Is that correct?
 
  • #14
BruceW
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epsilonjon said:
So if the particle is in the ground state of the Hamiltonian a competent measure of energy will definitely get E0. But at this moment with this energy measurement I only know that the magnitude of momentum is πℏ/a, I do not know its direction. However, if I then make a competent measure of the momentum, the wavefunction will collapse to one of either of the two momentum eigenfunctions and I will measure that corresponding eigenvalue (either +πℏ/a or −πℏ/a).
Yes. Full marks :) But I haven't heard of 'competent measure'. In textbooks I've read, they just call it a 'measurement', which means the state immediately after a measurement of some variable will be an eigenstate of that variable.

epsilonjon said:
So now I have a definite value for the momentum, but the particle is no longer in the ground state of the Hamiltonian so I no longer know the energy. And this all comes from the fact that Hamiltonian and the momentum operator don't share a common set of eigenfunctions, so they don't commute?
Essentially, yes. But strictly, you haven't said what the system is like after this 'measurement' occurs. For example, the particle might fly off into free space, so in that case it could be in an eigenstate of both momentum and energy.
 

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