Quantum Mechanics and Eigenfunctions Checks

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TFM
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Homework Statement



For each of the following wave functions check whether they are eigenfunctions of the momentum operator, ie whether they satisfy the eigenvalue equation:

[tex] \hat{p} \psi(x) = p\psi(x) with \hat{p} = i \hbar \frac{\partial}{\partial x} [/tex] and p is a real number.

For those that are eigenfunctions, calculate the eigenvalue p, the expectation value〈[tex] \hat{p} [/tex]〉, and the standard deviation Δ[tex]\hat{p}[/tex] ̂.

(a)

[tex] \psi(x) = \sqrt{\frac{2}{L}} cos(\frac{x\pi}{L}) for -L/2 \leq x \leqL/2[/tex]

0 for x > L/2, x < -L/2

(b)

[tex] \phi(x)= \frac{1}{\sqrt{L}}e^{ikx} for 0 \leq x \leq L with k is real [\tex]

0 for x > L, x < 0


(c)

[tex] \chi (x) = 2xe^{-x} for 0 \leq x [/tex]
0 for x < 0


Homework Equations



N/A

The Attempt at a Solution



Okay, I have gone some way through the first part.

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

[tex] -i\hbar \frac{\partial}{\partial x}(\sqrt{\frac{2}{L}}cos\frac{x\pi}{L}) [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} \frac{\partial}{\partial x}(cos\frac{x\pi}{L}) [/tex]

With Limits between L/2 and -L/2

[tex] -i\hbar \sqrt{\frac{2}{L}} (-\frac{\pi}{L}sin\frac{x\pi}{L})^{L/2}_{-L/2} [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} ([-\frac{\pi}{L}sin\frac{L\pi}{2L}]-[-\frac{\pi}{L}sin\frac{-L\pi}{2L}]) [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} ([-\frac{\pi}{L}sin\frac{\pi}{2}]-[-\frac{\pi}{L}sin\frac{-\pi}{2}]) [/tex]

But I am unsure where I am supoosed to go from here...

To show it is a Eigenfunction, I need to get rid of the i at the very begining, but I am unsure how to do this?

Any suggesstions where to rpoceed?

Thanks in advance,

TFM
 

Answers and Replies

  • #2
nrqed
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You do not need the boundary conditions to check if a function is an eigenstate so no need to replace x by L/2 or -L/2. These limits only enter when you calculate the expectation values.


You must ask yourself if the function you get after applying [itex] i \hbar \partial_x [/itex] to [itex] \psi [/itex] is proportional to the wavefunction itself. If yes, then the wavefunction is an eigenstate and the constant of proportionality is the eigenvalue. If not, the function is not an eigenstate.
 
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  • #3
TFM
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So do I need to differnitiate it then???
 
  • #4
nrqed
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So do I need to differnitiate it then???
Yes. Sorry, there was a typo in my post and the partial derivative did not show up. Yes, you must apply [itex] i \hbar \partial_x [/itex] to [itex] \psi [/itex] and check if the result is proportional to the wavefunction itself. Is it possible to write [itex] \sin(\pi x/l) [/itex] as a constant times [itex] \cos (\pi x/L) [/itex] in such a way that the result will be valid for any value of x ? If the answer is no, then this wavefunction (the cos) is not an eigenstate of the momentum operator
 
  • #5
TFM
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Okay so the original wavefunction is:

[tex] \psi(x) = \sqrt{\frac{2}{L}} cos(\frac{x\pi}{L}) [/tex] --- (1)

The [tex] \hat{p} [/tex] function is:

[tex] -i\hbar \sqrt{\frac{2}{L}} (-\frac{\pi}{L}sin\frac{x\pi}{L}) [/tex] Which can be simplified to:

[tex] i\hbar \sqrt{\frac{2}{L}} * \frac{\pi}{L}sin\frac{x\pi}{L} [/tex]

So now I have to see if this result is proportional to (1).

So I have to find how to write [tex] sin(\frac{\pi x}{L}) [/tex] as [tex] C cos(\frac{\pi x}{L}) [/tex]

Can this actually be done, though, since the only relation I know that will link sin and cos is [tex] sin^2(\theta) + cos^2(\theta) = 1 [/tex]?

Also, do I need to get rid of that i at the beginning, since p needs to be a real number?
 

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