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Quantum Mechanics and Eigenfunctions Checks

  • Thread starter TFM
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TFM

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1. 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


2. Homework Equations

N/A

3. 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
 

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.
 
Last edited:

TFM

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So do I need to differnitiate it then???
 

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
 

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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