Mean value and deviation of momentum in non-normalisable wave function

[castlemaster]

Homework Statement

Mean value and deviation of momentum for this wave function:

\Psi(x,0)=cos^2(kx/L)e^{2ikx/L}

Homework Equations

The Attempt at a Solution

I express the cos^2 in terms of exponencials:

\Psi(x,0)=(1/2)+(1/4)e^{2ikx/L}+(1/4)e^{4ikx/L}

the momentum of each plane wave is p=hk

How do I get the mean value and the deviation of the momentum?

 

[gabbagabbahey]

When in doubt, go back to the definitions. What are the definitions of mean value and deviation of any operator (otherwise known as 'expectation value' and 'uncertainty') that you have in your text/notes?

 

[castlemaster]

Hi,

Mean value is

<\psi,p,\psi>=\oint\psi^*(-ih)\frac{d\psi}{dx}dx

deviation

<\psi,p^2,\psi>=\oint\psi^*(-ih)^2\frac{d\psi^2}{dx^2}dx

This is for t=0, I guess I have to include e^{-iEh/t} in the wave function.

Is there any way to reach the solution without going through the integrals?

thanks

 

[gabbagabbahey]

Hi,

Mean value is

<\psi,p,\psi>=\oint\psi^*(-ih)\frac{d\psi}{dx}dx

deviation

<\psi,p^2,\psi>=\oint\psi^*(-ih)^2\frac{d\psi^2}{dx^2}dx

This is for t=0, I guess I have to include e^{-iEh/t} in the wave function.

Is there any way to reach the solution without going through the integrals?

thanks

I don't see any reason to include e^{-iEh/t} in the wave function unless you are asked to find the time evolution of the mean value and uncertainty. (In any case, because of the complex conjugation in the integrals it will cancel out anyways)

And yes, you have to go through and evaluate the integrals.

P.S. You can write Bras and Kets in \LaTeX more clearly by using the symbols \vert, \rangle and \langle and Planck's constant is given by the symbol \hbar and partial derivatives can be rendered using \partial...

\langle P\rangle=\langle\psi\vert P\vert\psi\rangle=\oint\psi^*(-i\hbar)\frac{\partial\psi}{\partial x}dx

 

[castlemaster]

When trying to solve the integrals I reach a point where the result gives infinity ... I guess it's because the original wave function is not normalisable.

How do I continue?

 

[gabbagabbahey]

Are you sure that the wave function is defined on the entire interval x\in (-\infty,\infty) and not just some section of it like say x\in (-\frac{L}{2},\frac{L}{2})?

 

[castlemaster]

The text doesn't say anything of boundary conditions. It says a free particle in one dimension and it says that I have to include the time dependence of the mean value and deviation.
Maybe I have to treat the particle as been in an infinite square well -L/2,L/2 and compute L->infinity to find the solution.

I was googling and found some comments about delta dirac normalization of wave functions for free particles, doyou know something about it? it's maybe part of the solution.

thanks

 

[gabbagabbahey]

Is this part of a larger problem? Was the first part of the problem to compute the wavefunction you gave in your problem statement?

 

[castlemaster]

No, it starts by giving <br /> \Psi(x,0)=Acos^2(kx/L)e^{2ikx/L}<br />

 

[gabbagabbahey]

Hmmm... that's an odd looking wavefunction for a free particle...could you post the exact wording of the original problem? (or, if the problem is straight from a textbook, post the name of the text and the problem number)

 

[castlemaster]

This is from a spanish university examination. This is a pretty fair translation:

A particle of mass m travels freely in one dimension. In t=0, the wave function (non-normalisable) of the particle is
<br /> <br /> \Psi(x,0)=Acos^2(x/L)e^{2ix/L}<br /> <br />

a) Find, for any given time, the expected value and the uncertainty of the lineal momentum of the particle

b) Write down explicitily the wave function of the particle \Psi(x,t) for any given time.


-------------

The fact that L is in the initial wave function is maybe a sign that I have to treat this as being in a infinite square well and compute L->infinity

 

[gabbagabbahey]

Were the 'k's in your first post just typos?

Anyways, I see what's going on here now...the reason your wave-function is non-normalizable is that it isn't in an energy/momentum eigenstate. Rather, it is in a state which is a superposition of 3 different momentum eigenstates.

So, your first approach (expressing the wavefunction as a linear combinatioon of plane waves) is the right way to go here. However,

\cos^2 (x/L)e^{2ix/L}\neq \frac{1}{2}+\frac{1}{4}e^{2ix/L}+\frac{1}{4}e^{4ix/L}

Your first two coefficients are incorrect.

The momentum eigenstates will be the individual plane waves e^{ipx/\hbar},,,and it should be clear, that a momentum measurement will yield either p=0,2h/L or 4h/L, and the probability of each value can be determined from the coefficient in front of each plane wave. Once you determine those probabilities, the mean value of the momentum, will just be the weighted average of each individual momentum value.

 

[castlemaster]

I didn't have the paper with me the first time I wrote the wave function, I thought the k's were there. Anyway what I want is to know how to proceed as I'm sure that the exact same problem won't be in the september test.

And uncertainty?

\langle P^2\rangle=|c1|^2(2\hbar/L)^2 + |c2|^2(4\hbar/L)^2

 

[gabbagabbahey]

Well, you basically have

\Psi(x,0)=\langle x\vert\Psi(t=0)\rangle=c_0e^{0ix}+c_1e^{2ix/L}+c_2e^{4ix/L}=c_0\langle x\vert p=0\rangle+c_1\langle x\vert p=\frac{2\hbar}{L}\rangle+c_2\langle x\vert p=\frac{4\hbar}{L}\rangle

Right?

So, this implies that \vert\Psi(t=0)\rangle=c_0\vert p=0\rangle+c_1\vert p=\frac{2\hbar}{L}\rangle+c_2\vert p=\frac{4\hbar}{L}\rangle

So you tell me, what will \langle P\rangle=\langle\Psi(t=0)\vert P\vert\Psi(t=0)\rangle and \Delta P=\left(\langle P^2\rangle-\langle P\rangle^2 \right)^{1/2}=\left(\langle\Psi(t=0)\vert P^2\vert\Psi(t=0)\rangle-\langle P\rangle^2 \right)^{1/2} be?

 

[castlemaster]

I think is this:

<br /> \langle P\rangle=|c1|^2(2\hbar/L) + |c2|^2(4\hbar/L)<br />

<br /> \langle P^2\rangle=|c1|^2(2\hbar/L)^2 + |c2|^2(4\hbar/L)^2<br />

You are right in the coefficients, I recalculated them.

 

[gabbagabbahey]

Looks good to me.

Just remember that the uncertainty/deviation is actually

\Delta P=\left(\langle P^2\rangle-\langle P\rangle^2 \right)^{1/2}\neq\langle P^2\rangle

 

[castlemaster]

<br /> \Delta P=\left(\langle P^2\rangle-\langle P\rangle^2 \right)^{1/2}\neq\langle P^2\rangle

yes I have it in mind

thank's a lot