Expectation value of momentum operator

[tanaygupta2000]

Homework Statement

Find the expectation value of momentum operator in a normalized real-valued wavefunction

Relevant Equations

momentum eigenstates = exp(ikx)

I know that the eigenstates of momentum operator are given by exp(ikx)
To construct a real-valued and normalized wavefunction out of these eigenstates,
I have,
psi(x) = [exp(ikx) + exp(-ikx)]/ sqrt(2)

But my trouble is, how do I find the expectation value of momentum operator <p> using this psi(x)?
On applying momentum operator, my integral is divergent.
I think that since there are equal probabilities in psi, <p> will be 0.
But how do I show it by calculations?
Please help!

 

[anuttarasammyak]

$$<p>=\int \psi^*(x) \frac{\hbar}{i}\frac{d}{dx} \psi(x) dx$$
is usual way of momentum calculation. Say $\psi(x)$ is real function ?

 

[PeroK]

Homework Statement:: Find the expectation value of momentum operator in a normalized real-valued wavefunction
Relevant Equations:: momentum eigenstates = exp(ikx)

View attachment 278349

I know that the eigenstates of momentum operator are given by exp(ikx)
To construct a real-valued and normalized wavefunction out of these eigenstates,
I have,
psi(x) = [exp(ikx) + exp(-ikx)]/ sqrt(2)

But my trouble is, how do I find the expectation value of momentum operator <p> using this psi(x)?
On applying momentum operator, my integral is divergent.
I think that since there are equal probabilities in psi, <p> will be 0.
But how do I show it by calculations?
Please help!

The problem says nothing about using the momentum eigenstates - which cannot be normalized in any case.

 

[tanaygupta2000]

I just took hint from this - https://physics.stackexchange.com/q...momentum-for-real-valued-wave-function/335213

 

[tanaygupta2000]

I just took hint from this - https://physics.stackexchange.com/q...momentum-for-real-valued-wave-function/335213

and normalized the eigenstates of momentum operator using dirac-delta function.
Since there will be equal and opposite probabilities, <p> will be 0.
Is my way of doing the solution correct?

 

[PeroK]

You are asked to calculate the integral in post #2:

$$<p>=\int \psi^*(x) \frac{\hbar}{i}\frac{d}{dx} \psi(x) dx$$
is usual way of momentum calculation. Say $\psi(x)$ is real function ?

This has nothing to do with momentum eigenfunctions.

 

[anuttarasammyak]

@tanaygupta If you are keen on momentum eigenfunction, you may expand the state by momentum eigenfunciton, say $\phi(p)$ instead of expansion of the state by coordinate eigenfunction $\psi(x)$.

Momentum expectation value is
$$<p>=\int \phi^*(p) p \phi(p) dp$$
The relation of $\phi(p)$ and $\psi(x)$ is
$$\phi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{-ipx/\hbar}\psi(x) dx$$
$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{ipx/\hbar}\phi(p) dp$$

You see the last equation tells $\phi(p)$ is a kind of amplitude for momentum eigenfunction component, i.e. $e^{ipx/\hbar}$, of the state.

As $\psi(x)$ is real you observe
$$\phi^*(p)=\phi(-p)$$
$$<p>=\int \phi(-p) p \phi(p) dp$$

How do you estimate this integral ?

 

[tanaygupta2000]

$$<p>=\int \psi^*(x) \frac{\hbar}{i}\frac{d}{dx} \psi(x) dx$$
is usual way of momentum calculation. Say $\psi(x)$ is real function ?

I think it will be easier to go with this one.

 

[tanaygupta2000]

I think it will be easier to go with this one.

What should I assume psi(x) to be?

 

[tanaygupta2000]

I just looked another question which was like this so I did like that.

 

[PeroK]

What should I assume psi(x) to be?

Any real-valued function!

 

[tanaygupta2000]

Any real-valued function!

Can I simply take sqrt(2/L) sin(n pi x/L)?
What should I get the value for <p>?

 

[PeroK]

By "any" we mean "arbitrary". You have to calculate the integral for any function. You can of course try a specific function to see what you get. But, the general solution is fairly easy.

 

[tanaygupta2000]

By "any" we mean "arbitrary". You have to calculate the integral for any function. You can of course try a specific function to see what you get. But, the general solution is fairly easy.

Can I simply take sin(x), cos(x) etc. ?
Please suggest me some easy function for calculating <p>.

 

[PeroK]

Can I simply take sin(x), cos(x) etc. ?
Please suggest me some easy function for calculating <p>.

Try $\psi(x) = e^{-x^2}$ and don't worry about the normalisation constant.