Finding eigenfunctions of the linear momentum operator

[t_n_p]

Homework Statement

http://img508.imageshack.us/img508/7199/46168034nt3.jpg

The Attempt at a Solution

I'm just totally lost with this question. The theory just eludes me totally. Just how do you determine whether it is/isn't an eighenfunction of the linear momentum operator?

 

[kdv]

Homework Statement

http://img508.imageshack.us/img508/7199/46168034nt3.jpg

The Attempt at a Solution

I'm just totally lost with this question. The theory just eludes me totally. Just how do you determine whether it is/isn't an eighenfunction of the linear momentum operator?

A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction).

So apply the operator p_x on that wavefunction and see if the result is a constant times the initial wavefunction

 

[Mindscrape]

In other words, one of the postulates of QM relates observables by eigenfunctions and eigenvectors. For example

\hat{H}\psi = E \psi

would be the hamiltonian operator giving energy eigenvalues. You could also have

\hat{x_o}\psi = x_n \psi

or more relevantly

\hat{p} \psi = p_n \psi

In the end, do exactly what kdv told you, see if when you apply the operator you get the function back with a constant.

 

[t_n_p]

pardon my lack of knowledge, but what is the linear momentum operator?

 

[Hootenanny]

pardon my lack of knowledge, but what is the linear momentum operator?

The linear momentum operator is a partial differential operator and in three dimensions has the form,

\hat{p} = -i\hbar\nabla

And in one dimension,

\hat{p} = -i\hbar\frac{d}{d x}

So for example, in 3D

\hat{p}\left(xyz\right) = -i\hbar\left(yz,xz,xy\right)

So as both Mindscrape & kdv have said, all you need to do is apply the operator to the wave function and see if you get the wave function (eigenfunction) multiplied by a constant (eigenvalue).

 

[t_n_p]

Ok, so I want to be using the one dimensional form. Pardon my lack of knowledge again, but what do you mean by apply?

 

[kdv]

Ok, so I want to be using the one dimensional form. Pardon my lack of knowledge again, but what do you mean by apply?

it means take the derivative with respect to x of the wavefunction then multiply by -i <br /> hbar

 

[t_n_p]

oh no, I havn't done calculus for a while...

Do I just treat A and i as constants?

If yes, by chain rule I get

http://img299.imageshack.us/img299/4323/15423623yp4.jpg

Then I multiple by i*h/2pi (although it's probably wrong, worth a try though)...?

 

[Hootenanny]

Yes, you have taken the derivative correctly, now just multiply it by -i\hbar and see if you can arrange it into the form,

\hat{p}\psi = p\psi

Where p is constant. If you can, then \psi is a linear eigenfunction of \hat{p}; if not, then \psi is not a linear eigenfunction of \hat{p}.

 

[t_n_p]

Just one question, is i the imaginary number? i.e. i^2=-1?

 

[malawi_glenn]

Just one question, is i the imaginary number? i.e. i^2=-1?

that is correct! :-)

 

[Mindscrape]

Sorry, I was looking at my help and I shouldn't have put p_n as eigenvalues because that would imply they are discrete, but really the momentum eigenvalues are continuous, as well as the position eigenvalues. I should have distinguished with maybe capitals as operators and lower case as eigenvalues.

<br /> \hat{X}\psi = x \psi<br />

<br /> \hat{P}\psi = p \psi<br />

We should really be talking about an eigenket, or state vector, that lives in Hilbert space (or its suburbs), but it would be bad to burden you with technicalities, so don't worry about it unless you want to.

 

[t_n_p]

Ok, so after all that I end up with...

http://img258.imageshack.us/img258/6994/20815713op7.jpg

So my constant is i*(hbar)*x + (hbar) ?

 

[Mindscrape]

Hold on there, is x a constant?

 

[t_n_p]

In that case, it isn't an eigenfunction of the linear momentum operator?

 

[Mindscrape]

Does it satisfy the postulate?

 

[t_n_p]

"A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction)."

x is not a constant, therefore applying the linear momentum operator on psi does not give the wavelength function back times some constant.

Therefore, psi is not an eigenfuction of the linear momentum operator.

(Well, that's what I think anyway...)

 

[Hootenanny]

Sounds good to me

 

[t_n_p]

Excellent! Thanks