Solving a Free Particle in 1D: Is \psi(X) an Energy Eigenstate?

[Epud]

Trying to get my head around this problem and would very much appreciate any suggestions.

Given a wavefunction \psi(X) i am asked if it is an energy eigenstate for a free particle moving in one dimension? Any suggestion on how I start a problem like this?

thanks,

Epud

 

[marlon]

Trying to get my head around this problem and would very much appreciate any suggestions.

Given a wavefunction \psi(X) i am asked if it is an energy eigenstate for a free particle moving in one dimension? Any suggestion on how I start a problem like this?

thanks,

Epud

Well, this is rather easy if you just know how to solve the Schrödinger equation in the case of a free particle (ie potential V(x) = 0)

regards
marlon

 

[sporkstorms]

As was pointed out, a free particle means V(x) = 0.
Furthermore, you'll be solving the one dimensional, time-independent S.E, since you're given psi(x).

Most undergrad texts work this out at one point or another. I especially like Griffith's explanations - and it should help you a lot (it's done in position space, in 1D).

In case you don't have it, to get you started:
Write the SE:
-\frac{\hbar ^2}{2m}\frac{d^2 \psi}{dx^2} = E\psi
Rearrange, then define
k \equiv \frac{\sqrt{2mE}}{\hbar}

Being able to just "see" that you should define k as such, to make it easier (or possible?) to solve isn't something I was able to do. It would have taken me ages to find that on my own.

hth.

 

[dextercioby]

Trying to get my head around this problem and would very much appreciate any suggestions.

Given a wavefunction \psi(X) i am asked if it is an energy eigenstate for a free particle moving in one dimension? Any suggestion on how I start a problem like this?

thanks,

Epud

The spectral problem

\hat{H}|\psi\rangle =E|\psi\rangle

in case of a free particle has a solution of the form

\psi (x)=\langle x|\psi\rangle

, where \langle x| is a tempered distribution and |\psi\rangle is a test function.

So you'll have to see whether your wavefunction can be obtained in this method: applying a linear functional on a vector from L^{2}\left(\mathbb{R}\right).


Daniel.