QM - Position/Momentum representation problem

  • Thread starter Thread starter knowlewj01
  • Start date Start date
  • Tags Tags
    Qm Representation
knowlewj01
Messages
100
Reaction score
0

Homework Statement



Write down the time independant Schrodinger eqn in the momentum representation for a particle with mass m when the potential is given by V(x) = \frac{1}{2} \gamma x^2
Given that a possible solution is given by \Phi(p) = e^{\frac{-Bp^2}{2}}

determine B and the corresponding energy eigenvalue.


Homework Equations



position / momentum

x \rightarrow i\hbar \frac{d}{dp}
-i\hbar \frac{d}{dx} \rightarrow p



The Attempt at a Solution



in position representation the full time independant schrodinger eqn is:

- \frac{\hbar^2}{2m}\frac{d^2}{dx^2} \Psi(x) + \frac{1}{2}\gamma x^2 \Psi(x) = E\Psi(x)

becomes

\frac{p^2}{2m}\Phi(p) - \frac{\hbar^2 \gamma}{2}\frac{d^2}{dp^2}\Phi(p) = E\Phi(p)
in the momentum representation, where Phi is the FT of Psi.

After plugging in the Trial Solution I get:

E = \frac{p^2}{2m} - \frac{\hbar^2 \gamma}{2}\left( B^2 p^2 - B\right)

Not sure what to do after this bit, I tried to normalise the wavefunction A = \left(\frac{B}{\pi}\right)^{\frac{1}{4}}

But I don't think that helps.

Any ideas on where i could get a second equation to find B and E?

Thanks
 
Physics news on Phys.org
The energy E needs to be a constant, so it can't depend on p. For what value of B will the dependence of E on p vanish?
 
ok, the only way i can see of doing this is, first rearrange to get:

p^2 \left(\frac{1}{2m} - \frac{B^2 \hbar^2 \gamma}{2}\right) + \frac{B\hbar^2 \gamma}{2} = E

choose B such that:

B^2 \frac{\hbar^2 \gamma}{2} = \frac{1}{2m}

which gives:

B = \frac{1}{\hbar}\sqrt{\frac{1}{\gamma m}}

So the energy eigenvalue would be:

\frac{\hbar}{2}\sqrt{\frac{\gamma}{m}}
 
Looks good.
 
thanks very much
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top