Energy eigen value

[Reshma]

The ground state of a one-dimensional Harmonic oscillator described by the Hamiltonian H = \frac{p^2}{2m} + \frac{kx^2}{2} is of the form, \psi = Ae^{-ax^2}. Determine 'A' and 'a' so that the wavefunction \psi is a normalized eigenstate of the Hamiltonian. What is the energy eigenvalue of the wavefunction?

Well, I was able to normalize the wavefunction and obtained the value of 'A'.
\int_{-\infty}^{\infty}\psi \psi^* dx =1

A^2\int_{-\infty}^{\infty}e^{-2ax^2}dx =1

A^2\sqrt{\frac{\pi}{2a}} =1

A = (\frac{2a}{\pi})^{1/4}

How do I determine 'a'? Any clues to obtain energy eigen value?

[Reshma]

Sorry, I could not correct my errors yesterday. I have rectified the LaTeX typos. Now can someone help me....?

[Hurkyl]

How do I determine 'a'? Any clues to obtain energy eigen value?
I would say use the definitions of "eigenstate" and "eigenvalue".

[Reshma]

I would say use the definitions of "eigenstate" and "eigenvalue".
You mean use the eigenfunction and obtain the eigenvalue?
i \hbar \frac{\partial}{\partial t} \psi = \mathcall H \psi

[inha]

Just apply the harmonic oscillators hamiltonian to the eigenfunction. And I don't think you can determine a but you can set some constraints on it. a just tells you how wide the gaussian is.

[Hurkyl]

You mean use the eigenfunction and obtain the eigenvalue?
The definitions are that \psi is an eigenfunction of H with eigenvalue \lambda if and only if H \psi = \lambda \psi.

[qtp]

yep you should just be able to operate on the wavefunction with the hamiltonian to obtain the eigenvalues which are the energy values

[Reshma]

Just apply the harmonic oscillators hamiltonian to the eigenfunction. And I don't think you can determine a but you can set some constraints on it. a just tells you how wide the gaussian is.

yep you should just be able to operate on the wavefunction with the hamiltonian to obtain the eigenvalues which are the energy values

Thank you for your time.

H = {p^2 \over 2m} + {1\over 2} m \omega^2 x^2

p = -i \hbar \partial / \partial x

{-\hbar^2\over 2m}{\partial^2 \psi \over \partial x^2} + {1\over 2} m \omega^2 x^2 \psi = E_n \psi

Looks familiar to me, energy eigen values given by:
E_n = \hbar \omega \left(n + {1\over 2}\right)