Calculating Probability for Harmonic Oscillator States

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
eku_girl83
Messages
89
Reaction score
0
I have the U(x) functions for the ground state and first excited state of the simple harmonic oscillator. I also have the psi (x,0) wave function for this situation. How do I find the probability the particle is in a particular state? Is it simply the integral of psi(x,0) * u(x) dx evaluated from -x0 to x0... and then the square of this value?

Or am I totally off base with this?
Thanks!
 
Physics news on Phys.org
Hi eku_girl,

You have the right idea except that you have to remember to integrate over all space, not just from -x0 to x0. All you're really doing is calculating the inner product or overlap integral between your state and the states of definite energy. This quantity tells you how much your state psi looks like the ground state or first the excited state or whatever you put in there.

Hope this helps.
 
The psi (x,0) is defined from -x0 to x0 in the problem. Do I still integrate over all space?
 
I see, your initial wavefunction vanishes outside of -x0 to x0? Sorry for the confusion; yes, you should integrate from -x0 to x0. In fact, you can think about integrating over all space except that the wavefunction is zero outside of -x0 to x0 so it all works out alright.