Energy probabilities of the harmonic oscillator

[xicor]

Homework Statement

A particl of mass m in the potential V(x) (1/2)*mω$^{2}$x$^{2}$ has the initial wave function ψ(x,0) = Ae$^{-αε^2}$.

a) Find out A.
b) Determine the probability that E$_{0}$ = hω/2 turns up, when a measuremen of energy is performed. Same for E$_{1}$ = 3hω/2
c) What energy values might turn up in an energy measurement?
d) Sketch the probability to measure hω/2 as a function of α and explain the maximum

Homework Equations

ψ$_{n}$ = (mω/πh)$^{1/4}$*[1/√(2$^{n}$*n!)]H$_{n}$(ε)e$^{-(ε^2)/2}$

H(0) =1, H(1) = 2ε, H(2) = 4ε$^{2}$ - 2

ε = √(mω/h)*x

The Attempt at a Solution

So far I have done the normalization and have got A = (2αmω/∏h)$^{1/2}$ but can't think my way through part b yet. My understanding so far is that you find ψ(x,t) and consider the fact that E$_{n}$ = (n+1/2)hω but that case was for when you was just a linear combination of wave functions and A is a numerical fraction. Are you suppose to use the ψ$_{n}$(x,0) formula to find the wave function at different excited states and find the probability based off the given H values? I still don't see how you would get a probability though since if I were to apply the c$_{n}$ terms, they would still have one of the parameters from a normalization.

 

[PhysicsGente]

Hey xicor. Try using the time independent Schrödinger equation to find the eigenvalue of ψ(x,0).

 

[xicor]

Further hints I was given to the problem suggested I don't need to solve the Schrödinger equation and instead the problem wanted me to get the probability values of C$_{n}$ from the Fourier's trick which states C$_{n}$ = ∫ψ$_{n}$(x)f(x)dx where in this case f(x) = ψ(x, 0). From this I was able to find the probability function for the ground state that with respect to α. The equation I got was 2(2σ)$_{1/2}$/(2α+1) which I had confirmed as being correct. However for the energy state E$_{1}$ = 3hω/2 I'm still trying to figure out what the correct solution is. If I plug ψ$_{1}$(x) into equation from the Fourier's trick, I get an integral of the form C∫xe$^{Dx^2}$ where a hint suggested you don't actually have to integrate and I interpreted this so far as saying you can consider the case of the derivative (d/dx)e$^{Dx^2}$ = Dxe$^{Dx^2}$ so the integral in form C∫xe$^{Dx^2}$ = (C/D)*e$^{Dx^2}$ where the direct integration will produce two values that are the same when squared and cancel out, therefore the direct integral is zero which applies to the case of C$_{1}$.

Therefore I got C$_{1}$ = 0 but I'm not sure how you would reason what would happen with higher energy probabilities. I was already told that there will be other energy states with energies probabilities that are greater then zero. Is it the case that the probability is 0 for odd energy states because of symmetries in the probability distributions? Also, it appears that for higher energy states, there would be a higher power of α in the dominator meaning that the probability will be smaller in that case then for the ground state.

I was also able to graph the ground state C$_{0}$(α) which had C$_{0}$(0) = 0 and increased until C$_{0}$(1/2) = 1 and then started dropping to 0 afterwards. I'm still not certain what the maximum C$_{0}$(1/2) = 1 means though, is this some special case in the harmonic oscillator?

 

[vela]

Further hints I was given to the problem suggested I don't need to solve the Schrödinger equation and instead the problem wanted me to get the probability values of C$_{n}$ from the Fourier's trick which states C$_{n}$ = ∫ψ$_{n}$(x)f(x)dx where in this case f(x) = ψ(x, 0). From this I was able to find the probability function for the ground state that with respect to α. The equation I got was 2(2σ)$_{1/2}$/(2α+1) which I had confirmed as being correct. However for the energy state E$_{1}$ = 3hω/2 I'm still trying to figure out what the correct solution is. If I plug ψ$_{1}$(x) into equation from the Fourier's trick, I get an integral of the form C∫xe$^{Dx^2}$ where a hint suggested you don't actually have to integrate and I interpreted this so far as saying you can consider the case of the derivative (d/dx)e$^{Dx^2}$ = Dxe$^{Dx^2}$ so the integral in form C∫xe$^{Dx^2}$ = (C/D)*e$^{Dx^2}$ where the direct integration will produce two values that are the same when squared and cancel out, therefore the direct integral is zero which applies to the case of C$_{1}$.

Therefore I got C$_{1}$ = 0 but I'm not sure how you would reason what would happen with higher energy probabilities. I was already told that there will be other energy states with energies probabilities that are greater then zero. Is it the case that the probability is 0 for odd energy states because of symmetries in the probability distributions? Also, it appears that for higher energy states, there would be a higher power of α in the dominator meaning that the probability will be smaller in that case then for the ground state.

You have the basic idea. Your integral for C₁ vanishes because you're integrating an odd function of x over a symmetric interval. Looking at the evenness and oddness of the Hermite polynomials, you should be able to convince yourself that the probability of finding the system in an n=odd state is zero.

I was also able to graph the ground state C$_{0}$(α) which had C$_{0}$(0) = 0 and increased until C$_{0}$(1/2) = 1 and then started dropping to 0 afterwards. I'm still not certain what the maximum C$_{0}$(1/2) = 1 means though, is this some special case in the harmonic oscillator?

You can express the state of the system as a linear combination of the eigenstates, so
$$\psi(x) = c_0\psi_0(x) + c_1\psi_1(x) + \cdots + c_n\psi_n(x) + \cdots$$ where $|c_n|^2$ is equal to the probability of finding the system in the state $\psi_n$. If $c_0=1$, what does that tell you about the other c's?