1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Energy probabilities of the harmonic oscillator

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

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

    a) Find out A.
    b) Determine the probability that E[itex]_{0}[/itex] = hω/2 turns up, when a measuremen of energy is performed. Same for E[itex]_{1}[/itex] = 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

    2. Relevant equations

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

    H(0) =1, H(1) = 2ε, H(2) = 4ε[itex]^{2}[/itex] - 2

    ε = √(mω/h)*x

    3. The attempt at a solution

    So far I have done the normalization and have got A = (2αmω/∏h)[itex]^{1/2}[/itex] 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[itex]_{n}[/itex] = (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[itex]_{n}[/itex](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[itex]_{n}[/itex] terms, they would still have one of the parameters from a normalization.
  2. jcsd
  3. Feb 28, 2013 #2
    Hey xicor. Try using the time independent Schrodinger equation to find the eigenvalue of ψ(x,0).
    Last edited: Feb 28, 2013
  4. Mar 10, 2013 #3
    Further hints I was given to the problem suggested I dont need to solve the Schrodinger equation and instead the problem wanted me to get the probability values of C[itex]_{n}[/itex] from the Fourier's trick which states C[itex]_{n}[/itex] = ∫ψ[itex]_{n}[/itex](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)[itex]_{1/2}[/itex]/(2α+1) which I had confirmed as being correct. However for the energy state E[itex]_{1}[/itex] = 3hω/2 I'm still trying to figure out what the correct solution is. If I plug ψ[itex]_{1}[/itex](x) into equation from the Fourier's trick, I get an integral of the form C∫xe[itex]^{Dx^2}[/itex] where a hint suggested you dont actually have to integrate and I interpreted this so far as saying you can consider the case of the derivative (d/dx)e[itex]^{Dx^2}[/itex] = Dxe[itex]^{Dx^2}[/itex] so the integral in form C∫xe[itex]^{Dx^2}[/itex] = (C/D)*e[itex]^{Dx^2}[/itex] 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[itex]_{1}[/itex].

    Therefore I got C[itex]_{1}[/itex] = 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[itex]_{0}[/itex](α) which had C[itex]_{0}[/itex](0) = 0 and increased until C[itex]_{0}[/itex](1/2) = 1 and then started dropping to 0 afterwards. I'm still not certain what the maximum C[itex]_{0}[/itex](1/2) = 1 means though, is this some special case in the harmonic oscillator?
  5. Mar 10, 2013 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted