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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

    vela

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    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?
     
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