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Quantum Harmonic oscillator, <T>/<V> ratio

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider an electron confined by a 1 dimensional harmonic potential given by ## V(x) = \dfrac{1}{2} m \omega^2 x^2##. At time t=0 the electron is prepared in the state

    [tex]\Psi (x,0) = \dfrac{1}{\sqrt{2}} \psi_0 (x) + \dfrac{1}{\sqrt{2}} \psi_4 (x)[/tex]

    with ## \psi_n (x) = \left( \dfrac{m \omega}{\pi \hbar}\right)^{0.25} \dfrac{1}{\sqrt{2^n n!}} H_n \left( \sqrt{\dfrac{m \omega}{\hbar}}x \right) exp \left( - \dfrac{m \omega}{2 \hbar} x^2 \right) ##

    the normalised solutions to the time independent Schrodinger equation, with ##H_n## being the Hermite Polynomials.

    (a) Express the square of the momentum operator as ladder operators.
    (b) Give the ratio of the expectation values of the kinetic and potential energies as a function of time.

    2. Relevant equations

    ## \hat{x} = \sqrt{\dfrac{\hbar}{2m \omega}} (a_+ + a_-) ##

    ## \hat{p} = i \sqrt{\dfrac{\hbar m \omega}{2}} (a_+ - a_-) ##

    3. The attempt at a solution

    For the first one I got:

    [tex]\hat{p}^2 = \left[ i \sqrt{\dfrac{\hbar m \omega}{2}} (a_+ - a_-) \right]^2 = -\dfrac{\hbar m \omega}{2} (a_+a_+ - a_-a_+ - a_+ a_- + a_-a_-) [/tex]


    For the second one, I attempted it using the operators
    [T = kinetic energy, V = potential energy]

    [tex]\displaystyle <T> = < \dfrac{\hat{p}}{2m} > = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_+a_+ - a_-a_+ - a_+ a_- + a_-a_-) \right] \Psi dx [/tex]

    [tex]\displaystyle <V> = <0.5 \omega ^2 \hat{x}^2> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_+a_+ + a_-a_+ + a_+ a_- + a_-a_-) \right] \Psi dx[/tex]


    Now each of them will split into 4 integrals. The a+a+ and a-a- operator integrals will evaluate to zero because of its orthogonality properties. That leaves us with:

    [tex]\displaystyle <T> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (- a_-a_+ - a_+ a_-) \right] \Psi dx [/tex]

    [tex]\displaystyle <V> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_-a_+ + a_+ a_-) \right] \Psi dx[/tex]


    That'll mean those terms disappear, and then... we end up with -1 as a ratio?? What did I do wrong?
     
  2. jcsd
  3. Jun 1, 2013 #2

    TSny

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    Did you drop a minus sign when substituting for ##\hat{p}##2?
     
  4. Jun 1, 2013 #3
    Seems to me you forgot a minus sing in ## \langle T \rangle ##
     
  5. Jun 1, 2013 #4
    Oh right, yea there should be a negative there.

    That just gives me 1 as the ratio, and it's independent of time.

    Does that mean that the ratio of kinetic energy to potential energy, no matter when it is, is always 1?
     
  6. Jun 1, 2013 #5

    TSny

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    Well, the expectation value of T equals the expectation value of V at any time. The values of T and V themselves are uncertain.
     
  7. Jun 1, 2013 #6
    Hmm, interesting. I cannot really visualise expectation values so it's sort of hard.

    So in a simple harmonic oscillator, I'm guessing the kinetic energy is sort of like a normal distribution, peaked at the center. Therefore the expectation value is at the middle.
    The potential energy is at both sides equally, and so its expectation value is in the middle as well.
    I guess that makes sense.
     
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