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Harmonic oscillator positive position expectation value?!

  1. Feb 21, 2016 #1
    So this is something that troubled me a bit- in Shankar's PQM, there's an exercise that asks you to find the position expectation value for the harmonic oscillator in a state [itex] \psi [/itex] such that
    [tex] \psi=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) [/tex]
    Where [itex] |n\rangle [/itex] is the [itex] n^{th} [/itex] energy eigenstate of the oscillator.

    Now, I calculated the expectation value of position in the following way:
    [tex] \langle X \rangle=\langle \psi| X|\psi\rangle=\frac{1}{2}((\langle 0|+\langle 1|)\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})(|0\rangle+|1\rangle)) [/tex]

    Where [itex]a[/itex] and [itex]a^{\dagger}[/itex] are the annihilation and creation operators, respectively. Using the properties of these operators, we quickly get to

    [tex]\langle X \rangle=\sqrt{\frac{\hbar}{2m\omega}} [/tex]

    (I also did it another slightly more complicated way, where I integrated in the position representation; this gave me the same result.)

    Now, what bothers me about this whole thing is the following:

    We know that the state is in a superposition of two positive energy eigenstates. Additionally, the expectation value of position for an energy eigenstate will be zero. Why is it that this superposition of eigenstates, this interference, brings out a positive expectation value for position? What kills me about this most is the fact that the expectation value is positive. This seems to do away with the apparent symmetry in the problem; why should the particle in a harmonic oscillator potential choose to be more at the "right" side? Why couldn't the expectation value be [itex]\pm[/itex] the above value? Or, even better, just zero?

    So essentially there's two questions here:
    1) why does the superposition of energy eigenstates here lead to a nonzero position expectation value? And,
    2) why is this value positive? Doesn't this violate the symmetry in the harmonic oscillator potential?
     
  2. jcsd
  3. Feb 21, 2016 #2

    blue_leaf77

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    Because the components of the superposition correspond to different parity signs. The ground state has even parity while the first excited state has the odd one.
    The superposition state you have there is not an energy eigenstate, therefore, its non-symmetric nature does not violate the Hamiltonian.
    The expectation value you have calculated is that at ##t=0##. If you let the system evolve in time, the expectation value of position will be a function of time. In particular, it will show some oscillatory motion between two fixed positions.
     
  4. Feb 22, 2016 #3
    Cool, thank you!
     
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