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Quantum Queries on the Harmonic Oscillator

  1. Nov 14, 2007 #1


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    Hi folks,

    I wonder if I could run a few things past the quantum gurus among you - I'm just not quite convinced of some of the results I've been deriving.

    1. The problem statement, all variables and given/known data

    Consider the ground state of the simple harmonic oscillator at t = 0 with the normalised wave function [tex]\sqrt[4]{\frac{m\omega}{\pi\hbar}}e^{-\frac{m\omega}{2\hbar} x^2}[/tex].

    If the position uncertainty is defined as [tex]\Delta x = \sqrt{<\widehat{x}^{2}>-<\widehat{x}>^{2}}[/tex] and the momentum uncertainty as [tex]\Delta p = \sqrt{<\widehat{p}^{2}>-<\widehat{p}>^{2}}[/tex], determine the product [tex]\Delta x . \Delta p[/tex]. Comment on the result.

    3. The attempt at a solution

    I begin by observing that [tex]<\widehat{x}>[/tex] and [tex]<\widehat{p}>[/tex] are both zero, by symmetry arguments. Thus [tex]\Delta x = \sqrt{<\widehat{x}^{2}>}[/tex] and [tex]\Delta p = \sqrt{<\widehat{p}^{2}>}[/tex].

    I go on to calculate [tex]<\widehat{x}^{2}>[/tex] and [tex]<\widehat{p}^{2}>[/tex].

    I find that [tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}[/tex], a result I am pretty confident of (it was an easy integration).

    My queries arise for [tex]<\widehat{p}^{2}>[/tex], which comes out as [tex]\hbar(m\omega - \frac{1}{2})[/tex], giving [tex]\Delta x . \Delta p = \sqrt{\frac{\hbar}{2m\omega}} . \sqrt{\hbar(m\omega - \frac{1}{2})} = \hbar\sqrt{1/2 - 1/4m\omega}[/tex] - not a very "nice" result. I'm suspicious.

    I make [tex]<\widehat{p}^{2}>[/tex] to be

    [tex]<\widehat{p}^{2}> = \sqrt{\frac{m\omega}{\pi\hbar}}\int^{+\infty}_{-\infty} e^{-\frac{m\omega}{2\hbar} x^2}.(-\hbar^{2}\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) dx = . . . = \sqrt{m^{3}\omega^{3}\hbar/\pi}.I_{2} - \sqrt{m^{3}\omega^{3}/\pi\hbar}.I_{1}[/tex]


    [tex](\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) = ... = -\frac{m\omega}{\hbar}.e^{-m\omega x^2/2\hbar} + (\frac{m\omega}{\hbar})^2.x^2.e^{-m\omega x^2/2\hbar}[/tex],


    [tex]I_{1} = \int^{+\infty}_{-\infty}x^2.e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar^{3}}{4m^{3}\omega^{3}}}[/tex]
    [tex]I_{2} = \int^{+\infty}_{-\infty}e^{-m\omega x^2/\hbar} dx = ... = \sqrt{\frac{\pi\hbar}{m\omega}}[/tex]


    determining finally that [tex]\hbar(m\omega - \frac{1}{2})[/tex]

    Has something gone wrong here, perhaps, or should I just accept the result?

    Last edited: Nov 14, 2007
  2. jcsd
  3. Nov 14, 2007 #2


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    there is something wrong here. Look below at your second derivative of x^2 times the exponential. The two terms you get from the second derivative have a relative factor of m omega/ hbar. and yet in the above expression th erelative factor is not m omega/hbar. Check that equation.

  4. Nov 15, 2007 #3


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    You are right. The expression should have read:

    [tex]<\widehat{p}^{2}> = \sqrt{\frac{m\omega}{\pi\hbar}}\int^{+\infty}_{-\infty} e^{-\frac{m\omega}{2\hbar} x^2}.(-\hbar^{2}\frac{\partial^2}{\partial x^2}).(e^{-\frac{m\omega}{2\hbar} x^2}) dx = . . . = \sqrt{m^{3}\omega^{3}\hbar/\pi}.I_{2} - \sqrt{m^{5}\omega^{5}/\pi\hbar}.I_{1}[/tex]

    This eventually leads to [tex]\hbar m\omega / 2[/tex], and then we do indeed derive Heisenberg's uncertainty relation in the form [tex]\hbar / 2 [/tex]

    Many thanks.
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