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Ground state of harmonic oscillator

  1. Jan 21, 2014 #1
    or6ja8.png

    Shouldn't the integrating factor be ##exp(\frac{m\omega x}{\hbar})##?
    [tex]\frac{\partial <x|0>}{\partial x} + \frac{m\omega x}{\hbar} <x|0> = 0 [/tex]

    This is in the form:

    [tex]\frac{\partial y}{\partial x} + P_{(x)} y = Q_{(x)} [/tex]

    Where I.F. is ##exp (\int (P_{(x)} dx)##
     
  2. jcsd
  3. Jan 21, 2014 #2

    Bill_K

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    And P(x) = mωx/ħ, and ∫P(x) dx = mωx2/2ħ, and the integrating factor is exp(mωx2/2ħ).
     
  4. Jan 21, 2014 #3

    vanhees71

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    You can as well solve the linear first-order ODE directly, i.e., with [itex]u_0(x)=\langle x|0 \rangle[/itex] given by the book to read
    [tex]u_0'=-\frac{x}{2l^2} u_0.[/tex]
    This you can write as
    [tex]\frac{u_0'}{u_0}=-\frac{x}{2l^2}.[/tex]
    This can be integrated
    [tex]\ln \left (\frac{u_0(x)}{u_0(0)} \right )=-\frac{x^2}{4l^2} \; \Rightarrow \; u_0(x)=u_0(0) \exp \left (-\frac{x^2}{4l^2} \right ).[/tex]
    Further the wave function should be normalized to 1, i.e.,
    [tex]\int_{\mathbb{R}} \mathrm{d}x \; |u_0(x)|^2=|u(0)|^2 \int_{\mathbb{R}} \mathrm{d} x exp \left (-\frac{x^2}{2l^2} \right )=\sqrt{2 \pi l^2} |u_0(0)|^2 \stackrel{!}{=}1.[/tex]
    From this we find, up to an irrelevant phase factor,
    [tex]u_0(0)=\left (\frac{1}{2 \pi l^2} \right )^{1/4}.[/tex]
    This completes the derivation for the ground state of the harmonic oscillator in position representation.
     
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