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446: The Variation Method in Momentum Space

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  • The following normalized trial wavefunction is proposed for a variational calculation on the harmonic oscillator.

    \[ \psi (x, a) := \sqrt{ \frac{1}{a}} exp( \frac{-|x|}{a})\]

    \[ \int_{- \infty}^{ \infty} \psi (x, a)^2 dx~~~assume,~a > 0 \rightarrow 1\]

    However, the graph below shows a cusp at x = 0, indicating that the wavefunction is not well‐behaved and therefore cannot be used for quantum mechanical calculations.

    Screen Shot 2019-02-14 at 1.23.02 PM.png

    Therefore, the wavefunction is Fourier transformed into the momentum representation.

    \[ \Phi (p, a) := \int_{- \infty}^{ \infty} \frac{exp(-ipx)}{ \sqrt{2 \pi}} \sqrt{ \frac{1}{a}} exp( \frac{-|x|}{a}) dx~|_{simplify}^{assume,~a>0} \rightarrow (-a^{ \frac{1}{2}}) \frac{2^{ \frac{1}{2}}}{(ipa-1) \pi ^{ \frac{1}{2}} (ipa +1)}\]

    Normalization is checked and the function is graphed.

    \[ \int_{- \infty}^{ \infty} \Phi (p, a)^2 dp~~~assume,~a > 0 \rightarrow 1\]

    Screen Shot 2019-02-14 at 1.23.08 PM.png

    The momentum wavefunction appears to be well‐behaved, so a variational calculation will be carried out in momentum space.

    Assuming a m = k =1 and h = 2π, we have for the harmonic oscillator in momentum space.

    • Momentum space integral: \( \int_{- \infty}^{ \infty} \blacksquare dp\)
    • Momentum operator: \( p \blacksquare\)
    • Kinetic energy operator: \( \frac{p^2}{2}\)
    • Position operator: \( i \frac{d}{dp} \blacksquare\)
    • Potential energy operator: \( \frac{-1}{2} \frac{d^2}{dp^2} \blacksquare\)

    Evaluate the energy integral in the momentum representation:

    \[ E(a) := \int_{- \infty}^{ \infty} \Phi (p, a) \frac{p^2}{2} \Phi (p, a) dp... + \int_{- \infty}^{ \infty} \Phi (p, a) \frac{-1}{2} \frac{d^2}{dp^2} \Phi (p, a) dp~|_{assume,~a >0}^{simplify} \rightarrow \frac{1}{4} \frac{2 + a^4}{a^2}\]

    Minimize energy with respect to the variational parameter:

    a := 1 a := Minimize (E, a) a = 1.189 E(a) = 0.707

    Display optimum wavefunction along with exact wavefunction:

    \[ Exact(p) := \frac{1}{ \pi ^{ \frac{1}{4}}} e^{ \frac{-1}{2} p^2}\]

    Screen Shot 2019-02-14 at 1.23.15 PM.png

    Naturally the agreement with the exact solution is not favorable because of the poor quality of the original coordinate space wavefunction.

    \[ \frac{E(a) - 0.5}{0.5} = 41.421\]