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Thread: Quantum Physics

  1. #1
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    Quantum Physics

    A system can be represented by $\displaystyle H = -\frac{d^2}{dx^2} + x^2$ (H is the Hamiltonian operator).

    Show $\displaystyle Axe^{\frac{-x^2}{2}}$ is an eigenfunction of $\displaystyle H$ and calculate the eigenvalue. Then, by normalizing, find $\displaystyle A$.

    I have the answer, but not sure how they get it. It is:

    Eigenvalue = 3

    $\displaystyle A=2^{\frac{1}{2}}\pi^{-\frac{1}{4}}$
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  2. #2
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    Quote Originally Posted by DiscreteW View Post
    A system can be represented by $\displaystyle H = -\frac{d^2}{dx^2} + x^2$ (H is the Hamiltonian operator).

    Show $\displaystyle Axe^{\frac{-x^2}{2}}$ is an eigenfunction of $\displaystyle H$ and calculate the eigenvalue. Then, by normalizing, find $\displaystyle A$.

    I have the answer, but not sure how they get it. It is:

    Eigenvalue = 3

    $\displaystyle A=2^{\frac{1}{2}}\pi^{-\frac{1}{4}}$
    The eigenvalue equation is
    $\displaystyle \left ( -\frac{d^2}{dx^2} + x^2 \right ) \psi = \lambda \psi$

    So you need to show that
    $\displaystyle \left ( -\frac{d^2}{dx^2} + x^2 \right ) \left [ Axe^{-x^2/2} \right ] = \lambda \left [ Axe^{-x^2/2} \right ]$
    for some constant value of $\displaystyle \lambda$. Really all you are doing here is using the differential operator on the trial wavefunction and solving for $\displaystyle \lambda$.

    To get a value for A, then
    $\displaystyle \int _{-\infty} ^{\infty} \psi ^* \psi~dx = 1$

    $\displaystyle \int _{-\infty} ^{\infty} A^2x^2e^{-x^2}~dx = 1$
    (You may assume that A is a real number.) This integral is easily done by integration by parts, or using a substitution to rewrite it as a gamma function, or simply looking it up in a table. It's a rather well known integral in Physics.

    -Dan
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