Quantum Physics

• Feb 24th 2008, 11:20 AM
DiscreteW
Quantum Physics
A system can be represented by $H = -\frac{d^2}{dx^2} + x^2$ (H is the Hamiltonian operator).

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

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

Eigenvalue = 3

$A=2^{\frac{1}{2}}\pi^{-\frac{1}{4}}$
• Feb 24th 2008, 08:43 PM
topsquark
Quote:

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

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

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

Eigenvalue = 3

$A=2^{\frac{1}{2}}\pi^{-\frac{1}{4}}$

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

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

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

$\int _{-\infty} ^{\infty} A^2x^2e^{-x^2}~dx = 1$
(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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