# Homework Help: Harmonic oscillator

1. Apr 20, 2010

### rayman123

1. The problem statement, all variables and given/known data
Can someone please give me some hints how to solve this problem.
Show that expected value for the kinetic energy is the same as the expected value for the potential energy for a harmonic oscillator in gound state.

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 20, 2010

### kuruman

One way would be to calculate the expectation values <T> and <V> separately and show that they are equal. You know what the ground state wavefunction for the harmonic oscillator is, so you have to do a couple of integrals.

3. Apr 20, 2010

### rayman123

hello! I have been trying to write my calculations by using 'latex' but then i get a problem, it only shows me the very first part of my solutions when i want to add more it does not work at all.....do you know what might be a problem?

4. Apr 20, 2010

### kuruman

Last edited: Apr 20, 2010
5. Apr 20, 2010

### rayman123

$$\psi_{0}= (\frac{\alpha}{\pi})^{\frac{1}{4}} e^{\frac{-y^2}{2}}$$

$$y= \sqrt{\frac{m\omega}{\hbar}}x\Rightarrow y=\sqrt{\alpha}x$$
$$\alpha= \frac{m\omega}{\hbar}$$

$$<|x^2|>=\int_{-\infty}^{\infty}dxx^2|{\psi_{0}}^2|=\sqrt{\frac{m\omega}{\pi \hbar}}\int_{-\infty}^{\infty}dxx^2e^{\frac{-m \omega x^2}{\hbar}}=I$$

$$\int_{-\infty}^{\infty}dxx^2e^{-\alpha x^2}=\frac{1}{2\alpha}\sqrt{\frac{\pi}{\alpha}}$$

$$I= \frac{1 \hbar}{2m \omega}$$

for $$<|p^2|>=\frac{m \hbar \omega}{2}$$

$$<|E_{k}| >= \frac{1}{2m}|<|p^2>|= \frac{\hbar \omega}{4}$$
$$<|E_{p}|> = \frac{m\omega^2}{2}<|x^2|>= \frac{\hbar \omega}{4}$$

can i calculate it this way?
I have problems with finding formulas for the expected value for kinetic and potential energy....

6. Apr 20, 2010

### kuruman

The <|x2|> expectation value looks correct. How did you do the <|p2/2m|> integral?

7. Apr 20, 2010

### rayman123

$$<|p^2|>=-\hbar^2 \int_{-\infty}^{\infty}dxe^{\frac{-m \omega x^2}{2\hbar}}\frac{\partial ^2}{\partial x^2}e^{\frac{-m \omega x^2}{2\hbar}}\sqrt{\frac{m\omega}{\pi \hbar}}= \hbar m \omega -m^2 \omega^2<|x^2|>=\frac{m \hbar \omega}{2}$$

8. Apr 20, 2010

### kuruman

Methinks you are done.

9. Apr 20, 2010

thank you!:)