# Kinetic energy of harmonic oscilator

1. Dec 6, 2007

### stunner5000pt

[SOLVED] Kinetic energy of harmonic oscilator

1. The problem statement, all variables and given/known data
Find the expectation value of the kinetic energy of the nth state of a Harmonic oscillator

2. Relevant equations
$$<T> = \frac{<p^2>}{2m}$$
$$p_{x} = \frac{1}{i} \sqrt{\frac{m\hbar\omega}{2}} (\hat{a} -\hat{a}^\dagger)$$
$$a^\dagger \psi_{n} = \sqrt{n+1} \psi_{n+1}$$
$$a\psi_{n} = \sqrt{n-1} \psi_{n-1}$$
3. The attempt at a solution
So $$p_{x}^2 = \left(\frac{1}{i} \sqrt{\frac{m\hbar\omega}{2}} (\hat{a} -\hat{a}^\dagger)\right)^2$$

So to calculate the <T> do i just do this:
$$<T> = <\Psi_{n}|\frac{p^2}{2m}|\Psi(n)>$$
$$<T> = -\frac{\hbar\omega}{4} \int \psi_{n}^* (aa - aa^\dagger - a^\dagger a + a^\dagger a^\dagger) \psi_{n} dx$$

$$<T> = -\frac{\hbar\omega}{4} \int\psi_{n}^* \sqrt{n(n-1)} \psi_{n-2} + \sqrt{n(n+1)}\psi_{n} + \sqrt{n^2} \psi_{n} + \sqrt{(n+1)(n+2)} \psi_{n+2} dx$$

only one term will survive, the nth state ones because the wave functions are orthogonal
$$<T> = -\frac{n\hbar\omega}{4}$$

Is this correct??

2. Dec 6, 2007

### malawi_glenn

Negative Kinetic energy is not allowed.

In your final integral, two of four terms will survive.

And also:
$$a\psi_{n} = \sqrt{n-1} \psi_{n-1}$$
should be:
$$a\psi_{n} = \sqrt{n} \psi_{n-1}$$

3. Dec 8, 2007

### stunner5000pt

thanks i got the required answer!
where do i mark this solved?