Quantum harmonic oscillator: ladder operators

[tjackson3]

Homework Statement

For a particle of mass m moving in the potential $$V(x) = \frac{1}{2}m\omega^2x^2$$ (i.e. a harmonic oscillator), it is often convenient to express the position and momentum operators in terms of the ladder operators $$a_{\pm}$$:

$$x = \sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)$$

$$p = i\sqrt{\frac{\hbar m \omega}{2}}(a_+ - a_-)$$

Evaluate $$<x>,<x^2>,\sigma_x,<p>,<p^2>,\sigma_p$$ for the stationary state $$\psi_n$$, where n = 0 is the ground state, and check that the uncertainty principle is obeyed.

Note that I'm not asking you guys to help with all six of those; I think just a nudge in the right direction on one of them is all I need. Also note that, according to my professor, it is entirely possible to fully complete this problem without evaluating a single integral.

Homework Equations

$$\psi_0 = A_0e^{-\frac{m\omega}{2\hbar}x^2}$$

$$\psi_n = A_n(a_+)^ne^{-\frac{m\omega}{2\hbar}x^2}$$

$$A_n = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{(-i)^n}{\sqrt{n!(\hbar\omega)^n}}$$

$$a_{\pm} = \frac{1}{\sqrt{2m}}\left(\frac{\hbar}{i}\frac{d}{dx} \pm im\omega x\right)$$

$$a_-a_+ - a_+a_- = \hbar\omega$$

$$a_-\psi_0 = 0$$

$$a_+\psi_n = i\sqrt{(n+1)\hbar\omega}\psi_{n+1}$$

$$a_-\psi_n = -i\sqrt{n\hbar\omega}\psi_{n-1}$$

And using this,

$$\psi_n = A_n(a_+)^ne^{-\frac{m\omega}{2\hbar}x^2}$$

The Attempt at a Solution

All I'm asking for help on is finding <x>. I think if I can do that, the rest will come fairly easily (if not quickly). At first I tried to do it in the most brute force way possible, using the last equation there in section 2, and literally taking $$\psi_n^*x\psi_n$$, but that went nowhere, due to the computational difficulties and the difficulty in taking the complex conjugate of that expression in a nice, closed form.

The more promising route seemed to be to find <x> for n = 0 and to use that somehow to generate <x> for some general n. <x> = 0 for n = 0 by parity. Unsure of where to go from there, I tried to use that to find <x> at n = 1. Thus I have (note that $$\psi^*_0 = \psi_0$$):

$$\[a_+\psi_0 = i\sqrt{\hbar\omega}\psi_1\]<br /> <br /> \[\psi_1 = \frac{a_+\psi_0}{i\sqrt{\hbar\omega}}\]\[\psi_1^*x\psi_1 = \frac{a^*_+\psi_0}{-i\sqrt{\hbar\omega}}\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)\frac{a_+\psi_0}{i\sqrt{\hbar\omega}}\]$$

I'm stuck here, though, for a couple of reasons. Since operators don't necessarily commute, I'm not sure exactly what liberties I can take in rearranging this into a more convenient form. Additionally, since $$a_+$$ is an operator, I don't know if it even makes sense to talk about its complex conjugate. Also, short of induction, I don't know how to use this to come up with a formula for a general n.

Thank you so much!

 

[betel]

Your second attempt is in the right direction. It is sensible to talk about the conjugate of an operator. You only have to be careful with the order
$$(a_+\psi)^*= \psi^* a_+^*$$
If you look at the definition of $$a_+$$ you will find it's conjugate. Then using the commutation relations for the a's you will find a formulat for general n.

By the way, did your professor mention bra-ket notation or hermitian conjugate in the lecture?