#### [SOLVED] How To Use Ladder Operators?

I'm studying for a test in quantum mechanics and I'm having a hard time understanding how to use ladder operators. There are no examples in my text book, only definitions that I can't understand how to use, so I hope you can help me instead.

So, there's an assignment where you're supposed to calculate first order correction to the ground state using ladder operators. In the right answer they say that

$$E_0^1 = \langle 0|H'|0\rangle$$

where does the zeros come from? I understand that they're coming from the wave function $\psi$. But I don't really understand the theory behind this.

Then, after some calculations it's written in the right answer

$$\langle 0|aa^\dagger a^\dagger|0\rangle +\langle0|aaa^\dagger a^\dagger|0\rangle =\langle0|aa^\dagger a|1\rangle +\langle0|aaa^\dagger|1\rangle$$

What are the mathematical rules of these ladder operators? How did these zeros become ones?

I would really appreciate if someone could help me understand this a little bit better. #### @bluesquare 2013-10-29 16:23:33

Well, $0$ is the ground state of the wave function and it's like the very first rung of the physical ladder and designated as $0$ in your case and when you operate lowering operator "$a$" on it, it will take you to the negative state but that is forbidden so you will get your answer as zero. And when you operate raising operator "$a^+$" on it, then it will simply move one step higher to the first excited state which was designated as $1$ in your case. So in principle by applying the raising operator you can go to infinitely higher excited states but by applying lowering operator you can go to lower states till you reach the last rung that means the ground state or the $0$th state and after that it's forbidden. Hope you can catch up now. #### @Kyle Kanos 2013-10-29 16:10:23

The $|0\rangle$ represents the ground state wavefunction. If we consider the harmonic oscillator, we know that the wavefunction takes the form $$\psi\left(x\right)=\frac{1}{2^n\sqrt{n}}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}e^{-m\omega x^2/2\hbar}\cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)$$ where $H_n(x)$ is the Hermite polynomial. The solution to the Schrdinger equation, $\hat{H}\psi=E_n\psi$, gives the energy as $$E_n=\hbar\omega\left(n+\frac12\right)$$

Since $\psi$ has a rather complicated form, it's easier just to represent it as a ket: $\psi\to|n\rangle$, since the energy really depends on $n$. The lowest value $n$ can be is 0, so the ground state is $|0\rangle$.

The ladder operators raise and lower the index $n$ in the eigenstate: $$a|n\rangle = \sqrt{n}|n-1\rangle \\ a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$ When you apply the bra $\langle n|$ to the above, we get $$\langle n|a|n\rangle = \sqrt{n+1}\,\langle n|n-1\rangle =0\\ \langle n|a^\dagger|n\rangle = \sqrt{n}\,\langle n|n+1\rangle=0$$ because each $|n\rangle$ is orthogonal to state $|m\rangle$ if $m\neq n$. If we did a raising & lowering, we'd have $$\langle n|aa^\dagger|n\rangle=\sqrt{n+1}\langle n|a|n+1\rangle=\sqrt{n+1}\cdot\sqrt{n+1}\langle n|n\rangle=n+1$$

For your assignment, you can separate the operators as $$aaa^\dagger a^\dagger|0\rangle=\left(aaa^\dagger\right)a^\dagger|0\rangle$$ And then using the relation above, we find $$a^\dagger|0\rangle=\sqrt{0+1}|0+1\rangle=|1\rangle$$ Now we can act the next operator on this, $$a^\dagger|1\rangle=\sqrt{1+1}|1+1\rangle=\sqrt{2}|2\rangle$$ The last two give us a result of $$aaa^\dagger a^\dagger|0\rangle=2|0\rangle$$ Closing this off with $\langle0|$, we get $$\langle0|aaa^\dagger a^\dagger|0\rangle=2\langle0|0\rangle=2$$