# Raising and lowering operators of the Hamiltonian

1. May 8, 2014

### CAF123

1. The problem statement, all variables and given/known data
a) The operators $a$ and $a^{\dagger}$ satisfy the commutation relation $[a,a^{\dagger}] = 1$. Find the normalization of the state $|\psi \rangle = C (a^{\dagger} )^2 |0\rangle$, where the vacuum state $|0\rangle$ is such that $a|0\rangle = 0$

b)A one dimensional simple harmonic oscillator has the Hamiltonian $$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}mw^2 \hat{x}^2.$$ The operator $\hat{a}$ is defined as $$\hat{a} = \left(\frac{mw}{2\hbar}\right)^{1/2}\hat{x} + \frac{i}{\sqrt{2mw\hbar}} \hat{p}$$

Suppose that the oscillator frequency is changed slightly to $w \rightarrow \sqrt{1+\epsilon} w$ What are the exact new energy eigenvalues? Now calculate the first order perturbation in the energy and compare with the exact answer.
2. Relevant equations
In a), perhaps that $a|n\rangle = \sqrt{n}|n-1\rangle$ and that $a^{\dagger} |n\rangle = \sqrt{n+1} |n+1\rangle$

3. The attempt at a solution
In a), write out the commutator explicitly $aa^{\dagger} - a^{\dagger}a = 1$. Now I tried various manipulations of this equation to get something that resembled the equation given. Multiply by $(a^{\dagger})^2$ on the left and $|0\rangle$ on the right would give $(a^{\dagger})^2 a a^{\dagger} |0\rangle - (a^{\dagger})^3 a|0\rangle = (a^{\dagger})^2 |0\rangle.$ But this simplifies to LHS=RHS, so it is of no use.

b)New energy eigenvalues would be $E_n = \hbar (\sqrt{1+\epsilon} w) (n+1/2)$ and the shift is $E' = \langle n | \hat{H'} | n\rangle$ where $\hat{H'}$ is the Hamiltonian above but with $w$ replaced with $\sqrt{1+\epsilon}w$ Do we treat $\epsilon$ as a constant here?

Many thanks.

2. May 8, 2014

### dextercioby

For a), just compute the vector itself, a†|0> = √0+1 |1> = |1>, etc.

3. May 9, 2014

### CAF123

Applying a† again would give √2 |2>, not some constant premultiplying the state |0>. So I think somehow we have to incorporate the commutation relation given, which I tried to do in the OP.

4. May 9, 2014

### vela

Staff Emeritus
Try using $\|C(a^\dagger)^2\vert 0 \rangle\|^2=1$.