Tessellate S.T.E.M.S - Physics - College - Set 1 - Problem 5

Consider the energy eigenstates for the harmonic oscillator, $\ket{n}$ . The vacuum state is given by $\ket{0}$ , such that $\hat{a}\ket{0} = \ket{0}$ .

Given the 'perturbed' Hamiltonian $\hat{H} = \hat{H}_{0} + \hat{H}^{\prime}$ , where $\hat{H}_{0}$ represents the harmonic oscillator Hamiltonian and $\hat{H}^{\prime} = \exp (\hat{a}\hat{a}^{\dagger} / \lambda)$ , we calculate the first order correction to the vacuum state energy as $E^{(1)}_{0} = \bra{0}\hat{H}^{\prime}\ket{0}$ . In this problem, it is a straightforward calculation,

$\displaystyle E^{(1)}_{0} = \bra{0}\hat{H}^{\prime}\ket{0} = \bra{0}\exp \left(\frac{\hat{a}\hat{a}^{\dagger}}{\lambda}\right)\ket{0} = \bra{0} \left[ \sum_{k=0}^{\infty} \frac{1}{k!}\left(\frac{\hat{a}\hat{a}^{\dagger}}{\lambda}\right)^{k} \right]\ket{0} = \bra{0} \left[ \sum_{k=0}^{\infty} \frac{(1/\lambda)^{k}}{k!}\left(\hat{a}\hat{a}^{\dagger}\right)^{k}\ket{0} \right]$ .

It can be easily shown that $\hat{a}\hat{a}^{\dagger}\ket{n} = (n+1)\ket{n}$ , i.e. the states $\ket{n}$ are eigenvectors of the $\hat{a}\hat{a}^{\dagger}$ operator. Therefore, it follows that $\left(\hat{a}\hat{a}^{\dagger}\right)^{k}\ket{n} = (n+1)^{k}\ket{n}$ . In particular, we have $\left(\hat{a}\hat{a}^{\dagger}\right)^{k}\ket{0} = (1)^{k}\ket{0} = \ket{0}$ . Applying this result in the expression above,

$\displaystyle E^{(1)}_{0} = \bra{0} \left[ \sum_{k=0}^{\infty} \frac{(1/\lambda)^{k}}{k!}\ket{0} \right] = \sum_{k=0}^{\infty} \frac{(1/\lambda)^{k}}{k!}\langle0\ket{0}$ .

Assuming the state is normalized, $\langle0\ket{0} = 1$ ,

$\displaystyle E^{(1)}_{0} = \sum_{k=0}^{\infty} \frac{(1/\lambda)^{k}}{k!}$ $\displaystyle \therefore \hspace{2pt} \boxed{E^{(1)}_{0} = e^{1/\lambda}}$ .