Let's now study the power method for estimating the ground-state energy, applied to the quantum harmonic oscillator. For this problem, we will work with the Hamiltonian

\[\begin{equation*}\hat{H}=\hbar \omega\left(1+\hat{a}^{\dagger} \hat{a}\right) \tag{10.120}\end{equation*}\]

where \(\hat{a}\) and \(\hat{a}^{\dagger}\) are the familiar lowering and raising operators, respectively. Note the slight difference between this Hamiltonian and what we would typically think of as the harmonic oscillator: all we've done here is shift the potential up by an amount \(\hbar \omega / 2\), which will make some of the mathematical manipulations simpler later. Importantly, this constant shift does not affect the eigenstates of the Hamiltonian, it just shifts the eigenvalues up by that same amount.

(a) Let's first calculate the inverse of this Hamiltonian, \(\hat{H}^{-1}\). One answer is, of course, simply

\[\begin{equation*}\hat{H}^{-1}=\frac{1}{\hbar \omega} \frac{1}{1+\hat{a}^{\dagger} \hat{a}} \tag{10.121}\end{equation*}\]

However, this isn't so useful for determining how this operator acts on states. Instead, we can express the inverse Hamiltonian as a sum over products of \(\hat{a}\) and \(\hat{a}^{\dagger}\) :

\[\begin{equation*}\hat{H}^{-1}=\frac{1}{\hbar \omega} \sum_{n=0}^{\infty} \beta_{n}\left(\hat{a}^{\dagger}\right)^{n} \hat{a}^{n} \tag{10.122}\end{equation*}\]

for some coefficients \(\beta_{n}\). Determine the coefficients \(\beta_{n}\) and thus the inverse Hamiltonian \(\hat{H}^{-1}\) by demanding that \(\hat{H} \hat{H}^{-1}=\mathbb{I}\), the identity operator.

This way of ordering the terms in \(\hat{H}^{-1}\) is called normal order: all raising operators \(\hat{a}^{\dagger}\) are to the left of all lowering operators \(\hat{a}\) in each term in the sum.

(b) Let's use this inverse Hamiltonian to estimate the ground-state energy of the harmonic oscillator in which we start with a coherent state \(|\chiangle\). Recall that \(|\chiangle\) satisfies \[\begin{equation*}\hat{a}|\chiangle=\lambda|\chiangle \tag{10.123}\end{equation*}\]
for some complex number \(\lambda\). From this coherent state, what is the expectation value of the Hamiltonian, \(\langle\chi|\hat{H}| \chiangle\) ?

(c) Now, let's use the inverse Hamiltonian to improve our estimate of the ground-state energy. What is your new estimate of the ground state after one application of \(\hat{H}^{-1}\) on the coherent state? Recall that this estimate is \[\begin{equation*}E_{0} \leq \frac{\left\langle\chi\left|\hat{H}^{-1} \hat{H} \hat{H}^{-1}\right| \chi\rightangle}{\left\langle\chi\left|\hat{H}^{-1} \hat{H}^{-1}\right| \chi\rightangle}=\frac{\left\langle\chi\left|\hat{H}^{-1}\right|\chi\rightangle}{\left\langle\chi\left|\hat{H}^{-1} \hat{H}^{-1}\right| \chi\rightangle} \tag{10.124}\end{equation*}\]
You should ultimately find \[\begin{equation*}E_{0} \leq \frac{\left\langle\chi\left|\hat{H}^{-1}\right| \chi\rightangle}{\left\langle\chi\left|\hat{H}^{-1} \hat{H}^{-1}\right| \chi\rightangle}=\hbar \omega \frac{e^{|\lambda|^{2}}-1}{\int_{0}^{|\lambda|^{2}} d x\frac{e^{x}-1}{x}} \tag{10.125}\end{equation*}\]
The integral that remains can't be expressed in terms of elementary functions.

 For the denominator factor, it might help to slide the identity \(\mathbb{I}\) in between the two inverse Hamiltonians, where \[\begin{equation*}\mathbb{I}=\sum_{n=0}^{\infty}\left|\psi_{n}\rightangle\left\langle\psi_{n}\right| \tag{10.126}\end{equation*}\]
where \(\left|\psi_{n}\rightangle\) is the \(n\)th energy eigenstate.

(d) Now, just to get a sense of this approximation, verify that you get the exact result for the ground-state energy if \(\lambda \rightarrow 0\). Also, evaluate the approximation of part

(c) for \(\lambda=1\). The remaining integral evaluates to \[\begin{equation*}\int_{0}^{1} d x \frac{e^{x}-1}{x}=1.31790215 \ldots \tag{10.127}\end{equation*}\]
How does this result compare to the initial estimate of part (b)?