We had shown that any state \(|\psiangle\) on the Hilbert space of the harmonic oscillator is described by some analytic function \(f\left(\hat{a}^{\dagger}\right)\) of the raising operator acting on the ground state \(\left|\psi_{0}\rightangle\) :

\[\begin{equation*}|\psiangle=f\left(\hat{a}^{\dagger}\right)\left|\psi_{0}\rightangle . \tag{6.130}\end{equation*}\]

Because both \(\left|\psi_{0}\rightangle\) and \(|\psiangle\) are on the Hilbert space and normalized, this might suggest that the resulting operator \(f\left(\hat{a}^{\dagger}\right)\) is unitary. However, we have only required \(f\left(\hat{a}^{\dagger}\right)\) to act on one special state in the Hilbert space, \(\left|\psi_{0}\rightangle\), and not a general state. So is it unitary?

(a) Consider a general state \(|\psiangle\) on the Hilbert space of the harmonic oscillator. Now, act the operator \(f\left(\hat{a}^{\dagger}\right)\) as defined in Eq. (6.56) on the state \(|\psiangle\); call the result \(|\xiangle\) :

\[\begin{equation*}|\xiangle=f\left(\hat{a}^{\dagger}\right)|\psiangle \tag{6.131}\end{equation*}\]

Is \(|\xiangle\) a physical state and on the Hilbert space, as well? That is, is \(f\left(\hat{a}^{\dagger}\right)\) unitary?

(b) If \(f\left(\hat{a}^{\dagger}\right)\) is not unitary, can you modify it so it is? Note that this modification cannot affect its action on the ground state \(\left|\psi_{0}\rightangle\). If \(f\left(\hat{a}^{\dagger}\right)\) is unitary, then it can be expressed as the exponential of a Hermitian operator \(\hat{T}\) :

\[\begin{equation*}f\left(\hat{a}^{\dagger}\right)=e^{i \hat{T}} \tag{6.132}\end{equation*}\]

What is \(\hat{T}\) ?

(c) For a coherent state, we had identified this analytic function as (up to nomalization)

\[\begin{equation*}f\left(\hat{a}^{\dagger}\right)=e^{\lambda \hat{a}^{\dagger}} \tag{6.133}\end{equation*}\]

Is this unitary? If not, can you modify it to make it unitary such that its action on the ground-state wavefunction \(\left|\psi_{0}\rightangle\) is unchanged?