We studied coherent states, which we had identified as eigenstates of the lowering operator, (hat{a}). A coherent
Question:
We studied coherent states, which we had identified as eigenstates of the lowering operator, \(\hat{a}\). A coherent state \(|\psiangle\) is defined by the eigenvalue equation\[\begin{equation*}\hat{a}|\psiangle=\lambda|\psiangle, \tag{6.122}\end{equation*}\]
where \(\lambda\) is a complex number. In terms of the harmonic oscillator's ground state \(\left|\psi_{0}\rightangle\), this coherent state can be expressed as\[\begin{equation*}|\psiangle=e^{-\frac{|\lambda|^{2}}{2}} e^{\lambda \hat{a}^{\dagger}}\left|\psi_{0}\rightangle . \tag{6.123}\end{equation*}\]
In this problem, we will study the properties of these coherent states.
(a) Calculate the variances of the position and momentum operators \(\hat{x}\) and \(\hat{p}\) on the general time-evolved coherent state \(|\psi(t)angle\). What is the Heisenberg uncertainty principle now?
Hint: Be sure to use the result of Eq. (6.118).
(b) The time evolution of a coherent state is eerily familiar to that of a classical particle. Classically, if you start at rest at a position \(\Delta x\) displaced from the center of the harmonic oscillator potential, what is your position \(x(t)\) and momentum \(p(t)\) as a function of time? How do these classical results compare to the time evolution of the expectation values of the position \(\hat{x}\) and momentum \(\hat{p}\) operators on the time-evolved state \(|\psi(t)angle\) ?
(c) Why don't we consider eigenstates of the raising operator, \(\hat{a}^{\dagger}\) ? What is wrong with a state \(|\chiangle\) that satisfies\[\begin{equation*}\hat{a}^{\dagger}|\chiangle=\eta|\chiangle \tag{6.124}\end{equation*}\]
for some complex number \(\eta\) ?
Step by Step Answer:
Quantum Mechanics A Mathematical Introduction
ISBN: 9781009100502
1st Edition
Authors: Andrew J. Larkoski