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### [1] pr. 2.19(c)

Show that \( \Abs{f(n)}^2 \) for a coherent state written as

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:561}

\ket{z} = \sum_{n=0}^\infty f(n) \ket{n}

\end{equation}

has the form of a Poisson distribution, and find the most probable value of \( n\), and thus the most probable energy.

### A:

The Poisson distribution has the form

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:581}

P(n) = \frac{\mu^{n} e^{-\mu}}{n!}.

\end{equation}

Here \( \mu \) is the mean of the distribution

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:601}

\begin{aligned}

\expectation{n}

&= \sum_{n=0}^\infty n P(n) \\

&= \sum_{n=1}^\infty n \frac{\mu^{n} e^{-\mu}}{n!} \\

&= \mu e^{-\mu} \sum_{n=1}^\infty \frac{\mu^{n-1}}{(n-1)!} \\

&= \mu e^{-\mu} e^{\mu} \\

&= \mu.

\end{aligned}

\end{equation}

We found that the coherent state had the form

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:621}

\ket{z} = c_0 \sum_{n=0} \frac{z^n}{\sqrt{n!}} \ket{n},

\end{equation}

so the probability coefficients for \( \ket{n} \) are

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:641}

\begin{aligned}

P(n)

&= c_0^2 \frac{\Abs{z^n}^2}{n!} \\

&= e^{-\Abs{z}^2} \frac{\Abs{z^n}^2}{n!}.

\end{aligned}

\end{equation}

This has the structure of the Poisson distribution with mean \( \mu = \Abs{z}^2 \). The most probable value of \( n \) is that for which \( \Abs{f(n)}^2 \) is the largest. This is, in general, hard to compute, since we have a maximization problem in the integer domain that falls outside the normal toolbox. If we assume that \( n \) is large, so that Stirling’s approximation can be used to approximate the factorial, and also seek a non-integer value that maximizes the distribution, the most probable value will be the closest integer to that, and this can be computed. Let

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:661}

\begin{aligned}

g(n)

&= \Abs{f(n)}^2 \\

&= \frac{e^{-\mu} \mu^n}{n!} \\

&= \frac{e^{-\mu} \mu^n}{e^{\ln n!}} \\

&\approx e^{-\mu – n \ln n + n } \mu^n \\

&= e^{-\mu – n \ln n + n + n \ln \mu }

\end{aligned}

\end{equation}

This is maximized when

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:681}

0

= \frac{dg}{dn}

= \lr{ – \ln n – 1 + 1 + \ln \mu } g(n),

\end{equation}

which is maximized at \( n = \mu \). One of the integers \( n = \lfloor \mu \rfloor \) or \( n = \lceil \mu \rceil \) that brackets this value \( \mu = \Abs{z}^2 \) is the most probable. So, if an energy measurement is made of a coherent state \( \ket{z} \), the most probable value will be one of

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:701}

E = \Hbar \lr{

\lceil\Abs{z}^2\rceil

+ \inv{2} },

\end{equation}

or

\begin{equation}\label{eqn:gradQuantumProblemSet2Problem1:721}

E = \Hbar \lr{

\lfloor\Abs{z}^2\rfloor

+ \inv{2} },

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.