## A curious proof of the Baker-Campbell-Hausdorff formula

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Equation (39) of [1] states the Baker-Campbell-Hausdorff formula for two operators $$a, b$$ that commute with their commutator $$\antisymmetric{a}{b}$$

\label{eqn:bakercambell:20}
e^a e^b = e^{a + b + \antisymmetric{a}{b}/2},

and provides the outline of an interesting method of proof. That method is to consider the derivative of

\label{eqn:bakercambell:40}
f(\lambda) = e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)},

That derivative is
\label{eqn:bakercambell:60}
\begin{aligned}
\frac{df}{d\lambda}
&=
e^{\lambda a} a e^{\lambda b} e^{-\lambda (a + b)}
+
e^{\lambda a} b e^{\lambda b} e^{-\lambda (a + b)}

e^{\lambda a} b e^{\lambda b} (a + b)e^{-\lambda (a + b)} \\
&=
e^{\lambda a} \lr{
a e^{\lambda b}
+
b e^{\lambda b}

e^{\lambda b} (a+b)
}
e^{-\lambda (a + b)} \\
&=
e^{\lambda a} \lr{
\antisymmetric{a}{e^{\lambda b}}
+
{\antisymmetric{b}{e^{\lambda b}}}
}
e^{-\lambda (a + b)} \\
&=
e^{\lambda a}
\antisymmetric{a}{e^{\lambda b}}
e^{-\lambda (a + b)}
.
\end{aligned}

The commutator above is proportional to $$\antisymmetric{a}{b}$$

\label{eqn:bakercambell:80}
\begin{aligned}
\antisymmetric{a}{e^{\lambda b}}
&=
\sum_{k=0}^\infty \frac{\lambda^k}{k!} \antisymmetric{a}{ b^k } \\
&=
\sum_{k=0}^\infty \frac{\lambda^k}{k!} k b^{k-1} \antisymmetric{a}{b} \\
&=
\lambda \sum_{k=1}^\infty \frac{\lambda^{k-1}}{(k-1)!} b^{k-1}
\antisymmetric{a}{b} \\
&=
\lambda e^{\lambda b} \antisymmetric{a}{b},
\end{aligned}

so

\label{eqn:bakercambell:100}
\frac{df}{d\lambda} = \lambda \antisymmetric{a}{b} f.

To get the above, we should also do the induction demonstration for $$\antisymmetric{a}{ b^k } = k b^{k-1} \antisymmetric{a}{b}$$.

This clearly holds for $$k = 0,1$$. For any other $$k$$ we have

\label{eqn:bakercambell:120}
\begin{aligned}
\antisymmetric{a}{b^{k+1}}
&=
a b^{k+1} – b^{k+1} a \\
&=
\lr{ \antisymmetric{a}{b^{k}} + b^k a
} b – b^{k+1} a \\
&=
k b^{k-1} \antisymmetric{a}{b} b
+ b^k \lr{ \antisymmetric{a}{b} + {b a} }
– {b^{k+1} a} \\
&=
k b^{k} \antisymmetric{a}{b}
+ b^k \antisymmetric{a}{b} \\
&=
(k+1) b^k \antisymmetric{a}{b}.
\end{aligned}

Observe that \ref{eqn:bakercambell:100} is solved by

\label{eqn:bakercambell:140}
f = e^{\lambda^2\antisymmetric{a}{b}/2},

which gives

\label{eqn:bakercambell:160}
e^{\lambda^2 \antisymmetric{a}{b}/2} =
e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)}.

Right multiplication by $$e^{\lambda (a + b)}$$ which commutes with $$e^{\lambda^2 \antisymmetric{a}{b}/2}$$ and setting $$\lambda = 1$$ recovers \ref{eqn:bakercambell:20} as desired.

What I wonder looking at this, is what thought process led to trying this in the first place? This is not what I would consider an obvious approach to demonstrating this identity.

# References

[1] Roy J Glauber. Some notes on multiple-boson processes. Physical Review, 84 (3), 1951.

## More on (SHO) coherent states

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

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

\label{eqn:gradQuantumProblemSet2Problem1:561}
\ket{z} = \sum_{n=0}^\infty f(n) \ket{n}

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

\label{eqn:gradQuantumProblemSet2Problem1:581}
P(n) = \frac{\mu^{n} e^{-\mu}}{n!}.

Here $$\mu$$ is the mean of the distribution

\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}

We found that the coherent state had the form

\label{eqn:gradQuantumProblemSet2Problem1:621}
\ket{z} = c_0 \sum_{n=0} \frac{z^n}{\sqrt{n!}} \ket{n},

so the probability coefficients for $$\ket{n}$$ are

\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}

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

\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}

This is maximized when

\label{eqn:gradQuantumProblemSet2Problem1:681}
0
= \frac{dg}{dn}
= \lr{ – \ln n – 1 + 1 + \ln \mu } g(n),

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

\label{eqn:gradQuantumProblemSet2Problem1:701}
E = \Hbar \lr{
\lceil\Abs{z}^2\rceil
+ \inv{2} },

or

\label{eqn:gradQuantumProblemSet2Problem1:721}
E = \Hbar \lr{
\lfloor\Abs{z}^2\rfloor
+ \inv{2} },

# References

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