Coherent states play two important roles. On the one hand they can be
used to illustrate how close a quantum system can be made to look like
the back and forth motion of a mass on a spring. On the other hand, they
play a critical role in the quantization of light, with coherent states
being the quantum representation of the electromagnetic fields in a
laser. We won’t be able to go into detail about that here, but we will
develop these states for the SHO. They are our connection to the
classical world. As a side note, I am currently working on generalizing
this idea into other systems than just the simple harmonic oscillator,
and it looks very promising.
There are many ways to motivate the coherent states, but here we de
so as the eigenstates of the lowering operator \(\hat{a} : |\alpha\rangle\) satisfies \(\hat{a}|\alpha\rangle=\alpha|\alpha\rangle\).
But wait, you say, \(\hat{a}\) is not
Hermitian, how can it have eigenstates? It turns out it can, but some of
our familiar properties do not hold. First, \(\alpha\) can be complex and need not be
real and second, the eigenstates are not orthogonal. They also are over
complete, which we will describe below.
But first, let’s refresh our memories about \(\hat{a}\). \[\begin{aligned}
& {[\hat{H}, \hat{a}]=\hbar \omega_{0}\left[\hat{a}^{\dagger}
\hat{a}+\frac{1}{2}, \hat{a}\right]=-\hbar \omega_{0} \hat{a}
\quad\left(\text { since }\left[\hat{a},
\hat{a}^{\dagger}\right]=1\right)} \\
& {\left[\hat{H}, \hat{a}^{\dagger}\right]=\hbar
\omega_{0}\left[\hat{a}^{\dagger}\hat{a}+\frac{1}{2},
\hat{a}^{\dagger}\right]=\hbar \omega_{0} \hat{a}^{\dagger}}
\end{aligned}\] These lead us to conclude that \(\hat{a}\) lowers the energy of an energy
eigenstate by \(\hbar \omega_{0}\),
while \(\hat{a}^{\dagger}\)raises it by
\(\hbar w_{0}\).
We also know \(\left[\hat{a} ,
\left(\hat{a}^{\dagger}\right)^{n}\right]=n\left(\hat{a}^{\dagger}\right)^{n-1}
\quad\text{and}\quad\left[\hat{a}^{\dagger},(\hat{a})^{n}\right]=-n
\hat{a}^{n-1}\) (you can use induction to show this). So \[\begin{aligned}
\hat{a}|n\rangle&=\hat{a}
\frac{\left(\hat{a}^{\dagger }\right)^{n}}{\sqrt{n!}}|0\rangle=\frac{\left(\hat{a}\left(\hat{a}^{\dagger}\right)^{n}-\left(\hat{a}^{\dagger}\right)^{n}
\overbrace{\hat{a}}^{\text{adding zero}}\right)}{\sqrt{n!}}|0\rangle\\
&=
\frac{\left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n}\right]}{\sqrt{n!}}|0\rangle=\frac{n}{\sqrt{n!}}\left(\hat{a}^{\dagger}\right)^{n-1}|0\rangle=\sqrt{n}|n-1\rangle.
\end{aligned}\] Similarly \(\hat{a}^{\dagger} |n \rangle = \sqrt{n+1}
|n+1\rangle\) (but this proof is much easier since no commutator
is needed, try it.)
Let’s find \(|\alpha\rangle . \text{ Let }
|\alpha\rangle=\sum_{n=0}^{\infty} c_{n}|n\rangle \quad\)
(possible since\(\{\ |n \rangle \}\) is
complete). Then \[\begin{aligned}
& \hat{a}|\alpha\rangle=\sum_{n=0}^{\infty} c_{n}
\hat{a}|n\rangle=\sum_{n=1}^{\infty} c_{n}
\underbrace{\sqrt{n}}_{\text{vanishes when n=0
}}|n-1\rangle=\sum_{n=0}^{\infty} c_{n+1} \sqrt{n+1}|n\rangle \\
& =\alpha|\alpha\rangle \Rightarrow c_{n+1} \sqrt{n+1}=\alpha c_{n}
\end{aligned}\] \[\Rightarrow \quad
c_{0}, \quad c_{1}=\frac{\alpha}{\sqrt{1}} c_{0}, \quad
c_{2}=\frac{\alpha}{\sqrt{2}} c_{1}=\frac{\alpha^{2}}{\sqrt{2 \cdot 1}}
c_{0}, ~~c_{3}=\frac{\alpha^{3}}{\sqrt{3 \cdot 2 \cdot 1}}
c_{0},\] and \(\quad
c_{n}=\frac{\alpha^{n}}{\sqrt{n!}} c_{0}\). The coefficient \(c_{0}\) is determined by normalization.
Let’s first compute the overlap \(\langle\beta
| \alpha\rangle\)
\[\begin{aligned}
\langle\beta| \alpha\rangle & =\sum_{n=0}^{\infty}
\frac{\left(\beta^{*}\right)^{n}}{\sqrt{n!}} c_{0}^{*}(\beta)\langle n|
\sum_{m=0}^{\infty} \frac{(\alpha)^{m}}{\sqrt{m!}}
c_{0}(\alpha)|m\rangle \\
&=\sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
\frac{\left(\beta^{*}\right)^{n}(\alpha)^{m}}{\sqrt{n!m!}}
c_{0}^{*}(\beta) c_{0}(\alpha) \underbrace{\langle n |
m\rangle}_{\delta_{nm}} \\
&=\sum_{m=0}^{\infty} \frac{\left(\beta^{*} \alpha\right)^{m}}{m!}
c_{0}^{*}(\beta) c_{0}(\alpha)=e^{\beta^* \alpha} c_{0}^{*}(\beta)
c_{0}(\alpha).
\end{aligned}\]
Set \(\alpha=\beta \Rightarrow\langle\alpha |
\alpha\rangle=e^{|\alpha|^{2}} | c_{0}\left(\alpha)
|^{2}=1\right.\)
\[\begin{aligned}
& \Rightarrow\left|c_{0}(\alpha)\right|=e^{-\frac{1}{2}|\alpha|^{2}}
\quad \text { (pick } c_{0}=\text { real and positive) } \\
& \boxed{|\alpha\rangle =\sum_{n=0}^{\infty}
\frac{\alpha^{n}}{\sqrt{n!}} e^{-\frac{1}{2}|\alpha|^{2}}|n\rangle
\quad\langle\alpha | \beta\rangle=e^{\beta^{*}
\alpha-\frac{1}{2}|\alpha|^{2}-\frac{1}{2}|\beta|^{2}}}
\end{aligned}\]
Note that the overlap is non zero, but as \(\alpha-\beta\) becomes large in magnitude,
\(\langle\alpha | \beta\rangle\) gets
small very quickly. Hence coherent states are not orthogonal!
You should think about why \(\hat{a}^{\dagger}\) has no eigenstate. Ask,
if you de not see why it cannot work.
Let us play with the representation of a coherent state
\[\begin{aligned}
|\alpha\rangle & =\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}}
e^{-\frac{1}{2}|\alpha|^{2}}|n\rangle=\sum_{n=0}^{\infty}
\frac{\alpha^{n}}{\sqrt{n!}} e^{-\frac{1}{2}|\alpha|^{2}}
\frac{\left(\hat{a}^{\dagger}\right)^{n}}{\sqrt{n!}}|0\rangle \\
& =e^{-\frac{1}{2}|\alpha|^{2}} \sum_{n=0}^{\infty}
\frac{\left(\alpha
\hat{a}^{\dagger}\right)^{n}}{n!}|0\rangle=e^{-\frac{1}{2}|\alpha|^{2}}
e^{\alpha \hat{a}^{\dagger}}|0\rangle
\end{aligned}\] but recall \(e^{-\alpha
\hat{\alpha}}|0\rangle=|0\rangle\) and recall \(\left(\alpha
\hat{a}^{+}\right)^{\dagger}=\alpha^{*} \hat{a}\)), so that \[|\alpha\rangle=e^{-\frac{1}{2}|\alpha|^{2}}
e^{\alpha \hat{a}^{\dagger}} e^{-\alpha^{*}
\hat{a}}|0\rangle\]
Now use BCH with \(\hat{A}=\alpha
\hat{a}^{\dagger} \quad \text{and}\quad \hat{B}=-\alpha^{*} \hat{a}
\quad\text{and}\quad[\hat{A}, \hat{B}]=|\alpha|^{2}\), \[\begin{aligned}
& |\alpha\rangle=e^{\left.\left.-\frac{1}{2} \right\rvert\,
\alpha\right|^{2}} e^{\alpha \hat{a}^{\dagger}-\alpha^{*}
\hat{a}+\frac{1}{2}|\alpha|^{2}}|0\rangle \\
& \boxed{|\alpha\rangle=e^{\alpha \hat{a}^{\dagger}-\alpha^{*}
\hat{a}}|0\rangle}.
\end{aligned}\]
We define the displacement operator \(\hat{D}(\alpha)\) to be \(\hat{D}(\alpha)=e^{\alpha
\hat{a}^{\dagger}-\alpha^{*} \hat{a}}\) \[\boxed{|\alpha\rangle=\hat{D}(\alpha)|0\rangle}.\]
But recall that \[\hat{a}=\sqrt{\frac{m
w_{0}}{2 \hbar}}\left(\hat{x}+\frac{i}{m \omega_{0}} \hat{p}\right)
\quad \hat{a}^{\dagger}=\sqrt{\frac{m \omega_{0}}{2
\hbar}}\left(\hat{x}-\frac{i}{m \omega_{0}} \hat{p}\right)\] So
that \(\alpha \hat{a}^{\dagger }-\alpha^{*}
\hat{a}=\sqrt{\frac{m \omega_{0}}{2 \hbar}}\left(\alpha \hat{x}-\frac{i
\alpha}{m \omega_{0}} \hat{p}-\alpha^{*} \hat{x}-\frac{i \alpha^{*}}{m
\omega_{0}} \hat{p}\right)\) \[\boxed{
\alpha \hat{a}^{\dagger}-\alpha^{\psi} \hat{a}=i \sqrt{\frac{2 m
\omega_{0}}{\hbar}} \operatorname{Im} (\alpha) \hat{x}-i
\sqrt{\frac{2}{\hbar m \omega_{0}}} \operatorname{Re(\alpha) }
\hat{p}}\] So, if \(\alpha\) is
real \(\alpha \hat{a}^{\dagger}-\alpha^{*}
\hat{a}=-i \frac{x_{0} \hat{p}}{\hbar} \quad x_{0}=\sqrt{\frac{2
\hbar}{m \omega_{0}}} \operatorname{Re} (\alpha)\) and
if \(\alpha\) is imaginary \(\alpha \hat{a}^{\dagger}-\alpha^{*} \hat{a}=i
\frac{p_{0} \hat{x}}{\hbar} \quad p_{0}=\sqrt{2 \hbar \pi \omega_{0}}
\operatorname{Im} (\alpha)\)
Hence, the displacement operator gives both a translation in space and a
translation in momentum to the ground-state energy eigenfunction.
Calculating with coherent states is easy because \[\hat{a}|\alpha\rangle=\alpha|\alpha\rangle \text
{ and }\langle\alpha| \hat{a}^{\dagger}=\langle\alpha|
\alpha^{*}.\] But note that we do not know what \(\hat{a}^\dagger|\alpha\rangle\) is, nor do
we know \(\langle\alpha|\hat{a}\). So
\[\langle\alpha|
\hat{x}|\alpha\rangle=\sqrt{\frac{\hbar}{2 m \omega_{0}}}\langle\alpha|
\hat{a}+\hat{a}^{\dagger}|\alpha\rangle=\sqrt{\frac{2 \hbar}{m
\omega_{0}}} \operatorname{Re} \alpha\] Then for a more
complicated calculation, we have \[\begin{aligned}
\langle\alpha| \hat{x}^{2}|\alpha\rangle & =\frac{\hbar}{2 m
\omega_{0}}\langle\alpha| \left (\hat{a}^{2}+\hat{a}
\hat{a}^{\dagger}+\hat{a}^{\dagger} \hat{a}+(\hat{a}^{\dagger})^2\right
)|\alpha\rangle \\
& =\frac{\hbar}{2 m \omega_{0}}\langle\alpha| \left
(\hat{a}^{2}+\hat{a}^{\dagger} \hat{a}+\left[\hat{a},
\hat{a}^{\dagger}\right]+\hat{a}^{\dagger}
\hat{a}+(\hat{a}^{\dagger})^{2}\right )|\alpha\rangle \\
& =\frac{\hbar}{2 m
\omega_{0}}\left(\alpha^{2}+2|\alpha|^{2}+1+\alpha^{* 2}\right)
\end{aligned}\] Note how we had to “normal-order” the operators
with daggers to the left, in order to evaluate the matrix element. This
then gives for uncertainty \[\begin{aligned}
(\Delta x)_{\alpha}^{2} & =\langle\alpha|
\hat{x}^{2}|\alpha\rangle-\langle\alpha| \hat{x}|\alpha\rangle^{2} \\
& =\frac{\hbar}{2 m
\omega_{0}}\left(\alpha^{2}+2|\alpha|^{2}+1+\alpha^{*^{2}}-\alpha^{2}-2|\alpha|^{2}-(\alpha^{*})^{2}\right)
\end{aligned}\] \[\boxed{
(\Delta x)_{\alpha}^{2}=\frac{\hbar}{2 m w_{0}}},\] which is
independent of \(\alpha\) (the coherent
state shifts the ground state but preserves the shape, and hence its
uncertainty).
Displacing the ground state does not change its
uncertainty.
Now, we compute the uncertainty for momentum in a coherent state. \[\begin{aligned} \langle\alpha| \hat{p}|\alpha\rangle & =-i \sqrt{\frac{\hbar m \omega_{0}}{2}}\langle\alpha| \hat{a}-\hat{a}^{\dagger}|\alpha\rangle=-i \sqrt{\frac{\hbar m \omega_{0}}{2}}\left(\alpha-\alpha^{*}\right) \\ & =\sqrt{2 \hbar m \omega_{0}} \operatorname{Im} \alpha \\ \langle\alpha| \hat{p}^{2}|\alpha\rangle & =-\frac{\hbar m \omega_{0}}{2}\langle\alpha| \left (\hat{a}^{2}-\hat{a} \hat{a}^{\dagger}-\hat{a}^{\dagger} \hat{a}-(\hat{a}^{\dagger})^{2}\right )|\alpha\rangle \\ & =-\frac{\hbar m \omega_{0}}{2}\left(\alpha^{2}-2|\alpha|^{2}-1-\alpha^{* 2}\right) \end{aligned}\]
So
\[\boxed{
(\Delta p)_{\alpha}^{2}=\frac{\hbar m \omega_{0}}{2}} \quad \Rightarrow
~~\boxed{(\Delta p)_{\alpha}(\Delta
x)_{\alpha}=\frac{\hbar}{2}}\] The uncertainty is unchanged
in a coherent state!
What about their time dependence? We have not discussed time
dependence in general yet, but one should see immediately that \[i \hbar \frac{\partial}{\partial
t}|\psi\rangle=\hat{H}|\psi\rangle\] is solved by \(|\psi(t)\rangle=\hat{U}(t)|\psi(0)\rangle=e^{-i
\frac{\hat{H} t}{\hbar}}|\psi(0)\rangle\).
Just check by taking the derivative. Because \(\hat{H}\) is independent of time, there is
no operator ordering issue. So, we have \[\begin{aligned}
& | \alpha(t)\rangle =e^{-i \frac{\hat{H} t}{\hbar}}|\alpha\rangle,
~
\end{aligned}\] But \(\hat{H}=\hbar
\omega_{0}\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)\) so
\[\begin{aligned}
|\alpha(t)\rangle & =e^{-i \omega_{0}
t\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)}|\alpha\rangle \\
& =e^{-i
\omega_{0}t\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)} e^{\alpha
\hat{a}^{\dagger}-\alpha^{*} \hat{a}}|0\rangle
\end{aligned}\]
How do we proceed? we need to move the two operators through each
other. This is the exponential reordering or the braiding identity—we
get \[\begin{aligned}
& |\alpha(t)\rangle=\exp \left[e^{-i \omega_{0}
t\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)}\left(\alpha
\hat{a}^{\dagger}-\alpha^{*} \hat{a}\right) e^{i \omega_{0}
t\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)}\right] \\
& \quad \underbrace{e^{-i \omega_{0}
t\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)}|0\rangle}_{e^{-i
\frac{\omega_{0}t}{2}}|0\rangle \text { since } \hat{a} | 0 \rangle = 0}
\end{aligned}\]
So we need to compute \(\hat{U}(t) \hat{a}
\hat{U}^{\dagger}(t)\) and \(\hat{U}(t)
\hat{a}^{+} \hat{U}^{\dagger}(t)\)
The easiest way to do this is by differentiating (note this does not
always work, but does so here because of the simplicity of \([\hat{H}, \hat{a}]\) and \([\hat{H}, \hat{a}^{\dagger}]\)). We have
\[\begin{aligned}
\frac{d}{d t}\left(\hat{U}(t) \hat{a} \hat{U}^{\dagger}(t)\right) &
=\left(\frac{d}{dt} \hat{U}(t)\right) \hat{a}
\hat{U}^{\dagger}(t)+\hat{U}(t) \hat{a}\left(\frac{d}{dt}
\hat{U}^{\dagger}(t)\right) \\
& =-i \omega_{0} \hat{U}(t)\left(\hat{a}^{\dagger} \hat{a}
\hat{a}-\hat{a} \hat{a}^{\dagger} \hat{a}\right) \hat{U}^{\dagger}(t) \\
& =-i \omega_{0} \hat{U}(t)\left[\hat{a}^{\dagger} \hat{a},
\hat{a}\right] \hat{U}^{\dagger}(t) \\
& =i \omega_{0} \hat{U}(t) \hat{a} \hat{U}^{\dagger}(t) \\
\Rightarrow & \boxed{\hat{U}(t) \hat{a} \hat{U}^{\dagger}(t) =e^{i
\omega_{0} t} \hat{a}}
\end{aligned}\] similarly \[\boxed{
\hat{U}(t) \hat{a}^{\dagger} \hat{U}^{\dagger}(t)=e^{-i \omega_{0} t}
\hat{a}^{\dagger}}\] because of the sign change in the
commutator.
Hence \[\begin{aligned}
|\alpha(t)\rangle & =\exp \left[\alpha e^{-i \omega_{0} t}
\hat{a}^{\dagger}-\alpha^{*} e^{i \omega_{0} t} \hat{a}\right]
e^{-i\frac{\omega_{0} t}{2}}|0\rangle \\
& =e^{-i \frac{\omega_{0} t}{2}}\left|\alpha e^{-i \omega_{0}
t}\right\rangle
\end{aligned}\]
This implies that the coherent state evolves in time simply via its
parameter \(\alpha\) being multiplied
by \(e^{-i \omega t}\) and being
multiplied by an overall phase! This is very simple.
Let’s try to understand by computing the expectation value of position as a function at time
\[\begin{aligned} \langle\alpha(t)| \hat{x}|\alpha(t)\rangle & =\sqrt{\frac{\hbar}{2 m \omega_{0}}}\left\langle\alpha e^{-i \omega_{0} {t}}\right| (\hat{a}+\hat{a}^{\dagger})\left|\alpha e^{-i \omega_{0} t}\right\rangle \\ & =\sqrt{\frac{\hbar}{2 m \omega_{0}}}\left(\alpha e^{-i \omega_{0} t}+\alpha^{*} e^{+i \omega_{0} t}\right) \\ & =\sqrt{\frac{2 \hbar}{m \omega_{0}}}\left(\operatorname{Re\alpha } \cos \omega_{0} t+\operatorname{Im} \alpha \sin \omega_{0} t\right) \\ & \boxed{\langle\alpha(t)| \hat{x}|\alpha(t)\rangle =x_{0} \operatorname{cos} \omega_{0} t+\frac{p_{0}}{m \omega_{0}} \sin \omega_{0} t} \end{aligned}\] This is the classical equation of motion for a spring!
Similarly \[\begin{aligned} \langle\alpha(t)| \hat{p} | \alpha(t)\rangle & =-i \sqrt{\frac{\hbar m \omega_{0}}{2}}\langle\alpha(t)| (\hat{a}-\hat{a}^{\dagger})|\alpha(t)\rangle \\ & =-i \sqrt{\frac{\hbar m \omega_{0}}{2}}\left(\alpha e^{-i \omega_{0} t}-\alpha^{\dagger} e^{i \omega_{0} t}\right) \\ & =-i \sqrt{2 \hbar m \omega_{0}}\left(-i \operatorname{Re} \alpha \sin \omega_{0} t+i \operatorname{Im} \alpha \cos \omega_{0} t\right) \\ & \boxed{\langle\alpha(t)| \hat{p}|\alpha(t)\rangle \left.=-m \omega_{0} x_{0} \sin \omega_{0} t+p_{0} \cos \omega_{0} t\right)} \end{aligned}\] Also the classical equation of motion!
You will show on a homework exercise that the uncertainty is
independent of time as well.
So we can think of the coherent states as being a "blob" whose
uncertainty remains the same for all time and it sloshes back and forth
as a classical mass on a spring.
This is as close to a classical image as we get with quantum
systems.
What are the probabilities to observe the system to have different
energies when it is prepared in a coherent state? The probability is
given by \[\begin{aligned}
|\langle n | \alpha\rangle|^{2} & =\left|\sum_{n=0}^{\infty}
\frac{\alpha^{m}}{\sqrt{m!}} e^{-\frac{1}{2}|\alpha|^{2}}\langle n |
m\rangle\right|^{2} \\
P(n) & =\frac{| \alpha | ^{2 n}}{n!} e^{-|\alpha|^{2}}
\end{aligned}\] This implies that for any nonzero \(\alpha\), there is a nonzero probability to
see any energy excitation.
By differentiating this, we find the maximum occurs when \(\frac{d}{d|\alpha|^{2}} \frac{|\alpha|^{2 n}}{n!}
e^{-|\alpha|^{2}} \Rightarrow \frac{n}{|\alpha|^{2}}-1=0\) or
\(|\alpha|^{2}=n\). Hence, this maximal
probability increases with \(n\). When
we later discuss photons, we will see we can think of \(n\) as being related to the number of
photons and classical sources of light (light bulbs and even lasers)
produce light in coherent (and incoherent) states. So no matter how dim
the light is, it is never a single photon—there is always the
possibility of photon bunching.
An experiment was done in the early days of quantum mechanics of a two
slit experiment with light dimmed so low that it took 3 months for
enough light to expose the photographic plate. It did show a two slit
interference pattern. But this experiment used an incoherent source not
a single photon source, so it could not rule out multiple photons being
in the apparatus at the same time. Indeed, it is certain that this did
occur due to photon bunching. But quantum theory was not well enough
developed at that time for anyone to know this was the case, and so the
experiment was influential about showing wave-particle duality of
photons, even if it had issues.