In the next two lectures, we will summarize quantum optics with the
goals of (1) establishing precisely what a photon is and (2) describing
how quantum optics principles are employed in the LIGO experiment to
improve the precision of the measurements. It will be a crash course.
You need to review ladder operators for the simple harmonic oscillator
and coherent states for this lecture.
We begin with classical description of an electric field given by \[\mathbf{E}(r,t)=\sum_l\mathbf{\varepsilon}_lE_l(t)e^{i\mathbf{k}_l\cdot\mathbf{r}}+\text{c.c.}\]
where we are describing a real traveling wave. Here \(l\) denotes the mode described by the
wavevector \(\mathbf{k}_l\) and
polarization \(\mathbf{\varepsilon}_l\). Here, we will
focus on linear polarization only. The function \(E_l(t)\) can be complex-valued.
We require the wave to satisfy Maxwell’s equations. \[\nabla\cdot\mathbf{E}(\mathbf{r},t)=0,\quad\nabla\times\mathbf{E}(\mathbf{r},t)=-\frac{\partial\mathbf{B}(\mathbf{r},t)}{\partial
t},\quad\nabla\cdot\mathbf{B}(\mathbf{r},t)=0,\quad\nabla\times\mathbf{B}(\mathbf{r},t)=\frac{1}{c^2}\frac{\partial\mathbf{E}(\mathbf{r},t)}{\partial
t}\] Using the fact that \(\nabla
e^{i\mathbf{k}_l\cdot\mathbf{r}}=i\mathbf{k}_le^{i\mathbf{k}_l\cdot\mathbf{r}}\)
and \(\nabla\cdot\mathbf{E}=0\), we can
say that \(\mathbf{k}_l\perp\mathbf{\varepsilon}_l\).
Also, \(\nabla\times\mathbf{E}=-\partial_t\mathbf{B}\implies\mathbf{B}(\mathbf{r},t)\propto\mathbf{k}_l\times\mathbf{\varepsilon}_l\).
Then \(\nabla\cdot\mathbf{B}=0\) is
automatically satisfied. We end up with \[\frac{\partial^2E_l(t)}{\partial
t^2}=-\frac{k_l^2}{c^2}E_l(t).\] Define the angular frequency
\(\omega_l=\frac{k_l}{c}\). Then \(E_l(t)=E_l(0)e^{-i\omega_l t}\). \[\implies\mathbf{B}(\mathbf{r},t)=\sum_l\frac{\mathbf{k}_l\times\mathbf{\varepsilon}_l}{\omega_l}E_l(t)e^{i\mathbf{k}_l\cdot\mathbf{r}}+\text{c.c.}\]
We define the single photon amplitude as \(\varepsilon_l^{(1)}=\sqrt{\frac{\hbar\omega_l}{2\epsilon_0V_l}}\)
and introduce \(V_l\) as the
quantization volume for mode \(l\). We
will be working in a volume \(L^3\)
with periodic boundary conditions so \(\mathbf{k}_l=\frac{2\pi}{L}(n_x,n_y,n_z)\)
for \(n_\alpha\in\mathbb{Z}\). The
polarization is one of the two directions perpendicular to \(k_l\). Note that we use \(\mathbf{k}_l,\mathbf{\varepsilon}_l,\) and
\(\mathbf{k}_l\times\mathbf{\varepsilon}_l\)
as a triad to define the coordinates of the three dimensions.
We also choose \(\mathbf{\varepsilon}_{-l}=\mathbf{\varepsilon}_l\)
when \(l:\frac{2\pi}{L}(n_x,n_y,n_z;\mathbf{\varepsilon}_l)\)
and \(-l:\frac{2\pi}{L}(-n_x,-n_y,-n_z;\mathbf{\varepsilon}_{-l})\).
This will help us with calculating the total energy. But before that, we
note two more things. We will be writing \(E_l(t)=\varepsilon_l\alpha_l(t)\) with
\(\alpha_l(t)\) possibly complex. The
we also define the two quadrature parameters \[\begin{aligned}
&Q_l(t)=\sqrt{\frac{\hbar}{2}}(\alpha_l(t)+\alpha_l^*(t)) \\
&P_l(t)=-i\sqrt{\frac{\hbar}{2}}(\alpha_l(t)-\alpha_l^*(t))
\end{aligned}\] The energy is given by \[E=\frac{\varepsilon_0}{2}\int d^3r \
(\mathbf{E}^2+c^2\mathbf{B}^2)\] In calculating this, we note
that \[\int d^3r \
e^{i\mathbf{k}_l\cdot\mathbf{r}-\mathbf{k}_{l'}\cdot\mathbf{r}}=\delta_{n_l,n_{l'}}V.\]
We end up with two types of terms: Those with \(l=l'\) and those with \(l=-l'\). Then (you should do this),
\[\begin{aligned}
E=& \
\frac{\varepsilon}{2}V\sum_l\left(2|E_l(t)|^2-E_l(t)E_{-l}(t)-E_l^*(t)E_{-l}^*(t)+2|E_l(0)|^2+E_l(t)E_{-l}(t)+E_l^*(t)E_{-l}^*(t)\right)
\\
=& \ 2\varepsilon_0V\sum_l(\varepsilon_l)^2|\alpha_l|^2 \\
=& \ \sum_l\hbar\omega_l|\alpha_l|^2
\end{aligned}\] Now we quantize: Let \(Q_l\rightarrow\hat{Q}_l\) and \(P_l\rightarrow\hat{P}_l\) with \([\hat{Q}_l,\hat{P}_{l'}]=i\hbar\delta_{ll'}\).
Then define \[\begin{aligned}
&\hat{a}_l=\frac{1}{\sqrt{2\hbar}}(\hat{Q}_l+i\hat{P}_l) \\
&\hat{a}^\dagger_l=\frac{1}{\sqrt{2\hbar}}(\hat{Q}_l-i\hat{P}_l).
\end{aligned}\] Written in terms of the quadratures, \[H=\sum_l\frac{\omega_l}{2}(\hat{Q}_l^2+\hat{P}_l^2),\]
which shows the relationship with the simple harmonic oscillator. Back
to the field, we now have an operator \[\hat{\mathbf{E}}(\mathbf{r})=\hat{E}^{(+)}+\hat{E}^{(-)}=i\sum_l\mathbf{\varepsilon}_l\hat{a}_le^{i\mathbf{k}_l\cdot\mathbf{r}}+\text{h.c.}\]
The eigenstates are labeled by number operator eigenstates \[\left|n_1,n_2,\ldots\right\rangle=\frac{(\hat{a}_1^\dagger)^{n_1}}{\sqrt{n_1!}}\frac{(\hat{a}_2^\dagger)^{n_2}}{\sqrt{n_2!}}\ldots\left|0\right\rangle\]
and \(E_{n_1\cdots}=\sum_l\hbar\omega_l\left(n_l+\frac{1}{2}\right)\).
Usually, we only include nonzero \(n_l\)’s in the labeling. Note the vacuum
has “infinite energy” given by \(\sum_l\hbar\omega_l\frac{1}{2}\). But we
avoid this energy by focusing on the excitation energy with respect to
the vacuum given by \(E^{ex}=\sum_l\hbar\omega_ln_l\).
We will be working in the Heisenberg representation where \(\hat{a}_l(t)=\hat{a}_le^{-i\omega_lt}\) and
\(\hat{a}^\dagger_l(t)=\hat{a}^\dagger_le^{i\omega_lt}\).
Now consider a single mode state \(\left|n_l\right\rangle\). The average of
the electric field vanishes \[\left\langle
n_l\middle|\hat{\mathbf{E}}(\mathbf{r},t)\middle|n_l\right\rangle=\left\langle
n_l\middle|\hat{a}_le^{ik\cdot r-i\omega_lt}-\hat{a}_l^\dagger
e^{-ik\cdot r+i\omega_lt}\middle|n_l\right\rangle=0\] since \(\hat{a}\) and \(\hat{a}^\dagger\) operators are unbalanced.
We calculate the fluctuations from \[\left\langle
n_l\middle|\hat{\mathbf{E}}\cdot\hat{\mathbf{E}}\middle|n_l\right\rangle=-\varepsilon_l\cdot\varepsilon_l(\varepsilon_l^{(1)})^2\left\langle
n_l\middle|\hat{a}_l^2e^{i(2k_l\cdot
r-2\omega_lt)}-\hat{a}_l\hat{a}_l^\dagger-\hat{a}_l^\dagger\hat{a}_l+\hat{a}_l^{\dagger
2}e^{-i(2k_l\cdot r-2\omega_lt)}\middle|n_l\right\rangle\] \[=(\varepsilon_l^{(1)})^2(2n_l+1)\] so
\(\Delta\mathbf{E}=\varepsilon_l^{(1)}\sqrt{2n_l+1}\).
In particular, the vacuum \((n_l=0)\)
has fluctuations. This is real and responsible for spontaneous emission,
Lamb shift, g-2, Casimir effect, etc. A similar calculation shows that
\[\begin{aligned}
&\left\langle
0\middle|\hat{P}_l\middle|0\right\rangle=\left\langle
0\middle|\hat{Q}_l\middle|0\right\rangle=0 \\
&\left\langle
0\middle|\hat{P}_l^2\middle|0\right\rangle=\left\langle
0\middle|\hat{Q}_l^2\middle|0\right\rangle=\sqrt{\frac{\hbar}{2}}
\end{aligned}\] so \(\Delta Q_l\Delta
P_l=\frac{\hbar}{2}\), which is a minimum uncertainty state (same
as the ground state).
Our next step is to describe photodetection. We will describe a
photomultiplier tube which uses the photoelectric effect and a
cascade.
A single photon releases an electron which is accelerated and leads to
huge amplification of electrons which can be measured as an electron
pulse. The important aspects are that these detectors are efficient and
fast (there are more efficient fast detectors, but we focus on these
which is all we need).
Suppose light is traveling with a areal profile of \(S\) and impinging on detectors 1 and 2. The
probability to detect one photon is \(dP(\mathbf{r},t)=W^{(1)}(r,t)dS \ dt\) with
\(W^{(1)}(r,t)=s|\hat{E}^+(\mathbf{r},t)\left|\psi\right\rangle|^2\),
which is the single-photon detection probability. \(s\) is the sensitivity of the detector and
\(\left|\psi\right\rangle\) is the
photon state.
The probability to detect two photons is \(dP(\mathbf{r}_1, t_1,
\mathbf{r}_2,t_2)=W^{(2)}(\mathbf{r}_1, t_1, \mathbf{r}_2,t_2)dS \
dt\) with \[W^{(2)}(\mathbf{r}_1, t_1,
\mathbf{r}_2,t_2)=S^2|\hat{E}^+(\mathbf{r}_2,t_2)\hat{E}^+(\mathbf{r}_1,t_1)\left|\psi\right\rangle|^2\]
and \[\hat{E}_l^+(\mathbf{r},t)=i\varepsilon_l\varepsilon_l^{(1)}\hat{a}_le^{i(\mathbf{k}_l\cdot\mathbf{r}-i\omega_l
t)}\] Consider a single photon state. We have \(\left|1_l\right\rangle=\hat{a}_l^\dagger\left|0\right\rangle\).
Then, \[W^{(1)}(\mathbf{r},t)=s|i\varepsilon_l\varepsilon_l^{(1)}e^{i(\mathbf{k}_l\cdot\mathbf{r}-i\omega_l
t)}\hat{a}_l\hat{a}_l^\dagger\left|0\right\rangle|^2\] But \[\hat{a}_l\hat{a}_l^\dagger\left|0\right\rangle=[\hat{a}_l,\hat{a}_l^\dagger]\left|0\right\rangle=\left|0\right\rangle\]
so \(W^{(1)}(\mathbf{r},t)=s\).
Similarly, we have that \[W^{(2)}(\mathbf{r}_1,t_1,\mathbf{r}_2,t_2)=s^2|i\varepsilon_l\varepsilon_l^{(1)}e^{i(\mathbf{k}_l\cdot\mathbf{r}-i\omega_l
t)}i\varepsilon_l\varepsilon_l^{(1)}e^{i(\mathbf{k}_l\cdot\mathbf{r}-i\omega_l
t)}\hat{a}_l\hat{a}_l\hat{a}_l^\dagger\left|0\right\rangle|^2=0,\]
since \(\hat{a}_l\hat{a}_l\hat{a}_l^\dagger\left|0\right\rangle=0\).
So a single photon can only be measured once! Measuring one photon
alters the quantum state so it cannot be measured again. This is, in
many respects, one of the critical aspects of what a photon is.
We now describe how one makes a single photon source. Here is the two
photon excitation process, and a two photon decay. The two photons
emerge within a few nanoseconds. So, by observing \(\omega_H\), one is sure a photon \(\omega_0\) is emitted within the next 10-15
ns.
Collect the atoms at the focus of a parabolic mirror. Then, the photon
will “live” in a volume given by \(S\subset
T\) with \(T\) being the
lifetime of the excited state. Recall \[\frac{dP^{(1)}}{dt \
dS}=s|\varepsilon_l^{(1)}|^2=\frac{s\hbar\omega_l}{2\varepsilon_0V_l}=\frac{s\hbar\omega_l}{2\varepsilon_0ScT}\]
So \[\frac{dP^{(1)}}{dt}=\int dS \
\frac{dP^{(1)}}{dt \ S}=\frac{s\hbar\omega_l}{2\varepsilon_0
cT_l}\] and \[\int_T\frac{dP^{(1)}}{dt}=\frac{s\hbar\omega_l}{2\varepsilon
c}=1\] for a perfect detector. So \(S_{\text{perfect}}=\text{perfect
efficiency}=\frac{2\varepsilon_0}{\hbar\omega_l}\). We define the
quantum efficiency as \[s=\eta
S_{\text{perfect}}\] Then, \(\boxed{\omega^{(1)}=\frac{\eta}{ST}}\). The
real photon emitted is in a wavepacket \(\left|\psi\right\rangle=\sum_lc_l\left|1_l\right\rangle\)
with \[c_l=\frac{\kappa e^{i\omega
t_0}}{\omega_l-\omega_0+i\frac{\Gamma}{2}}=\text{Lorentzian
lineshape}\] \(\kappa=\text{normalization constant}\),
\(\omega_0=\text{frequency of the excited
state}\), \(\Gamma=\text{lifetime}=\sqrt{\frac{c\pi}{L}}\).
Assume the photon is emitted at time \(t_0\). It must travel a distance \(z\) to reach the detector. So, \[W^{(1)}(z,t)=s|\hat{E}^+(z,t)\left|\psi(t_0)\right\rangle|^2=s|\sum_l\varepsilon_l\varepsilon_l^{(1)}\hat{a}_le^{-i\omega_l(-\frac{z}{2}+t)}\sum_{l'}\frac{\sqrt{\frac{c\Gamma}{L}}e^{i\omega_{l'}t_0}}{(\omega_{l'}-\omega_0)+i\frac{\Gamma}{2}}\hat{a}_{l'}^\dagger\left|0\right\rangle|^2\]
But \(\hat{a}_l\hat{a}_{l'}^\dagger\left|0\right\rangle=\delta_{ll'}\left|0\right\rangle\)
so \[=s|\sum_li\varepsilon_0\varepsilon_l^{(1)}\sqrt{\frac{c\Gamma}{L}}\frac{e^{-i\omega_l(t-t_0-\frac{z}{c})}}{(\omega_l-\omega_0)+i\frac{\Gamma}{2}}\left|0\right\rangle|^2.\]
Assuming \(\varepsilon_l^{(1)}\) is
small, since \(\Gamma\) is small, and
transform the sum to an integral. \[i\int_{-\infty}^\infty d\omega_l\frac{L}{2\pi
c}\left(\frac{c\Gamma}{L}\right)^{1/2}\frac{e^{-i\omega_l(t-t_0-\frac{z}{c})}}{\omega_l-\omega_0+i\frac{\Gamma}{2}}\left|0\right\rangle=i\frac{L}{2\pi
c}\left(\frac{c\Gamma}{L}\right)^{1/2}e^{-\omega_0(t-t_0-\frac{z}{c})}(-2\pi
i\Theta(t-t_0-\frac{z}{c}))e^{-\frac{\Gamma}{2}(t-t_0-\frac{z}{c})}\left|0\right\rangle\]
and \[W^{(1)}(z,t)=s\frac{\hbar\omega_0}{2\varepsilon_0V}\frac{L\Gamma}{c}e^{-\Gamma(t-t_0-\frac{z}{c})}\Theta\left
(t-t_0-\frac{z}{c}\right )\] Recall \(V=sL\), \(s=\frac{2\varepsilon_0S}{\hbar\omega_0}\).
So, \[W^{(1)}(z,t)=\eta\frac{\Gamma}{c}e^{-\Gamma(t-t_0-\frac{z}{c})}\Theta\left
(t-t_0-\frac{z}{c}\right ).\] The probability exponentially
decays away from the initial time which has been measured
experimentally! These single photon sources can be verified by measuring
them on a beam splitter. We analyze this next:
The beam splitter is a partially silvered mirror that reflects the
amplitude with strength \(r\) (or \(-r\)) depending on which side (silvered or
not) and with strength \(t\). We have
\[\hat{E}_3=r\hat{E}_1+t\hat{E}_2\ \text{and} \
\hat{E}_4=t\hat{E}_1-r\hat{E}_2.\] The probability in 3 and 4 for
single-photon detection is \(s|E_3+(r_3,t_3)\left|\psi_{in}\right\rangle|^2\)
and \(s|E_4^+(r_4,t_4)\left|\psi_{in}\right\rangle|^2\)
where \[\left|\psi_{in}\right\rangle=(\gamma\left|1\right\rangle_1+\sqrt{1-\gamma^2}\left|0\right\rangle_1)\otimes\left|0\right\rangle_2\]
where the efficiency to detect the heralded photon is \(\gamma\) (not all photons are detected).
One finds \[\begin{aligned}
&\frac{dP_3(t)}{dt}=\eta_3|\gamma|^2|r|^2\Theta\left
(t_3-t_0-\frac{z_3}{c}\right)e^{-\Gamma(t_3-t_0-\frac{z_3}{c})} \\
&\frac{dP_4(t)}{dt}=\eta_4|\gamma|^2|r|^2\Theta\left
(t_4-t_0-\frac{z_4}{c}\right )e^{-\Gamma(t_4-t_0-\frac{z_4}{c})}.
\end{aligned}\] Integrating over a few \(1/\Gamma\)’s yields \[N_3=\eta_3|\gamma|^2|r|^2\ \text{and} \
N_4=\eta_4|\gamma|^2|t|^2.\] Coincidences are of course zero, but
in a real experiment, we see coincidence due to dark background current
and more than one atom spontaneously emitting in the measurement window.
We define: \[P_3=\frac{N_3}{N_H}, \
P_4=\frac{N_4}{N_H},\ \text{and} \ P_c\frac{N_c}{N_H}\] where
\(N_H\) is the number of heralded
photons and \[\alpha=\frac{P_c}{P_3P_4}=\frac{N_cN_H}{N_3N_4}.\]
For a quantum system, we have \(\alpha<1\). A classical system must have
\(\alpha>1\), since \[P_C=\langle W^{(2)}\rangle\ge\langle
W^{(1)}\rangle^2=P_3P_4.\] We can characterize the single-photon
quantum nature by observing \(\alpha<1\).
In summary, a one photon state, given by \(\left|1\right\rangle=\sum_lc_l\left|1_l\right\rangle\)
is an eigenstate of \(\hat{N}\), but
not necessarily \(\hat{H}\) (it has
definite particle number but not necessarily definite energy). When we
measure it, it can be observed only once. It has an extent in time given
by some small multiple of \(1/\Gamma\)
with high probability. Sources of single photons are not just very dim
light, as we see next.
We end the chapter with a discussion of semiclassical states of
light, described by our old friend the coherent state, which satisfies
\[\hat{a}_l\left|\alpha_l\right\rangle=\alpha_l\left|\alpha_l\right\rangle\]
where \[\left|\alpha_l\right\rangle=D(\alpha_l)\left|0\right\rangle=e^{\alpha_l\hat{a}_l+\alpha^*\hat{a}}\left|0\right\rangle.\]
For a semiclassical state, we have \[\begin{aligned}
W^{(1)}(\mathbf{r},t)=& \
s|\hat{E}^+(\mathbf{r},t)\left|\alpha_l\right\rangle|^2 \\
=& \ s(\varepsilon_l^{(l)})^2|\alpha_l|^2 \\
W^{(2)}(\mathbf{r},t,\mathbf{r}',t')=& \
s^2|\hat{E}^\dagger(\mathbf{r}',t')\hat{E}^+(\mathbf{r},t)\left|\alpha_l\right\rangle|^2
\\
=& \
s^2(\varepsilon_l^{(2)})^4|\alpha_l|^4=W^{(1)}(\mathbf{r},t)W^{(1)}(\mathbf{r}',t')
\end{aligned}\] This is the classical result for a classical
field as well.
Note that it shows that even for very dim light, with much less than one
photon in each measurement interval, we will sometimes observe two
photons in one interval. Furthermore the \(\alpha\) for semiclassical systems is 1.
The experiment has been done and verified. This clarifies an old result
where it was believed that classical light becomes a single photon
source when very dim. But it never does.
Incandescent light, LEDs, lasers, are all sources of semiclassical
light. Essentially all light sources we commonly use are semiclassical.
Single photon light sources are much more difficult to make. Next time,
we discuss squeezed light, measuring quadratures and how they improve
LIGO.