Phase-Phase Interactions – Quantum Information Science & Technology

(1) $\begin{equation*} \underbrace{ \overbrace{\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})}^{photon A}\overbrace{\ket{1}}^{B}}_{input} \implies \underbrace{\overbrace{\frac{1}{\sqrt{2}}(\ket{0}+e^{-i\pi} \ket{1})}^{photon A}\overbrace{\ket{1}}^{B}}_{output} \end{equation*}$

It is common knowledge that light consists of “photons,” but what exactly does this mean? What are “single photons” and how exactly do they relate to the electric field? And what about quantum mechanics being about waves? How did we go from classical waves to quantum waves to particles?

In the Stony Brook QIT lab, we can measure and characterize the quantum states of light. And we can not just detect and measure these “single photons,” but we can observe more complicated quantum states. These quantum states can be in quantum mechanical superpositions of individual states, so, for example, we can not just distinguish between “single photons” and “two photons,” but we can measure coherent superpositions of the two states. We can observe a “single photon state” and a “two photon state,” but we can also characterize a state that is a \textit{wave} of half “single photon” and half “two photon.” And we can distinguish that this \textit{wave of probability} is fundamentally different than a state that is simply having a mixture of the individual states (e.g. a box containing a coherence of single photon and two photon states is not as that same box containing simply a mixture of half single photons and half two photon states).

We will discuss in detail how a modern lab measures these things. More specifically, we will show how oscilloscope measurements of a homodyne detector allow us to reconstruct electric field quadrature data; and how we can use a maximum likelihood algorithm to fit this quadrature data to a set of wavefunctions to create the most likely density matrix. We will pay special attention to appealing to intuition, so that at the end of the day the reader will hopefully be able to understand what it means to be in a wavelike superposition of a “single photon” state and a “two photon” state.

\section{Homodyne Detector}

To characterize the quantum state of light you need a computer, a coherent light source (a laser), an oscilloscope, and a homodyne detector.
In order to understand precisely what a homodyne detector is doing, it is helpful first to take a step back and understand how a homodyne is useful for classical systems.

Before applying any quantum mechanics to the electric field, let’s first consider it as a simple wave:

$E = A cos(\omega t)$

As you can see in the figure, radio-frequencies have frequencies on the order of kilohertz, and optical light has frequencies in the hundreds of terrahertzs. While radio frequencies can be easily measured on a detector, optical light oscillates at too high of a frequency when compared to the clock speed of computers and response time of detectors. (To give a rough idea, the clock speed of modern processors is a few gigaHertz, which is around a hundred thousand times slower than the frequency of optical light.)

Does that mean we can’t see high-frequency of light on our detectors at all? Most modern detectors work by giving a change in photocurrent that is proportional to a change in intensity. We can then see that the voltage we’d read off of an oscilloscope is:

$V_{oscilloscope} \propto \Delta I \propto E^2 = (A cos(\omega t))^2$

And using $cos^2(A)= \frac{1}{2}(1+cos(2A))$

$V_{oscilloscope} \propto (A cos(\omega t))^2 = \frac{A^2}{2}(1+cos(2\omega t)$

When the $\omega$ in $cos(\omega t)$ is very large, our detector can only see its time-averaged value, zero.

Therefore we see that when we put an quickly-oscillating electric field in front of a detector detector, we end up seeing a constant value:

%HERE IS A HELPFUL SECOND QUANTIZATION LINK <3
%https://www.eng.fsu.edu/~dommelen/quantum/style_a/qft.html#eq:ex36

%http://hitoshi.berkeley.edu/221a/coherentstate.pdf

%http://www.waltervansuijlekom.nl/wp-content/uploads/2014/05/thesisMoniek.pdf

%http://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf

%http://pages.uoregon.edu/svanenk/solutions/NotesBS.pdf

%http://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf

$<V_{oscilloscope}> \implies \frac{A^2}{2}$

This means that if we put some light in front of our detector, we cannot see the time dependence of our E-field. This means if you put two different frequencies with the same amplitude in front of your detector, you would see the same result on your oscilloscope; and thus, you would not be able to figure out which frequency is which by looking at your oscilloscope.

Is there a way that we can be able to see this time dependence without having to get faster detectors or computers? Imagine instead of sending a single light field alone into the detector, we combine it with a second light field (additionally, we add the ability to change the phase of this second light field, which is easily attainable with a movable mirror):

$E_1 = A_1 cos(\omega_1 t)$

$E_2 = A_2 cos(\omega _2 t+ \delta (t))$

$\begin{align*} V_{oscilloscope} \propto (E_1 + E_2)^2 & = E_1^2 + E_2^2 + 2 E_1 E_2 \\ %& = (A_1 cos(\omega_ t))^2 + (A_2 cos(\omega _2 t + \delta (t)))^2 + 2 (A_1 cos(\omega_ t)) (A_2 cos(\omega _2 t + \delta (t)) \\ & = E_1^2 + E_2^2 + 2 (A_1 cos(\omega_ t)) (A_2 cos(\omega _2 t + \delta (t))) \end{align*}$

%An optical heterodyne detector can see these differences.

Now we see that our detector picks up interference between the two fields! We showed earlier that $E_1^2$ appears as a constant on the oscilloscope (due to the frequency being too fast for the detector to pick up). Likewise if $\omega_2$ is fast, and we don’t change the phase too quickly (i.e. $\dot{\delta}<< \omega$ ), then $E_2^2$ also appears as a constant. So now we see that we can recover the time dynamics of our oscillating beam:

$\begin{align*} V_{oscilloscope} \propto \frac{A_1^2}{2} + \frac{A_2^2}{2} + 2 A_1 A_2 cos(\omega_ t)cos(\omega _2 t + \delta (t)) \end{align*}$

This interference term is typically referred to as a “beat note.” We can get a better understanding of what this looks like by using a simple trig identity:

$Cos(A)Cos(B) = Cos(A+B) - Cos(A-B)$

\interfootnotelinepenalty=10000

$\begin{align*} V_{oscilloscope} \propto \frac{A_1^2}{2} + \frac{A_2^2}{2} + 2 A_1 A_2 cos((\omega _1 +\omega _2) t + \delta (t))- 2 A_1 A_2 cos((\omega _1 -\omega _2 ) t - \delta (t)) \end{align*}$

So now we have two constants and two cosine functions. One of the cosine functions has a frequency of $(\omega _1 +\omega _2)$ , and since $\omega _1$ was too high a frequency, and $\omega _2$ was too high of a frequency, then their sum frequency will also be too high; and thus this cosine term will average to zero. But their difference frequency is not necessarily too fast. If we measure frequencies such that $\omega_1-\omega_2 << \omega_1$ (meaning that our two fields aren’t that different in frequency), then this cosine term will not appear to be a constant value on our detector. So now we can see the time dynamics of our quickly oscillating light on our slow detector:

$\begin{align*} V_{oscilloscope} \propto Constant - 2 A_1 A_2 cos((\omega _1 -\omega _2 ) t - \delta (t)) \end{align*}$

We can use this setup to measure the frequency and amplitude of an unknown source. If we have a laser of known amplitude, $A_1$ , and frequency, $\omega _1$ , we can interfere it with an unknown source. Then when we look at a detector (if we hold our $\delta (t)$ to a constant), \textit{we can find the frequency of this “beat”} and since we know that $\omega_{beat} = \omega_1 - \omega_{unknown}$ , this implies $\omega_{unknown} = \omega_1 - \omega_{beat}$ . Likewise the amplitude of this beat is $2 A_1 A_2$ , so we can also solve for the amplitude of the unknown source. So we’ve managed to be able to come up with a scheme to detect light that was initially too high of a frequency for our detector to see it. But we intend on using this type of detector to detect single photons. Can our detector measure the light of a single photon? Is the light of a single photon too small to be seen by a detector? We can do an incredibly important trick by exploiting the fact that our beat signal is proportional to both electric field amplitudes. We can magnify the signal of our unknown source by increasing the amplitude of our known source. If our signal of a single photon is 100 times too small, we simply need to make our known laser source to be 100 times larger to compensate. This allows us to be able to detect the light of a single photon even with a detector that is not sensitive enough to see it on its own. Additionally there are a couple of experimental techniques that clean things up a bit. For starters, it would be nice if our oscilloscope voltage was proportional to just the interference beat note without the constant. The constant can be a little bit of a pain, particularly because our known source being large ( $A_{known}>>A_{unknown}$ ) implies that the constant value associated with our known field, $\frac{A_{1}^2}{2}$ , is much larger than the amplitude of our beat note. Graphing something like “ $cos(\omega t)$ + 10000 ” is a pain to see because it either requires you to look at our oscilloscope with a huge DC-offset value, or additional electronics are needed to filter this DC signal out.

Consider the following two equations:

$\begin{align*} (E_1 + E_2)^2 &= E_1^2 + E_2^2 + 2 E_1 E_2 \\ (E_1 - E_2)^2 &= E_1^2 + E_2^2 - 2 E_1 E_2 \end{align*}$

if we subtract these two equations from each other we obtain:

$(E_1 + E_2)^2 - (E_1 - E_2)^2 = 4 E_1 E_2$

Therefore, if we can detect the two signals, $(E_1 + E_2)$ and $(E_1 - E_2)$ , we can then subtract these two signals, and we will also subtract out the constant DC signal we don’t want. So if we combine $E_1$ and $E_2$ in a beam splitter, not only is there an output that returns $(E_1 + E_2),$ but there is a second output that returns $(E_1 - E_2)$ . In our heterodyne detector, we can take measurements of both outputs at the same time. Then we can send both signals to an electronic subtractor that subtracts the two signals, and finally we can obtain a signal that doesn’t include that annoying constant DC component. In our lab we use slightly different equipment that gives the same 2 outputs, $(E_1 + E_2)$ and $(E_1 - E_2)$ . Instead of using a normal beam splitter, we use two \textit{polarizing} beam splitters and a halfwave plate. First we combine two oppositely polarized beams (one H and one V), but since the combined beam has two E-fields of orthogonal polarization they will not interfere\footnote{The detector detects the norm, $|E|^2$ . So if E is a vector, the norm of that vector is $E_x^2+E_y^2+E_z^2$ . Before we when we summed up two waves we assumed they were oscillating in the same direction. Putting it more precisely, we could have included that polarization in the math: $E_1cos(\omega_1 x)\hat{x}+E_2cos(\omega_2 x)\hat{x} = (E_1cos(\omega_1 x)+E_2cos(\omega_2 x))\hat{x}$ This means that if the $E_x$ and $E_y$ are separate waves, our detector sees $E_x^2+E_y^2$ \textit{without} the interference term. E-fields only get their interference term when their polarizations are the same. (i.e. when they wiggle up and down in the same direction.}.

Since the two E-fields coming out of our PBS (polarized beam splitter), we need to come up with a scheme to line up these two E-fields. By putting a half-wave plate at 22.5 degrees, and then splitting that light with a second polarizing beam splitter, that second beam splitter produces two outputs: $E_1-E_2$ and $E_1 + E_2$ . You might ask what the heck was the point of doing this complicated setup when we simply could have used a single beam splitter to get $E_1-E_2$ and $E_1 + E_2$ . In essence, having the half-wave plate gives us one extra “nob” to turn, which lets us have slightly more control over our experimental imperfections. If our beam splitters do not split our light at a perfect 50/50 ratio, we can compensate that by over or under-rotating our halfwave plate such that the projection projects slightly different ratio’s to our second PBS to get back to a 50/50 ratio of $E_1$ and $E_2$ .