Interferometry and metrology

Interferometry and metrology

Suppose you are given a phase shifter of unknown $\phi$:

\noindent How accurately can you determine $\phi$, given a certain time, and laser power? In this section, we consider this basic measurement problem, and show how the usual shot noise limit can be exceeded by using quantum states of light, reaching a quantum limit determined by Heisenberg's uncertainty principle. This limit is achieved using entanglement, which can be realized using entangled multi-mode photons, or by a variety of squeezed states. The physics behind such quantum measurement techniques generalizes to a wide range of metrology problems, but a common challenge the need to reduce loss.

Shot noise limit

The Poisson distribution of photon number in coherent (laser) light contributes an uncertainty of $\Delta n\sim {\sqrt {n}}$ to optical measurements. It is therefore reasonable to anticipate that with $n$ photons, the uncertainty $\Delta \phi$ with which an unknown phase $\phi$ can be determined might be bounded below by $\Delta \phi \geq 1/{\sqrt {n}}$, based on the heuristic that $\Delta n\Delta \phi \geq 1$. Such a limit is known as being due to shot noise, arising from the particle nature of photons, as we shall now see rigorously. Consider a Mach-Zehnder interferometer constructed from two 50/50 beamsplitters, used to measure $\phi$:

\noindent Let us analyze this interferometer, first by using a traditional quantum optics approach in the Heisenberg picture, and second by using single photons in the Schrodinger picture. Previously, we've defined the unitary transform for a quantum beamsplitter as being a rotation about the ${\hat {y}}$ axis, so as to avoid having to keep track of factors of $i$. For variety, let's now use a different definition; nothing essential will change. Let the 50/50 beamsplitter transformation be

$B={\frac {1}{\sqrt {2}}}\left[{\begin{array}{cc}{1}&{-i}\\{-i}&{1}\end{array}}\right]\,.$

This acts on $[a,b]^{T}$ to produce operators describing the output of the beamsplitter; in particular,

$\left[{\begin{array}{c}{a-ib}\\{b-ia}\end{array}}\right]=B\left[{\begin{array}{c}{a}\\{b}\end{array}}\right]\,.$

Similarly, the phase shifter acting on the mode operators performs

$P=\left[{\begin{array}{cc}{1}&{0}\\{0}&{e^{i\phi }}\end{array}}\right]\,.$

The Mach-Zehnder transform is thus

${\begin{array}{rcl}U&=&BPB\\&=&{\frac {1}{2}}\left[{\begin{array}{cc}{1}&{-i}\\{-i}&{1}\end{array}}\right]\left[{\begin{array}{cc}{1}&{0}\\{0}&{e^{i\phi }}\end{array}}\right]\left[{\begin{array}{cc}{1}&{-i}\\{-i}&{1}\end{array}}\right]\\&=&-ie^{i\phi /2}\left[{\begin{array}{cc}{\sin(\phi /2)}&{\cos(\phi /2)}\\{\cos(\phi /2)}&{-\sin(\phi /2)}\end{array}}\right]\,.\end{array}}$

The way we have defined these transformations here, the output modes of the interferometer, $c$ and $d$, are

$\left[{\begin{array}{c}{c}\\{d}\end{array}}\right]=U\left[{\begin{array}{c}{a}\\{b}\end{array}}\right]\,.$

We are interested in the difference between the photon numbers measured at the two outputs, $M=(d^{\dagger }d-c^{\dagger }c)/2$, where the extra factor of two is introduced for convenience. We find

${\begin{array}{rcl}c^{\dagger }c&=&\sin ^{2}{\frac {\phi }{2}}a^{\dagger }a+\cos ^{2}{\frac {\phi }{2}}b^{\dagger }b+\sin {\frac {\phi }{2}}(a^{\dagger }b+b^{\dagger }a)\\d^{\dagger }d&=&\cos ^{2}{\frac {\phi }{2}}a^{\dagger }a+\sin ^{2}{\frac {\phi }{2}}b^{\dagger }b-\sin {\frac {\phi }{2}}(a^{\dagger }b+b^{\dagger }a)\end{array}}$

The measurement result $M$ is thus

$M=(a^{\dagger }a-b^{\dagger }b)\cos \phi -(a^{\dagger }b+b^{\dagger }a)\sin \phi \,.$

Define $X=a^{\dagger }a-b^{\dagger }b$, and $Y=a^{\dagger }b+b^{\dagger }a$. Recognizing that $X$ is the difference in photon number between the two output arms, and recalling that this is the main observable result from changing $\phi$, we identify the signal we wish to see as being $X$. Ideally, the output signal should go as $\cos \phi$. The signal due to $Y$ goes as $\sin \phi$, and we shall see that this is the noise on the signal. The average output signal $\langle X{\rangle }$, as a function of $\phi$, looks like this:

\noindent Note that if our goal is to maximize measurement sensitivity to changes in $\phi$, then the best point to operate the interferometer at is around $\phi =\pi /2$, since the slope $d\langle x{\rangle }/d\phi$ is largest there. At this operating point, if the interferometer's inputs have laser light coming into only one port, then the outputs have equal intensity; thus, the interferometer is sometimes said to be "balanced" when $\phi =\pi /2$. What is the uncertainty in our measurement of $\phi$, derived from the observable $M$? By propagating uncertainties, this is

${\langle }\Delta \phi ^{2}\rangle ={\frac {{\langle }\Delta M^{2}{\rangle }}{\left|{\frac {\partial \langle M{\rangle }}{\partial \phi }}\right|^{2}}}\,,$

where

${\frac {\partial \langle M{\rangle }}{\partial \phi }}=-X\sin \phi -Y\cos \phi \,.$

Let $\phi =\pi /2$, such that $M=Y$, and $\partial M/\partial \phi =-X$. For a coherent state input, $|\psi _{in}\rangle =|\alpha {\rangle }|0{\rangle }$, we find

${\begin{array}{rcl}\langle X\rangle &=&\langle 0|{\langle }\alpha |(a^{\dagger }a-b^{\dagger }b)|\alpha {\rangle }|0{\rangle }\\&=&|\alpha |^{2}\\&=&n\,,\end{array}}$

if we define $n=|\alpha |^{2}$ as the input state mean photon number. Also,

${\begin{array}{rcl}\langle Y\rangle =\langle 0|{\langle }\alpha |(a^{\dagger }b+b^{\dagger }a)|\alpha {\rangle }|0\rangle =0\,.\end{array}}$

This is consitent with our intuition: the signal should go as $\sim n$, and the undesired term goes as $Y$, so it is good that is small on average. However, there are nontrivial fluctuations in $Y$, because

$\langle Y^{2}\rangle =\langle a^{\dagger }ba^{\dagger }b+a^{\dagger }bb^{\dagger }a+b^{\dagger }ab^{\dagger }a+b^{\dagger }aa^{\dagger }b{\rangle }\,,$

and $\langle a^{\dagger }bb^{\dagger }a\rangle =\langle a^{\dagger }a(1+b^{\dagger }b){\rangle }$ is nonzero for the coherent state! Specifically, the noise in $Y$ is

$\langle Y^{2}\rangle =|\alpha |^{2}=n\,,$

and thus the variance in the measurement result is

${\langle }\Delta M^{2}\rangle =\langle Y^{2}\rangle -\langle Y{\rangle }^{2}=|\alpha |^{2}=n\,.$

From Eq.(\ref{eq:l7-dphi}), it follows that the uncertainty in $\phi$ is therefore

${\langle }\Delta \phi \rangle ={\frac {\sqrt {{\langle }\Delta M^{2}{\rangle }}}{\left|{\frac {\partial \langle M{\rangle }}{\partial \phi }}\right|}}={\frac {\sqrt {n}}{n}}={\frac {1}{\sqrt {n}}}\,.$

This is a very reasonable result; as the number of photons used increases, the accuracy with which $\phi$ can be determined increases with ${\sqrt {n}}$. The improvement arises because greater laser power allows better distinction between the signals in $c$ and $d$. Another way to arrive at the same result, using single photons, gives an alternate interpretation and different insight into the physics. As we have seen previously, acting on the $|01{\rangle }$, $|10{\rangle }$ "dual-rail" photon state, a 50/50 beamsplitter performs a $R_{y}(\pi /2)$ rotation, and a phase shifter performs a $R_{z}(\phi )$ rotation. The Mach-Zehnder interferometer we're using can thus be expressed as this transform on a single qubit:

\noindent where the probability of measuring a single photon at the output is $P$. Walking through this optical circuit, the states are found to be

${\begin{array}{rcl}|\psi _{1}\rangle &=&{\frac {|0{\rangle }+|1{\rangle }}{\sqrt {2}}}\\|\psi _{2}\rangle &=&{\frac {|0{\rangle }+e^{i\phi }|1{\rangle }}{\sqrt {2}}}\\|\psi _{3}\rangle &=&{\frac {1-e^{i\phi }}{2}}|0\rangle +{\frac {1+e^{i\phi }}{2}}|1{\rangle }\,,\end{array}}$

such that

$P={\frac {1+\cos \phi }{2}}\,.$

Repeating this $n$ times (so that we use the same average number of photons as in the coherent state case), we find that the standard deviation in $P$ is

$\Delta P={\sqrt {\frac {p(1-p)}{n}}}={\frac {\sin \phi }{2{\sqrt {n}}}}\,.$

Given this, the uncertainty in $\phi$ is

$\Delta \phi ={\frac {\Delta \phi }{\left|{\frac {dP}{d\phi }}\right|}}={\frac {1}{\sqrt {n}}}\,.$

This is the same uncertainty as we obtained for the coherent state input, but the physical origin is different. Now, we see the noise as being due to statistical fluctuations of a Bernoulli point process, one event at a time. The ${\sqrt {n}}$ noise thus comes from the amount of time the signal is integrated over (assuming a constant rate of photons). The noise is simply shot noise.

Heisenberg limit: entanglement

The shot noise limit we have just seen, however, is not fundamental. Here is a simple argument that something better should be possible. Recall that the desired signal at the output of our Mach-Zehnder interferometer is $X=a^{\dagger }a-b^{\dagger }b$, and the noise is $Y=a^{\dagger }b+b^{\dagger }a$. If the inputs have $a^{\dagger }a=n$ and $b^{\dagger }b=0$, and if $Y$ were zero, then the measured signal would be $n\cos \phi$. And at the balanced operating point $\phi =\pi /2$,

${\frac {|\Delta m|}{\Delta \phi }}=n\sin \phi \leq n\,.$

Thus, if the smallest photon number change resolvable is $\Delta m=1$, then $n\Delta \phi \geq 1$, from which it follows that

$\Delta \phi \geq {\frac {1}{n}}\,.$

This is known as the "Heisenberg limit" on interferometry. There are some general proofs in the literature that such a limit is the best possible on interferometry. It governs more than just measurements of phase shifters; gyroscopes, mass measurements, and displacement measurements all use interferometers, and obey a Heisenberg limit. The argument above only outlines a sketch for why $1/n$ might be an achievable limt, versus $1/{\sqrt {n}}$; it assumes that the noise $Y$ can be made zero, however, and does not provide a means for accomplishing this in practice. Many ways to reach the Heisenberg limit in interferometry are now known. Given the basic structure of a Mach-Zehnder interferometer,

\noindent one can consider changing the input state $|\psi _{in}{\rangle }$, changing the beamsplitters, or changing the measurement. Common to all of these approaches is the use of entangled states. How entanglement makes Heisenber-limited interferometry possible can be demonstrated by the following setup. Let us replace the beamsplitters in the Mach-Zehnder interferometer with entangling and dis-entangling devices:

\noindent Conceptually, the unusual beamsplitters may be the nonlinear Mach-Zehnder interferometers we discussed in Section~2.3. They may also be described by simple quantum circuits, using the Hadamard and controlled-{\sc not} gate; for two qubits, the circuit is

\noindent Note how the output is one of the Bell states. For three qubits, the circuit is

\noindent This output state, $|000{\rangle }+|111{\rangle }$ (suppressing normalization) is known as a GHZ (Greenberger-Horne-Zeilinger) state. Straightforward generalization leads to larger "Schrodinger cat" states $|00\cdots 0{\rangle }+|11\cdots 1{\rangle }$, using one Hadamard gate and $n$ controlled-{\sc not} gates. Note that the reversed circuit unentangles the cat states to produce computational basis states. The important feature of such $n$-qubit cat states, for our purpose, is how they are transformed by phase shifters. A single qubit $|0{\rangle }+|1{\rangle }$ becomes $|0{\rangle }+e^{i\phi }|1{\rangle }$. Similarly, two entangled qubits in the state $|00{\rangle }+|11{\rangle }$, when sent through two phase shifters, becomes $|00{\rangle }+e^{2i\phi }|11{\rangle }$, since the phases add. And $n$ qubits in the state $|00\cdots 0{\rangle }+|11\cdots 1{\rangle }$ sent through $n$ phase shifters becomes $|00\cdots 0{\rangle }+e^{ni\phi }|11\cdots 1{\rangle }$. When such a phase shifted state is un-entangled, using the reverse of the entangling circuit, the $n$ controlled-{\sc not} gates leave the state $|00\cdots 0{\rangle }[|0\rangle +e^{ni\phi }|1\rangle ]$, where the last $n-1$ qubits are left in $|0{\rangle }$, and the first qubit (the qubit used as the control for the {\sc cnot} gates) is

${\frac {|0\rangle +e^{ni\phi }|1{\rangle }}{\sqrt {2}}}\,.$

Compare this state with that obtained from the single qubit interferometer, Eq.(\ref{eq:l7-1qubitphase}); instead of a phase $\phi$, the qubit now carries the phase $n\phi$. This means that the probability $P$ of measuring a single photon at the output becomes

$P={\frac {1+\cos(n\phi )}{2}}\,.$

The standard deviation, from repeating this experiment, on average, would be

$\Delta P={\sqrt {P(1-P)}}={\frac {\sin(n\phi )}{2}}\,.$

Using $dP/d\phi =-n\sin(n\phi )/2$, we obtain for the uncertainty in $\phi$,

$\Delta \phi ={\frac {\Delta \phi }{\left|{\frac {dP}{d\phi }}\right|}}={\frac {1}{n}}\,,$

which meets the Heisenberg limit.

Squeezed light interferometry

Heisenberg-limited interferometry can also be accomplished using a variety of states of light, including squeezed states we studied in Section~2.2. Let us explore three configurations here. Vacuum squeezed state input. Because squeezed states can move noise between the $x$ and $p$ quadratures, it is intuitively reasonable that a state with low phase noise could be used to provide more accurate measurements of $\phi$ than is possible with a coherent state, which has equal noise in the two quadratures. It might seem counter-intuitive, however, that we can get to the Heisenberg limit by replacing not the coherent state input, but rather, the vacuum state, in the Mach-Zehnder interferometer. This works because at the balanced operating point, the noise in the output is due to fluctuations entering in at the vacuum port. Before, we used $|\alpha {\rangle }|0{\rangle }$ as input. Let us now replace this by

$|\psi _{in}=|\alpha \rangle |0_{r}{\rangle }\,,$

where $|0_{r}\rangle =S(r)|0{\rangle }$ is a squeezed vacuum state. Recall that for the balanced interferometer, the final uncertainty in the phase measurement is

${\langle }\Delta \phi \rangle ={\frac {\sqrt {{\langle }\Delta Y^{2}{\rangle }}}{|\langle X{\rangle }|}}\,,$

where $X=a^{\dagger }a-b^{\dagger }b$, and $Y=a^{\dagger }b+b^{\dagger }a$. For the squeezed vacu