Finding interferometer phase

Setup

When making the Pauli Matrix Measurement ${\sigma }_X$ on an interferometer in the Quantum state $\cos (\theta )|\downarrow ⟩+\sin (\theta )|\uparrow ⟩$, we get $\uparrow$ with probability ${\sin }^2(\theta )$ and $\downarrow$ with probability ${\cos }^2(\theta )$, i.e.

$$\{\begin{array}{r}\mathbb{P}[\uparrow |\theta ]={\sin }^2(\theta )\\ \mathbb{P}[\downarrow |\theta ]={\cos }^2(\theta )\end{array}$$

Depending on the value of $\theta$ we can get more/less information per experiment, based on the derivatives of ${\sin }^2(\theta )$ and ${\cos }^2(\theta )$. Since the underlying value of $\theta$ is uniform on the interval $[0,2\pi ]$, we can attempt to maximise the information obtained by applying different phase shifts before applying the ${\sigma }_X$ Measurement. As such, we will consider our observations to be the set $\{(a_i,{\theta }_i):a_i\in \{\uparrow ,\downarrow \},{\theta }_i\in \mathbb{R}\}$, obtained with the underlying probabilities

$$\{\begin{array}{r}\mathbb{P}[\uparrow |\theta ,{\theta }_i]={\sin }^2(\theta +{\theta }_i)\\ \mathbb{P}[\downarrow |\theta ,{\theta }_i]={\cos }^2(\theta +{\theta }_i)\end{array}$$

Our goal is to find the most likely $\theta$ given our observations.

Solution

We will apply the Maximum Likelihood Estimation framework. Up to a me, the probability of our observations equals $L:={\prod }_{i,a_i=\uparrow }{\sin }^2(\theta +{\theta }_i){\prod }_{i,a_i=\downarrow }{\cos }^2(\theta +{\theta }_i)$ . Since $\log$ is Monotonic, $\underset{\theta }{\mathrm{argmax}}\{L\}=\underset{\theta }{\mathrm{argmax}}\{\log (L)\}$, allowing us to simplify the latter expression:

$$\begin{array}{rlc}\log (L)& =\log \left({\prod }_{i,a_i=\uparrow }{\sin }^2(\theta +{\theta }_i){\prod }_{i,a_i=\downarrow }{\cos }^2(\theta +{\theta }_i)\right)& \text{As log(xy)=log(x)+log(y)}\\ & =\log \left({\prod }_{i,a_i=\uparrow }{\sin }^2(\theta +{\theta }_i)\right)+\log \left({\prod }_{i,a_i=\downarrow }{\cos }^2(\theta +{\theta }_i)\right)& \text{As log(∏xi)=∑log(xi)}\\ & ={\sum }_{i,a_i=\uparrow }\log \left({\sin }^2(\theta +{\theta }_i)\right)+{\sum }_{i,a_i=\downarrow }\log \left({\cos }^2(\theta +{\theta }_i)\right)& \end{array}$$

In order to maximise $L$ we need to have

$$\begin{array}{rll}0& =\frac{\partial }{\partial \theta }\log (L)& \\ & =\frac{\partial }{\partial \theta }\left[{\sum }_{i,a_i=\uparrow }\log \left({\sin }^2(\theta +{\theta }_i)\right)+{\sum }_{i,a_i=\downarrow }\log \left({\cos }^2(\theta +{\theta }_i)\right)\right]& \text{Using the linearity of ∂θ∂}\\ & ={\sum }_{i,a_i=\uparrow }\frac{\partial }{\partial \theta }\left[\log \left({\sin }^2(\theta +{\theta }_i)\right)\right]+{\sum }_{i,a_i=\downarrow }\frac{\partial }{\partial \theta }\left[\log \left({\cos }^2(\theta +{\theta }_i)\right)\right]& \text{By the chain rule}\\ & ={\sum }_{i,a_i=\uparrow }2{\sin }^{-2}(\theta +{\theta }_i)\sin (\theta +{\theta }_i)\cos (\theta +{\theta }_i)-{\sum }_{i,a_i=\downarrow }2{\cos }^{-2}(\theta +{\theta }_i)\sin (\theta +{\theta }_i)\cos (\theta +{\theta }_i)& \text{Removing 2 as it does not impact the final result}\\ & ={\sum }_{i,a_i=\uparrow }\arctan (\theta +{\theta }_i)-{\sum }_{i,a_i=\downarrow }\tan (\theta +{\theta }_i)& \end{array}$$

While finding the closed formula solution for the equation above might not be possible, finding the roots can be done relatively quickly via the Newton-Raphson method.

Fisher Information

We can calculate the Fisher Information for this setup by:

$$\begin{array}{rll}{\mathcal{I}}_X(\theta )& =\mathbb{E}[(\frac{\partial }{\partial \theta }\log (f_{\theta }(X){)}^2|\theta ]& \text{Since ∂θ∂logfθ(x)=∂θ∂logL, with L as above}\\ & =\mathbb{E}[{\left({\sum }_{i,a_i=\uparrow }\arctan (\theta +{\theta }_i)-{\sum }_{i,a_i=\downarrow }\tan (\theta +{\theta }_i)\right)}^2|\theta ]& \text{Taking the expectation over the 2N possible outcomes, with α∈{↑,↓}N }\\ & =2^{-N}{\sum }_{\alpha }{\left({\sum }_{i,a_i=\uparrow }\arctan (\theta +{\theta }_i)-{\sum }_{i,a_i=\downarrow }\tan (\theta +{\theta }_i)\right)}^2& \text{Rearranging into a single sum }\\ & =2^{-N}{\sum }_{\alpha }{\left({\sum }_i^N\tan (\theta +{\theta }_i)(-1{)}^{1_{\{{\alpha }_i=\downarrow \}}}\right)}^2& \text{Splitting the i sum as 2 sums}\\ & =2^{-N}{\sum }_{\alpha }\left({\sum }_i^N\tan (\theta +{\theta }_i)(-1{)}^{1_{\{{\alpha }_i=\downarrow \}}}\right)\left({\sum }_j^N\tan (\theta +{\theta }_j)(-1{)}^{1_{\{{\alpha }_j=\downarrow \}}}\right)& \text{Pulling the sums out}\\ & =2^{-N}{\sum }_{\alpha }{\sum }_{i,j\in \{\mathrm{1...}N{\}}^2}\tan (\theta +{\theta }_i)(-1{)}^{1_{\{{\alpha }_i=\downarrow \}}}\tan (\theta +{\theta }_j)(-1{)}^{1_{\{{\alpha }_j=\downarrow \}}}& \\ & =2^{-N}{\sum }_{i,j\in \{\mathrm{1...}N{\}}^2}{\sum }_{\alpha }\tan (\theta +{\theta }_i)(-1{)}^{1_{\{{\alpha }_i=\downarrow \}}}\tan (\theta +{\theta }_j)(-1{)}^{1_{\{{\alpha }_j=\downarrow \}}}& \text{As (−1)1{αi=↓}(−1)1{αj=↓}=(−1)1{αi=αj}}\\ & =2^{-N}{\sum }_{i,j\in \{\mathrm{1...}N{\}}^2}{\sum }_{\alpha }\tan (\theta +{\theta }_i)\tan (\theta +{\theta }_j)(-1{)}^{1_{\{{\alpha }_j={\alpha }_i\}}}& \text{The i=j terms cancel out 🎉}\\ & =2^{-N}{\sum }_{i=1}^N{\sum }_{\alpha }\tan (\theta +{\theta }_i{)}^2& \text{Since there are 2N possible configurations for α}\\ & =2^{-N}{\sum }_{i=1}^N{2}^N\tan (\theta +{\theta }_i{)}^2& \\ & ={\sum }_{i=1}^N\tan (\theta +{\theta }_i{)}^2& \end{array}$$

• Replace mentions of $\arctan$ above with ${\tan }^{-1}$.

Open Questions

Some questions remain. Help me ❓

1. Is the root $\theta$ unique, up to translations of $2\pi$?
2. Can we find the Variance of the estimated value $\bar{\theta }$?
3. $\underset{\theta +{\theta }_i\to \pm \pi }{\lim }\tan (\theta +{\theta }_i)=\pm \infty =\underset{\theta +{\theta }_i\to 0}{\lim }\arctan (\theta +{\theta }_i)$.
   1. Could this cause problems in the equality above due to an absurd sensitivity to “edge” cases?
   2. Could we get rid of this infinite information explosion by adding a possibly Normally distributed error term to the measurement of $\theta +{\theta }_i$?