Compute the output of a quantum circuit with classically controlled operations

Question

Consider the following circuit :

where $|\psi\rangle$ is a qubit in $\mathbb{C}^2$, $|0\rangle= \begin{pmatrix}1 \\ 0 \end{pmatrix}$, $T= \begin{pmatrix}1 & 0\\ 0 & e^{i\pi/4} \end{pmatrix}$ is the $\pi/8$ gate, $H= \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\ 1 & -1 \end{pmatrix}$ is the Hadamard gate, $X= \begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}$ and $P= \begin{pmatrix}1 & 0\\ 0 & i \end{pmatrix}$.

In this circuit, the $PX$ gate is only applied when the measurement of the top wire outcome is $1$. This measurement is done with respect to the computational basis $|0\rangle, |1\rangle$.

The task is to verify the resulting state from this circuit, but it doesn't seem clear how we can retrieve $T|\psi\rangle$ exactly.

Note that: $TH|0\rangle= \frac{1}{\sqrt{2}}T\begin{pmatrix}1\\ 1 \end{pmatrix}= \frac{1}{\sqrt{2}}\begin{pmatrix}1\\ e^{i\pi/4} \end{pmatrix}$ then after the CNOT gate:

$\begin{align} \text{CNOT}(TH|0\rangle|\psi\rangle)&= \begin{pmatrix}1 & 0& 0& 0 \\ 0 & 1& 0& 0 \\ 0 & 0& 0& 1 \\ 0 & 0& 1& 0 \\ \end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ e^{i\pi/4} \end{pmatrix}\otimes \begin{pmatrix}\psi_0\\ \psi_1 \end{pmatrix}\\ &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 0& 0& 0 \\ 0 & 1& 0& 0 \\ 0 & 0& 0& 1 \\ 0 & 0& 1& 0 \\ \end{pmatrix}\begin{pmatrix}\psi_0\\ \psi_1 \\ e^{i\pi/4}\psi_0 \\ e^{i\pi/4}\psi_1 \end{pmatrix}\\ &= \frac{1}{\sqrt{2}} \begin{pmatrix}\psi_0\\ \psi_1 \\ e^{i\pi/4}\psi_1 \\ e^{i\pi/4}\psi_0 \end{pmatrix} \end{align}$

is not necessarily a separable state.

How do you go from here? The measurement is done on the top qubit only.

Answer 1

Consider the following circuit :

where $|\psi\rangle$ is a qubit in $\mathbb{C}^2$, $|0\rangle= \begin{pmatrix}1 \\ 0 \end{pmatrix}$, $T= \begin{pmatrix}1 & 0\\ 0 & e^{i\pi/4} \end{pmatrix}$ is the $\pi/8$ gate, $H= \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\ 1 & -1 \end{pmatrix}$ is the Hadamard gate, $X= \begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}$ and $P= \begin{pmatrix}1 & 0\\ 0 & i \end{pmatrix}$.

I'll get you part way there, and I'll work with the kets rather than the matrix representations (for the most part).

The input hits the first H and goes to: $$ |0\rangle\otimes|\psi\rangle \to \frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)\otimes|\psi\rangle $$ then hits the T gate and becomes: $$ \to \frac{1}{\sqrt{2}}\left(|0\rangle+e^{i\phi}|1\rangle\right)\otimes|\psi\rangle\;, $$ which I'll write as $e^{i\phi}$, but this all works for $\phi=\pi/4$ too.

This hits the CNOT gate and becomes: $$ \to \frac{1}{\sqrt{2}}\left\{|0\rangle\otimes|\psi\rangle+e^{i\phi}|1\rangle\otimes \left(\psi_1|0\rangle+\psi_0|1\rangle\right) \right\} $$ $$ \frac{1}{\sqrt{2}}\left\{|0\rangle\otimes\left(\psi_0|0\rangle+\psi_1|1\rangle\right)+e^{i\phi}|1\rangle\otimes \left(\psi_1|0\rangle+\psi_0|1\rangle\right) \right\} $$ $$ =\frac{1}{\sqrt{2}} \left\{ \left(\psi_0|0\rangle+e^{i\phi}\psi_1|1\rangle\right)\otimes|0\rangle + \left(\psi_1|0\rangle+e^{i\phi}\psi_0|1\rangle\right)\otimes|1\rangle \right\}\;. $$

If I measure 0 for the top ket (the rightmost in the direct product), then the states of the bottom key is (since no PX is applied): $$ \left(\psi_0|0\rangle+e^{i\pi/4}\psi_1|1\rangle\right) = T|\psi\rangle $$

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In order to show that the same is true when $1$ is measured for the top ket, use the fact that $\phi$ is really $\pi/4$ (the general case of $\phi$ won't work, it only works with $T$) and use the fact that $PX$ can be written as: $$ PX = \left( \begin{matrix}1 & 0 \\ 0 & e^{i\pi/2} \end{matrix} \right) \left( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} \right)\;, $$ where I have written $i=e^{i\pi/2}$.

And remember that you can always factor out an overall phase.