Austin Adams

Proof of the Expectation Value of Z for a Qubit State

April 16, 2022

It has been a few years since I wrote a post on this blog, so I wanted to share a fun proof, even if it’s a little trivial.$ \newcommand{bra}[1]{\langle #1 |} \newcommand{ket}[1]{| #1 \rangle} \newcommand{bramket}[2]{\langle #1 | #2 \rangle} \newcommand{bramket}[3]{\langle #1 | #2 | #3 \rangle} $

For many quantum applications, you care about $\bramket{\psi}{Z}{\psi}$. That is, the expectation value of the Pauli operator $Z$. I’m a computer scientist, so $\ket{\psi}$ is a qubit state for me, but this is defined for any quantum state $\ket{\psi}$. One way to calculate it may be common knowledge, but I couldn’t find a proof online (only some code), so here is my little proof.

Examples for One and Two Qubits

First, let’s consider the case where $\ket{\psi}$ is a single-qubit state. It’s useful to know that

\[ \begin{align} Z &= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix} \\
&= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} - \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} \\
&= \ket{0}\bra{0} - \ket{1}\bra{1} \end{align} \]

thus

\[ \begin{align} \bramket{\psi}{Z}{\psi} &= \bramket{\psi}{\left[\ket{0}\bra{0} - \ket{1}\bra{1}\right]}{\psi} \\
&= \bramket{\psi}{\left[\ket{0}\bra{0}\right]}{\psi} - \bramket{\psi}{\left[\ket{1}\bra{1}\right]}{\psi} \\
&= \bramket{\psi}{M_0}{\psi} - \bramket{\psi}{M_1}{\psi} \\
&= \bramket{\psi}{M_0^\dagger M_0}{\psi} - \bramket{\psi}{M_1^\dagger M_1}{\psi} \\
&= P(0) - P(1) \end{align} \]

where $M_0$ and $M_1$ are the measurement operators for 0 and 1, respectively, and $P(0)$ and $P(1)$ are the probabilities of measuring 0 and 1, respectively.

Now, let’s suppose $\ket{\psi}$ is a two qubit state. Do we get something similar? First, let’s observe something similar to the above about $Z \otimes Z$: \[ \begin{align} Z \otimes Z &= \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\
&= \ket{00}\bra{00} - \ket{01}\bra{01} - \ket{10}\bra{10} + \ket{11}\bra{11} \end{align} \]

Now, we can see

\[ \begin{align} \bramket{\psi}{Z \otimes Z}{\psi} &= \bramket{\psi}{\left[\ket{00}\bra{00} - \ket{01}\bra{01} - \ket{10}\bra{10} + \ket{11}\bra{11} \right]}{\psi} \\
&= \bramket{\psi}{\left[\ket{00}\bra{00}\right]}{\psi} - \bramket{\psi}{\left[\ket{01}\bra{01}\right]}{\psi} \\ &\quad - \bramket{\psi}{\left[\ket{10}\bra{10}\right]}{\psi} + \bramket{\psi}{\left[\ket{11}\bra{11}\right]}{\psi} \\
&= \bramket{\psi}{M_{00}}{\psi} - \bramket{\psi}{M_{01}}{\psi} - \bramket{\psi}{M_{10}}{\psi} + \bramket{\psi}{M_{11}}{\psi} \\
&= \bramket{\psi}{M_{00}^\dagger M_{00}}{\psi} - \bramket{\psi}{M_{01}^\dagger M_{01}}{\psi} \\ &\quad - \bramket{\psi}{M_{10}^\dagger M_{10}}{\psi} + \bramket{\psi}{M_{11}^\dagger M_{11}}{\psi} \\
&= P(00) - P(01) - P(10) - P(11) \end{align} \]

A pattern is emerging here. In general, if $p(x)$ is the bit parity of bitstring $x$, is $\bramket{\psi}{Z}{\psi} = \sum_{x\in\{0,1\}^n} (-1)^{p(x)}P(x)$? Yes!

Proof

We need to show that $\underbrace{Z \otimes Z \otimes \cdots \otimes Z}_{n\text{ times}} = \sum_{x\in\{0,1\}^n} (-1)^{p(x)} \ket{x}\bra{x}$. To help with the proof, we should also assert that this is a diagonal matrix. Let’s use a proof by induction.

Base case

Consider $n=1$. We already showed above that $Z$ is a diagonal matrix and $Z = \ket{0}\bra{0} - \ket{1}\bra{1}$, so the base case holds.

Inductive step

For the inductive step, we assume that for $n-1$, the proposition holds. Thus, $Z' = \underbrace{Z \otimes Z \otimes \cdots \otimes Z}_{n-1\text{ times}}$ is a diagonal matrix and $Z' = \sum_{x\in\{0,1\}^{n-1}} (-1)^{p(x)} \ket{x}\bra{x}$. We want to show that $Z' \otimes Z = \sum_{x\in\{0,1\}^n} (-1)^{p(x)} \ket{x}\bra{x}$.

In general, we can observe that for some $2^n\times2^n$ diagonal matrix $D$

\[ D = \begin{bmatrix} d_{00\cdots00} & & & 0 \\ & d_{00\cdots01} & & \\ & & \ddots & \\ 0 & & & d_{11\cdots11} \end{bmatrix} \]

we have

\[ \begin{align} D \otimes Z &= \begin{bmatrix} d_{00\cdots00} & & & & & & 0 \\ & -d_{00\cdots00} & & \\ & & d_{00\cdots01} & \\ & & & -d_{00\cdots01} \\ & & & & \ddots \\ & & & & & d_{11\cdots11} \\ 0 & & & & & & -d_{11\cdots11} \end{bmatrix} \\
&= \begin{bmatrix} d'_{00\cdots000} & & & & & & 0 \\ & d'_{00\cdots001} & & \\ & & d'_{00\cdots010} & \\ & & & d'_{00\cdots011} \\ & & & & \ddots \\ & & & & & d'_{11\cdots110} \\ 0 & & & & & & d'_{11\cdots111} \end{bmatrix} \end{align} \]

Above, $d_x$ denotes the diagonal entry on row/column $x$ in $D$, and $d'_x$ denotes the diagonal entry on row/column $x$ in $D \otimes Z$. From above, we can see that we have $d'_{xb} = (-1)^bd_x$. Henceforth, let’s assume that $D = Z'$.

Consider two cases for the bit parity of $x$:

  1. If $x$ has even bit parity, then $d_{x} = 1$ by the inductive hypothesis. Thus, for $b=0$, the $d'_{xb}$ equation above gives $1$, which aligns with our goal to show that $d'_{xb} = 1$ since $xb$ would have even bit parity. For $b=1$, however, $xb$ would have odd bit parity; the $d'_{xb}$ equation above gives $-1$, which matches our goal to show that $d'_{xb} = -1$.

  2. If $x$ has odd bit parity, then $d_{x} = -1$ by the inductive hypothesis. For $b=0$, $xb$ still has odd bit parity, and the $d'_{xb}$ equation above gives $-1$, which is good because it aligns with our goal. For $b=1$, observe that $xb$ now has even bit parity! The $d'_{xb}$ equation above yields $1$, which matches our goal to show that $d'_y = 1$ for any $y \in \{0,1\}^n$ with even bit parity.

This proves that if the proposition holds for $n-1$, it holds for $n$.

Wrapping up

By induction, then, we’ve shown that $\underbrace{Z \otimes Z \otimes \cdots \otimes Z}_{n\text{ times}} = \sum_{x\in\{0,1\}^n} (-1)^{p(x)} \ket{x}\bra{x}$.

Now let’s plug this in. If $\ket{\psi}$ is an $n$-qubit state, then

\[ \begin{align} \bramket{\psi}{\underbrace{Z \otimes Z \otimes \cdots \otimes Z}_{n\text{ times}}}{\psi} &= \bramket{\psi}{\left[\sum_{x\in\{0,1\}^n} (-1)^{p(x)} \ket{x}\bra{x} \right]}{\psi} \\
&= \bramket{\psi}{\left[\sum_{x\in\{0,1\}^n} (-1)^{p(x)} M_x \right]}{\psi} \\
&= \sum_{x\in\{0,1\}^n} (-1)^{p(x)} \bramket{\psi}{M_x}{\psi} \\
&= \sum_{x\in\{0,1\}^n} (-1)^{p(x)} \bramket{\psi}{M_x^\dagger M_x}{\psi} \\
&= \sum_{x\in\{0,1\}^n} (-1)^{p(x)} P(x) \end{align} \]

Done!