Measurement
A projective measurement is a measurement characterized by a set $M=\{P_i\}$ of $n$ projectors $P_i$ which satisfy the completeness relation $\sum_iP_i=1$. Each projector $P_i$ is associated with an outcome $i$ where $i\in S=\{1,...,n\}\subset R$. This measurement $M=\{P_i\}$ is naturally associated with the actual measurement of a physical observable $O$ which is characterized by eigenvalues equal $i$ with eigenspaces defined by the projectors $P_i$. In other words, the observable $O$ can be rewritten in the form
\begin{eqnarray}
O=\sum_i i P_i.
\end{eqnarray}
The expected value $\langle O\rangle$ of the physical observable $O$ in a state of the system given by a density matrix $\rho$ is given by the usal formula
\begin{eqnarray}
\langle O\rangle=Tr \rho O.
\end{eqnarray}
The expected value is the value found on average in the measurement $M=\{P_i\} (or more precisely it is the average of the values found in many identical measurements).
Spin and Qubit
We start by stating the eigenvalues and the eigenvectors associated with the measurement of the angular momentum operator of a spin $1/2$ particle in the rest frame in an arbitrary direction $\hat{n}$ given by
\begin{eqnarray}
\hat{n}=\sin\theta\cos\phi \hat{i}+\sin\theta\sin\phi \hat{j}+\cos\theta \hat{k}.
\end{eqnarray}
The eigenvalues of the measurement of the spin component $S_n=\hat{S}.\hat{n}=\sin\theta\cos\phi S_x+\sin\theta\sin\phi S_y+\cos\theta S_z$ are given by $s_n=\pm 1/2$ with corresponding eigenvectors given respectively by
\begin{eqnarray}
|\hat{n}+\rangle=\cos\frac{\theta}{2}e^{-i\frac{\phi}{2}}|+\rangle+ \sin\frac{\theta}{2}e^{i\frac{\phi}{2}}|-\rangle. \label{def01}
\end{eqnarray}
\begin{eqnarray}
|\hat{n}-\rangle=-\sin\frac{\theta}{2}e^{-i\frac{\phi}{2}}|+\rangle+ \cos\frac{\theta}{2}e^{i\frac{\phi}{2}}|-\rangle. \label{def02}
\end{eqnarray}
We have then (we have obviously set $\hbar=1$)
\begin{eqnarray}
S_n |\hat{n}\pm\rangle=\pm\frac{1}{2} |\hat{n}\pm\rangle~,~\vec{S}^2 |\hat{n}\pm\rangle=\frac{1}{2}(\frac{1}{2}+1) |\hat{n}\pm\rangle.
\end{eqnarray}
As our first example we consider measurement of the spin in the $z-$direction, i.e. $\hat {n}=\hat{k}$ or equivalently $\theta=0$ and $\phi=0$. This measurement is defined by the set of projectors $M=\{|+\rangle\langle +|, |-\rangle\langle -|\}$ corresponding to the two outcomes: A spin up $s_z=+1/2$ with eigenvector $|+\rangle$ and spin down $s_z=-1/2$ with eigenvector $|-\rangle$. The spin observable $S_z$ is given explicitly by
\begin{eqnarray}
S_z=\frac{1}{2}|+\rangle\langle +|-\frac{1}{2}|-\rangle\langle -|.
\end{eqnarray}
Besides the physical basis $\{|+\rangle, |-\rangle\}$ we introduce the computational basis $\{|0\rangle, |1\rangle\}$ (where $|0\rangle$ denotes the quantum bit or qubit $0$ and $|1\rangle$ denotes the qubit $1$) by the relations
\begin{eqnarray}
|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)~,~|-\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle).\label{def1}
\end{eqnarray}
Or equivalently
\begin{eqnarray}
|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)~,~|1\rangle=\frac{1}{\sqrt{2}}(|+\rangle-|-\rangle).\label{def2}
\end{eqnarray}
The spin observable $S_z$ in the computational basis takes the form
\begin{eqnarray}
S_z=\frac{1}{2}(|0\rangle\langle 1|+|1\rangle\langle 0|)=\frac{1}{2}X.
\end{eqnarray}
The $X$ matrix defines our first example of quantum gates. Indeed, $X$ is the quantum NOT gate. It is given explicitly by the matrix
\begin{equation}
X=|0\rangle\langle 1|+|1\rangle\langle 0|=
\left( \begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right).
\end{equation}
From the definitions (\ref{def1}) and (\ref{def2}) it is seen that the computational basis vectors $|0\rangle$ and $|1\rangle$ are obtained from the physical basis vectors $|+\rangle$ and $|-\rangle$ by a complex reflection (not a complex rotation but still a unitary transformation) about the line $\theta=\pi/8$ where the $|+\rangle-$axis is transformed to the $45-$degree line (the $|0\rangle-$line) whereas the $|-\rangle-$axis is transformed to the $-45-$degree line (the $1\rangle-$line).
We remark that if we substitute $\theta=\pi/2$ and $\phi=0$ in (\ref{def01}) and (\ref{def02}) and then compare with (\ref{def2}) we obtain immediately
\begin{eqnarray}
|+\rangle_x=|0\rangle~,~|-\rangle_x=-|1\rangle.
\end{eqnarray}
This means that the measurement of the spin component along the $x-$direction is effectively equivalent to (up to an irrelevant phase factor) the measurement in the computational basis $\{|0\rangle, |1\rangle\}$ with outcomes given precisely by the usual values $s_x=\pm 1/2$.
The measurement in the computational basis given by $M=\{|0\rangle\langle 0|, |1\rangle\langle 1|\}$ corresponds therefore to another quantum gate (the so-called $Z$ gate) given explicitly by
\begin{equation}
Z=|0\rangle\langle 0|-|1\rangle\langle 1|=
\left( \begin{array}{cc}
1 & 0\\
0 &-1
\end{array}\right).
\end{equation}
Let us point out that observables are hermitian operators whereas quantum gates are unitary transformations acting on operators which are two quite different things in general. It just happens that the $X$ and $Z$ gates are at the same time measurable observables (hermitian operators) and unitary matrices.
Another example is the Hadamard gate which is perhaps the single most important quantum gate. It is the unitary transformation which takes the physical basis $\{|+\rangle, |-\rangle\}$ to the computational basis $\{|0\rangle, |1\rangle\}$ or vice versa. Thus it is given by
\begin{eqnarray}
H&=&\frac{1}{\sqrt{2}}
\left( \begin{array}{cc}
1 & 1\\
1 &-1
\end{array}\right)\nonumber\\
&=& \frac{1}{\sqrt{2}}(|+\rangle\langle +|+|+\rangle\langle -|+|-\rangle\langle +|-|-\rangle\langle -|)\nonumber\\
&=& \frac{1}{\sqrt{2}}(|0\rangle\langle 0|+|0\rangle\langle 1|+|1\rangle\langle 0|-|1\rangle\langle 1|).
\end{eqnarray}
The Hadamard gate is also an observable with eigenvalues $\pm 1$ and corresponding eigenvectors given by
\begin{eqnarray}
|+\rangle_H=\cos\frac{\pi}{8}|+\rangle+\sin\frac{\pi}{8}|-\rangle.
\end{eqnarray}
\begin{eqnarray}
|-\rangle_H=-\sin\frac{\pi}{8}|+\rangle+\cos\frac{\pi}{8}|-\rangle.
\end{eqnarray}
So, the Hadamard gate corresponds to the measurement of the spin component along the $45-$degree line in the $\phi=0-$ plane (the $x-z$ plane).
Another observable of importance to us here is the conjugated Hadamard gate $H^{\prime}=HZ$ given explicitly by
\begin{eqnarray}
H^{\prime}&=&\frac{1}{\sqrt{2}}
\left( \begin{array}{cc}
1 & -1\\
1 &1
\end{array}\right)\nonumber\\
&=& \frac{1}{\sqrt{2}}(|+\rangle\langle +|-|+\rangle\langle -|+|-\rangle\langle +|+|-\rangle\langle -|)\nonumber\\
&=& \frac{1}{\sqrt{2}}(|0\rangle\langle 0|-|0\rangle\langle 1|+|1\rangle\langle 0|+|1\rangle\langle 1|).
\end{eqnarray}
The conjugated Hadamard gate is also an observable with eigenvalues $\pm 1$ and corresponding eigenvectors given by
\begin{eqnarray}
|+\rangle_{H^{'}}=\cos\frac{\pi}{8}|+\rangle-\sin\frac{\pi}{8}|-\rangle.
\end{eqnarray}
\begin{eqnarray}
|-\rangle_{H^{\prime}}=\sin\frac{\pi}{8}|+\rangle+\cos\frac{\pi}{8}|-\rangle.
\end{eqnarray}
So, the conjugated Hadamard gate corresponds to the measurement of the spin component along the $45-$degree line in the $\phi=\pi-$plane.
We have now $4$ operators of interest given by
\begin{eqnarray}
A_0=Z~,~A_1=X~,~B_0=H~,~B_1=H^{\prime}.
\end{eqnarray}
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