Pages

Saturday, 30 November 2013

Constructing an AND gate as a quantum operation

Here we consider operations that map the two-bit state $\ket{a,b}$ to the one-qubit state $\ket{a\wedge b}$, for all $a,b\in\{0,1\}$, where $a\wedge b$ represents the binary logical AND operation of $a$ and $b$. We will construct  $k$ matrices $A_1,\dots, A_k$ (where $k\leq 4$, and each $A_j$ is a $2 \times 4$ matrix) such that $\sum_
 {j=1}^{k}A^\dagger_jA_j=I$, and whose quantum operation $\Phi$ computes the above mapping. In other words, for all $a,b\in \{0,1\}$, when $\rho=\ket{a,b}\bra{a,b}$,
\[
 \Phi(\rho):=\SUM{j=1}{k}A_j\rho A^\dagger_j=\ket{a\wedge b}\bra{a\wedge b}.
\]

 Define the following matrices:
\[
 A_1=\begin{pmatrix} 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}, \
 A_2=\begin{pmatrix} 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}, \
 A_3=\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \end{pmatrix}, \
 A_4=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \end{pmatrix},
\]
 and observe that
\[ \begin{align*}
 A^\dagger_1A_1=\begin{pmatrix} 1&0&0&0\\0&0&0&0\\0&0&0&0\\
0&0&0&0 \end{pmatrix} , \ &  A^\dagger_2A_2=\begin{pmatrix} 0&0&0&0\\0&1&0&0\\0&0&0&0\\
0&0&0&0 \end{pmatrix} \\
 A^\dagger_3A_3=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&1&0\\
0&0&0&0 \end{pmatrix} , \ &  A^\dagger_4A_4=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&0\\
0&0&0&1 \end{pmatrix}.
 \end{align*}\]
 Therefore $\sum_{j=1}^{4}A^\dagger_j A_j=I$ is the $4\times 4$ identity matrix, which is necessary to define a valid channel in the Krauss representation

 Now consider the four computational basis states $\ket{a,b}$, for $a,b\in\{0,1\}$, with corresponding density operators given by $\rho_i:=\ket{a,b}\bra{a,b}$ where the index can be represented in terms of $a$ and $b$ as $i=2a+b+1$. More explicitly,

 \[ \begin{align*}
\rho_1:=\ket{0,0}\bra{0,0}=\begin{pmatrix} 1&0&0&0\\0&0&0&0\\0&0&0&0\\
0&0&0&0 \end{pmatrix} , \ &  \rho_2:=\ket{0,1}\bra{0,1}=\begin{pmatrix} 0&0&0&0\\0&1&0&0\\0&0&0&0\\
0&0&0&0 \end{pmatrix} \\
\rho_3:=\ket{1,0}\bra{1,0}=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&1&0\\
0&0&0&0 \end{pmatrix} , \ &  \rho_4:=\ket{1,1}\bra{1,1}=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&0\\
0&0&0&1 \end{pmatrix}.
 \end{align*}\]

 It then follows that,
\[
 A^\dagger_j \rho_i A_j=\left\{
 \begin{array}{rl} \begin{pmatrix}0 &0 \\
 0 &0 \end{pmatrix}  \ \ &\text{if} \ \ i\neq j,  \\
 \\
\begin{pmatrix}1 &0 \\
 0 &0 \end{pmatrix}  \ \ &\text{if} \ \ i= j<4, \\
 \\
\begin{pmatrix}0 &0 \\
 0 &1 \end{pmatrix}  \ \ &
 \text{if} \ \ i= j=4.
 \end{array}
 \right.
\]

 Therefore,
\[
 \Phi(\rho_i)=\SUM{j=1}{k}A_j\rho_i A^\dagger_j\left\{
 \begin{array}{rl}
\begin{pmatrix}1 &0 \\
 0 &0 \end{pmatrix}  \ \ &\text{if} \ \ i<4, \\
 \\
 \begin{pmatrix}0 &0 \\
 0 &1 \end{pmatrix}  \ \ &
 \text{if} \ \ i=4.
  \end{array}
 \right.
\]

Now, for $a,b\in\{0,1\}$ the logical AND operation is such that $a\wedge b=1$ if and only if $a=b=1$, and $a\wedge b=0$ otherwise. Moreover, 
\[
\ket{a\wedge b}\bra{a\wedge b}=\left\{
\begin{array}{rl}
\ket{0}\bra{0}=\begin{pmatrix}1 &0 \\
 0 &0 \end{pmatrix}  \ \ &\text{if} \ \ a\wedge b=0, \\
 \\
  \ket{1}\bra{1}=\begin{pmatrix}0 &0 \\
 0 &1 \end{pmatrix}  \ \ &\text{if} \ \ a\wedge b=1.
 \end{array}
 \right.
\]

Hence, the quantum operation defined in this manner does indeed compute the logical AND operation
\[
  \Phi(\rho_i)=\SUM{j=1}{k}A_j\rho_i A^\dagger_j=\ket{a\wedge b}\bra{a\wedge b},
\]
 where, as originally defined $\rho_i:=\ket{a,b}\bra{a,b}$ with $i=2a+b+1$ for $a,b\in\{0,1\}$.
 
This operation maps all basis states to pure states, but does it map all pure input states to pure output states?

 Consider the pure state $\ket{\psi}:=\frac{1}{\sqrt{2}}(\ket{0,0}+\ket{1,1})$, with density matrix given by
\[ \begin{align*}
 \rho:=\ket{\psi}\bra{\psi}&=\frac{1}{2}(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})\\
 &=\frac{1}{2}(\ket{0,0}\bra{0,0}+\ket{0,0}\bra{1,1}+\ket{1,1}\bra{0,0}+\ket{1,1}\bra{1,1}) \\
 &=\frac{1}{2}\begin{pmatrix}
 1&0&0&1 \\
 0&0&0&0 \\
 0&0&0&0 \\
 1&0&0&1
 \end{pmatrix}.
 \end{align*} \]

Recall that $\Phi(\ket{0,0}\bra{0,0})=\ket{0}\bra{0}$ and $\Phi(\ket{1,1}\bra{1,1})=\ket{1}\bra{1}$. Moreover it can be seen that $\Phi(\ket{0,0}\bra{1,1})=\Phi(\ket{1,1}\bra{0,0})=\mathbf{0}_2$ (where $\mathbf{0}_2$ is the $2\times 2$ zero matrix) since $\Phi$ only acts nontrivially on  a matrix which has some nontrivial coefficient along its main diagonal.That is, for $a,b,c,d\in\{0,1\}$, $\Phi(\ket{a,b}\bra{c,d})\neq \mathbf{0}_2$ only if $a=c$ and $b=d$. Then by the linearity of the quantum operation $\Phi$,
\[ \begin{align*}
 \Phi(\rho)&=\frac{1}{2}(\Phi(\ket{0,0}\bra{0,0})+\Phi(\ket{0,0}\bra{1,1})+\Phi(\ket{1,1}\bra{0,0})+\Phi(\ket{1,1}\bra{1,1}))\\
 &=\frac{1}{2}(\ket{0}\bra{0}+\ket{1}\bra{1}) \\
 &=\frac{1}{2}\begin{pmatrix}1 &0 \\
 0 &0 \end{pmatrix} + \frac{1}{2}\begin{pmatrix}0 &0 \\
 0 &1 \end{pmatrix} \\
 &=\frac{1}{2}\begin{pmatrix}1 &0 \\
 0 &1 \end{pmatrix}
 \end{align*}\]
 
  Thus, since $\Phi(\rho)$ in this case is a mixed state, but $\rho$ was a pure state, the quantum operation $\Phi$ does not always take pure input states to pure output states.

No comments :

Post a Comment