## Sunday, 8 September 2013

### Bell States

Vector spaces generally admit many choices for a suitable basis to represent its elements. For a Hilbert space describing a register of qubits $\mathcal{H}= \bigotimes_{i=1}^n\mathcal{H}^2$  the  canonical choice of basis is the computational basis given by the states expressed in the decimal representation as $\left|0\right>, \left|1\right>, . . . ,\left|N-1\right>$, where $N=2^n$, or equivalently, in the binary representation as states of the form $\left|b_1b_2...b_n\right>$ for $b_i\in\{0,1\}^n$.

A particularly important set of basis states for the Hilbert space $\mathcal{H}^2\otimes\mathcal{H}^2$ of two qubits consists of the $4$ Bell states. The Bell states can be defined in terms of the computational basis states $\left|0\right>$ and $\left|1\right>$ as
$\left| \beta_{xy}\right>\equiv \frac{1}{\sqrt2}(\left|0y\right>+(-1)^x\left|1\overline y\right>),$
where here $\overline y=(y+1)mod2$ denotes the NOT operation which simply flips the state of the qubit to the opposite value. Therefore, this change of basis maps the computational basis as follows
$\begin{array}{r l} \left|00\right> &\mapsto \frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)=\left| \beta_{00}\right>, \\ \left|01\right> &\mapsto \frac{1}{\sqrt2}(\left|01\right>+\left|10\right>)=\left| \beta_{01}\right>, \\ \left|10\right> &\mapsto \frac{1}{\sqrt2}(\left|00\right>-\left|11\right>)=\left| \beta_{10}\right>, \\ \left|11\right> &\mapsto \frac{1}{\sqrt2}(\left|01\right>-\left|10\right>)=\left| \beta_{11}\right> \end{array}$
An explicit circuit implementing this transformation of basis states using a single Hadamard gate followed by a controlled$-NOT$ gate, or $c_X$, is given in the figure shown below

(A circuit for creating Bell states. A change of basis from the computational basis states $\left|ij\right>$ to the Bell basis $\left|\beta_{ij}\right>$ is accomplished using a single Hadamard gate $H$ followed by a controlled$-NOT$ gate $c-X$.)