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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 \(c-X\).)

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