Editing Bell state (section)

=== Creating Bell states via quantum circuits ===
[[File:The Hadamard-CNOT transform on the zero-state.png|thumb|right|400px|Quantum circuit to create Bell state <math>|\Phi^+\rangle</math>.]]Although there are many possible ways to create entangled Bell states through [[Quantum circuit|quantum circuits]], the simplest takes a computational basis as the input, and contains a [[Hadamard gate]] and a [[Cnot gate|CNOT gate]] (see picture). As an example, the pictured quantum circuit takes the two qubit input <math>|00\rangle</math> and transforms it to the first Bell state <math>|\Phi^+\rangle.</math> Explicitly, the Hadamard gate transforms <math>|00\rangle</math> into a [[Quantum superposition|superposition]] of <math>(|0\rangle|0\rangle + |1\rangle|0\rangle) \over \sqrt{2}</math>. This will then act as a control input to the CNOT gate, which only inverts the target (the second qubit) when the control (the first qubit) is 1. Thus, the CNOT gate transforms the second qubit as follows <math>\frac{(|00\rangle + |11\rangle)}{\sqrt{2} } = |\Phi^+\rangle</math>.

For the four basic two-qubit inputs, <math>|00\rangle, |01\rangle, |10\rangle, |11\rangle</math>, the circuit outputs the four Bell states ([[#Bell basis|listed above]]). More generally, the circuit transforms the input in accordance with the equation

<math display="block">|\beta(x,y)\rangle = \left ( \frac{|0,y\rangle + (-1)^x|1,\bar{y}\rangle}{\sqrt{2}} \right ),</math>

where <math>\bar{y}</math> is the negation of <math>y</math>.<ref name="Nielsen-2010" />