Editing Quantum logic gate (section)

==== Application on entangled states ====
If two or more qubits are viewed as a single quantum state, this combined state is equal to the tensor product of the constituent qubits. Any state that can be written as a tensor product from the constituent subsystems are called ''[[separable states]]''. On the other hand, an ''[[Quantum entanglement|entangled state]]'' is any state that cannot be tensor-factorized, or in other words: ''An entangled state can not be written as a tensor product of its constituent qubits states.'' Special care must be taken when applying gates to constituent qubits that make up entangled states.

If we have a set of ''N'' qubits that are entangled and wish to apply a quantum gate on ''M'' < ''N'' qubits in the set, we will have to extend the gate to take ''N'' qubits. This application can be done by combining the gate with an [[identity matrix]] such that their tensor product becomes a gate that act on ''N'' qubits. The identity matrix {{nowrap|(<math>I</math>)}} is a representation of the gate that maps every state to itself (i.e., does nothing at all). In a circuit diagram the identity gate or matrix will often appear as just a bare wire.

[[File:Shows_the_application_of_a_hadamard_gate_on_a_state_that_span_two_qubits.png|thumb|right|upright=1.8|The example given in the text. The Hadamard gate <math>H</math> only act on 1 qubit, but <math>|\psi\rangle</math> is an entangled quantum state that spans 2 qubits. In our example, <math>|\psi\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}</math>.]]
For example, the Hadamard gate {{nowrap|(<math>H</math>)}} acts on a single qubit, but if we feed it the first of the two qubits that constitute the [[quantum entanglement|entangled]] [[Bell state]] {{nowrap|<math>\frac{|00\rangle + |11\rangle}{\sqrt{2}}</math>,}} we cannot write that operation easily. We need to extend the Hadamard gate <math>H</math> with the identity gate <math>I</math> so that we can act on quantum states that span ''two'' qubits:

:<math>K = H \otimes I = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1\end{bmatrix}</math>

The gate <math>K</math> can now be applied to any two-qubit state, entangled or otherwise. The gate <math>K</math> will leave the second qubit untouched and apply the Hadamard transform to the first qubit. If applied to the Bell state in our example, we may write that as:

:<math>K \frac{|00\rangle + |11\rangle}{\sqrt{2}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 0 \\ 1\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ 1 \\ -1 \end{bmatrix} = \frac{|00\rangle + |01\rangle + |10\rangle - |11\rangle}{2}</math>