Quantum Country exercise 3

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 3. Consider a quantum circuit in which first the Hadamard gate H is applied to a quantum state \vert \psi \rangle and then the X gate is applied to the output of the first gate. Explain why the output from this circuit is XH \vert \psi \rangle and not

Answer: Applying the gate H to the quantum state \vert \psi \rangle is equivalent to multiplying the state vector \begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix} (where \psi_1 and \psi_2 are complex values) by the 2 by 2 matrix \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.

By the rules of matrix multiplication this multiplication occurs from the left as H \vert \psi \rangle and produces a two element column vector as a result, representing the output quantum state \vert \psi' \rangle.

Applying the second gate X to that result again requires multiplying that two-element column vector by the matrix \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} from the left as X \vert \psi' \rangle. That produces another two-element column vector representing the final quantum state \vert \psi'' \rangle output from the quantum circuit.

We thus have

\vert \psi'' \rangle = X \vert \psi' \rangle = X (H \vert \psi \rangle) = XH \vert \psi \rangle

In contrast, the expression HX \vert \psi \rangle amounts to first applying the gate X to \vert \psi \rangle and then applying the Hadamard gate H afterward, the opposite of the given circuit.

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