# Unitary Matrix

Recently, I was trying to get the hang of quantum computing. I found myself in a position where I forgot most of the linear algebra stuff I’ve learned in past semesters. So again, I decide to put them down in hope that some of the knowledge here will stay in my memory a bit longer.

## General single-qubit Gates

Trying to understand unitary matrix in the context of pure linear algebra is, I must admit, rather boring. Perhaps that is one reason why I brushed them off so quickly and so easily. However, explaining it in the context of quantum computing feels a lot more fun. Maybe it’s because I can associate a unitary matrix with a quantum gate, which is something a bit more concrete, or simply because the term ‘‘quantum computing’’ makes me sound smarter.

Speaking of something concrete, here ara two example unitary matrices: the NOT gate ($$X$$) and Hadamard gate ($$H$$):

$X =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} ;\ H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$

For example, if we take the Hadamard gate ($$H$$) and compute its adjoint $$H^{\dagger}$$:

$H^{\dagger} = \begin{pmatrix} \begin{pmatrix}\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \end{pmatrix}^T \end{pmatrix}^{*}$

We know the transpose of $$H$$ is still $$H$$, and taking the complex conjugate of $$H^T$$ doesn’t do anything since $$H^T$$ is a real matrix. Thus, we can verify that $$H^{\dagger}H = I$$.

There are other single-qubit quantum gates such as the $$Y$$ and $$Z$$ matrices (Pauli matrices) introduced by physicist Wolfgang Pauli. It’s a good exercise to verify they are also unitary matrices.

## What does it mean for a matrix to be unitary

The most important property of unitary matrices is that they preserve the length of inputs. It means that given a quantum state, represented as vector $$|\psi\rangle$$, it must be that $$\left\lVert U|\psi\rangle \rangle \right\rVert = \left\lVert |\psi\rangle \right\rVert$$.

Proving unitary matrix is length-preserving is straightforward. We wanna show that $$\left\lVert U |\psi\rangle \right\rVert_2 = \left\lVert |\psi\rangle \right\rVert_2$$:

\begin{aligned} \left\lVert U |\psi\rangle \right\rVert_2^2 &= (U |\psi\rangle)^H(U |\psi\rangle) \\ &= |\psi\rangle^H U^H U |\psi\rangle \\ &=|\psi\rangle^H |\psi\rangle \\ &= \left\lVert |\psi\rangle \right\rVert_2^2 \end{aligned}

## Why are unitaries the only matrices that preserve length

Previously, we use the ket notation for quantum state vectors. We can extend the two-dimensional quantum state vectors to more general vectors and the properties of unitary matrix will still hold.

Putting our questions in formal terms, we want to show that if $$A \in \mathbb{C}^{m \times m}$$ preserves length ($$\left\lVert A x \right\rVert_2 = \left\lVert x \right\rVert_{2}\ \forall x \in \mathbb{C}^m$$, then $$A$$ is unitary).

We first prove that $$(Ax)^H(Ay) = x^Hy$$ for all $$x$$, $$y$$ by considering that $$\left\lVert x - y \right\rVert_2^2 = \left\lVert A(x - y) \right\rVert_2^2$$. Then we will the result to evaluate $$e_i^H A^HAe_j$$.

Let $$x$$, $$y \in \mathbb{C}^m$$, then we can use the alternative definition for the matrix 2-norm (e.g. $$\left\lVert y \right\rVert_2 = y^Hy$$) for $$\left\lVert x - y \right\rVert_2^2 = \left\lVert A(x - y) \right\rVert_2^2$$,

$(x-y)^H(x-y) = (A(x-y))^HA(x-y)$

Based on that fact that the hermitian transpose rule that $$(Ax)^H = x^HA^H$$, we get

$(x-y)^H(x-y) = (x-y)^HA^HA(x-y)$

Multiplying the above formula out,

\begin{align} x^Hx - y^Hx - x^Hy + y^Hy &= x^HA^HAx - y^HA^HAx \\ &\quad - x^HA^HAy + y^HA^HAy \end{align}

The alternative definition for $$y^Hx$$ is $$\overline{x^Hy}$$, so we apply the definition here,

\begin{align} x^Hx - (\overline{x^Hy} + x^Hy) + y^Hy &= x^HA^HAx - (\overline{x^HA^HAy} + x^HA^HAy) \\ &\quad + y^HA^HAy \end{align}

We know that $$A$$ preserves length, and that $$\frac{\alpha + \overline{\alpha}}{2} = Re(\alpha)$$. so we can simplify the above formula as:

$Re(x^Hy) = Re((Ax)^H(Ay))$

We know that $$A$$ preserves length, and thus we need to show that $$A^HA = I$$ by using the fact that the standard basis vectors have the property that

$\begin{equation} e_i^H e_j = \begin{cases} 1 & \text{if $$i = j$$}\\ 0 & \text{otherwise} \end{cases} \end{equation}$

Therefore, $$e_i M e_j$$ will essentially extract the $$i,\ j$$th entry in matrix $$M$$. So we know that

$e_i A^HA e_i = \left\lVert Ae_i \right\rVert^2 = \left\lVert e_i \right\rVert^2 = 1$

We can conclude that all the diagonal elements of $$A^HA$$ are $$1$$.

A side question remains, how do we prove that all the off-diagonal elements in $$A^HA$$ are $$0$$? Turns out it very straightforward to illustrate the process if we resort back to the two-dimensional quantum vector state matrix.

Suppose we have $$|\psi\rangle = |e_i\rangle + |e_j\rangle$$, we already know that $$\left\lVert A |\psi\rangle \right\rVert^2 = \left\lVert |\psi\rangle \right\rVert^2 = 1 + 1 = 2$$, and we know we can expand $$\left\lVert A |\psi\rangle \right\rVert^2$$ to $$1 + e_i A^HA e_j + e_j A^HA e_i + 1$$, we would get $$e_i A^HA e_j + e_j A^HA e_i = 0$$.

Then, suppose instead we have $$|\psi\rangle = |e_i\rangle + i|e_j\rangle$$, following the same process, we would get $$e_i A^HA e_j - e_j A^HA e_i = 0$$. Combining with the fact that $$e_i A^HA e_j + e_j A^HA e_i = 0$$, we’ve proven that the off-diagonal elements in $$A^HA$$ are all $$0$$. We can extend the vector $$\psi$$ to higher-dimensional vectors and the proof will be similar.