# Unitary matrices – some examples

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Exercises 1.6.3 – 1.6.6.

Here are a few more results about unitary operators.

Shankar defines a unitary operator ${U}$ as one where

$\displaystyle UU^{\dagger}=I \ \ \ \ \ (1)$

From this we can derive the other condition by which they can be defined, namely that a unitary operator preserves the norm of a vector:

$\displaystyle \left|Uv\right|=\left|v\right| \ \ \ \ \ (2)$

This follows, for if we define the effect of ${U}$ by

$\displaystyle \left|v_{1}^{\prime}\right\rangle =U\left|v_{1}\right\rangle \ \ \ \ \ (3)$

then

 $\displaystyle \left\langle v_{1}^{\prime}\left|v_{1}^{\prime}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle Uv_{1}\left|Uv_{1}\right.\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle v_{1}\left|U^{\dagger}Uv_{1}\right.\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle v_{1}\left|v_{1}\right.\right\rangle \ \ \ \ \ (6)$

Thus ${\left|v_{1}^{\prime}\right|^{2}=\left|v_{1}\right|^{2}}$.

Theorem 1 The product of two unitary operators ${U_{1}}$ and ${U_{2}}$ is unitary.

Proof: Using Shankar’s definition 1, we have

 $\displaystyle \left(U_{1}U_{2}\right)^{\dagger}U_{1}U_{2}$ $\displaystyle =$ $\displaystyle U_{2}^{\dagger}U_{1}^{\dagger}U_{1}U_{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U_{2}^{\dagger}IU_{2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U_{2}^{\dagger}U_{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I \ \ \ \ \ (10)$

$\Box$

Theorem 2 The determinant of a unitary matrix ${U}$ is a complex number with unit modulus.

Proof: The determinant of a hermitian conjugate is the complex conjugate of the determinant of the original matrix, since ${\det U=\det U^{T}}$ (where the superscript ${T}$ denotes the transpose) for any matrix, and the hermitian conjugate is the complex conjugate transpose. Therefore

$\displaystyle \det\left(UU^{\dagger}\right)=\left[\det U\right]\left[\det U\right]^*=\det I=1 \ \ \ \ \ (11)$

Therefore ${\left|\det U\right|^{2}=1}$ as required.$\Box$

Example 1 The rotation matrix ${R\left(\frac{\pi}{2}\mathbf{i}\right)}$ is unitary. We have

$\displaystyle R\left(\frac{\pi}{2}\mathbf{i}\right)=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (12)$

By direct calculation

 $\displaystyle RR^{\dagger}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{array}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]=I \ \ \ \ \ (14)$

Example 2 Consider the matrix

$\displaystyle U=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & i\\ i & 1 \end{array}\right] \ \ \ \ \ (15)$

By calculating

 $\displaystyle UU^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cc} 1 & i\\ i & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -i\\ -i & 1 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cc} 2 & 0\\ 0 & 2 \end{array}\right]=I \ \ \ \ \ (17)$

Thus ${U}$ is unitary, but because ${U\ne U^{\dagger}}$ it is not hermitian. Its determinant is

$\displaystyle \det U=\left(\frac{1}{\sqrt{2}}\right)^{2}\left(1-i^{2}\right)=1 \ \ \ \ \ (18)$

This is of the required form ${e^{i\theta}}$ with ${\theta=0}$.

Example 3 Consider the matrix

 $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cc} 1+i & 1-i\\ 1-i & 1+i \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle UU^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left[\begin{array}{cc} 1+i & 1-i\\ 1-i & 1+i \end{array}\right]\left[\begin{array}{cc} 1-i & 1+i\\ 1+i & 1-i \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left[\begin{array}{cc} 4 & 0\\ 0 & 4 \end{array}\right]=I \ \ \ \ \ (21)$

Thus ${U}$ is unitary, but because ${U\ne U^{\dagger}}$ it is not hermitian. Its determinant is

 $\displaystyle \det U$ $\displaystyle =$ $\displaystyle \left(\frac{1}{2}\right)^{2}\left[\left(1+i\right)^{2}-\left(1-i\right)^{2}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i \ \ \ \ \ (23)$

This is of the required form ${e^{i\theta}}$ with ${\theta=\frac{\pi}{2}}$.