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}}.

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