All 2x2 Hermitian and Unitary Matrices (Check My Proof)

[RJLiberator]

Homework Statement

Find all 2x2 Matrices which are both hermitian and unitary.

Homework Equations

Conditions for Matrix A:
A=A^†
A^†A=I
I = the identity matrix
† = hermitian conjugate

The Attempt at a Solution

1. We see by the conditions that A^† = A and by the second condition, we see that AA=I. So we write:

(\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\begin{pmatrix}
a & b\\
c & d
\end{pmatrix})= \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}

Which sets us up with:
2. Solve
\begin{pmatrix}
a^2+bc & ab+bd\\
ca+dc & cb+d^2
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}

3. So our system:
cb+d^2=1
a^2+bc=1
ab+bd=0
ca+dc=0

4. In solving, we see that a= +/1, b=+/-1 and c,d = 0.

So we get 4 total matrices that are diagonal and all possible values +/-1.
Is this correct? Am I forgetting anything? Any complex elements am I forgetting to check?

 

[fzero]

3. So our system:
cb+d^2=1
a^2+bc=1
ab+bd=0
ca+dc=0

4. In solving, we see that a= +/1, b=+/-1 and c,d = 0.

The 3rd and 4th equations require $d=-a$ if either $b$ or $c$ is nonzero, so there are more solutions, which include imaginary components .

 

[RJLiberator]

Damn, I was afraid of that.

Well, if d=-a. Then bc=1.

There seems to be a lot of solutions...

I'll have to write the general answer down then.

a=-d, and bc=1

 

[fzero]

Damn, I was afraid of that.

Well, if d=-a. Then bc=1.

There seems to be a lot of solutions...

I'll have to write the general answer down then.

a=-d, and bc=1

Since the matrix is Hermitian, $c =\bar{b}$, and $a$ is real, so you should have a condition $a^2 + |b|^2 =1$. You can solve this with trig functions.