Are Pauli Matrices Unitary and Do Their Complex Conjugates Match?

[shounakbhatta]

Hello,

I am new to this:

Taking the first Pauli Matrix:

σ1=[0 1
1 0]

Doing the transpose it becomes:

[0 1
1 0]

So is it a unitary matrix?

Similarly
σ2= [0 -i
i 0]

Doing a transpose

=[0 i
[-i 0]

Does it mean the complex conjugates are the same?

-- Shounak

 

[tom.stoer]

A matrix M is unitary iff

$$M^\dagger M = MM^\dagger = 1$$

A matrix M is hermitian iff

$$M^\dagger = M$$

with

$$M^\dagger = {M^\ast}^t$$

So for the Pauli matrix σ² you have

$$M = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$$

$$M^\dagger = {\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}^\ast}^t = {\begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}}^t = {\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}} = M$$

$$MM^\dagger = M^2 = 1$$

The same applies to other Pauli matrices, so every a Pauli matrix σⁱ is both hermitian and unitary

 

[Fredrik]

Since Tom answered the question, I'll just make a comment about something that I find useful to know.

Note that the diagonal elements of a hermitian matrix must be real. The Pauli matrices are also traceless, i.e the sum of the diagonal elements is 0. Every complex 2×2 traceless hermitian matrix can be written in the form
$$\begin{pmatrix} x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix},$$ where the $x_i$ are real numbers, and this can clearly can also be written as $\sum_i x_i\sigma_i$. So the Pauli matrices are basis vectors for the vector space of complex 2×2 traceless hermitian matrices. If you can remember this, it's pretty easy to remember what they look like.

 

[tom.stoer]

If you add the 2*2 identity matrix you get a basis for 2*2 hermitean matrices (no longer traceless)

These matrices are related to 4-dim. Minkowski space in SR which you see immediately by calculating the determinant

 

[shounakbhatta]

Thanks for the help.

Just a small question to Tom"

(0 -i
i 0) when transposed becomes (0 i
i 0) but why again= (0 -i
i 0)?

Thank you.

-- Shounak

 

[Fredrik]

(0 -i
i 0) when transposed becomes (0 i
i 0) but why again= (0 -i
i 0)?

I don't quite understand what you're asking, but maybe this will clear it up:
$$\begin{align}
\sigma_2 &=\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}\\
\sigma_2^T &=\begin{pmatrix}0 & i\\ -i & 0\end{pmatrix}\\
\sigma_2^\dagger &= \begin{pmatrix}0^* & i^*\\ (-i)^* & 0^*\end{pmatrix} =\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}=\sigma_2
\end{align}$$

 

[shounakbhatta]

Thank you. Yes, it clears up.

I have one more question:

For a spin 1 for Pauli matrix it follows:

Jx=hbar/√2 (0 1 0
1 0 1
0 1 0)


Now the above, does it follows from below?

σ1=σx= (0 1
1 0)

But Jz=hbar( 1 0 0
0 0 0
0 i 0)

whereas σ3=σz= (1 0
0 -1)

is it somehow related to each other?

 

[tom.stoer]

the 3*3 matrices for spin 1 have partially different properties and are NOT Pauli matrices!