Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pauli Matrices

  1. Oct 22, 2012 #1
    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
     
  2. jcsd
  3. Oct 23, 2012 #2

    tom.stoer

    User Avatar
    Science Advisor

    A matrix M is unitary iff

    [tex]M^\dagger M = MM^\dagger = 1[/tex]

    A matrix M is hermitian iff

    [tex]M^\dagger = M[/tex]

    with

    [tex]M^\dagger = {M^\ast}^t[/tex]

    So for the Pauli matrix σ² you have

    [tex]M = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} [/tex]

    [tex]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 [/tex]

    [tex]MM^\dagger = M^2 = 1[/tex]

    The same applies to other Pauli matrices, so every a Pauli matrix σi is both hermitian and unitary
     
  4. Oct 23, 2012 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
    Last edited: Oct 23, 2012
  5. Oct 23, 2012 #4

    tom.stoer

    User Avatar
    Science Advisor

    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
     
  6. Oct 23, 2012 #5
    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
     
  7. Oct 23, 2012 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't quite understand what you're asking, but maybe this will clear it up:
    $$\begin{align}
    \sigma_2 &=\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}\\
    \sigma_2^T &=\begin{pmatrix}0 & i\\ -i & 0\end{pmatrix}\\
    \sigma_2^\dagger &= \begin{pmatrix}0^* & i^*\\ (-i)^* & 0^*\end{pmatrix} =\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}=\sigma_2
    \end{align}$$
     
  8. Oct 23, 2012 #7
    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?
     
  9. Oct 23, 2012 #8

    tom.stoer

    User Avatar
    Science Advisor

    the 3*3 matrices for spin 1 have partially different properties and are NOT Pauli matrices!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pauli Matrices
  1. Pauli matrices (Replies: 7)

  2. Pauli matrices (Replies: 10)

  3. Pauli matrices (Replies: 3)

  4. Pauli Matrices (Replies: 6)

  5. Pauli matrices? (Replies: 1)

Loading...