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Simple matrix algebra

  1. Jul 24, 2007 #1
    1. The problem statement, all variables and given/known data
    This is from Ryder's QFT book, second ed. page 37. At the bottom of the page it says that the commutation relations (eqn 2.68?) are satisfied by:
    [itex]K = \pm i\frac{\sigma}{2}[/itex]
    However, I do not find this to be so. What am I missing?

    2. Relevant equations
    Here is one of the commutation relations that I think he means.
    [itex][K_x,K_y] = -iJ_z[/itex]

    3. The attempt at a solution
    Using [itex]K = i\frac{\sigma}{2}[/itex], I get:
    [itex][K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = \frac{-1}{4}[\sigma_x,\sigma_y] = -\frac{1}{2}\sigma_z = iK_z \neq -iJ_z[/itex]
     
    Last edited: Jul 24, 2007
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  3. Jul 24, 2007 #2

    Dick

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    Isn't there an i in the commutation relations of the Pauli matrices as well?
     
  4. Jul 24, 2007 #3

    olgranpappy

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    yep,
    [tex]
    \left[\sigma_j,\sigma_k\right]=2i\epsilon_{jkl}\sigma_l
    [/tex]
     
  5. Jul 24, 2007 #4
    Thanks Dick. Here is the corrected attempt. I still don't get the right commutation relation.

    [itex][K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = \frac{-1}{4}[\sigma_x,\sigma_y] = -i\frac{\sigma_z}{2} = -K_z \neq -iJ_z[/itex]
     
  6. Jul 24, 2007 #5

    olgranpappy

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    yes, you do get the right relation. Ryder is talking about (2-component) Pauli spinors for which [tex]J_z=\frac{\sigma_z}{2}[/tex].

    Look at equation (2.74). That is a boost and a rotation of a 2-component spinor where the rotation generator is
    [tex]\frac{\vec \sigma}{2}[/tex] and the boost generator is [tex]i\frac{\vec \sigma}{2}[/tex].
     
  7. Jul 25, 2007 #6
    Thanks olgranpappy, your reply is what I needed. If I make the substitutions [tex]K = i\frac{\sigma}{2}[/tex] and [tex]J = \frac{\sigma}{2}[/tex], then I get:

    [tex][K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = -i\frac{\sigma_z}{2} = -iJ_z[/tex] just as in (2.68)

    I have also verified the other relations in (2.68). I came close to solving it this morning as I was driving to work. It occured to me that there might be a typo in the book and that the author meant J instead of K in eqn (2.69). If I had followed that thought a while longer, I might have come up with the solution on my own. Thanks again for your help.
     
  8. Jul 25, 2007 #7

    olgranpappy

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    no problem. cheers.
     
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