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

Commutator of charges in QFT

  1. May 4, 2009 #1

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Consider the SUSY charge

    [tex] Q= \int d^3y~ \sigma^\mu \chi~ ~\partial_\mu \phi^\dagger~ [/tex]

    The SUSY transformation of fields, let's say of the scalar field, can be found using the commutator

    [tex] i [ \epsilon \cdot Q, \phi(x)] = \delta \phi(x) [/tex]

    using the equal time commutator

    [tex] [\phi(\vec{x},t), \dot{\phi}^\dagger(\vec{y},t) ] = \delta^3(\vec{x} - \vec{y}) [/tex]


    Everything works fine at the condition of assuming that the time in [tex] \phi^\dagger(y) [/tex] is equal to the time in in [tex] \phi(x) [/tex] .

    However, I don't see why we need to assume this. The charge is time independent so we should be able to use whatever time we like to calculate the commutator. But if we pick a different time than the time of [tex] \phi(x) [/tex], we get zero for the transformation of the scalar field, which is incorrect.

    So why do we need to set the two times equal?

    Thanks in advance,


    Patrick
     
  2. jcsd
  3. May 5, 2009 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    The reason for it is called the equal time commutator is simply that it is defined to be like that. At least that is the impression I have by reading 7 introductory books on QFT. Nothing deeper than "in the spirit of non Rel QM"...

    But the reason for choosing equal time then is that one includes time dependence by Heisenberg picture, i.e we start with time INDEP. fields, and then have the commutator:
    [tex]
    [\phi(\vec{x}), \dot{\phi}^\dagger(\vec{y}) ] = \delta^3(\vec{x} - \vec{y})
    [/tex]

    Then we add the time dependence of the fields, which are operators in QFT, by Heisenberg picture.

    See e.g. page 41 in srednicki's textbook

    http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf
     
  4. May 5, 2009 #3

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi Malawi Glenn,

    thanks for your reply.

    Sorry if my question was not clear. I was not asking about the equal time commutation relation itself but about using a charge to generate the transformation of a field. When we use the commutator of a charge with a field to generate the transformation of that field, we must assume that the time in the fields of thecharge must be equal to the time of the field that we are varying. I wonder why this is the case.

    Thanks!
     
  5. May 5, 2009 #4

    StatusX

    User Avatar
    Homework Helper

    It's not strictly necessary to use the charge in terms of fields at the same time, but it is easier. The commutation relations between fields at different times can be worked out (and are not zero), and using them here would give the same answer, but that way is more complicated.
     
  6. May 5, 2009 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Interesting! Thanks for clarifying this. It seemed to me at first sight that it would give zero. It does not give zero because there are derivatives acting on the fields?
     
  7. May 5, 2009 #6

    StatusX

    User Avatar
    Homework Helper

    It doesn't give zero when the times are different because it doesn't give zero when they're the same, and these two commutators are related by a simple differential equation. Namely, with a little work you can show that this commutator is basically the Green's function for whatever equations of motion the fields satisfy. One way to compute this for the free field is show in equation 4.11 in Srednicki.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutator of charges in QFT
Loading...