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Verifying a simple QFT derivation in Peskin's

  1. Sep 9, 2008 #1
    Hiya,

    just stumbled upon this forum searching out 'Peskin errata' when trying to figure out a simple QFT calculation in the textbook. Apparently, there is no mention of the simple derivation that I'm struggling with, so there must be something wrong with my own working. I would really appreciate any help with this.

    Anyway, here's the derivation -

    basically, the I'd like to ask how (2.38) results from (2.35) and (2.37).

    [tex]|\textbf{p}\rangle = \sqrt{2E_{\textbf{p}}}a^{t}_{\textbf{p}}|0\rangle[/tex]
    (2.35)​

    [tex]U(\Lambda)|\textbf{p}\rangle = |\Lambda\textbf{p}\rangle[/tex]
    (2.37)​

    [tex]U(\Lambda)a^{t}_{\textbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}}}a^{t}_{\Lambda\textbf{p}}[/tex]
    (2.38)​

    It looks like something that you could hardly go wrong with, but I get the following instead:

    [tex]U(\Lambda)a^{t}_{\textbf{p}} = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}}}a^{t}_{\Lambda\textbf{p}}[/tex]

    Thanks in advance for any help rendered :)
     
  2. jcsd
  3. Sep 9, 2008 #2

    samalkhaiat

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  4. Sep 9, 2008 #3

    Avodyne

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    You can't actually get from (2.35) to (2.38), and P&S don't claim to. They are using the general framework of unitary transformations in QM. All QM statements (eg, commutation relations, eigenvalue equations, etc) are unchanged if every operator [itex]A[/itex] is replaced by [itex]U\!\!AU^{-1}[/itex], and every state [itex]|\psi\rangle[/itex] by [itex]U|\psi\rangle[/itex], where [itex]U[/itex] is a unitary operator.
     
  5. Sep 9, 2008 #4
    But where does [tex]U(\Lambda)a^{t}_{\textbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}} }a^{t}_{\Lambda\textbf{p}} [/tex] come from?
     
  6. Sep 9, 2008 #5

    Avodyne

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    P&S are simply postulating the existence of such a unitary operator, since symmetries in QM are, in general, implemented by unitary operators. (Find a mind-numbingly detailed exposition of this, see Weinberg volume 1.)
     
  7. Sep 10, 2008 #6

    haushofer

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  8. Sep 11, 2008 #7

    haushofer

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  9. Sep 11, 2008 #8

    Avodyne

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    The vacuum is rotation invariant and boost invariant; rotate it or boost it, and you get the same state back again. That's the meaning of [itex]U(\Lambda)|0\rangle=|0\rangle[/itex].
     
  10. Sep 12, 2008 #9

    haushofer

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    Yes, I can understand that for inertial observers. But according to the Unruh-effect an accelerating observer will measure that the vacuum |0> as observed by inertial observers will be a heath bath with a certain temperature T. So, as far as I can understand, the statement

    U|0> = |0>

    only is true for those U which consists of Lorentz transformations transforming inertial observers to inertial observers.

    Am I getting something wrong?
     
  11. Sep 12, 2008 #10

    samalkhaiat

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    The existence of a unique, Poincare' invariant vacuum and its cyclicity is one of the postulates of the (special) relativistic quantum field theory.
    I do understand your concerns regarding non-inertial observers and their choices of vacuum! However, I cannot produce a "short" and "easy" answer in here. A very readable account on the issues involved can be found in Sec. 3.3 of Birrell & Davies book "Quantum Fields in Curved Space".


    regards

    sam
     
  12. Sep 12, 2008 #11

    strangerep

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    That's correct. The Hilbert/Fock space of standard QFT is constructed to carry a
    representation of the Poincare group. If one then tries (naively) to represent a larger
    group thereon (eg a group also containing accelerations) one gets a new Fock space which
    is disjoint from the original. (Well, that's what the maths says anyway. Whether this is an
    experimentally-real physical effect remains to be seen.)
     
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