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

Lorentz Transformation and Creation Operators

  1. Jun 7, 2009 #1
    Hi

    Suppose [itex]\Lambda[/itex] is a Lorentz transformation with the associated Hilbert space unitary operator denoted by [itex]U(\Lambda)[/itex]. We have

    [tex]U(\Lambda)|p\rangle = |\Lambda p\rangle[/tex]

    and

    [tex]|p\rangle = \sqrt{2E_{p}}a_{p}^{\dagger}|0\rangle[/tex]

    Equivalently,

    [tex]U(\Lambda)|p\rangle = U(\lambda)\sqrt{2E_{p}}a_{p}^{\dagger}|0\rangle[/tex]

    Now, by definition,

    [tex]|\Lambda p\rangle = \sqrt{2E_{\Lambda p}}a_{\Lambda p}^{\dagger}|0\rangle[/tex]

    Therefore it follows that

    [tex]\sqrt{2E_{p}}U(\Lambda) a_{p}^{\dagger}|0\rangle = \sqrt{2E_{\Lambda p}} a_{\Lambda p}^{\dagger}|0\rangle[/tex]

    or

    [tex]U(\Lambda)a_{p}^{\dagger} = \sqrt{\frac{E_{\Lambda p}}{E_p}}a_{\Lambda p}^{\dagger}[/tex]

    But apparently the correct expression is

    [tex]U(\Lambda)a_{p}^{\dagger}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_p}}a_{\Lambda p}^{\dagger}[/tex]

    Can someone please point out my mistake?

    Thanks.
     
  2. jcsd
  3. Jun 7, 2009 #2
    When performing a Lorentz transformation (or any other symmetry transformation for that matter) you should keep in mind that states transform according to:

    [tex] |p\rangle \longrightarrow |p'\rangle = U|p\rangle[/tex]

    while operators transform according to:

    [tex] a_p \longrightarrow a_p' = Ua_pU^{-1}[/tex]

    To see how this works out in your derivation you have to insert [tex]\mathbf{1} = U^{-1}U[/tex] (= identity operator) in the following step:
    [tex]
    \sqrt{2E_{p}}U(\Lambda) a_{p}^{\dagger}|0\rangle = \sqrt{2E_{p}}U(\Lambda) a_{p}^{\dagger}(U(\Lambda)^{-1}U(\Lambda))|0\rangle
    [/tex]

    Next, we make the assumption that the vacuum state [tex]|0\rangle[/tex] is unique and the same for all observers. In other words, it doesn't transform under a Lorentzboost:
    [tex] U(\Lambda)|0\rangle = |0\rangle [/tex]

    What remains in the expression above is precisely the transformation rule of the operator.

    One way to see why operators transform this way, is to consider a state made up out of multiple creation operators, i.e. [tex]|p_1,p_2\rangle[/tex]. You'll find that under a Lorentz boost the transformation rule you derived will not work, while the other one does.
     
  4. Jun 7, 2009 #3
    Thanks xepma.

    Now, I see that this assumption is the key. It seems intuitive, but what really is the basis for this assumption?
     
  5. Jun 7, 2009 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Shouldn't the transformation laws follow directly from the way Fock space is constructed? They look like the things you typically see in (multi)linear algebra.
     
  6. Jun 7, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi maverick280857! :smile:

    (have a lambda: Λ :wink:)
    Short answer … you can't divide by |0> (and you did :wink:)

    Slightly longer answer … |0> is Lorentz invariant, ie U(Λ)|0> = |0>, so U(Λ)|0> = U(Λ')|0> for any Λ and Λ', but that doesn't mean that U(Λ) = U(Λ') ! :smile:
     
  7. Jun 7, 2009 #6
    It means that vacuum stays vacuum (a state without particles) whatever reference frame you choose.

    Or a ground state remains the grounds state in another reference frame.

    Bob.
     
  8. Jun 7, 2009 #7
    What I meant to ask was the following: We are explicitly enforcing this condition, viz. the ground state is Lorentz invariant. Can it ever happen that this condition is false?
     
  9. Jun 7, 2009 #8
    Lol, I feel dumb :tongue2:. But yeah, I got it..thanks.
     
  10. Jun 8, 2009 #9

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    We always assume/postulate symmetries in physics, like the global phase transformation invariance, rotational invariance, translation invariance. Ask yourself and try to construct a framework where the vacuum is not a Lorentz Invariant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lorentz Transformation and Creation Operators
  1. Creation operator (Replies: 2)

  2. Creation operator (Replies: 1)

Loading...