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Lorentz Transformation and Creation Operators

  1. Jun 7, 2009 #1

    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]


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


    [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]


    [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?

  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:
    \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

    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


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    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


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    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.

  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


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    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.
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