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Vacuum energy of free scalar field

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I have the following task:

    In quantum free scalar field theory find commutators of creation and anihilation operators with total four-momentum operator, starting with commutators for fields and canonical momenta. Show that vacuum energy is zero.

    2. Relevant equations

    [itex]\hat{p}^\nu=\int d^3\vec{k} k^\nu \hat{a}^\dagger(k)\hat{a}(k)[/itex]
    [itex][\hat{a}(p),\hat{a}^\dagger(q)] = \delta^{(3)}(\vec{p} - \vec{q})[/itex]

    3. The attempt at a solution

    I managed to find commutators:
    [itex][\hat{a}(q),\hat{p}^\nu] = q^\nu \hat{a}(q)[/itex]
    [itex][\hat{a}^\dagger(q),\hat{p}^\nu] = - q^\nu \hat{a}^\dagger(q)[/itex]

    Then I used this result to show that [itex]\hat{a}(q)|p>[/itex] and [itex]\hat{a}^\dagger(q)|p>[/itex] are eigenstates of the four-momentum operator [itex]\hat{p}^\nu[/itex] with eigenvalues [itex](p^\nu - q^\nu)[/itex] and [itex](p^\nu + q^\nu)[/itex] respectively.

    However i seem to be stuck now as I have no ide how to show that there is a lower boundary for energy values and that this value is indeed 0.
     
  2. jcsd
  3. Jul 9, 2012 #2

    gabbagabbahey

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    Homework Helper
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    Hi amgo100, welcome to PF! :smile:

    I think the general idea is to:

    (1) Compute the Hamiltonian [itex]H[/itex] in terms of the creation & annihilation operators from its expression in terms of the the free-scalar field and momentum density, then

    (2) compute the commutation relationships between the Hamiltonian and the creation and annihilation operators, and

    (3) use those commutation relations to look at the spectrum of [itex]H[/itex] and show that the vacuum state is the state [itex]|0\rangle[/itex] such that [itex]\hat{a}(\mathbf{p})|0\rangle = 0[/itex] for all [itex]\mathbf{p}[/itex] and show that it has zero energy (expectation value of [itex]H[/itex] is zero)

    most QFT text I've seen do at least the first of these step for you (or give the result at any rate)
     
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