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Vacuum in qft, what are we refering to?

  1. May 13, 2012 #1
    When we talk about the vacuum in qft, what are we refering to? The lowest state vector of the Fock space or the lowest energy field configuration that minimize the Lagrangian?

    Also related, when we sandwich the free field between two vacuum states, we get zero plus quantum fluctuations. But the same applies, of course, when we sandwich the lowest field configuration (which is zero) between two vacuum states. So does the vacuum consist of unmeasurable/ unphysical free fields plus fluctuations, or zero field plus fluctuations?


    thanks in advance for any replies
     
    Last edited: May 13, 2012
  2. jcsd
  3. May 14, 2012 #2
    Re: vacuum

    <0|psi(x,t)|0> = 0

    The vacuum expectation value of the (scalar) quantum field is zero. But for what psi(x,t)? It is certaintly zero for the free field. But the free field is not the field configuration that minimizes the Hamiltonian, that is psi(x,t) = zero.

    So can we say, the quite obvious, that psi(x,t)= 0 is the vacuum?

    I ask since I am confused because every qft book says |0> is the vacuum state and that the vacuum expectation value of the free field is zero. Furthermore, we can renormalized interacting fields, which then give zero expectation values for the renormalized interacting fields in the so-called physical vacuum.

    As it seems, classically only psi(x,t) = 0 is the lowest energy configuration, but quantum physically in the vacuum the expectation values of every field (zero, free, interacting) gives per defintion zero.

    Is that correct? If not, what got me confused?
     
  4. May 14, 2012 #3

    Bill_K

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    Re: vacuum

    The vacuum state is defined to be the state in which there are no particles present, i.e. nothing available to be destroyed. In momentum representation, a(k)|0> = 0. In position representation define the positive frequency part ψ(+)(x) = ∫eik·x a(k) d3k. Then ψ(+)(x)|0> = 0.
     
  5. May 14, 2012 #4
    Re: vacuum

    Thanks for answering, Bill!

    What you wrote is certaintly true, and something I also found in the textbooks.

    But what the textbooks also say, especially in those sections that treat spontaneous symmetry breaking, is that the vacuum is calculated to be the state of least energy. Here ψ is treated as classical field that minimizes the Hamiltonian, and by that giving us the ground energy state of the theory. For example, for a scalar field theory with a quartic interaction term, the minimum energy is achieved for ψ = 0, and if the sign of the mass term is flipped, we have two vacuum states, achieved by ψ = v and ψ = -v, where v is some constant.

    I assume that the ground state energy of the free Klein-Gordon scalar-field Lagragian is also achieved by the configuration ψ = 0.

    But why then is the vacuum expectation value <0|ψ|0> for a free field also zero? Does it also minimize the energy of the Hamiltonian of the theory?

    How is the Fock space ground state ket|0> and the ground state energy of a Hamiltonian/ Lagrangian connected?
     
  6. May 15, 2012 #5

    Bill_K

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    Re: vacuum

    I think that's generally right, for a self-interacting field you find the vacuum states by treating Ψ as a classical field and looking for the values that minimize the energy.

    This is unrelated to <0|Ψ|0> = 0, I think. The field operator Ψ always creates or annihilates a particle, i.e. it changes the number of particles. So in general for any state |λ> with a finite number of particles, <λ|Ψ|λ> = 0.
     
  7. May 15, 2012 #6
    thanks, Bill! much appreciated
     
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