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Interpretation of the creation and annihilation operators of a (free) scalar field

  1. Jun 4, 2009 #1
    Hi everyone

    The Hamiltonian of the Klein Gordon field can be written as

    [tex]H = \frac{1}{2}\int d^{3}E_p \left[a^{\dagger}(p)a(p) + a(p)a^{\dagger}(p)\right][/tex]

    and we have

    [tex][H,a(p')] = -E_{p'}a(p')[/tex]
    [tex][H,a(p')] = +E_{p'}a^{\dagger}(p')[/tex]

    The book I'm reading states that

    What does this mean?

    Thanks.
     
  2. jcsd
  3. Jun 5, 2009 #2
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    [tex][H,a^{\dagger}(p')] = +E_{p'}a^{\dagger}(p')[/tex]

    [tex]|p'>=a^{\dagger}(p')|0>[/tex]

    has energy [tex] E_{p'} [/tex] because:

    [tex]H|p'>=Ha^{\dagger}(p')|0>=[H,a^{\dagger}(p')]|0>+a^{\dagger}(p')H|0>[/tex]

    The 2nd term is zero because the vacuum |0> has no energy so is a zero eigenvalue of H. The 2nd term is equal to [tex]E_{p'}|p'>[/tex]

    and hence the state |p'> has that energy, and was created from the vacuum |0> by the creation operator 'a dagger (p')'.
     
  4. Jun 5, 2009 #3
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    The quantum field looks like

    [tex] \phi(x) = \int d^3p a^{\dag}(p) e^{ipx -iE(p)t} + a(p) e^{-ipx +iE(p)t} [/tex]

    So, creation operator is associated with "negative" energy exp(-iEt) and annihilation operator is associated with "positive" energy exp(iEt).
     
  5. Jun 5, 2009 #4
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    Thanks meopemuk, this is a cool way of looking at it.

    Thanks RedX, this also clarifies my doubt about the operators.

    ----

    Now, a question to both of you: what does creation and annihilation really mean here, physically? I understand how these operators work mathematically and how this is a generalization of the nonrelativistic simple harmonic oscillator in QM. But what I want to understand more deeply is their physical meaning, in QFT.

    Now lets say I have a scalar field [itex]\phi[/itex] in a region, described by the expression meopemuk has given.

    (a) What is the (physical) interpretation of the linear superposition of the two exponentials, suitably weighted by the creation and annihilation operators, in the context of QFT?

    (b) Is [itex]\phi(x)[/itex] so defined, an operator or an operator valued distribution?
     
  6. Jun 5, 2009 #5
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    Hi maverick,

    I'll give you my personal opinion, which is not necessarily in accord with most textbooks.

    Creation/annihilation operators do not have any specific physical meaning. They are just convenient symbols for compact writing of physically relevant operators of observables (see e.g. the total momentum and energy discussed in a parallel thread).


    Quantum fields do not have physical interpretation either. They are just convenient building blocks for relativistically invariant and cluster separable interactions in the Hamiltonian. This point of view was expressed most forcefully in S. Weinberg "The quantum theory of fields", vol. 1.


    This is an operator function depending on 4 real parameters. So, "operator valued distribution" is an appropriate name.
     
  7. Jun 5, 2009 #6
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    It is easy to understand if you know how to describe a system of identical particles in the non relativistic QM. Learn first that. You will see how one goes from the (anti) symmetrized product of the usual wave functions to the representation in population numbers.

    The difference between non relativistic and relativistic QM of many particles is in larger number of possible states: in the latter case there are also the states with negative frequency = the antiparticle states. The rest is quite similar to the non-relativistic case. You will find all answers there.

    Bob_for_short.
     
    Last edited: Jun 5, 2009
  8. Jun 5, 2009 #7
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    Thank you.
     
  9. Jun 5, 2009 #8
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    You can view the creation and annihilation operators as operators that creaet/destroy the energyquanta of the system. All states are build up from the vacuum using the creation operators.

    In the non-interacting case the spectrum is continous. But remember for instance the hydrogen atom: in that case there is a discrete spectrum of energy levels (the bound states). This is in essence the "quantum" in quantum theory: systems tend to develop a smallest energy quanta.

    As for the field operator: in non-relativistic quantum mechanics the field operator is in fact nothing but the wavefunction. It has a clear meaning, since it's absolute square stands for the probability distribution of the particle at hand. In relativistic QFT this picture breaks down, and the field operator can strictly no longer be viewed as such a distribution. It should be viewed as an operator which contains all the information possibly present in the physical state. But only by acting on the state with this field operator can we extract this information. By itself the field operator is "just" a mathematical construct - it has no physical meaning, yet.
     
  10. Jun 7, 2009 #9
    Re: Interpretation of the creation and annihilation operators of a (free) scalar fiel

    I did that, re-read the section on Fock states and went back to QFT. I think I have a better understanding of it now. Thank you :smile:
     
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