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Regarding the creation and annhilation operators in QFT

  1. Jan 27, 2014 #1
    I'm trying to understand QFT for the moment and have a question regarding the basic.

    So we have a vectorspace (Hilbertspace) of our states. The operator \phi(x) measures the amplitude at point x, whereas the operator \pi(x) measures the momentum density..
    The ladder operator
    creates a state with momentum p. But it seems here that there are many different states with momentum p. Is the momentum density always homogenously distributed in this case?
  2. jcsd
  3. Jan 27, 2014 #2


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    In what sense? What exactly is the ambiguity you're referring to in say the following: ##a^{\dagger}_{k}|n_k \rangle =(n_k + 1)^{1/2} |n_k + 1\rangle ##?
  4. Jan 28, 2014 #3
    Hmm.. In the case of the SHO in QM, this seems alright. There is a 1-1 correspondence between the energy states and the spatial wavefunctions that solves the Schrödinger equation. I thought that in this case, for the field, we integrate over all of space in order to look at the momentum, thus the excitation can either correspond to a plane wave or a 'more localized bump'. There is nothing that is preferable.

    But. If we are looking at the free Klein-Gordon theory, I guess that the plane wave is the preferable choice.

    Would you agree if I said that, when acting on the vacuum state, ##a^\dagger_p## corresponds to an excitation such that our QFT state has a homogenously distributed momentum density?
  5. Jan 28, 2014 #4


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    I think, we should clarify some misunderstandings first. As an example we take the free charged Klein-Gordon field with the Lagrangian
    [tex]\mathcal{L}=(\partial_{\mu} \phi^*)(\partial^{\mu} \phi)-m^2 \phi^* \phi.[/tex]
    The canonical field momenta (which have nothing to do with momentum!) are
    [tex]\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}^*, \quad \Pi^*=\frac{\partial \mathcal{L}}{\partial \dot{\phi}^*}=\dot{\phi}.[/tex]
    This determines the canonical equal-time commutators to be
    [tex][\phi(t,\vec{x}),\dot{\phi}^{*}(t,\vec{y})]=\mathrm{i} \delta^3(\vec{x}-\vec{y}).[/tex]
    All other combinations vanish.

    The equations of motion for the field operators read
    [tex](\Box + m^2) \phi=(\Box+m^2) \phi^*=0.[/tex]
    The solutions in terms of annihilation and creation operators, implementing the correct time dependence (Feynman-Stückelberg trick) are given by
    [tex]\phi(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega(\vec{p})}} \left [a(\vec{p}) \exp(-\mathrm{i} p \cdot x)+b^{\dagger}(\vec{p}) \exp(+\mathrm{i} x \cdot p ) \right ]_{p^0=\omega(\vec{p})}[/tex]
    with [itex]\omega(\vec{p})=+\sqrt{\vec{p}^2+m^2}[/itex]. Then the operators [itex]a(\vec{p})[/itex] and [itex]b(\vec{p})[/itex] are annihilation operators for a particle and an antiparticle with momentum [itex]\vec{p}[/itex], respectively. They obey the commutator relations for creation and annihilation operators independent harmonic oscillators,
    with all other combinations in the commutator vanishing.

    The energy and momentum density are given via Noether's theorem as the temporal components of the energy-momentum tensor, [itex]\Theta^{\mu 0}[/itex], i.e., after normal ordering by
    [tex](P^{\mu})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \begin{pmatrix} \omega(\vec{p}) \\ \vec{p} \end{pmatrix} [a^{\dagger}(\vec{p}) a(\vec{p})+b^{\dagger}(\vec{p})].
    A basis of the corresponding Hilbert space, the Fock space, is then given by the occupation-number basis
    [tex]|\{N(\vec{p},\bar{N}(\vec{p})) \}_{\vec{p}} \rangle=\prod_{\vec{p}} \frac{1}{\sqrt{N(\vec{p})! \bar{N}(\vec{p})!}} [a^{\dagger}(\vec{p})]^{N(\vec{p})} b^{\dagger}(\vec{p})]^{\bar{N}(\vec{p})} |\Omega \rangle.[/tex]
    The product here goes over any (finite!) set of momenta and [itex]N(\vec{p}),\overline{N}(\vec{p}) \in \mathbb{N}_0[/itex]. Further [itex]|\Omega \rangle[/itex] is defined as the "vacuum state", fulfilling
    [tex]a(\vec{p}) | \Omega \rangle= b(\vec{p}) | \Omega \rangle=0[/tex]
    for all [itex]\vec{p}[/itex]. It is assumed (!) that there is only one such state (non-degenerate ground state with total energy [itex]E=0[/itex]).
    Last edited: Jan 28, 2014
  6. Jan 28, 2014 #5
    Thanks for the answers!
    Could you give me a reference for futher reading on the Feynman-Stückelberg trick? :)
  7. Jan 28, 2014 #6


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    There are two meanings of "vacuum state".

    The creation and annhilation operators are defined relative to some "reference vacuum state". However, the "reference vacuum state" that is used in the definition of the creation and annihilation operators is not the ground state of an interacting theory of many-particles.

    You can see this from second quantization, in which some convenient basis states for single particles are used to make basis states for the many-particle Hilbert space. The annihilation and creation operators are defined so that they create basis states of the many-particle Hilbert space from some "reference vacuum state". But for an interacting system, the "true ground state" is different from the "reference vacuum state", and the "true ground state" will be a complicated superposition of basis states of the many-particle Hilbert space.

    Whereas the reference vacuum state is the state that appears in the definition of the creation and annihilation operators, in the LSZ formula, it is the true ground state that appears. For example, take a look at Srednicki's derivation of the LSZ formula, especially the discussion starting with "However, our derivation of the LSZ formula relied on the supposition that the creation operators of free field theory would work comparably in the interacting theory. This is a rather suspect assumption, and so we must review it." http://web.physics.ucsb.edu/~mark/qft.html (p51).
    Last edited: Jan 28, 2014
  8. Jan 28, 2014 #7


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    I'm not sure what you mean by this. Take for example a (free) real scalar field ##\varphi##. What does it mean for ##\theta^{0i} = \partial_0 \varphi \partial^i \varphi## to be "homogenously distributed"?

    As you already know, the 3-momentum itself is given by ##P^{i} = \int d^{3}x :\theta^{0i}: ## which for ##\varphi## is the usual ##P^{i} = \int \frac{d^{3}p}{(2\pi)^3}p^{i}a^{\dagger}_p a_{p}## of which ##a^{\dagger}_p|0\rangle ## is an eigenstate of momentum ##p^i## so again I'm not sure what you mean when you say ##\theta^{0i}## is "homogenously distributed".
  9. Jan 29, 2014 #8
    Hmm, maybe this is the surface of a much bigger misunderstanding. I will try to write my (so called) logic, and hopefully you have the time to follow my mental picture. :)

    Starting with a classical scalar field, ##\phi##, in free space, the field has a amplitude and momentum density assigned to each point. Since the physics is translation invariant we know that we have conservation of momentum,
    \partial_\nu T^{\nu i} = 0.
    Where ##T^{0 i} = \dot{\phi}\partial^i \phi## which allows me the calculate ##T^{0 i}## for any given space-time event, assuming I know the configuration and dynamics of my field.
    Futhermore the total charge of this current (in other words the total momentum of the field ) is
    P^i = \int d^3 x T^{0i} = \int d^3x \;\dot{\phi}\partial^i \phi

    Then we quantize these fields. This gives rise to states that are really messy to work with. Instead we focus on the operators the act on these states. The amplitude in point x is related to an operator ##\hat{\phi}(x)## - just as we do with our dynamical variables in QM. (Lets use hats so we can tell the difference between FT and QFT.)

    The problem for me is that: In field theory there are many different field configurations that has the same total momentum. They doesn't necessarily represent free fields or plane waves but is seems as if I can picture a bunch of them (I just need to make sure the that the time evolution and spatial "bumps" amounts to the same vector ##P^i## in the equation above).. However it is only the infinitely smeared out configurations that has a homogenous distribution of the momentum density in space.

    The same goes for QFT. It seems as if we can construct many states that at one given moment in time has the same momentum. After all we have infinitely many degrees of freedom.

    However I suspect that we always talk about the free particle solutions since we are doing free theory. :/

    Was this understandable at all?
    Last edited: Jan 29, 2014
  10. Jan 29, 2014 #9


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    Your total-momentum integral becomes, of course, an operator in QFT, the momentum operator that generates spatial translations as it should according to Noether's theorem. Of course, you have to normal order the integrand, because otherwise you get an undefined expression. See also my previous posting!
  11. Jan 29, 2014 #10

    Yes, but what about my argument that there are many states with the same total momentum? :/
  12. Jan 29, 2014 #11


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    I think so.

    We know that ##\varphi## has to satisfy ##(\partial_{\mu}\partial^{\mu} + m^2)\varphi = 0##. So at the least the dynamical evolution of any given configuration is constrained to the space of configurations that evolve according to the free KG equation. So are you talking about different field configurations ##\varphi##, with different ##\theta^{\mu\nu} = \partial^{\mu}\varphi\partial^{\nu}\varphi -\eta^{\mu\nu}\mathcal{L}## (of course ##\mathcal{L}## will be the same if we be as simple as possible), all solving the free KG equation such that ##P^{i} =\int d^{3}x :\theta^{0i}:= \int d^{3}x :\partial_0\varphi\partial^i \varphi:## is the same for all of them? In other words are you talking about different field configurations solving the free KG equation with different momentum densities such that ##a^{\dagger}_p |0\rangle ## for all of them yields states with the same momentum ##p##? I suppose by "infinitely smeared out" you're referring to a plane wave solution in order to contrast with localized solutions; textbooks seem to usually only treat the creation and annihilation operators with respect to the former.
  13. Jan 29, 2014 #12
    This helped alot. I will continue my reading of Peskin and Schroeder. I think I graps it.

    Once again, thanks for your time and dedication. :)
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