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What is propagator

  1. Jul 10, 2007 #1
    Here's how I've learned propagators, and how far I felt I understood them. For a shrodinger's equation

    [tex]
    i\hbar\partial_t\Psi = -\frac{\hbar^2}{2m}\nabla^2 \Psi
    [/tex]

    the propagator is

    [tex]
    K(t,\boldsymbol{x},\boldsymbol{y}) =\int\frac{d^3p}{(2\pi\hbar)^3}\exp\Big(-\frac{i}{\hbar}\Big( \frac{|p|^2}{2m}t - \boldsymbol{p}\cdot(\boldsymbol{y}-\boldsymbol{x})\Big) \Big)
    [/tex]

    The way to use this propagator is this. For a given initial wave function [tex]\psi(\boldsymbol{x})[/tex] we can define a wave function into the future spacetime like this

    [tex]
    \Psi(t,\boldsymbol{y}) = \int d^3x\; K(t,\boldsymbol{x},\boldsymbol{y})\psi(\boldsymbol{x})
    [/tex]

    There are few remarks that should be made of this propagator. It satisfies following two equations.

    [tex]
    K(0,\boldsymbol{x},\boldsymbol{y})=\delta^3(\boldsymbol{x}-\boldsymbol{y})
    [/tex]

    [tex]
    K(t_1+t_2,\boldsymbol{x},\boldsymbol{z}) = \int d^3y\; K(t_1,\boldsymbol{x},\boldsymbol{y}) K(t_2,\boldsymbol{y},\boldsymbol{z})
    [/tex]

    These are very pleasant features. The first one means that the wave function doesn't change in zero time when the time evolution is defined with the propagator. The second one means that the time evolution is associative in some sense. From the second equation it also follows, that if we want to study evolution from time [tex]t[/tex] to time [tex]t+\Delta t[/tex], we don't need to do any stuff with inverse propagations ([tex]t\to 0[/tex] and then [tex]0\to t+\Delta t[/tex]), but instead we can just take the direct propagation ([tex]t\to t+\Delta t[/tex]).

    We can solve the differential equation satisfied by the wave function [tex]\Psi(t,y)[/tex] by studying the propagation with linear approximation of t. The result turns out to be precisly the Shrodinger's equation.

    To get the propagator for Klein-Gordon equation it suffices to replace the energy term with a relativistic energy [tex]E_{\boldsymbol{p}}=\sqrt{|p|^2c^2 + (mc^2)^2}[/tex]. It is easy to see that the propagator

    [tex]
    K(t,\boldsymbol{x},\boldsymbol{y}) = \int\frac{d^3p}{(2\pi\hbar)^3}\exp\Big(-\frac{i}{\hbar}\Big( E_{\boldsymbol{p}}t - \boldsymbol{p}\cdot(\boldsymbol{y}-\boldsymbol{x})\Big) \Big)
    [/tex]

    also satisfies the two important equations. It is a small exercise to check, that studying time evolution of a wave function given by definition

    [tex]
    \Phi(t,\boldsymbol{y}) = \int d^3x\; K(t,\boldsymbol{x},\boldsymbol{y}) \phi(\boldsymbol{x})
    [/tex]

    gives a Klein-Gordon equation.

    So far everything is fine, to me.

    Then I've encounterd QFT and there this propagator of KG-field is modified by adding a factor [tex]1/(2E_{\boldsymbol{p}})[/tex]. I can immidiately make an important remark. This factor ruins the two equations, which were true for the previous propagators. The propagator does not become a delta function for zero time, so it allows propagations in zero time, and the time evolution of the wave function cannot be defined in the same way as it could be defined with the earlier propagators. Also this propagator is not associative in the same way as the earlier propagators were. It cannot be used in series.

    Now knowing how to not use this propagator, I would like to know, how is it supposed to be used. What does this propagator even mean? All QFT-texts I have encountered somehow assume, that it is trivial that propagation amplitude is just propagation amplitude, but what is this propagation amplitude really? At least it is not the same thing it is with SE or with the KG equation (or in way I presented it with KG equation).

    Some people have answered me, that the [tex]1/(2E_{\boldsymbol{p}})[/tex] factor has to be there to make the propagator Lorentz's invariant. Well it is so nice it is Lorentz's invariant, but what is this quantity that is now Lorentz's invariant?

    The calculations with propagators in QFT are usually very abstract and complicated, and the matter is not improved by the fact, that I have no clue of what the quantity is that these calculations deal with.

    This matter is also not improved by a fact that the mainstream view of the propagator without [tex]1/(2E_{\boldsymbol{p}})[/tex] factor seems to be, that this propagator is imcompatible with relativistic QM, because it would violate causality. (At least P&S explain it like this). Since the propagator gives solutions of Lorentz's invariant KG equation, it obviously is not violating causality.

    I think this post is already too long like this, so I was forced to leave out some details. If you have problems with some specific mathematical claims I made, I'm ready to post more details. Just mention about it.
     
    Last edited: Jul 11, 2007
  2. jcsd
  3. Jul 10, 2007 #2
    Great post. I think you raised very relevant and important questions. Let me give you my take on them. I am sure some people will disagree.

    Particle propagators in QFT are not supposed to describe the amplitude for a particle to move from point x to point y. Some textbooks may try to convince you that propagators are doing just that. But you should take these claims with scepticism. After all, these claims have never been compared with experiment. QFT is just not good at calculating how things are propagating in space and time. QFT is designed for a completely different purpose. The main goal of QFT is to calculate the S-matrix. This quantity contains important information about things that can be directly compared with experiment - scattering amplitudes, decay rates, energies of bound states, etc. QFT is great for calculating these things.

    S-matrix calculations in QFT are normally done within the Feynman-Dyson perturbation theory. Individual terms look like this

    [tex] \langle in| \int \limits_{-\infty}^{\infty} dt_1 \int \limits_{-\infty}^{\infty} dt_2
    T V(t_1) V(t_2) |out \rangle[/tex] (1)

    where [itex]|in> [/itex] is "in" state in the remote past, [itex]|out> [/itex] is "out" state in the remote future. These states are obtained by applying creation operators to the vacuum state [itex] |0 \rangle [/itex]. For example, for a two-particle state one can write something like

    [tex] |out \rangle = a^{\dag}_{\mathbf{p}}b^{\dag}_{\mathbf{q}} |0 \rangle [/tex]

    T is the chronological ordering sign, and [itex] V(t) [/itex] is interaction part of the Hamiltonian. The interaction is normally written as a product of quantum fields

    [tex] V(t) \propto \int d^3x \phi_1(\mathbf{x}, t) \phi_2(\mathbf{x}, t) ... [/tex]

    Looking at these formulas one can easily convince oneself (for details see any QFT textbook) that S-matrix amplitudes like (1) can be eventually expressed as a multiple 4-dimensional integral of products of several propagators like

    [tex] \langle 0 | T \phi(x) \phi(y) | 0 \rangle [/tex]

    or their Fourier transforms (momentum space propagators).

    My point is that the only reason for introducing and evaluating propagators in QFT is to assist S-matrix calculations outlined above. I consider propagators as useful formal expressions and don't even try to interpret them as true propagation amplitudes of free particles.
     
  4. Jul 10, 2007 #3
    I have, in fact, not taken any course on QFT yet. I've tried to read about perturbation theory and Feynman's digrams on my own, but I've found the derivations too confusing (I manage with the non-interacting theory somehow). I'm probably taking a course this fall, and then I'll be forced to do lot of exercises. I'm not yet convinced I'll learn to understand anything there, but at least I'll get used to this stuff.

    I remember asking here at physics forums that is QFT good for anything else than calculating particle collisions, and I received a firm answer that sure, for example it is used also for solid state physics. But now this is starting to look like, that even though this QFT is called "a theory", it seems to be somehow in pieces. Some techniques used for some purposes, and other techniques for other purposes, and it is just called one theory. Or is it like this?
     
  5. Jul 10, 2007 #4
    In my understanding, QFT used in solid state physics is quite different from QFT in particle physics, e.g., QED. One important difference is that in QED there is no well-defined Hamiltonian. For example, interation operators [itex] V(t) [/itex] that I mentioned in the previous post are actually infinite (they contain infinite mass and charge renormalization counterterms). So, in fact, it is not possible to do rigorous calculations of time-dependent processes in QED. At least, I haven't seen any calculations of that sort. However, somewhat miraculously, all these infinities cancel out when the S-matrix is calculated. This is the idea of the Tomonaga-Schwinger-Feynman renormalization approach. Luckily for theoreticians, it is impossible to measure time dependence in collisions of high energy particles, so S-matrix methods are sufficient for comparison with experiment. So, when learning QFT in particle physics, keep in mind that the ultimate goal is scattering, and all the machinery (quantum fields, propagators, etc.) may not have its alleged physical meaning.
     
  6. Jul 10, 2007 #5

    olgranpappy

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    The main difference is that in solid state physics we deal for the most part with non-relativistic particles. It's amusing that non-relativistic QFT was developed after relativistic QFT... but this is not too surprising because, analogously, the KG equation was written down before the Schrodinger equation.

    The other difference is that high-energy theorists do their experiements in vacuum and are interested in things like "vacuum self-energy" etc. Whereas solid state physicists really want to have a finite amount of stuff (finite density) around. We also work at finite temperature (but so do some high-energy people). Because we have a finite amount of stuff around the self-energies etc occuring in solid state are mainly due to interaction of the particle with the real medium (i.e. other electrons, phonons, plasmons).

    The other great thing about QFT in condensed matter is that the only interaction is the well-known and friendly electromagnetic interaction--there's none of this "electroweak" and "strong" interaction crap.:biggrin:
     
  7. Jul 11, 2007 #6
    Yes, and this is related to another important difference and similarity. In both particle and solid state QFT there is effect of renormalization. For example, an electron moving through crystal lattice polarizes atoms around itself. In QFT language this can be described as creation of virtual phonons. This results in modification of the electron's mass. This mass change (renormalization) is perfectly physical and finite. The finiteness comes from the fact that phonons with small (less than the interatomic distance) wavelengths cannot exist in the lattice.

    A similar, but more sinister effect occurs in the traditional version of QED. It appears that a free electron "polarizes vacuum" around itself by creating virtual photons and electron-positron pairs. The bad news is that the electron mass (and charge) modification induced by this polarization is infinite. Indeed, there is no "interatomic distance" in vacuum that could limit the wavelength of virtual photons, like in the case of crystal. Tomonaga, Schwinger, and Feynman found a way how to do meaningful calculations with these infinite-mass and infinite-charge electrons. However, the fundamental problem of infinities remains unresolved until now.

    The most popular idea for dealing with infinities is "effective field theory". Basically it says that vacuum (or space-time) *is* sort of like a crystal. It says that there is some structure of space at very short distances, and these distances (like interatomic distances in crytals) provide the necessary limit for wavelengths of virtual photons. Then the renormalization of electron's mass and charge, instead of being infinite, becomes finite, but huge. Needless to say that nobody knows what is this space-time structure. The range of wild speculations is very broad. To see some of them you can visit http://www.arxiv.org/hep-th on any given day.
     
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