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Particle number and resonance with Klein-Gordon waves.

  1. Aug 13, 2012 #1
    Hi all,

    I'm hoping this will be a quickly solved question. In Peskin and Schroeder (2.66), when dealing with source terms in the Klein-Gordon equation, ##(\partial^2+m^2)\phi(x) = j(x)##, they have
    $$\int d N =\int \frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p}|\tilde{j}(p)|^2\quad \text{where}\quad\tilde{j}(p)=\int d^4y \ e^{ipy}\ j(y)$$
    and ##N## is the particle number. They go on to state
    I'm a little confused by this, and was wondering if someone could explain

    1. what is meant by "resonance"
    2. where being "on mass" comes into it. (I assume is to do with ##1/2E_p## being large, or some such, but I haven't figured it out.)

  2. jcsd
  3. Aug 13, 2012 #2
    They're doing the same sneaky trick here that they've been doing up until now in the Fourier expansion of the free field equations (the integral over the creation/annihilation ops which define the field). Namely, they're starting with a momentum 3-vector (in the [itex]dp^3[/itex] of the integral), and converting that into a momentum four-vector (the [itex]p[/itex] in the [itex]\widetilde{j}(p)[/itex]). The conversion occurs by giving the four-vector a time component that causes its total magnitude to be equal to the mass of the particle, i.e. [itex]p^2 = m^2[/itex] (in the book, they first do this in 2.46 and 2.47). This is called putting the particle "on the mass shell".

    Since we only care about these values of [itex]\widetilde{j}(p)[/itex] for which [itex]p^2 = m^2[/itex], that means there are a whole lot of values from the function that we ignore completely. Therefore, what they mean by #1 is that only those few values actually contribute to the amplitude for particle creation--if you have some big spike in [itex]\widetilde{j}(p)[/itex] that's centered around some value that is not on-shell, it won't have any effect on [itex]j(x)[/itex]'s ability to create particles.
  4. Aug 13, 2012 #3
    In classical mechanics, your first approximation to some free oscillating system will be the simple harmonic oscillator, which has a characteristic frequency [itex]\omega[/itex]. If you now try and drive this system (say you're pushing a child on a swing), energy is transferred into the system most efficiently if your force varies periodically with the same natural frequency [itex]\omega[/itex] of the system. An oscillating system being driven at it's characteristic frequency is said to be in resonance, and the frequency is often called the resonant frequency.

    In the mathematical analogy between the Klein-Gordan particle and the SHO, remember that E= [itex]\omega[/itex] in particle physics units. The "natural frequency" of the system is just the energy that is related to the 3-momentum and mass of the particle by E^2=p^2 +m^2. The statement that the energy, 3-momentum and mass of a particle are related in this way is the same as saying that the Klein-Gordan equation is satisfied, and these equivalent conditions are referred to as "the on-shell" condition. Adding a "source" term
    is analogous (in a formal way) to driving your unperturbed classical oscillator.

    If you want to understand this analogy better, try looking at Zee's book "Quantum Field Theory in a Nutshell". The mathematical treatment isn't really robust enough for you to learn QFT from it without referring to Peskin (or another good book like Srednicki), but I found it really useful for getting some kind of intuitive feeling for QFT, whereas the opening chapters of Peskin left me in a total mental fog even though I could follow the maths.
  5. Aug 13, 2012 #4
    Ah yes, silly me! I just read it again, and right after PS define ##\tilde{j}(p)## they state "evaluated at 4-momenta ##p## such that ##p^2=m^2##". That being said, why do we choose to ignore the off-shell terms completely? As far as I can tell, ##\tilde{j}(p)## is valid (as an equation) for any ##p##, so I'm not sure why it must vanish for off-shell momenta. Unless one defines:
    $$\tilde{j}(p) ={\Bigg\{}\begin{array}{cl} \int d^4y\ e^{ipy}\ j(y) & if\ p^2=m^2 \\ 0 & otherwise \end{array}$$
    I guess it's still a Fourier transform of sorts.

    I understand the classical sense, but not this application of "resonance". I'd understand what is meant if ##j(x)## had the same frequency as ##e^{ipx}## (i.e. ##p^0 = E_p##), but the problem is there is nothing to mention ##j(x)## has a frequency, or how to determine it.

    Thanks for the reference to Zee. I'll have a look at his mattresses.
  6. Aug 13, 2012 #5
    It's not a choice. We've worked out the formula for how many particles are produced, and it turned out only to depend on the on-shell components of ##\tilde{j}(p)##.

    In general ##\tilde{j}(p)## won't vanish for off-shell 4-momenta; however our calculation shows that these momentum components of the source don't produce any particles.

    If you have a classical SHO with a characteristic frequency, and you perturb it with a source that contains many frequencies, only the perturbations with the same frequency as the oscillator are effective at producing oscillations. That is, only the frequency component of the source that is resonant with the oscillator matters.

    We can think of a free quantum field as being an infinite set of quantum SHO's, one oscillator for each possible 3-momentum. The frequency of the oscillator for a given 3-momentum is proportional to the energy corresponding to that 3-momentum: ##E = \sqrt{\vec{p}^2 + m^2}##. If we perturb this oscillator with a source, we will only produce oscillations if the source has a 3-momentum matching the oscillator's 3-momentum, and a frequency matching the oscillator's frequency. Again only the resonant (on-shell) frequencies of the source produce oscillations (particles).
  7. Aug 13, 2012 #6
    That's kind of asking my question again: how do our formula show particle number only depends on the on-shell components?

    I understand that --- I guess I was assuming a source to be constant, like a charge. That is ##\partial_t j(x) =0##. Hence it has no frequency, so resonance didn't make sense. I take it then that for resonance we require ##iE_p = \partial_t j(x)##, or something of that kind?
  8. Aug 13, 2012 #7
    In Eq. (2.63) we write out a formula for ##\phi(x)## in terms of j. This equation involves the Fourier expansion of the retarded propagator D_R(x - y). This Fourier expansion of the propagator involves only those 4-momenta p which are on-shell. Then we are changing the order of the integrals over p and y, and noting that the integral over y gives us the Fourier transform of j, ##\tilde{j}(p)##. But again, in the p integral we are only integrating over the on-shell 4-momenta. So ##\phi(x)## only depends on the value of ##\tilde{j}(p)## for values of p such that p is on-shell. In the following steps, wherever p appears it is still an on-shell 4-momentum. So in the final result, we have an integral over all possible spatial components of the 4-momentum p, with the time component of the 4-momentum set to whatever energy it needs to be to make p on-shell (just as in the Fourier expansion of the retarded propagator).

    Ah. No, the source depends on both space and time. When P&S write ##j(x)##, x is a spacetime point, and thus j in general is time-dependent. If j were time-independent, and only depended on spatial position, they would probably write it as ##j(\mathbf{x})##. A time-independent source wouldn't produce any particles. As with the classical SHO, you need a time-dependent source jiggling around to excite oscillations: a force that is constant for all time won't cause the oscillator to oscillate.

    Something like that. A source ##j(x) = \Re \exp [i \mathbf{p} \cdot \mathbf{x} - i E t]## will be resonant with the mode of the field with spatial momentum ##\mathbf{p}## if ##E = E_\mathbf{p} \equiv \sqrt{\mathbf{p}^2 + m^2}##. This source will create particles with 4-momentum ##(E, \mathbf{p})##. Then we can superpose many on-shell plane waves with different momenta to get a source that will create various particles with different momenta. We could also add non-resonant off-shell plane waves to the source, but they won't create any particles.
  9. Aug 13, 2012 #8
    Ok, now I see what you mean. So most of these equations should really have ##|_{p^0=E_p}## on them? I know it might be tedious for the authors, but it'd be clearer for me anyway.

    That's what I was looking for. A nice analogy too.

    That's neat --- I like that.

    Thanks for clearing it all up,

  10. Aug 13, 2012 #9
    Oh, if you're feeling extra helpful, PS just pull the number probability density ##|\tilde{j}(p)|^2/2E_p## out of thin air, as far as I can tell. It is somewhat mentioned in (2.65) but nothing really identifies this term in terms of particle number. Could you explain why?
  11. Aug 13, 2012 #10
    Right. It might not be too much of an exaggeration to say that your progress in QFT can be measured by how much suppressed notation you can restore in your head when you read an equation. It's not too uncommon for equations to contain four different kinds of invisible indices that you're supposed to know about implicitly.

    Denote the expected number of particles whose momentum lies in a small volume ##d^3 p## around the momentum p by ##N(p) \frac{d^3 p}{(2 \pi)^3}##. So N(p) is the expected number density of particles in momentum space. Then the expected energy of the particles in this small volume of momentum space is ##E_p N(p) \frac{d^3 p}{(2 \pi)^3}##. And so the total expected energy is

    ##\int \frac{d^3 p}{(2 \pi)^3} E_p N(p) ##

    Equating this to Eq. (2.65) which gives the expected total energy after the source has acted, we have:

    ##\int \frac{d^3 p}{(2 \pi)^3} E_p N(p) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{2} |\tilde{j}(p)|^2 ##

    So we must have ## N(p) = \frac{1}{2 E_p} |\tilde{j}(p)|^2 ##
  12. Aug 13, 2012 #11
    Oh dear!

    Righto, that was very clear and helpful. Cheers
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