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Magnitude of the scattering ampliudes in QFT

  1. Feb 19, 2010 #1
    I have yet another question...

    I was always thinking that the scattering amplitudes one computes in QFT are complex numbers of modulus between 0 and 1. And I was thinking that because it is supposed to be related to the probability of some transition between states happening. And then I tried to find out some support to this idea of mine but without success. Does anyone know if the scattering amplitudes are of modulus less than 1? And, most importantly, any references where I could find support for the claim, if it's true?

    Thanks!
     
  2. jcsd
  3. Feb 19, 2010 #2

    strangerep

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    Aren't they distributions? E.g., for trivial scattering, where the S operator is just 1,
    we have
    [tex]
    \langle p| q \rangle ~=~ \delta(p - q) ~.
    [/tex]
    Less trivial scattering amplitudes still involve a momentum-conservation delta function.
    One then integrates to get cross-sections, as in (eg) Peskin & Schroeder section 4.5.

    Or were you thinking about something else?
     
  4. Feb 20, 2010 #3
    strangerep -> Yes and no. One usually defines the invariant scattering amplitudes [tex]\mathcal{M}[/tex] by [tex] \langle out| in \rangle =: (2 \pi)^4 \delta^{(4)}(P_{in} - Q_{out}) \mathcal{M}(in \to out) [/tex]. Now both the RHS and LHS are distributional, but the [tex]\mathcal{M}[/tex] is a smooth complex function of the momenta. I was talking about the latter. It's the quantity that one computes with Feynman diagrams.
     
  5. Feb 20, 2010 #4
    [tex] \mathcal{M}(in \to out) [/tex] is a complex function of "out" particles momenta (if "in" momenta are fixed). There are no restrictions on the values of this function. It can be even singular as in the case of Coulomb scattering. However if you take square of the modulus of this probability amplitude and integrate over all possible "out" momenta you should get a positive number less than 1 (the total probability of scattering). This is guaranteed by the unitarity of the S-operator.

    Eugene.
     
  6. Feb 20, 2010 #5
    meopemuk ->I suspect unitarity of S has something to do with it, but I'm looking for a clear argument. Have to point out though that your argument is flawed. [tex]\mathcal{M}(in \to out)[/tex] is definitely a smooth function of all the momenta, at least when you compute it in QFT and after renormalization (don't wanna go into renormalization at all, let's just agree we have somehow made it finite). It cannot be singular because it's essentially given by an integral over some "internal" momenta and the integrand is not singular - product of Feynman propagators, which are not singular for real momenta.

    Most importantly, however, what I'm puzzled about is the next point. If, as you claim, one has [tex]\int |\mathcal{M}(P)|^2 dP \le 1[/tex], then [tex]|\mathcal{M}| \le 1[/tex]. This is simply because [tex]|\mathcal{M}|^2 \ge 0[/tex] being the modulus squared of a complex number. And by integrating positive quantities you are "summing positive numbers", so to say. And if the overall result is less than 1, than each "summand" must be less than one. Incidentally, this is more ore less the kind of argument I have in my mind why it should be [tex]|\mathcal{M}| < 1[/tex], but it's not a rigorous argument. Do you have any references?
     
  7. Feb 20, 2010 #6
    The summand is [tex]|M(p)|^2dp[/tex], which is emphatically less than one no matter what the value of |M(p)|, since dp is infintesimal. Revise what you know about probability density functions and all should be clear.
     
  8. Feb 20, 2010 #7
    peteratcam -> I realized my argument was not correct a minute after posting it. Still, your argument that the integrand is always less than 1 because pd is infinitesimal doesn't help much. An integral can be thought of as an area, so if the total area under a curve is to be less than one, the curve better not be bigger than one "too much". This is again heuristic and hand wavy, but the idea is that the function should not have to many "spikes" above one if the integral is to be less than one. Or one should only integrate over a sufficiently small interval. But in the above case the integral is over the real line and so the integrand might at best fluctuate wildly around one, but the overall contribution of the "above one spikes" should be "small". (Again, the argument is heuristic, but enough to make point.) And that is what I'm trying to understand better, the behaviour of [tex]|\mathcal{M}|[/tex].
     
  9. Feb 20, 2010 #8
    A counterexample: the photon propagator [tex]\propto 1/k^2[/tex] is singular at small values of transferred momentum k. This leads to the [tex]\propto 1/k^2[/tex] singularity of the scattering amplitude for the Coulomb scattering.


    A counterexample: Function [tex]\log(x)[/tex] is singular at x=0. However the integral [tex]\int_0^1 \log(x) dx = -1[/tex] is finite.


    Eugene.
     
  10. Feb 20, 2010 #9
    meopemuk -> I suspect it's massless particles that cause problems. Do you know of any such "problem" for massive theories?

    About the integral, I already said my argument was incorrect. A crucial difference is that the scattering amplitudes now gets integrated on the entire real line - unless in your integral above you meant something else. So this does constrain the integrand somewhat. Admittedly, not much. But precisely how much is what I'm asking for.
     
  11. Feb 20, 2010 #10

    If all involved particles are massive, then I would guess that scattering amplitudes are finite everywhere. However, I don't think that conservation of probability sets any limit on the amplitude values. The only limitation is that the integral of the squared amplitude is less than 1. This is a very soft restriction, which does not limit amplitude values at all.

    Eugene.
     
  12. Feb 4, 2012 #11
    Sorry to resurrect this, but I've started banging my head against this question after a row with my supervisor.

    I'm quite firmly of the opinion that the invariant amplitude [itex]\mathcal{M}[/itex] should be less than one. If you look at a derivation of the relationship between the cross-section and the scattering amplitude, one always uses the modulus square of the amplitude as a probability to infer the fraction of incident particles that actually get scattered.
    Regarding the point about probability density functions: In NRQM, the wavefunction has to have modulus less than one everywhere, so that the probability of finding a particle in any finite region is less than one. I believe an analagous statement holds in QFT, albeit one that's more complicated to phrase (probability for scattering to any number of on-shell particles with arbitrary momenta must be smaller than one, etc).
    As for the unconstrained growth of nearly-on-shell propagators, I think it's important that propagators are only components in diagrams, and diagrams themselves are only approxmations to the full amplitude. I think I'm right in saying that the sum of some set of diagrams exceeding modulus one means that you're making an approximation that breaks down.
     
  13. Feb 4, 2012 #12

    strangerep

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    Maybe I didn't understand what you meant, but consider this example. Suppose
    [itex] \psi(x) = \sqrt{2}[/itex] for x in [0,1/4], and [itex]\sqrt{2/3}[/itex] for x in (1/4, 1], and zero elsewhere. Then
    [tex]
    \int_{-\infty}^\infty dx |\psi(x)|^2
    ~=~ 2 \times \frac{1}{4} ~+~ \frac{2}{3} \times \frac{3}{4} ~=~ 1 ~.
    [/tex]
    but [itex]|\psi(x)| = \sqrt{2} > 1[/itex] in [0,1/4].

    (Or did I misunderstand the question?)
     
    Last edited: Feb 4, 2012
  14. Feb 5, 2012 #13
    Thanks for your reply, strangerep. You've forced me to clarify an important point. The wavefunction of NRQM isn't dimensionless, as a probability density; its numerical values depend on the units in which you measure volume. The probability of finding the particle in any finite volume in the above example is less than one, but because we're taking the whole volume of our system to define the unit volume, large values of [itex]|\psi|^2[/itex] can be suppressed by fractional volumes. In general, the only constraint on the value of the wavefunction is that the probability of finding the particle in unit volume must be less than or equal to one, with equality attained only in the limit where the entire system has unit volume. (So if, for example, you're talking about the hydrogen atom in an infinite and otherwise empty universe, and using SI units, or measuring length in Angstroms, or whatever, then your wavefunction should have numerical values less than one.)

    In my case, I'm calculating dimensionless quantities, [itex]\mathcal{M}[/itex] for 2->2 scattering amplitudes, so the overall scale of my amplitudes is actually meaningful.
     
  15. Feb 5, 2012 #14
    So you believe that the solution of the ground state of the quantum harmonic oscillator is not a suitable wave function, at large [itex]\omega[/itex]?

    Ref: http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

    If I am understanding your statements correcty, I think you have misunder stood something about probability densities.

    There is no upper limit on the target space [itex]\mathbb{R}^+[/itex] of a probability distribution. Only the integral of it across the sample space, e.g. [itex]\mathbb{R}[/itex], must be one. E.g. consider the normal distribution centered at the origin. Its integral is 1 and independent of the variance. But for small variances, it is very large near the origin. The total prob. within any region of the target space is still less than one. These are all dimensionless quantities.
     
  16. Feb 5, 2012 #15

    strangerep

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    I don't think that makes any difference. M is still a (generalized) function over some variables, and some sort of (multiple) integral over these variables is required for normalization of total probability to 1.

    Writing a scattering matrix as [itex]S=1+iT[/itex], and looking at Peskin & Schroeder's eq(4.73):
    [tex]
    \def\<{\langle}
    \def\>{\rangle}
    \def\to{\rightarrow}
    \<p_1 p_2 \dots | iT | k_A k_B\>
    ~=~ (2\pi)^4 \; \delta^{(4)}\Big(k_A + k_B - \sum p_f\Big)
    ~ i {\mathcal M}(k_A,k_B \;\to p_f)
    [/tex]
    we see that both the LHS and RHS must be interpreted as distributions over many-body momentum space. For the [itex]2\to 2[/itex] case, this is 4-body momentum space, but the Dirac delta removes one effective degree of freedom by constraining the momenta.

    If we specialize further and regard the initial momenta as fixed, then we have a distribution over the difference [itex]q := p_1 - p_2[/itex] of the 2 final state momenta. This still means that we must normalize by integrating the square of this quantity over all possible values of q. On some parts of the q axis, |M| could be >1, but <1 on others -- such that the overall normalization integral is satisfied.

    [Edit: strictly speaking, I should use the full S in the normalization, not just iT, but I don't think this changes the underlying point.]
     
    Last edited: Feb 5, 2012
  17. Feb 6, 2012 #16
    Thank you both for your helpful replies.
    I understood the idea of probability density functions, but I was deceived by the dimensional arguments in applying them to physics. Your example works of course because the extremely narrow width of the distribution is well confined within "unit volume", so one is effectively adding the product of a fractional volume and the appreciable values of [itex]|\psi|[/itex] to zero. In my argument above, I shouldn't have taken the "volume of the system" to be infinite, but to be what any sane person would use to characterise such a system- something like the standard deviation in position, or the Bohr radius of the hydrogen atom. If the unit of length is chosen to be small compared to such "characteristic scales", then the wavefunction should have modulus less than one.

    This was extremely helpful in understanding the relativistic case, thank you.

    In my problem, the amplitudes are typically of order 100 at q=0, and fall off like 1/q^4 at large q, so I think I still have a problem! I think I should also point out that you'd expect this distribution as a function of q to be normalised to something smaller than one, as it only corresponds to the total probability of 2->2 scattering, and in general in a relativistic problem you'd expect there to be nonzero probabilities associated with 2->n processes. We're also always using units that are comparable to the scales in the problem, the amplitudes perturbatively fall off like inverse powers of q (2 for every propagator) and q is bounded from above by the centre-of-mass energy, so I think it's still reasonable to start worrying if your amplitudes are very much bigger than one. Fair comment?
     
  18. Feb 6, 2012 #17

    strangerep

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    Do you mean a logarithmic divergence in M at high q? (If so, isn't this just a regularization+renormalization problem, as occurs so often in QFT?)

    Typically, I'd expect that those 2->n processes would involve higher order perturbation. I.e., (connected) Feynman diagrams with more vertices. So you've got to decide what perturbation order you're working at, and construct a normalization constant accordingly. In general, the normalization constant at (say) order-N perturbation does not contain contributions from order-(N+1) and higher. But if the integral over q diverges, then you've probably got a familiar regularization+renormalization task to deal with first. (It's hard to be more specific without knowing the details of your case.)

    More sophisticated QFT calculations (at order-N, say) therefore typically try to estimate the magnitude of order-(N+?) contributions, even if one can't calculate the latter explicitly. This is why one often sees "error bars" (parenthesized numbers at the end of a value) on theoretical predictions. E.g., some documents on the Particle Data Group's website talk about how anomalous magnetic moments are calculated, and try to estimate the magnitude of contributions from higher order terms (while also keeping an eye on renormalization).
     
  19. Feb 7, 2012 #18
    No, I mean that for increasing q the amplitude decays like 1/q^4. (Long story.)

    So, this is the crux of my problem, now it's established that the integral over q has to be some number of modulus less than one. :smile:

    I've been scouring my textbooks looking for the right way to go about normalising amplitudes, and as far as I can tell there are essentially two things that need to be normalised. The vacuum-vacuum amplitude needs to be set to one, but this is trivially accounted for by not including disconnected diagrams when calculating the amplitude. The other factor is the overall field-strength normalization [itex]Z= \langle p | \phi (0) | \Omega\rangle [/itex]. If you include the appropriate factors of [itex]1/\sqrt{Z}[/itex] then your distribution should automatically be normalised such that it integrates to less than one; if it isn't (e.g. as you increase the centre-of-mass energy), that's a sign that you need to include higher-order corrections. Right?
     
  20. Feb 8, 2012 #19

    strangerep

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    Still doesn't sound quite right to me. Can't you post the integral?

    With a decent renormalization scheme, I would have expected things to make sense at each order of perturbation.
     
  21. Feb 9, 2012 #20
    For the general principle under discussion, does the specific form of the integral matter? I took the following post to indicate that the 2->2 scattering amplitude was essentially a probability distribution in the Mandelstam variable t:
    What you don't specify here is the overall scale. My reasoning was that there is a finite probability for 2->n scattering for each value of n, which must collectively sum to one and will in general depend on energy (e.g. below the threshold energy for the production of some particle, the corresponding amplitude must vanish). It seems then that this overall scale will be roughly one at energies less than the rest energy of the lightest particle in the theory, and in general less than one.
     
  22. Feb 9, 2012 #21

    strangerep

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    Only in that I'm still not entirely sure whether I've been answering the question that's actually in your mind, or some distorted version of it that I've synthesized from your posts. :-)

    A more concrete question could eliminate that ambiguity...
     
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