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Polarization vector in Peskin & Schroeder QFT

  1. Jul 19, 2012 #1
    Hi all,

    A friend and I are working through Peskin and Schroeder, and we're both stumped with only the fourth equation! The interaction in question is [itex]e^+ + e^- \to \mu^+ + \mu^-[/itex] with a virtual photon as the inner branch. P&S state that
    [tex]\mathcal{M}\propto \langle \mu^+\mu^- | H_I | \gamma \rangle ^\nu \langle \gamma |H_I|e^+ e^-\rangle_\nu[/tex]
    where [itex]H_I[/itex] is the interaction part of the Hamiltonian. They then try to "guess the form of [itex]\langle \gamma |H_I|e^+ e^-\rangle_\nu[/itex], and conclude that if, in this case, the electron and muon particles are right-handed then that gives one unit of angular momentum in the [itex]+z[/itex] direction. They the say that
    where [itex]e[/itex] is electric charge.

    I have some questions:

    1. I've never come across a polarization four vector, and can't find descriptions online...could someone provide an explanation?

    2. How is this form of polarization vector determined, and could someone explain the "Thus we have..." statement?

  2. jcsd
  3. Jul 20, 2012 #2
    Photons have three key properties: they are described by vector fields, A^μ, (so their polarisation tensors, ζ, have one Lorentz index, ζ^μ), they are massless (so that the momentum four-vector squared vanishes), and they are transversely polarised with respect to their momentum vector (due to gauge invariance, which manifests as invariance under ζ --> ζ+k),
    k^2=0,\qquad k\cdot \zeta = 0.
    Normally, one normalises the polarisation tensor such that:
    \zeta^*\cdot \zeta =1.
    (This latter equation is consistent with the following non-ambiguous requirement for normalisation but not necessary: for a single-particle wavefunction, the total energy of the universe is .. lo and behold .. that of a single particle.) Now, the idea is to solve these conditions in a particular simple case, for instance when:
    k^{\mu} = (k^0,0,0,k^z),
    and then use the underlying Lorentz invariance to Lorentz boost to a more general frame (if necessary). One choice of polarisation tensor is the one you quote:
    \zeta^{\mu} = \frac{1}{\sqrt{2}}
    which you can check to convince yourself that it satisfies the above conditions. Notice also that both ζ and ζ* satisfy the above conditions, so photons have two polarisation states.

    If you want a deep understanding, the details are quite intricate and you should delve into a good textbook such as Hammermesh's book on group theory (I think the ~last? section of this book where he talks about little groups in particular, I don't have it at hand) and Weinberg's book on QFT.

    In general, photon or electron wavefunctions are suppose to transform like one-particle states under Lorentz transformations, and the irreducible representations of the full Poincar ́e group SO(3,1) are determined from the irreducible representations of the little group, SO(2) for massless states (e.g. photons) and SO(3) for massive states (e.g. electrons).
    Last edited: Jul 20, 2012
  4. Jul 20, 2012 #3
    By the way, if I remember correctly there should be a whole section on e+e- --> mu+mu- scattering in P&S. If I remember correctly, they should go through it in a lot of detail, but I don't have the book at hand. I would suspect that you were reading the introduction where most details are omitted.
  5. Jul 20, 2012 #4
    OK, so in the case I've given, the momentum is in the z-direction, and since the polarization vector satisfies those conditions you've given, I can see how the result is obtained. However, I do not yet see what the polarization vector is, so finding a book on gauge invariance might help.

  6. Jul 20, 2012 #5
    Indeed I am reading the introduction...so hopefully you're right, and things are cleared up!
  7. Jul 20, 2012 #6
    A polarisation vector (or tensor) more generally tells you the directions in which the field is fluctuating. There can be redundant fluctuations (in the sense that they can be removed by performing a symmetry transformation), or there can be physical fluctuations, the number of which corresponds to the number of physical degrees of freedom of your field. For instance, for gravitons the polarisation tensor has two indices (it's a rank two symmetric traceless tensor), and this leads to the more elaborate quadrupole fluctuations. You might like to check out Weinberg's (very expensive..) cosmology and gravity book for a great description of polarisation tensors, and the associated physical degrees of freedom. (If I remember correctly he also describes photon polarisation tensors there.)
  8. Jul 20, 2012 #7
    It might also be useful to note that you can trace the origin of the photon polarisation vector it back to Maxwell's equations, which are equations of motion for the vector field A^μ(x). The solution to these equations of motion may always be written in the form:
    A^{\mu}(x) = \frac{1}{\sqrt{2k^0V}}\Big(\zeta^{\mu}(x)+\zeta^{*\mu}(x)\Big),
    which ensures that A^μ(x) is real, while preserving its vector-like character. For example, for plane waves of a single wavelength,
    A^{\mu}(x) = \frac{1}{\sqrt{2k^0V}}\Big(\zeta^{\mu}e^{ik\cdot x}+\zeta^{*\mu}e^{-ik\cdot x}\Big),
    with, with k^μ the momentum, ζ^μ the polarisation vector and,
    k^2=0,\qquad k\cdot \zeta=0,\qquad {\rm and}\qquad \zeta^*\cdot \zeta =1,
    the normalisation being such that the the total energy in volume V, call it H_V, is that of a single particle:
    \frac{H_V}{k^0} = 1,\qquad {\rm with}\qquad k^0=\sqrt{{\bf k}^2}.
    H_V is in turn derived from the Hamiltonian density:
    H_V = \int_V d^3{\bf x}\mathcal{H}(x),
    which is in turn derived from your Maxwell Lagrangian by a Legendre transform. H_V is evaluated on the specific solution to the classical equations of motion (as given above).
    Last edited: Jul 20, 2012
  9. Jul 20, 2012 #8
    OK, you mean the potential formulation of Maxwell's equations: http://en.wikipedia.org/wiki/Mathem...lectromagnetic_field#Potential_field_approach ?

    In "Introductoin to elementary particles" by Griffiths, page 240, he says that the plane wave solution of Maxwell's equations with zero current density, in the Coulomb gauge (i.e. [itex]\Box A^\mu = 0[/itex]) are
    $$A^\mu (x) = ae^{-(i/\hbar)p\cdot x}\epsilon^\mu (p)$$
    where [itex]p[/itex] and [itex]x[/itex] are the relevant 4-vectors, [itex]\epsilon^\mu[/itex] is the polarization vector, and [itex]a[/itex] is a normalization factor. In the Coulomb gauge, one has [itex]p^\mu \epsilon^\mu = 0[/itex], [itex]\epsilon^0 = 0[/itex], so [itex]\mathbb{\epsilon}\cdot \mathbb{p} = 0[/itex], where the boldface symbols are for the spatial 3-vectors. If [itex]\mathbb{p}[/itex] is in the [itex]z[/itex] direction, then there are two linearly independent possibilities:
    which represent the two spin states.

    I think this is similar to what you have mentioned, though I fail to see two things:

    1. Where the complex numbers come from (though later on, he does mention that [itex]\epsilon^{\mu*}\epsilon_\mu = 1[/itex] and [itex]\epsilon_\mu^{(1)*}\epsilon^{(2)\mu} = 0[/itex], and I don't see where these conditions come from either.

    2. Why Griffiths has given two polarization vectors, and you only one (and it's conjugate). (I suspect that
    work, as they are orthogonal to [itex]\mathbb{p}[/itex] and satisfy [itex]\epsilon^{\mu*}\epsilon_\mu = 1[/itex] and [itex]\epsilon_\mu^{(1)*}\epsilon^{(2)\mu} = 0[/itex].)

    3. Why complex conjugate has all of a sudden entered into the dot product, i.e. [itex]\zeta^*\cdot \zeta =1[/itex]? (Griffiths introduces this too, but I don't see an explanation...).

  10. Jul 21, 2012 #9
    1) It is best to think of A(x) as being real; after all, from it follow the electric and magnetic fields, which satisfy Gauss' law, etc .. (You wouldn't, presumably, want to have to interpret the notion of complex charge densities and so on ..) So, if (when discussing plane waves) you like to think in terms of complex exponentials, e^{ik.x}, the most general possibility that solves the equations of motion and leading to real A(x) is that the ζ be complex. Finally, we always need the most general possibility if we want to make manifest all degrees of freedom.

    2) Precisely! As I mentioned above, the fact that ζ, ζ* are distinct while satisfying the above physical state conditions means that there are at least two (and in fact in this case only two) polarisation states. That the photon can only fluctuate in 2 of the 3 spatial dimensions is closely related to the fact that it's massless (or, equivalently, that its little group is SO(2) and not SO(3) as it would be for massive states in 4 flat spacetime dimensions).

    3) The ζ are complex and so one needs to specify three quantities:
    \zeta^*\cdot \zeta,\qquad \zeta\cdot \zeta,\qquad {\rm and}\qquad \zeta^*\cdot \zeta^*.
    The reason why we need to specify these is that they will naturally appear when we try to actually compute something.
    The available choices are largely arbitrary (as long as they span all directions orthogonal to the momentum k), and so it is convenient to choose:
    \zeta^*\cdot \zeta=1,\qquad \zeta\cdot \zeta=\zeta^*\cdot \zeta^*=0.
    Notice that we can't set all of them to zero because then our wavefunctions will have zero norm ...
  11. Jul 21, 2012 #10
    By the way, if you want to start thinking quantum, then you might want to start thinking about annihilation and creation operators. It's better however (in my opinion) to study the Feynman path integral quantisation where everything (apart from Grassmann variables and matrices) commutes and is (again in my opinion) much more intuitive. (Although most textbooks or field theory courses start off with canonical quantisation, most of them will admit that when one starts thinking about gauge theories or more elaborate theories one is better off with the path integral formulation.)
  12. Jul 21, 2012 #11
    Ah, that's actually quite obvious, so I feel a little silly for missing that.

    Oh, and the important point (if anyone else reads this) I forgot to mention was that, for the vectors I gave, [itex]\epsilon^{(2)} = \epsilon^{(1) *}[/itex], so that [itex]\zeta = \epsilon^{(1)},\ \zeta^* = \epsilon^{(2)}[/itex]. So the reason you only gave one, is that it's conjugate is like the additional one, since they're orthogonal.

    True, but again, I was stuck on a silly mistake: the dot product for complex numbers is [itex]a^{\mu*}b_\mu = \overline{a}\cdot b[/itex]. So the conditions I gave are simply saying that the polarisation vectors are normalised and orthogonal.

    Now I'm just not 100% sure on the link with the Hamiltonian and amplitudes etc, but I'm willing to fluff over that, as I've got the main problem solved. Cheers w4k4b4lool4 !

  13. Jul 21, 2012 #12


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    I just want to say that you shouldn't worry about being caught up on the fourth equation. The first chapter of P&S is really just a preview of what's to come, presented in a way that you won't understand everything yet, and the real meat starts in the second chapter.
  14. Jul 21, 2012 #13
    Yup, I've started the second chapter, and you're right.
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