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  • Whenever Feynman rules are stated they are always without any mention of the helicities - this I find to be very confusing. How does one introduce and account for that?

  • Is there an intuitive/simple argument for why massless particles should have "helicities" (and not polarizations) and they can only be of the form $\pm\text{ some positive integer}$? (..i have seen some very detailed arguments for that which hinge on the representation theory for the little group of massless particles and various other topological considerations - i am here looking for some "quick" explanation for that..)

  • Is there some reason why polarized gluon scattering amplitudes at the tree-level can somehow "obviously" be written down? Like for example, consider a process where two positive helicity gluons of momenta $p_1$ and $p_2$ scatter into two negative helicity gluons of momenta $p_3$ and $p_4$ then at tree level the scattering amplitude is,

$A(p_1^+,p_2^+,p_3^-,p_4^-)= \frac{ig^2}{4p_1.p_2} \epsilon_2^+ \epsilon_3^-(-2p_3.\epsilon_4^-)(-2p_2.\epsilon_1^+)$

where $\epsilon^{\pm}_i$ is the polarization of the $i^{th}$ particle.

I have at places seen this expression being almost directly written down. Is the above somehow obvious?

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helicity ($h = \vec{S} \cdot \hat{p}$) is only useful for massless particles because for them it is Lorentz Invariant. For particles with mass you can find a Lorentz transformation (just travel in the same direction as the particle but faster) such that $\hat{p} \rightarrow - \hat{p}$. So the $h$ for massive particles is frame dependent. Helicities are $\pm$ some (half) integer for the same reason the spin takes (half) integer values. For that you need to understand the representations of $SU(2)$. That should answer the second point at least. –  Kyle Feb 23 '12 at 3:51
    
@Kyle I guess my question about the numerical restriction on the values of helicity was not clear. It seems to work unlike the representation theory of $SU(2)$. A particle in the spin $s$ representation of $SU(2)$ can have all the spin states from $-s$ to $s$ in steps of $1$ where $s$ is restricted to be a (half)integer. BUT that is NOT how helicity works - helicities can only be $\pm (half) integer$. I wanted to know if there is a "quick" explanation for that. –  user6818 Feb 25 '12 at 0:49
    
I think that is how helicity works.... if you are thinking about massless spin $1/2$ particles then clearly the helicity will only by $\pm 1/2$. For massless spin 1 particles you could in principle think that the helicity could be $0, \pm 1$ but the $0$ mode turns out break gauge invariance so the helicity is the $\pm 1$. If you were to look at helicity states for massless spin $3/2$ particles, you would have $\pm 3/2, \pm 1/2$. –  Kyle Feb 25 '12 at 4:06
    
@Kyle Can you kindly explain as to why does helicity work this way? I want to know of a "quick proof" of this - that for massless spin $\frac{1}{2}$ the helciity has to be $\pm \frac{1}{2}$ and for massless spin $\frac{3}{2}$ the helicity has to be $\pm \frac{3]{2}$ etc.. –  user6818 Feb 25 '12 at 19:28
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2 Answers

up vote 3 down vote accepted

The argument for the first question goes as follows:

Consider the Pauli-Lubanski vector $ W_{\mu} = \epsilon_{\mu\nu\rho\sigma}P^{\nu}M^{\rho\sigma}$. Where $P^{\mu}$ are the momenta and $M^{\mu\nu}$ are the Lorentz generators. (The norm of this vector is a Poincare group casimir but this fact will not be needed for the argument.)

By symmetry considerations We have $W_{\mu} P^{\mu} = 0$. Now, in the case of a massless particle, a vector orthogonal to a light-like vector must be proportional to it (easy exercise). Thus $ W^{\mu} = h P^{\mu}$, ($ h = const.$). Now, the zero component of the Pauli-Lubanski vector is given by:

$ W_{0} = \epsilon_{0\nu\rho\sigma}P^{\mu}M^{\mu\nu} = \epsilon_{abc}P^{a}M^{bc} = \mathbf{P}.\mathbf{J}$, (where the summation after the second equality is on the spatial indices only, and $\mathbf{J}$ are the rotation generators ).

Therefore the proportionality constant $ h = \frac{W^{0}}{P^{0}}= \frac{\mathbf{P}.\mathbf{J}}{|\mathbf{P}|}$ is the helicity.

Now, on the quantum level, if we rotate by an angle of $2 \pi$ around the momentum axis, the wave function acquires a phase of: $exp(2 \pi i\frac{\mathbf{P}}{|\mathbf{P}|}.\mathbf{J}) = exp(2 \pi i h)$. This factor should be $\pm 1$ according to the particle statistics thus $h$ must be half integer.

As for the second question, a very powerful method to construct the gluon amplitudes is by the twistor approach. Please see the following article by N.P. Nair for a clear exposition.

Update:

This update refers to the questions asked by user6818 in the comments:

For simplicity I'll consider the case of a photon and not gluons.

The strategy of the solution is based on the explicit construction of the angular momentum and spin of a free photon field (which depend on the polarization vectors) and showing that the above relations are satisfied for the photon field. The photon momentum and the angular momentum densities can be obtained via the Noether theorem from the photon Lagrangian. Alternatively, it is well known that the photon linear momentum is given by the Poynting vector (proportional to) $\vec{E}\times\vec{B}$, and it is not difficult to convince onself that the total angular momentum density is (proportional to) $\vec{x}\times \vec{E}\times\vec{B}$.

Now, the total angular momentum can be decomposed into angular and spin angular momenta (please see K.T. Hecht: quantum mechanics (page 584 equation 16))

$\vec{J} = \int d^3x (\vec{x}\times \vec{E}\times\vec{B}) =\int d^3x (\vec{E}\times\vec{A} + \sum_{i=1}^3 E_j \vec{x} \times \vec{\nabla} A_j )$

The first term on the right hand side can be interpreted as the spin and the second as the orbital angular momentum as it is proportional to the position.

Now, Neither the spin nor the orbital angular momentum densities are gauge invariant (only their sum is). But, one can argue that the total orbital angular momentum is zero because the position averages to zero, thus the total spin:

$ \vec{S} =\int d^3x (\vec{E}\times\vec{A})$

is gauge invariant:

Now, we can obseve that in canonical quantization: $[A_j, E_k] = i \delta_{jk}$, we get $[S_j, S_k] = 2i \epsilon_{jkl} S_l$. Which are the angular momentum commutation relations apart from the factor 2.

Now, by substituting the plane wave solution:

$A_k = \sum_{k,m=1,2} a_{km} \vec{\epsilon_m}(k) exp(i(\vec{k}.\vec{x}-|k|t)) +h.c.$

(The condition $\vec{\epsilon_m}(k).\vec{k} = 0$, is just a consequence of the vanishing of the sources).

We obtain:

$\vec{S} = \sum_{k,m=1,2}(-1){m} a^\dagger_{km}a_{km} \hat{k} = \sum_{k}(n_1-n_2)\hat{k}$

(where $n_1$, $n_2$ are the numbers of right and left circularly polarized photons). Thus for a single free photon, the total spin, thus the total angular momentum are aligned along or opposite to the momentum, which is the same result stated in the first part of the answer.

Secondly, the photon total spin operators exist and transform (up to a factor of two) as spin 1/2 angular momentum operators.

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Thanks for your illuminating reply. Is there a typo in your definition of $W_\mu$? Like shouldn't it be $W_\mu = \epsilon_{\mu \nu \rho \sigma}M^{\nu \rho}P^{\sigma}$? Can you kindly explain how your definition of "helicity" corresponds to the "other" way of thinking as the $4$-vectors $\epsilon_\mu(p)$ that is required to describe a photon of momentum $p$ as $A_\mu(p) = \epsilon_\mu(p) e^{ip_\nu x^\nu}$? (..and one can choose gauge such that $\epsilon_0 = 0$ and $\epsilon_\mu p^\mu = 0$..) And if you could explain as to why there does not exist a notion of "spin" for massless particles? –  user6818 Feb 25 '12 at 1:01
    
I do know of the explanation about why there is no spin for massless particles coming from thinking of what their little group is - its $SO(3)$ massive particles and $SO(2)$ massless and hence spin for the former and helicity for the later. But I am asking if there is a "quick" intuitive explanation and if you can reconcile these different pictures - like see my comments on Sidious Lord's answer. –  user6818 Feb 25 '12 at 1:20
    
I have added an update to the answer, in which I have corrected the index error and added an explanation on the relation of the helicity to the polarization vectors (not based on Wigner's classification). Also, I have added an explanation about the photon spin. –  David Bar Moshe Feb 26 '12 at 16:20
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Whenever Feynman rules are stated they are always without any mention of the helicities - this I find to be very confusing. How does one introduce and account for that?

In QFT you can represent the state of a gauge quanta by its momentum and helicity. You can also do it in gauge dependent way by specifying the momentum and a polarization vector $\epsilon_\mu(p)$. This is null $\epsilon(p)^2 = 0$ and is subject to a gauge equivalence $\epsilon_\mu(p) \sim \epsilon_\mu(p) + p_\mu$. When you compute a scattering amplitude by using Feynman rules the way you describe the states of the external particles is by using polarization vectors. This is standard textbook material so I don't see what's confusing you.

Is there an intuitive/simple argument for why massless particles should have "helicities" (and not polarizations) and they can only be of the form ± some positive integer? (..i have seen some very detailed arguments for that which hinge on the representation theory for the little group of massless particles and various other topological considerations - i am here looking for some "quick" explanation for that..)

Mathematically, particles are in correspondence with irreducible representations of the Lorentz group. The representation theory of the Lorentz group is a bit delicate because it is non-compact. But the non-compactness is easy to understand physically: it's because one can boost by an arbitrary amount. So let's forget about the boosts and look only at the Lorentz transformations which preserve the direction of momentum. You should be able to show that these transformations form a $SO(2) \simeq U(1)$ subgroup. They act on the states by multiplying them by a phase. The charge under this $U(1)$ group is just the helicity.

Is there some reason why polarized gluon scattering amplitudes at the tree-level can somehow "obviously" be written down?

What do you mean by obviously? It's easy to write it down at tree level, it's a typical QFT exercise in using Feynman rules. And by the way, it seems to me the formula you wrote down can't be right since it breaks gauge invariance. It should be invariant under $\epsilon_\mu(p) \sim \epsilon_\mu(p) + p_\mu$.

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Thanks for the reply. I am aware of the definition of helcity that you give using representation theory. But this definition is hard to reconcile with the one given through $\epsilon_\mu$ or the one in terms of the Pauli-Lubansky vector (like the answer above). It would be great if you can help connect these 3 "different" pictures of helicity. And also if you can give a "quick" explanation from your picture of why helicity should be $\pm (half)integer$. –  user6818 Feb 25 '12 at 1:11
    
About the Feynman rules with helicity - I guess my question was not clear. Let us say that I have a 3-gluon vertex where the three gluons with all incoming momenta are labelled by the triplet $(colour\text{ }index,Lorentz\text{ }index,momentum)$ as $(a,\mu,p_1), (b,\nu,p_2), (c,\lambda,p_3)$ then the vertex factor for this is $-igf^{abc}[\eta^{\mu \nu}(p_1-p_2)^\lambda + \eta^{\nu \lambda}(p_2-p_3)^\mu + \eta ^{\lambda \mu}(p_3-p_1)^\nu]$ - this is standard. Now can you tell me as to how in the above do I incorporate helicities of the three gluons if that is also specified? –  user6818 Feb 25 '12 at 1:17
    
@user6818 Your question about the relation between polarization vector $\epsilon_\mu$, Pauli-Lubansky vector and the helicity is much more precise than your initial question so I can also be more precise in my answer. The quickest way I know to connect helicity and polarization vectors uses the spinor helicity language. For any vector in 4D we can form the following quantity $v_{\alpha \dot{\alpha}} = v_\mu \sigma^\mu_{\alpha \dot{\alpha}}$, where $\sigma^\mu = (1, \mathbf{\sigma})$ and $\mathbf{\sigma}$ are the Pauli matrices. [see next...] –  Sidious Lord Feb 25 '12 at 8:11
    
[...continued] Then a light-like momentum can be written in a factorized form $p_{\alpha \dot{\alpha}} = \lambda_\alpha \tilde{\lambda}_{\dot{\alpha}}$ (prove this!). The little group acts by multiplication by a phase: $\lambda \to e^{i \phi} \lambda$ and $\tilde{\lambda} \to e^{-i \phi} \tilde{\lambda}$. Then suppose I want to write some wave-function for a massless spin $\tfrac 1 2$ particle. In momentum space this will be just $\lambda$ for helicity $\tfrac 1 2$ and $\tilde{\lambda}$ for helicity $-\tfrac 1 2$ (or the opposite, depending on conventions) [see next...] –  Sidious Lord Feb 25 '12 at 8:17
    
[...continued] The rule is that for a particle of helicity $h$ the wave-function transforms by a phase $e^{2 i h \phi}$ under a rotation by $\phi$ which preserves the momentum. Now, for a particle of spin $1$ things get more interesting. If you think a bit about it you can write a wave-function which transforms as $e^{2 i \phi}$ but it's going to be of the form $\lambda_\alpha \lambda_\beta$. This is not of vector type (vectors have one undotted and one dotted index) but in factor corresponds to the anti-selfdual part of a rank two tensor. [see next...] –  Sidious Lord Feb 25 '12 at 8:24
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