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How can it be shown without using the little group formalism?

Let's have the Wigner's classification for the irreducible represetation of the Poincare group. For the massless case the eigenvalues of two Casimir operators of the group, the squares of Pauli-Lubanski operator and momentum operator, $\hat {W}_{\alpha}W^{\alpha}, \hat {P}_{\alpha}\hat {P}^{\alpha}$, is equal to zero.

Together with $\hat {W}_{\alpha}\hat {P}^{\alpha} = 0$ it leads to an expression $\hat {W}_{\alpha} = \hat {h}\hat {P}_{\alpha}$, where eigenvalues of $\hat {h}$ has dimension like angular momentum. It is called helicity. I want to get it "properties" without using small groups formalism (by the other words, not as Weinberg).

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  • $\begingroup$ @WetSavannaAnimalakaRodVance . Let's have the Wigner's classification for the irreducible represetation of the Poincare group. For the massless case the eigenvalues of two Casimir operators of the group, the squares of Pauli-Lubanski operator and momentum operator, $\hat {W}_{\alpha}W^{\alpha}, \hat {P}_{\alpha}\hat {P}^{\alpha}$, is equal to zero. $\endgroup$ – user8817 Aug 12 '13 at 1:09
  • $\begingroup$ Sorry, I slightly misunderstood the point of your question. But it is good to have a fuller characterization. $\endgroup$ – WetSavannaAnimal Aug 12 '13 at 1:10
  • $\begingroup$ Together with $\hat {W}_{\alpha}\hat {P}^{\alpha} = 0$ it leads to an expression $\hat {W}_{\alpha} = h\hat {P}_{\alpha}$, where $h$ has dimension like angular momentum. It is called helicity. I want to get it "properties" without using small groups formalism (by the other words, not as Weinberg). $\endgroup$ – user8817 Aug 12 '13 at 1:11
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    $\begingroup$ @PhysiXxx: For a massless particle, helicity and chirality are the same thing. Chirality is determined by whether the particle transforms in a right or left-handed representation (of the Lorentz group). That is : representations $(h,0)$ have chirality $h$, while representations $(0,h)$ have chirality $-h$. $\endgroup$ – Trimok Aug 12 '13 at 8:34
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    $\begingroup$ @PhysiXxx : For massive particle, as electrons, helicity depends on the frame. Suppose you electron has some speed $v_z$.If you take a new frame which speed relatively to the electron which is positive, you will have an opposite helicity than if you take a frame which speed relatively to the electron is negative. Now, for massless particles, you cannot have a frame which speed is greater than $c$, so the helicity/chirality is constant, and in fact is coming from the representations of the Lorentz group. $\endgroup$ – Trimok Aug 13 '13 at 8:29
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  1. Construction of the helicity formula using 3-vector notation

The zero component of the pauli Lubanski vector

$W^0 = \epsilon^{0 ijk}J_{ij}p_k = \epsilon^{ijk}J_{ij}p_k $

The angular momentum genrerators

$ j^k = \epsilon^{ijk}J_{ij}$

Thus

$W^0 = j^k p_k = \vec{j}.\vec{p} $

The orbital angular momentum

$ \vec{l} = \vec{x} \times \vec{p}$

is orthogonal to the momentum:

$ \vec{l}.\vec{p} = 0$

And since the total angular momentum is the vector sum of the orbirtal and the spin angular momenta

$ \vec{j} = \vec{l} + \vec{\Sigma}$

Thus

$W^0 = j^k p_k = \vec{j}.\vec{p} = (\vec{j-l}).\vec{p} = \vec{\Sigma}.\vec{p} $

Now, since

$W^0 = \hat{h} p_0$

and for a massless particle

$ p_0 = p$

We obtain:

$ \hat{h} = \frac{\vec{\Sigma}.\vec{p}}{p_0} = \vec{\Sigma}.\hat{p}$

  1. The helicity operator

$$\hat{h} = \Sigma.\hat{p}$$

where $\Sigma$ is the spin operator and $\hat{p}$ is the momentum unit vector is a projection along the axis $\hat{p}$ of a spin operator, thus one might expect it to have for a helicity $\lambda$ the eigenvalues $\lambda$, $\lambda-1$, ..., $-\lambda$.

However, the eigenvectors corresponding to all eigenvalues except $\pm \lambda$ are not physical, because they describe longitudinal polarizations which do not exist in free massless particles.

Here is an example of the massless spin-1 case (photon). In this case, we may choose the spin operators as:

$ \Sigma_x = \begin{bmatrix} 0 & 0& 0\\ 0 & 0& -i\\ 0 & i& 0 \end{bmatrix}$

$\Sigma_y = \begin{bmatrix} 0 & 0& i\\ 0 & 0& 0\\ -i & 0& 0 \end{bmatrix}$

$\Sigma_z = \begin{bmatrix} 0 & -i & 0\\ i & 0& 0\\ 0 & 0& 0 \end{bmatrix}$

The action of the Helicity operator on (say), the electric field in the momentum representation is:

$$\hat{h} \vec{E} = i\begin{bmatrix} 0 & -\hat{p}_z & \hat{p}_y\\ \hat{p}_z & 0& -\hat{p}_x\\ -\hat{p}_x & \hat{p}_x& 0 \end{bmatrix}\begin{bmatrix} E_x\\ E_y\\ E_z \end{bmatrix} = i \hat{p}\times \vec{E}$$

Thus:

$$\hat{h}^2 \vec{E} = - \hat{p} \times ( \hat{p}\times \vec{E}) = \vec{E} -\hat{p}(\hat{p}. \vec{E})$$

But, since for a free electromagnetic field:

$$\hat{p}. \vec{E} = 0$$

We get:

$$\hat{h}^2 = 1$$,

and the only admissible eigenvalues are $\pm 1$

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  • $\begingroup$ "...The helicity operator $\hbar = \Sigma \hat {p}...$, from where did you take this definition? $\endgroup$ – user8817 Aug 12 '13 at 11:33
  • $\begingroup$ "... the eigenvectors corresponding to all eigenvalues except ±λ are not physical, because they describe longitudinal polarizations which do not exist in free massless particles...", - can it be generalized without using examples? $\endgroup$ – user8817 Aug 12 '13 at 11:35
  • $\begingroup$ @PhysiXxx I have added an update showing the computation of the helicity in 3-vector notation. The generalization of the case of the electromagnetic field (that I know of) requires the use of spinor notation of the massless field equations. It requires more than a few lines. I'll try to find a good reference. $\endgroup$ – David Bar Moshe Aug 12 '13 at 12:15
  • $\begingroup$ Thank you! "...I'll try to find a good reference...". I would be grateful. $\endgroup$ – user8817 Aug 12 '13 at 23:07
  • $\begingroup$ But there is a question about derivation an expression of helicity operator. How did you avoid problem with $\hat {p}_{0}$ (by the other words - interpreting it as a scalar) when passing from $$ \hat {W}^{0} = \hat {h}\hat {p}^{0} $$ to $$ \hat {h} = \frac{\hat {W}^{0}}{p_{0}}? $$ You used something like $$ \hat {W}^{0}\psi = \hat {h}\hat {p}^{0}\psi = \hat {h}p^{0}\psi \Rightarrow \hat {h}\psi = \frac{\hat {W}^{0}}{p^{0}}\psi ? $$ $\endgroup$ – user8817 Aug 12 '13 at 23:09

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