Chirality can be interpreted as a property of Lorentz group - Lorentz transformation of field through representation $(s, 0)$ or representation $(0, s)$. For the massless particles one says, that chirality and helicity are the same (helicity values are a chirality values multiplied by $ \hbar $). So it means, that chirality values are $\pm s$.

The questions:

  1. How can I get these values for chirality? I.e., how right- and left-handed representation of the Lorentz group are connected with plus-minus signs of the chirality value?

  2. Equivalence of helicity and chirality in the massless case is little not obvious for me. One uses idea that for the massless case helicity of given particle is Lorentz-invariant, the same as chirality. But it seems to me that this is not enough to identify these values. So how to identity them?

  3. Also, in my previous question on the subject (Why helicity is proportional to the spin of particle and has two values?), there was a comment which give an idea of getting an answer on the question, using an equivalence of helicity and chirality. Can the helicity operator properties for massless case be given from the equivalence between helicity and chirality? Or at the beginning we get the properties of helicity operator, and only then we can identify them?


1 Answer 1


In the general case of a representation $(j_-, j_+)$ of Lorentz, chirality is defined as $j_+ - j_-$. The chiralities $\pm s$ for $(s,0)$ and $(0,s)$ are a particular case of this definition.

So then I assume the answer to your first question is "by definition", and the next question is, "How can this definition of chirality be related to helicity ?".

To find a relation between chirality and helicity, let's consider the wave function $$ \langle 0 | O_{(m_-,m_+)}(0) | \bar{p}, \sigma\rangle $$ where $O_{(m_-, m_+)}$, with $m_- = - j_-, ..., j_-, m_+ = -j_+, ..., j_+$ denotes an operator transforming in the $(j_-, j_+)$ representation of Lorentz, and $| \bar{p}, \sigma \rangle$ is a massless state in the reference $\bar{p}^\mu=(p,0,0,p)$ with helicity $\sigma$.

We can show, by group theoretic arguments (proof below), that this matrix element is non-vanishing only in the case where $m_- + m_+ = - \sigma$ and $j_+ - j_- = \sigma$. The second relation establishes the relation between chirality and helicity. More precisely, the statement is that massless particles of helicity $\sigma$ can only be interpolated by operators of chirality $j_+ - j_- = \sigma$.

Proof of the statement

Defining the Pauli-Lebanski vector $ W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} J_{\nu \rho} P_\sigma $, and denoting $J_\pm^i$ the generators in the $j_\pm$ $SU(2)$ irreps respectively, we have the following relations / definitions :

$$ \begin{cases} W^{\pm} \equiv W^1 \pm i W^2 \\ J_{\pm}^i = \frac{1}{2} (J^i \pm i K^i) \\ W^1 = K^2 - J^1 \\ W^2 = - K^1 - J^2 \end{cases} $$

After some easy manipulations, we find $$ W^+ = - 2 \sqrt{2} J_+^+ \\ W^- = - 2 \sqrt{2} J_-^- $$ where, $J^+_+$ is the raising operator in the $j_+$ rep, and $J_-^-$ is the lowering operator in the $j_-$ rep.

Since in the massless case, the $W^\pm$ operators should be 0, we have $$ J_-^- | \bar{p}, \sigma \rangle = J_+^+ | \bar{p}, \sigma \rangle = 0 $$

Remember the action of the $J_-^-$, $J_+^+$, $J^3$ operators : $$ [J^3,O_{(m_-,m_+)}] = [J_-^3, O_{(m_-,m_+)}] + [J_+^3,O_{(m_-,m_+)}] = (m_- + m_+) O_{(m_-,m_+)} \\ [J_-^-,O_{(m_-,m_+)}] \propto O_{(m_- - 1,m_+)} \\ [J_+^+,O_{(m_-,m_+)}] \propto O_{(m_-,m_+ + 1)} $$

First note that the first relation implies

$$ - \sigma \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle = \langle 0 | [J^3,O_{(m_-, m_+)}] (0) | \bar{p}, \sigma \rangle = (m_- + m_+) \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$

which implies $-\sigma = m_- + m_+$ for the matrix element to be non-zero.

Now, assume that $j_- > m_-$, so that that there is a state $O_{m_- + 1, m_+}$. In this case, the second equation implies $$ 0 = \langle 0 | [J_-^-, O_{m_- + 1, m_+}] | \bar{p}, \sigma \rangle \propto \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$ The matrix element would be vanishing. For it not to vanish, we thus need $m_- = j_-$ to be the highest state. Similarly, assume $m_+> - j_+$ so that there is an operator $O_{m_-, m_+ - 1}$. Then $$ 0 = \langle 0 | [J_+^+, O_{m_- , m_+ - 1}] | \bar{p}, \sigma \rangle \propto \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$ Thus for the matrix element not to vanish, $m_+ = - j_+$. Comining these conditions with the first one $$ j_+ - j_- = - m_+ - m_- = \sigma $$ Otherwise, the matrix element vanishes.


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