# Helicity under rotation

Suppose that the state $$|p,\sigma\rangle$$ (for a massless particle) has 3 momentum $${\bf p}=p_3$$ (that is the momentum is in the $$z$$ direction) and that $$J_3|p,\sigma\rangle=\sigma|p,\sigma\rangle$$ where $${\mathbf{P }}=(P_1,P_2,P_3)$$ is the momentum operator and $${\mathbf{J }}=(J_1,J_2,J_3)$$ the angular momentum operator. Than

$$\frac{{\bf J} \cdot {\mathbf{P }}}{{|\mathbf{P}|}}|p,\sigma\rangle=\frac{{\bf J_3} {\mathbf{P_3 }}}{{|\mathbf{P}|}}|p,\sigma\rangle =\sigma|p,\sigma\rangle$$

Now suppose that we make a rotation around the $$x$$ axis than for a massless particle the state should transform like this $$U(\Lambda)|p,\sigma\rangle=e^{i\theta\sigma}| \Lambda p,\sigma\rangle.$$

Since $$\mathbf{J_2}|\Lambda p,\sigma\rangle=0$$, how can we proof that $$\frac{{\bf J} \cdot {\mathbf{P }}}{{|\mathbf{P }|}}|\Lambda p,\sigma\rangle=\frac{p'_3 \mathbf{J_3 }}{|\mathbf{P }|}|\Lambda p,\sigma\rangle=\frac{p'_3 \sigma}{|\mathbf{P }|}|\Lambda p,\sigma\rangle.$$

is equal to

$$\sigma|\Lambda p,\sigma\rangle$$

or since rotation leaves the norm of a vector invariant, that is the same as to proof that
$$\frac{p'_3}{|\mathbf{P }|}|\Lambda p,\sigma\rangle=\frac{p'_3}{p_3}|\Lambda p,\sigma\rangle$$

is equal to

$$|\Lambda p,\sigma\rangle$$

• Why is $\boldsymbol J\cdot\boldsymbol P=J_3P_3$? – AccidentalFourierTransform Aug 15 '19 at 15:17
• Because i am assuming that the $|p,\sigma\rangle$ has momentum $p_3$ (in the $z$ direction) and angular momentum $\sigma=\sigma_3$ (in the $z$ direction) – amilton moreira Aug 15 '19 at 16:56
• if $|p,\sigma\rangle$ has momentum in the $z$ direction, then $\frac{1}{|P|}|p,\sigma\rangle=\frac{1}{p_3}|p,\sigma\rangle$, and the factors of $p_3$ disappear. – AccidentalFourierTransform Aug 15 '19 at 19:04
• You are right but the norm of a vector is rotation invariant and so $\frac{1}{|P|}|\Lambda p,\sigma\rangle=\frac{1}{p_3}|\Lambda p,\sigma\rangle$ and then we would have $\frac{p'_3}{|\mathbf{P }|}|\Lambda p,\sigma\rangle=\frac{p'_3}{p_3}|\Lambda p,\sigma\rangle$. But the factor $\frac{p'_3}{p_3}$ does not disapear – amilton moreira Aug 16 '19 at 6:54

Your mistake is overlooking the fact that the parameter $$\sigma$$ measures the helicity referring to the actual direction of $${\bf p}$$.
Therefore, following again the first (correct) part of your reasoning where $$p$$ was directed along $$x_3$$, simply replacing the unit vector $${\bf e}_3$$ for the rotated axis $$\Lambda {\bf e}_3$$ you find that $$|\Lambda p, \sigma\rangle$$ satisfies $$\frac{{\bf J}\cdot {\bf P}}{|{\bf P}|}|\Lambda p, \sigma\rangle = (\Lambda {\bf e}_3)\cdot {\bf J}|\Lambda p, \sigma\rangle = \sigma |\Lambda p, \sigma\rangle\:.$$ As $$\Lambda {\bf e}_3$$ generally admits components along both $$x_2$$ and $$x_3$$, you cannot argue that $$J_2 |\Lambda p, \sigma\rangle = 0,$$ $$J_3 |\Lambda p, \sigma\rangle =\sigma |\Lambda p, \sigma \rangle,$$ and your claim does not follows.
• Suppose that $\Lambda {\bf e}_3=\cos\theta{\bf e}_3+\sin\theta{\bf e}_2$ i am not seeing why $(\cos\theta J_3+\sin\theta J_2)|\Lambda p,\sigma\rangle=(\sigma \cos\theta +\sin\theta J_2)|\Lambda p,\sigma\rangle$ is equal to $\sigma |\Lambda p,\sigma\rangle$ – amilton moreira Aug 16 '19 at 9:29
• By definition. It is the definition of the states $|p,\sigma\rangle$. Your second identity above is wrong $J_3|\Lambda p,\sigma\rangle \neq \sigma|\Lambda p,\sigma\rangle$. That is the same mistake you took in the post. – Valter Moretti Aug 16 '19 at 9:41
• If states do transform like this $U(\Lambda)|p,\sigma\rangle=e^{i\theta\sigma}| \Lambda p,\sigma\rangle.$ that means that the $z$ component of angular momentum does not change so we would have $J_3|\Lambda p,\sigma\rangle =\sigma|\Lambda p,\sigma\rangle$ or am i understanding it wrong? – amilton moreira Aug 16 '19 at 9:48
• $|p,\sigma\rangle$ has the property that $\sigma$ is the component along ${\bf p}$ of ${\bf J}$. If ${\bf p}$ is directed along ${\bf e}_3$, then $\sigma$ is the value of the component $J_3$. Therefore, keeping ${\bf p}$ as said, if $\Lambda$ is a rotation $\Lambda p$ has not spatial components directed along ${\bf e}_3$ and consequently $\sigma$ cannot coincide to the component $J_3$ any more. – Valter Moretti Aug 16 '19 at 9:49