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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

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

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    $\begingroup$ Why is $\boldsymbol J\cdot\boldsymbol P=J_3P_3$? $\endgroup$
    – AccidentalFourierTransform
    Commented Aug 15, 2019 at 15:17
  • $\begingroup$ 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) $\endgroup$
    – amilton moreira
    Commented Aug 15, 2019 at 16:56
  • $\begingroup$ 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. $\endgroup$
    – AccidentalFourierTransform
    Commented Aug 15, 2019 at 19:04
  • $\begingroup$ 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 $\endgroup$
    – amilton moreira
    Commented Aug 16, 2019 at 6:54

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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.

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  • $\begingroup$ 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 $\sigma |\Lambda p,\sigma\rangle$ $\endgroup$
    – amilton moreira
    Commented Aug 16, 2019 at 9:29
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    $\begingroup$ 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. $\endgroup$
    – Valter Moretti
    Commented Aug 16, 2019 at 9:41
  • $\begingroup$ 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? $\endgroup$
    – amilton moreira
    Commented Aug 16, 2019 at 9:48
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    $\begingroup$ $|p,\sigma\rangle$ 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. $\endgroup$
    – Valter Moretti
    Commented Aug 16, 2019 at 9:49
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    $\begingroup$ Please, stop insisting on your erroneous claims and try to focus on what I am trying to communicate you. Otherwise this discussion is useless. $\endgroup$
    – Valter Moretti
    Commented Aug 16, 2019 at 9:51

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