# Helicity states

On page 71 of Weinberg's book The Quantum Theory of Fields: Volume I, he defines the operators $$A=J_2+K_1$$and $$B=-J_1+K_2$$ where $${\mathbf{J }}=(J_1,J_2,J_3)$$ are the rotation generators and $${\mathbf{K }}=(K_1,K_2,K_3)$$ are the boost generators. These operators have the following commutation relationships: $$[J_3,A]=iB$$ $$[J_3,B]=-iA$$ $$[A,B]=0$$
Assume that $$k=[\omega,0,0,\omega]$$

For physical reasons it is assumed that $$A|k,\sigma\rangle=B|k,\sigma\rangle=0$$

so states are defined by the eigenvalue of the operator $$J_3$$ :$$J_3|k,\sigma\rangle=\sigma|k,\sigma\rangle$$ Then he was able to show that under a Lorentz transformation, a massless particle state should transform like this: $$U(\Lambda)|k,\sigma\rangle=e^{i\theta\sigma}| \Lambda k,\sigma\rangle.$$ My question is, for this case ($$k=[\omega,0,0,\omega]$$) should we have $$J_3e^{i\theta\sigma}| \Lambda k,\sigma\rangle= \sigma e^{i\theta\sigma}| \Lambda k,\sigma\rangle$$ ?

Weinberg uses $$k=[\omega,0,0,\omega]$$ as a fiducial 4-momentum of a massless particle moving along the z or $$x^{3}$$ axis (this answer uses the Minkowski metric $$\eta^{00}=+1,\eta^{11}=-1,...$$) . Weinberg uses the notation $$p$$ for a general 4-momentum which is made by transforming the fiducial momentum by a standard Lorentz transformation. $$$$p=L(p)k$$$$ Here, $$L(p)$$ is the standard Lorentz transformation that takes the fiducial momentum $$k$$ into the general momentum $$p$$.

Weinberg writes (equation (2.5.39) of his QFT text), $$$$J_{3}|k,\sigma\rangle=\sigma|k,\sigma\rangle$$$$ Here $$\sigma$$ is the particle's helicity (spin measured along the direction of the particle's momentum). The questioner wrote this equation in terms of the general momentum $$p$$ instead of the fiducial momentum $$k$$. Notice that this equation relates to the fiducial momentum $$k$$ that points along the z axis. The above equation would not be true if the fiducial momentum $$k$$ were replaced with the general momentum $$p$$, $$J_{3}|p,\sigma\rangle\neq\sigma|p,\sigma\rangle$$ because the particle is no longer moving along the z axis.

The question asks whether or not $$J_{3}|\Lambda p,\sigma\rangle=\sigma|\Lambda p,\sigma\rangle$$? Since $$\Lambda p$$ is just some arbitrary momentum, we might as well replace $$\Lambda p$$ by $$p$$. The equation is now $$J_{3}|p,\sigma>=\sigma|p,\sigma\rangle$$ which we've already said is false.

Edit to answer version 2 of the question.

The questioner writes that Weinberg was able to show that $$U(\Lambda)|k,\sigma\rangle=e^{i\theta\sigma}|\Lambda k,\sigma\rangle$$. In fact, Weinberg was not restricted to acting on the fiducial state $$k$$ with a Lorentz transformation $$\Lambda$$. Equation (2.5.42) of his QFT textbook gives the response of a general momentum $$p$$ to a Lorentz transformation, $$$$U(\Lambda)|p,\sigma\rangle=e^{i\sigma\theta}|\Lambda p,\sigma\rangle$$$$ The above formula assumes a Lorentz-invariant measure (Haar measure) is used to sum over momentum states and this removes the need for the square root in the textbook formula.

The question then asks if the following equation is true. $$$$J_3 e^{i\theta\sigma}| \Lambda k,\sigma\rangle= \sigma e^{i\theta\sigma}| \Lambda k,\sigma\rangle$$$$ Firstly, the factors $$e^{i\theta\sigma}$$ are complex numbers and so the one on the LHS can be taken through the angular momentum operator $$J_{3}$$ and then it cancels out the same number on the RHS. The question is now about the truth of the following equation, $$$$J_3 | \Lambda k,\sigma\rangle= \sigma| \Lambda k,\sigma\rangle$$$$ $$\Lambda k$$ is the action of an arbitrary Lorentz transformation $$\Lambda$$ on the fiducial momentum $$k$$. The result will be some arbitrary momentum $$p=\Lambda k$$. So, the question is about the truth of $$J_{3}|p,\sigma\rangle=\sigma|p,\sigma\rangle$$ which is false because the momentum is, in general, not along the z axis.

I think the question is trying to find a situation in which a general momentum state is an eigenstate of an angular momentum operator. If we take a general state $$|p,\sigma\rangle$$ and measure the component of spin along the momentum axis with operator, $$$$\frac{\vec{J}\bullet\vec{p}}{|\vec{p}|}|p,\sigma\rangle=\sigma|p,\sigma\rangle$$$$

• i will update my question – amilton moreira Aug 17 '19 at 4:01
• @amiltonmoreira : I've updated my answer. – Stephen Blake Aug 17 '19 at 7:21