3
$\begingroup$

In Weinberg QFT section 2.5.5, he defines the states of momentum $p$ by

$$\Psi_{p,\sigma}=U\bigl(L(p)\bigr)\Psi_{k,\sigma}$$

up to some irrelevant normalisation, and $L(p)$ is the Lorentz transformation that takes us from $k^\mu=(m,0,0,0)$ to any $p^\mu$ with $p^2=-m^2$ and $p^0>0$.

From hindsight I know that $\sigma$ labels the spin of the particle along the z-axis, and that $L(p)$ is a boost. So by the definition above this boost should not change our observed spin along the z-axis.

I have $J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma}$, so by definition I should have $$J_3\bigl[U\bigl(L(p)\bigr)\Psi_{k,\sigma}\bigr]=\sigma \bigl[U\bigl(L(p)\bigr)\Psi_{k,\sigma}\bigr]$$ But this is inconsistent with Poincare algebra. Indeed, if I only boost along the $z$-axis then everything is okay, since $J_3$ and $K_3$, the generator of boosts along the $z$-axis, commute. But I cannot reach arbitrary $p^\mu$ by only boosting along the $z$-axis. I can only get to $p^\mu=(p^0,0,0,p^3)$.

I cannot hope to fix this by rotation around $z$-axis either, I must rotate around the $x$ or $y$ axis. But these also do not commute with $J_3$, so I still have a problem.

How can I reach states of arbitrary $p^\mu$?

As an example suppose I start with a definite spin state in the rest frame, so I have $$J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma}$$ Now I perform an infinitesimal boost in the $y$ direction to get to a new state $$e^{iK_2p_2}\Psi_{k,\sigma}=(1+iK_2p_2)\Psi_{k,\sigma}$$ Now I measure my spin. I find $$J_3(1+iK_2p_2)\Psi_{k,\sigma}=\sigma (1+iK_2p_2)\Psi_{k,\sigma}-iK_1p_2\Psi_{k,\sigma}$$ The first term is what I want but the second term ruins everything. How can I resolve this issue?

$\endgroup$
  • 1
    $\begingroup$ Why should you think that spin operator $J_{3}$ commutes with $U(L, p)$? By the definition, the spin operator is the component of antisymmetric Lorentz group generator $\hat{M}_{\mu \nu}$, and thus it transforms as the component of tensor. $\endgroup$ – Name YYY Mar 12 '16 at 15:59
  • $\begingroup$ @NameYYY Because $\sigma$ index doesn't change under the transformation, so the transformed state must also be a $J_3$ eigenstate with the same eigenvalue $\sigma$. As in the example I gave, this cannot be the case unless $J_3$ commutes with $U(L(p))$. $\endgroup$ – qftey Mar 12 '16 at 16:37
  • $\begingroup$ "...Because σ index doesn't change under the transformation, so the transformed state must also be a J3 eigenstate with the same eigenvalue σ..." It is the number, but corresponding eigenstate is labelled by $\sigma$: $$ J_{3}|\mathbf k , \sigma\rangle = \sigma $$ From the general thinking the Lorentz transformation $U(\Lambda)$ which acts on $|p, \sigma\rangle$ converts it in the superposition of states $|\Lambda p , \sigma{'}\rangle$ with matrix weights $C_{\sigma {'}\sigma}$. This again shows that your statement is incorrect. $\endgroup$ – Name YYY Mar 12 '16 at 17:12
  • $\begingroup$ I do not understand. I agree that a general Lorentz transformation will produce a superposition of states with different $\sigma$ with weights $C_{\sigma'\sigma}$. But the transformation $U(L(p))$ is defined by Weinberg to not change the $\sigma$ index, he then goes on to say that it is specifically a boost, and even gives a specific matrix representation for it. $\endgroup$ – qftey Mar 12 '16 at 17:25
  • $\begingroup$ Related (possible duplicate?): physics.stackexchange.com/questions/240746/… $\endgroup$ – Art Brown Mar 13 '16 at 4:13
2
$\begingroup$

It seems that you're right, and I was wrong in the comments section. The answer on your question is simple: $\sigma$ doesn't mean the polarization for arbitrary momenta $p^{\mu}$, although it coincides with the polarization at rest. So, how to get the interpretation of the $\sigma$ label?

By the definition, $$ \tag 1 U(L(p))|k,\sigma\rangle = |p , \sigma \rangle, $$ so $\sigma$ quantity of one-particle state, defined at rest, is unchanged under the standard Lorentz boost. The physical meaning of $\sigma$ for arbitrary momenta is thus not simple. In general, it doesn't coincide with the spin $\hat{\mathbf J}_{3}$ eigenvalue.

Realize first that the Little group of massive orbit of the Lorentz group is isomorphic to $SO(3)$ group, whose irreducible representations generators are $\hat{\mathbf J}_{i}^{\sigma \sigma{'}}$ with $\sigma = -s,...,s$). So that it seems that $\sigma$ is just the eigenvalue of $\hat{\mathbf J}_{3}$. But there is the problem.

We know that $\hat{\mathbf J}_{3}$ is changed under the general Lorentz boost as the component of antisymmetric tensor; in the result we come to the statement that $\sigma$ is defined as the eigenvalue of some operator, the action of which on one-particle state with $k_{\mu} = (m, \mathbf 0)$ coincides with the action of $\hat{\mathbf J}_{3}$. For the general momentum $$ \tag 2 P_{\mu} = (E,\mathbf P), \quad P^{2} = m^{2}, \quad P_{\mu} = L_{\mu}^{\ \nu}k_{\nu} = (Lk)_{\mu}, $$ however, this correspondence is not true, i.e., $\sigma$ isn't the eigenvalue of $\hat{\mathbf J}_{3}$ and thus for general momenta it isn't the spin.

Maybe the operator which defines the $\sigma$ label is Pauli-Lubanski operator $\hat{\mathbf W}_{\mu}$ multiplied on $L^{-1}(p)$: $$ \tag 3 \hat{\mathbf V}_{\mu} \equiv L_{\mu}^{\ \nu}(p)\hat{\mathbf W}_{\nu} = \frac{1}{2m}L_{\mu}^{\ \nu}(p)\epsilon_{\nu \alpha \beta \gamma}\hat{p}^{\alpha}\hat{J}^{ \beta \gamma} $$ Really, note that due to invariance of $\epsilon$ tensor we have that $$ L_{\mu}^{\ \nu}\epsilon_{\nu \alpha \beta \gamma} = \epsilon_{\mu \rho \delta \epsilon}L^{\rho}_{\ \alpha}L^{\delta}_{\ \beta}L^{\epsilon}_{\ \gamma}, $$ so that, by using relation $$ (L^{-1}(p) p)_{\mu} = m\delta_{\mu 0} $$ we have that $$ \hat{\mathbf{V}}_{\mu} = \begin{cases} 0, \quad \mu = 0 \\ \frac{1}{2}\epsilon^{0\alpha\beta l}L_{\alpha}^{\ \delta}L_{\beta}^{\ \kappa}\hat{\mathbf J}_{\delta\kappa} = \hat{\mathbf J}_{l}, \quad \mu = l\end{cases} $$ Thus you have that independently on the value of general 4-momentum $p$ operator $\hat{\mathbf V}_{3}$, by acting on the state $|p, \sigma\rangle$, will always give $\sigma$ value as eigenvalue of $\hat{\mathbf J}_{3}^{\sigma \sigma{'}} = \sigma \delta^{\sigma \sigma{'}}$, with $\sigma = -s,...,s$.

$\endgroup$
  • $\begingroup$ I'm afraid I'm still unconvinced. At the start of chapter 3 Weinberg says; "To label the one-particle states we use their four-momenta $p^\mu$, spin-z component $\sigma$." There is nothing to allude that this spin must be measured in the rest frame, in fact it seems that it is measured in the frame where the particle has momenta $p^\mu$, unless I am completely misinterpreting the sentence. $\endgroup$ – qftey Mar 13 '16 at 13:37
  • $\begingroup$ @qftey : You may interpret $\hat{\mathbf V}_{\mu} = (0, \hat{\mathbf J})$ as the definition of the the quantity $\hat{\mathbf J}$. You may invert Eq. $(3)$ to get the expression of the spin: $$ \hat{\mathbf J}_{i} = \frac{1}{2}\epsilon_{ijk}\left(L^{-1}(p)\right)^{j}_{\ \mu}\left(L^{-1}(p)\right)^{k}_{\ \nu}\hat{\mathbf J}^{\mu \nu} $$ Thus the spin label $\sigma$ is interpreted as the angular momentum of particle with momentum $p$, measured in the frame at which this particle is at rest (corresponding transformation is given by $L^{-1}(p)$). $\endgroup$ – Name YYY Mar 13 '16 at 13:53
  • $\begingroup$ There is no other possibility if $\sigma$ is unchanged under boosts. $\endgroup$ – Name YYY Mar 13 '16 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.