0
$\begingroup$

Imagine we are studying the spin quantization along the same axis as the momentum. What if I have a Dirac spinor with a spin up but no definite helicity ($\psi_L,\psi_R\neq0$):

$$ u(p)= \left(\begin{array}{c} \sqrt{p ·\sigma}\xi_{\uparrow} \\ \sqrt{p ·\bar\sigma}\xi_{\uparrow} \end{array}\right) = \left(\begin{array}{c} \psi_L \\ \psi_R \end{array}\right)$$

How should I understand this? If the spin along the $p$-direction is positive, shouldn’t the particle have definite right-hand helicity and have 0 left-hand component? In other words, shouldn’t we have $ u(p) \sim \left(\begin{array}{c} 0 \\ \rm something \end{array}\right)$ ?

$\endgroup$
  • $\begingroup$ Yes for a massless particle. Are you asking a question about the partial overlap of helicity and chirality as a function of the mass? $\endgroup$ – Cosmas Zachos Dec 1 '17 at 19:35
  • $\begingroup$ Not at all. Nor am I asking about massless particles in particular. If the spin along the $p$-direction is positive, shouldn’t the particle have definite right-hand helicity and have 0 left-hand component? I.e. shouldn’t $u(p) \sim \left(\begin{array}{c} 0 \\ something \end{array}\right)$ ? $\endgroup$ – Y2H Dec 1 '17 at 20:12
  • $\begingroup$ Yes, spin aligned with momentum denotes Positive helicity... I'm not sure why your are bringing up handedness.... $\endgroup$ – Cosmas Zachos Dec 1 '17 at 20:32
  • $\begingroup$ Positive helicity? Don’t you mean right-handed helicity? Also I am sorry that I cannot make my question clear but I am not sure what you need me to elaborate on? $\endgroup$ – Y2H Dec 1 '17 at 20:39
  • $\begingroup$ Right. Well a right-handed (resp. left-handed) particle means that it’s a positive (resp. negative) eigenvalue of the helicity operator. $\endgroup$ – Y2H Dec 1 '17 at 20:42
2
$\begingroup$

OK, I suspect I understand your puzzlement...You are simply faked out by the inappropriate sloppy language "right handed helicity". There is no such thing.

For simplicity, take the spin up to be in the z direction, and the momentum in the z direction, $p^\mu= (\sqrt{m^2+p^2},0,0,p)$, and inspect your spinor in the Weyl (chiral) basis, $\psi_L=\frac12(1-\gamma^5)\psi=\begin{pmatrix} I_2 & 0 \\0 & 0 \end{pmatrix}\psi,\quad \psi_R=\frac12(1+\gamma^5)\psi=\begin{pmatrix} 0 & 0 \\0 & I_2 \end{pmatrix}\psi$, $$ u(p)= \left(\begin{array}{c} \sqrt{p ·\sigma}\xi_{\uparrow} \\ \sqrt{p ·\bar\sigma}\xi_{\uparrow} \end{array}\right) = \left(\begin{array}{c} \psi_L \\ \psi_R \end{array}\right).$$

Now $ \sqrt{p ·\sigma}\xi_{\uparrow}= \sqrt{E-p} ~\xi_{\uparrow}$, and $ \sqrt{p ·\bar\sigma}\xi_{\uparrow}= \sqrt{E+p}~\xi_{\uparrow}$, so that $$ u(p)= \left(\begin{array}{c} \sqrt{E-p} ~\xi_{\uparrow} \\ \sqrt{E+p}~\xi_{\uparrow}\end{array}\right), $$ of course both left- and right-chiral. The upper component does not vanish, in general, for positive helicity.

Inspect 3 limits. For p=0, E=m and $$ u(0)= \sqrt{m} \left(\begin{array}{c} \xi_{\uparrow} \\ \xi_{\uparrow}\end{array}\right), $$ completely even-handed--helicity is undefined.

For m=0, the upper two components are projected out, and only the lower two (actually only the third) survive(s), $$ u(p)= \sqrt{2p} \left(\begin{array}{c} 0 \\ \xi_{\uparrow}\end{array}\right), $$ so truly right-handed: the origin of sloppily referring to positive helicity as "R", as helicity and chirality are identical.

For $p\gg m$, $$ u(p)\longrightarrow \sqrt{2p} \left(\begin{array}{c}m/2p ~ \xi_{\uparrow} \\ \xi_{\uparrow}\end{array}\right), $$ so there is a left-chiral piece, as well, but wildly subdominant to the right-chiral piece. Again, this is a positive helicity spinor not divorced from its left chiral piece.

The obverse statement is provided in Wikipedia: For massive particles, distinct chirality states (e.g., as occur in the weak interaction charges) have both positive and negative helicity components, in ratios proportional to the mass of the particle.

The textbook of Itzykson & Zuber, 2-2-1, provides a serviceable definition of the helicity operator; in our case, $$ \hat{h}= \tfrac{1}{2} \begin{pmatrix} \sigma^3 & 0 \\0 & \sigma^3 \end{pmatrix}, $$ comporting with the above: it reads off an eigenvalue of +1/2 for both the upper, $\psi_L$, and the lower, $\psi_R$, component, as it should.

Unless one were safely clear of the language pitfall to avoid, or one were a sucker for paradoxical statements to keep the audience on edge, it is an untoward idea to be tossing around chirality terms to quantify helicity...

  • PS. A more apposite title would have been "Helicity with indefinite chirality"!
$\endgroup$
  • $\begingroup$ Thank you for your efforts. It may be to my lack of intelligence but I do not feel that this answers my question at all. Simply, if spin is aligned along $p$ shouldn’t the upper spinor in $u(p)$ be $0$? Incidently I think you could check out this other question I posted also physics.stackexchange.com/questions/371978/…, it’s kind of related. $\endgroup$ – Y2H Dec 2 '17 at 9:13
  • $\begingroup$ No. No. No. I show you why and how. Please review your chirality and do not confuse it with helictity. $\endgroup$ – Cosmas Zachos Dec 2 '17 at 11:14
  • $\begingroup$ Again with chirality. I thank you very much for trying to help me and I do not mean to sound rude at all but you keep insisting on 1 thing and you refuse to understand what I am saying. It’s like you’re telling me ‘‘no no this is not your question, your question is this’’. No, I know what I want to ask. $\endgroup$ – Y2H Dec 2 '17 at 11:18
1
$\begingroup$

There are two things I think you might possibly be misunderstanding. One is that the spinor doesn't just contain information about the spin, it also contains information about the energy and momentum. $$\bar{u}\gamma^\mu u= 2(E,\vec{p})$$You need to use all 4 components of the spinor in order to describe this.

The other thing is calling the components of a spinor $\psi_L$ and $\psi_R$ only makes sense for massless particles (or particles that are moving very fast).

If you are at rest, $p=(m,0,0,0)$ and the 0th component of $\sigma$ is the identity matrix, so your spinor is just $$u(p_{rest frame})= \left(\begin{array}{c} \sqrt{m}\xi_{\uparrow} \\ \sqrt{m}\xi_{\uparrow} \end{array}\right) $$ And both $\psi_L=\psi_R$ no matter if it's pointing up or down (or anything in between). As you boost the components in the direction of the spin, one of $\psi_L$ or $\psi_R$ will get smaller and the other bigger. In the limit of very large but finite momentum, one of these two components will be very small depending on the helicity. This is the limit in which the names $\psi_R$ and $\psi_L$ make sense. But the 'wrong' component still needs to be nonzero to give you information on the momentum.

$\endgroup$
0
$\begingroup$

So the problem is that I did not understand that $\psi_L$and $\psi_R$ are chirality eigenstates not helicity eigenstates.

In the high energy limit where helicity and chirality are basically the same thing, the state does in fact become an eigenstate (I did the math but it’s too long to post here unless someone asks for it.)

$\endgroup$

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.