I'm studying the Dirac equation for free particles and read that spin doesn't commute with the hamiltonian and one has to define the helicity operator to find a third good quantum number. What's the physical meaning behind that?
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$\begingroup$ Helicity is only a good quantum number for massless particles, if you are working with massive particles it is frame dependant. $\endgroup$– TriatticusJan 5, 2021 at 0:10
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$\begingroup$ I fully agree with your intuition that spin should be a "good" quantum number. $\endgroup$– my2ctsJan 5, 2021 at 0:40
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2$\begingroup$ @Triatticus "Good quantum number" does not require frame independence, only conservation. Helicity is conserved for massive particles too. $\endgroup$– nanomanJan 5, 2021 at 9:38
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$\begingroup$ @nanoman I was thinking of Lorentz a invariance for some reason $\endgroup$– TriatticusJan 5, 2021 at 21:24
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2 Answers
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The Dirac Hamiltonian for free particles doesn't commute with the $z$ component of the spin operator $\hat S_z$, but it does commute with the helicity operator $\hat h=S⋅p$. This means one can know simultaneously the energy and helicity of a particle.
And since the Hamiltonian does not commute with $\hat S_z$, so the same is not true.
From this Wiki article, take note that:
Systems which can be labelled by good quantum numbers are actually eigenstates of the Hamiltonian. They are also called stationary states. They are so called because the system remains in the same state as time elapses, in every observable way.
That is to say, good quantum numbers correspond to conserved quantities. Helicity is good, while $\hat S_z$ is not so good.
Side note: Helicity can be considered to be the most fundamental quantum number for massless particles since it distinguishes between the two inequivalent representations of the Poincaré group.
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$\begingroup$ Thank you for your answer. The mathematical reasoning is clear to me, but I was looking for the physical intuition behind that. $\endgroup$ Jan 5, 2021 at 9:07
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1$\begingroup$ I think the physical intuition is clear in this answer : good quantum numbers are conserved, spin is not generally. $\endgroup$– FrotaurJan 5, 2021 at 10:39
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1$\begingroup$ @Frotaur thanks, I was missing the link given by nanoman's aswer $\endgroup$ Jan 5, 2021 at 13:49
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There is generally no reason for spin to be conserved (i.e., a good quantum number) if it isn't the only form of angular momentum in a system. Total angular momentum ($\mathbf{J}$), consisting of orbital ($\mathbf{L} = \mathbf{r} \times \mathbf{p}$) plus spin ($\mathbf{S}$), is conserved (corresponding to rotation invariance).
It happens that the Schrödinger equation (the nonrelativistic limit) for a particle with spin conserves orbital and spin angular momentum separately, because spin appears as a separate quantum number uncoupled with motion. However, the Dirac equation couples spin with motion via the $\gamma$ matrices.
Helicity is manifestly conserved when written as $\mathbf{p} \cdot \mathbf{J}$, which is equivalent to $\mathbf{p} \cdot \mathbf{S}$ because $\mathbf{L}$ is always orthogonal to $\mathbf{p}$. (Why use helicity and not $J_z$? While $J_z$ is conserved, it doesn't commute with $\mathbf{p}$. We would like to use eigenstates of $\mathbf{p}$, i.e., plane waves.)
