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I am reading Zee's QFT book and he is developing the field theory of photons without introducing gauge invariance. He's putting a small photon mass into the Lagrangian which he will later let go to zero. He is assuming I just finished my courses on EM and QM, and that I should know what he's talking about in some brief comments, but I am confused. Zee writes:

A massive spin-1 particle has three degrees of polarization for the obvious reason that in its rest frame the spin vector can point in three different directions. The three polarization vectors $\varepsilon^{(a)}_\lambda$ are simply the three unit vectors pointing along the $x$, $y$, and $z$ axes.

Here I am embarrassingly confused already. Can't any spin vector, spin-1/2 for example, point in three different spatial directions? When I think of three possibilities for spin-1, I think $\{+1,0,-1\}$. When I think of a spin-1 "vector state," I think the the three positions in the vector represent $\{+1,0,-1\}$ and not $\{x,y,z\}$. What am I missing here? Why would the spin be able to point in three spatial directions for spin-1 exclusively?

I tried to just read past it but I quickly became even more lost. Zee writes:

The amplitude for a particle with momentum $k$ and polarization $a$ to be created at a source is proportional to $\varepsilon^{(a)}_\lambda(k)$, and the amplitude for it to be absorbed at the sink is proportional to $\varepsilon^{(a)}_\nu(k)$.

Here I understand that due to special relativity, $\varepsilon$ is a function of $k$, but since I don't see the connection to the polarization states, I am missing the relationship to the amplitude. I believe Zee when he cites this dependence of the amplitude, but where does it come from?

A little further down on the page (p34 in Zee's QFT book, 2nd Ed), Zee writes

Now we understand the residue of the pole in the spin -1 propagator $$D_{\nu\lambda}=\dfrac{g_{\nu\lambda}-\frac{k_\nu k_\lambda}{m^2}}{k^2-m^2}.$$ It represents $\sum\varepsilon^{(a)}_\nu(k)\varepsilon^{(a)}_\lambda(k)$. To calculate this quantity, note that by Lorentz invariance it can only ber a linear combination of $g_{\nu\lambda}$ and $k_\nu k_\lambda$. The condition $h^\mu \varepsilon^{(a)}_\mu=0$ fixes it to be proportional to $g_{\nu\lambda}-\frac{k_\nu k_\lambda}{m^2}$. [sic] Thus $$\sum \varepsilon^{(a)}_\nu(k)\varepsilon^{(a)}_\lambda(k)=-\left(g_{\nu\lambda}-\frac{k_\nu k_\lambda}{m^2}\right)$$

Here I am confused again. How Zee is able to conclude immediately that it is "fixed proportional to $g_{\nu\lambda}-\frac{k_\nu k_\lambda}{m^2}$? Also, the residue of $D$, I believe, should have a denominator like $$\text{Res}_D(k_0^-)=\dfrac{g_{\nu\lambda}-\frac{k_\nu k_\lambda}{m^2}}{k-k_0^+}$$ due to the Laurent series representation of $D$. I forgot as much from my complex analysis course as I did from my EM and QM courses. Am I wrong about the residue? Thanks for looking at my long question!!!

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  • $\begingroup$ This isn't a full answer, but to your first question, maybe he's just contrasting it with the massless case, like photons. Photons have no rest frame so saying that their polarization has two degrees of freedom described by helicity makes sense. For massive particles however, this cannot make sense since it has a rest frame and in which the polarization should just be a vector in a 3 dimensional space. $\endgroup$ – yankyl Sep 13 at 19:04
  • $\begingroup$ @yankyl Can't a spin-1/2 vector also point in any of the three spatial directions? Why does spin-1 have uniquely the three spatial directions for its polarization directions? $\endgroup$ – hodop smith Sep 13 at 19:07
  • $\begingroup$ They can, but he might be contrasting massless spin 1 where polarization can be two dimensional with massive spin 1 where it cannot $\endgroup$ – yankyl Sep 13 at 19:11
  • $\begingroup$ @yankyl The reason I am confused is that even in the massless case, the spin can point in (1) the direction of the momentum, or (2) oppositely to the momentum, and the momentum can be in any of three spatial directions. Indeed, the massive case adds the transverse spin possibility, but that adds two spatial vectors beyond the massless case. After he cites these three spin-1 polarization vectors, he goes on to talk about the five polarization spin tensors for spin-2. So certainly there must be something important about the 3 spin-1 spatial vectors since he builds the spin-2 analogy on top of it $\endgroup$ – hodop smith Sep 13 at 19:18
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    $\begingroup$ I've said this before in an answer to another question: you are not always supposed to see why everything he says is true, and he doesn't bother to tell you when that happens. Zee is best read like a novel, not like a "real" textbook, IMO. He claims that if you do all the math you can follow everything he says, and technically that's true, but it's not true in practice. $\endgroup$ – Javier Sep 13 at 23:28
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A vector representing the spin state of a spin-1/2 particle is two-dimensional. Spin is associated with the group $SU(2)$. The Pauli sigma matrices form a basis for a two-dimensional representation of this group. For example, if we let the operator that measures spin along the $z$-axis be proportional to the Pauli matrix $$\sigma_z=\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)$$ (as is conventional), then the spin eigenstates are the two-dimensional vectors $(1, 0)$ and $(0,1)$. We can have eigenstates of the Pauli matrices corresponding the $y$-axis and the $z$-axis, or take linear combinations to get any arbitrary axis. However, this is due to the superposition principle. Any arbitrary state can be expanded as a sum of eigenstates of $\sigma_z$. This is only a two-dimensional space.

A vector representing the spin state of a spin-1 particle is three-dimensional. This is because spin-1 is associated with the three-dimensional representation of $SU(2)$. Now in Quantum Field Theory, spin is a frame-dependent quantity. However, we know that in the rest frame of a spin-1 particle, its spin vector lives in a three-dimensional space. Therefore, we should associate spin with three degrees of freedom. I believe Zee is associating these three degrees of freedom with the three degrees of freedom of the polarization vector in the rest frame.

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  • $\begingroup$ Thank you for your answer. What prevents the spin-1/2 vector from pointing in any of the three dimensions of space? Maybe I'm not understanding a difference between a spin vector and a spin polarization vector? For instance, the dimensions of the Pauli matrices have nothing to do with the directions a spin-1/2 vector could point, so I don't see why the 3 spin-1 polarization states "should associate" with the 3 spatial directions but the spin-1/2 polarization states should not. Could you say a little more about when you mean by saying they "should associate?" This is where I fail to understand. $\endgroup$ – hodop smith Sep 13 at 22:10
  • $\begingroup$ The distinction is confusing. When I say "spin vector" for a spin-1/2 particle, I am referring to the two-dimensional vector that describes the particle's spin state. The spin can lie along any direction in physical space. However, for example, spin state $(1,1)$ (the spin-up state along the $x$-direction) doesn't represent a new degree of freedom. It's just a superposition of the states with spin up and spin down along the $z$-axis. That's why we use a two-dimensional vector to describe spin-1/2 instead of a three-dimensional one. $\endgroup$ – JoshuaTS Sep 13 at 22:49
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    $\begingroup$ "Spin-1 is associated with the group $SU(3)$" No! All spin comes from different dimensional representations of $SU(2)$. Sorry to be emphatic, but this is very important. $\endgroup$ – Javier Sep 13 at 23:32
  • $\begingroup$ @Javier Thanks for the catch! I corrected the error, $\endgroup$ – JoshuaTS Sep 14 at 0:14
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The Lagrangian for a massive vector field (without sources) has the form

$$ L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}M^{2}A_{\mu}A^{\mu}$$

and is not gauge invariant due to the mass term. The eq.s of motion are

$$\partial _{\mu}F^{\mu\nu} + M^{2}A^{\nu} = 0$$ and by deriving a second time

$$\partial_{\mu} A^{\mu} = 0$$

that is not a gauge-fixing. Instead it arises dynamically and reduces the d.o.f. from 4 to 3. In the electromagnetic case (massless vector fields) once you fix the gauge you have an additional residual gauge condition that reduces the d.o.f. from 3 to 2. The e.o.m. can be solved in the momentum space $$A^{\mu}(x) = \frac{1}{4\pi^{2}} \int \frac{d^{3}k}{2\omega}(e^{ik\dot x}\epsilon^{\mu}(\mathbf{k}) + c.c.)$$

What I think Zee wants to say is that in the massive case you can always choose a reference frame on the particle and so the time component of the polarization vector $\epsilon^{\mu}$ became redundant.

When I think of three possibilities for spin-1, I think {+1,0,βˆ’1}. When I think >of a spin-1 "vector state," I think the three positions in the vector represent {+1,0,βˆ’1}and not {π‘₯,𝑦,𝑧}

Maybe you are talking about the $z$-component of the spin. For $\frac{1}{2}$-spin it can also point in 3 directions, in fact there is the quantum number $helicity$ that is the projection on the direction of motion of the spin.

Here I understand that due to special relativity, πœ€ is a function of π‘˜, but since >I don't see the connection to the polarization states, I am missing the >relationship to the amplitude. I believe Zee when he cites this dependence of the amplitude, but where does it come from?

In an analogous way to the scalar field, from the expression of the $A^{\mu}(x)$ you can see that the creation and absorption amplitude are proportional to the polarization vectors.

How Zee is able to conclude immediately that it is "fixed proportional to $𝑔_{πœˆπœ†}βˆ’\frac{π‘˜_{𝜈}π‘˜_{πœ†}}{π‘š^{2}}$?

For the Lorentz invariance $$\sum \epsilon_{\nu}^{(a)}(k) \epsilon_{\lambda}^{(a)}(k) = -g_{\nu\lambda}A(k^{2})+k_{\nu}k_{\lambda}B(k^{2}) $$ You can multiply the previous for $k^{\nu}$ and applying $k^{\mu}\epsilon_{\mu}^{(a)} = 0$ you find $$0 = -k_{\lambda}(-A(k^{2})+k^{2}B(k^{2})) \iff B(k^{2}) = \frac{A(k^{2})}{m^{2}}$$ Also in the case $\nu = \lambda$ for the completeness relation $$\sum \epsilon_{\nu}^{(a)}(k) \epsilon_{\nu}^{(a)}(k) = -1 = A(k^{2})$$

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  • $\begingroup$ When you write "For the Lorentz decomposition," where does that come from? Are you just supposing that the tensor indices on both sides can match up if the answer is a linear combination of those two terms? $\endgroup$ – hodop smith Sep 15 at 18:30
  • $\begingroup$ Yes. I was meaning for Lorentz invariance. I have corrected the answer. $\endgroup$ – Pipe Sep 15 at 20:41

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