Spin-1 polarization vectors (massive particle)

Question

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!!!

Answer 1

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.

Answer 2

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,−1}. When I think >of a spin-1 "vector state," I think the three positions in the vector represent {+1,0,−1}and not {𝑥,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, 𝜀 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?

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 $𝑔_{𝜈𝜆}−\frac{𝑘_{𝜈}𝑘_{𝜆}}{𝑚^{2}}$?

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 $$\sum \epsilon_{\nu}^{(a)}(k) \epsilon_{\nu}^{(a)}(k) = -1 = A(k^{2})$$