# Photons have Spin 1 - Franz Gross' Relativistic Quantum mechanics and Field Theory

I've got a question regarding the derivation of spin 1 for photons in Franz Gross' Relativistic Quantum Mechanics and Field Theory.

From pages 50 to 56 he attempts to derive "how this [spin 1] comes about naturally in the particle picture." He takes the angular momentum operator for the EM field, and separates it into two parts, one associated with the transformation of the field indices (spin), and the other with the transformation of the field arguments (orbital).

He then expresses this in terms of creation and annihilation operators in a polarization basis (helicity basis) and claims that the expression alone shows that the photon has spin one. I simply don't understand how the form of the function is supposed to entail that result. I'll repeat his steps below for posterity:

Using normal ordered product notation (the : : terms), the angular momentum operator is first reduced (via some un-noteworthy expansions) to:

(2.57)$$\Omega^i = -i\int d^3r: \frac{\partial A^b}{\partial t} (L^iA^b-i\epsilon_{ibl}A^l):$$

Which is followed by a brief explanation of the resulting commutation relations (done to show that this operator is indeed the generator of infinitesimal rotations for the EM field theory). More importantly, Gross then breaks down the operator into its spin and orbital components, as mentioned above:

(2.60) $$\Omega^i_{spin} = -\int d^3r: \frac{\partial A^j}{\partial t} \epsilon_{ijk}A^k):$$ $$\Omega^i_{L} = -i\int d^3r: \frac{\partial A^j}{\partial t} L^iA^j:$$

This spin operator is then expressed "in terms of the a's and a$^{\dagger}$'s, using real polarization vectors for simplicity,"

$$\Omega^i_{spin} = -\int d^3r \frac{\epsilon_{ijk}}{2L^3}\sum_{n,\alpha ; n',\alpha'}^{} -\frac{i\omega_n}{\sqrt{\omega_n\omega_{n'}}} \epsilon^{\alpha j}_{n}\epsilon^{\alpha' k}_{n'} \times \Big\{a_{n,\alpha}a_{n',\alpha'}e^{-i(k_n+k_{n'})\cdot x}-a^{\dagger}_{n,\alpha}a^{\dagger}_{n',\alpha'}e^{i(k_n+k_{n'})\cdot x}-a^{\dagger}_{n,\alpha}a^{}_{n',\alpha'}e^{i(k_n-k_{n'})\cdot x}+a^{\dagger}_{n',\alpha'}a^{}_{n,\alpha}e^{-i(k_n-k_{n'})\cdot x}\Big\}$$

$$= i\sum_{n,\alpha ,\alpha'}^{} \frac{\epsilon_{ijk}}{2} \Big\{\epsilon^{\alpha j}_{n}\epsilon^{\alpha' k}_{-n}\Big[a_{n,\alpha}a_{-n,\alpha'}e^{-2i\omega_nt}-a^{\dagger}_{n,\alpha}a^{\dagger}_{-n,\alpha'}e^{2i\omega_nt}\Big]+\Big[\epsilon^{\alpha j}_{n}\epsilon^{\alpha' k}_{n}-\epsilon^{\alpha' j}_{n}\epsilon^{\alpha k}_{n}\Big]a^{\dagger}_{n,\alpha'}a_{n,\alpha}\Big\}.$$

"The first term in the {} bracket is symmetric in j and k. This can be seen by changing n to -n and $\alpha$ to $\alpha'$. Hence it is zero when contracted with $\epsilon_{ijk}$. The second term is clearly antisymmetric in j and k and requires $\alpha \neq \alpha'$. Hence"

$$\Omega_{spin} = i\sum_{n,\alpha\neq\alpha'}^{} \Big( \epsilon^{\alpha}_n\times\epsilon^{\alpha'}_n\Big)a^{\dagger}_{n,\alpha'}a_{n,\alpha}$$

"Recall that $\epsilon^1\times\epsilon^2 = \hat{k}$. Hence, carrying out the sum over $\alpha, \alpha'$ gives"

$$\Omega_{spin} = i\sum_{n}^{}\hat{k}_n\Big[a^{\dagger}_{n,2}a_{n,1} -a^{\dagger}_{n,1}a_{n,2}\Big]$$

"This is not a convenient form because it is not given in terms of number operators. We can express $\Omega_{spin}$ in terms of number operators by introducing a new polarization basis referred to as the circular, or helicity, basis. If $\epsilon^1 \to \hat{x}$,$\epsilon^2 \to \hat{y}$, and $\hat{k} \to \hat{z}$, then the circular polarization basis, in which the states have a definite spin projection along $\hat{z}$, is defined by"

(2.61) $$\epsilon^+ = spin in +\hat{k}direction = -\frac{1}{\sqrt{2}}(\epsilon^1+i\epsilon^2)$$ $$\epsilon^- = spin in -\hat{k}direction = \frac{1}{\sqrt{2}}(\epsilon^1-i\epsilon^2)$$

"Note the appearance of the minus sign in the definition of $\epsilon^+$; this is a standard phase convention used in the construction of the spherical harmonics $Y_{lm}$ for $l = 1$ and $m = \pm 1$ from $\hat{x}$ and$\hat{y}$. Then we define $a_{\pm}$, the annihilation operators corresponding to these circularly polarized states, by the relation (suppress n for now)..."

$$\epsilon^1a_1+\epsilon^2a_2 =\epsilon^+a_++\epsilon^-a_-$$ This gives $$a_1=-\frac{1}{\sqrt{2}}(a_+-a+-)$$ $$a_2=-\frac{i}{\sqrt{2}}(a_++a+-)$$ $$[a_+,a_+^{\dagger}]=[a_-,a_-^{\dagger}] =1$$ $$[a_+,a_-^{\dagger}]=[a_-,a_+^{\dagger}] =0$$ "Hence," $$a_2^{\dagger}a_1-a_1^{\dagger}a_2 = -i[a_+^{\dagger}a_+-a_-^{\dagger}a_-]$$ "And, restoring n"

(2.62) $$\Omega_{spin} =\sum_{n}^{}\hat{k}[a_{n,+}^{\dagger}a_{n,+}-a_{n,-}^{\dagger}a_{n,-}]$$

"The spin operator has now been expressed in terms of number operators for photons with a definite helicity. Note that it is a vector sum of terms which point in the $+\hat{k}$ direction for positive helicity and in the $-\hat{k}$ direction for negative helicity.

In general, the helicity of a particle is the projection of its spin along the direction of its motion, and it a massive particle has spin s, its helicity can take on any integer value between s and -s. The direction of motion is simply one special direction in space, and a massive particle of spin s has 2s+1 states which can always be expanded in terms of states having a definite spin projection along an chosen axis. However, Eq(2.62) shows that photons do not have this property. It shows that the photon has spin 1, but that out of three posible states ($\pm1 or 0$), only helicity states +1 and -1 can occur. The absence of helicity zero is due to the transverse nature of the field, which is due in turn to the absence of a photon rest mass."

Phew, that was long. Ok. So with the substitution of the helicity basis I can appreciate that the final operator expansion in terms of creation and annihilation operators is only summed over $\hat{k}$, and that therefore there are only 2 helicity values allowed. My issue is that Gross directly states that the form of eq 2.62 should tell me that these helicity values are 1 and -1, and that s = 1. I can't see this. From my limited viewpoint it looks as though the equation limits us to two helicities, but I don't see the definitive s = 1 moment.

Do any of you see this? Can you explain it? I hope that by providing the reference work up front I am making it easy to answer my question. I do not understand the deduction of the spin angular momentum from the form of the operator. I would appreciate any and all help.

• There seem to be some brackets or parenthesis missing. It's been a while since I've gone through any of these derivations but as I recall one approach is to square the operator and see if it gives the identity, if so your eigenvalues are restricted to be roots of unity. Whether this is "obvious" is a matter of experience and/or opinion. – ggcg Sep 4 '18 at 1:53

It's simpler than you think. As a familiar example, take the harmonic oscillator Hamiltonian, for simplicity setting $$\hbar \omega = 1$$, $$H = a^\dagger a + \text{const}.$$ This tells us the energy levels are spaced by $$\hbar \omega$$, because $$a^\dagger a$$ counts the number of excitations. Now you might ask how we know it's $$\hbar \omega$$ instead of $$2 \hbar \omega$$, because there appears to be no explicit "$$1$$" in the expression, but that's just because we don't write $$1$$ as a multiplicative factor.
Similarly, in this case we essentially have $$L \sim a_{+}^\dagger a_{+} - a_{-}^\dagger a_{-}$$ This tells us each $$+$$ excitation contributes $$+1 \hbar$$ angular momentum, while each $$-$$ excitation contributes $$-1 \hbar$$ angular momentum. The $$1$$ is not written, because $$1x = x$$. If we were dealing with gravitons, we would instead have found $$L \sim 2 a_{+}^\dagger a_{+} - 2 a_{-}^\dagger a_{-}$$ indicating that gravitons have spin 2.
• Hey knzhou! Thanks for responding! This may be (probably is) naive on my part. But I can still see my issue in your answer. What I mean is, you say that the form tells us how the levels are spaced, but if say, s = $\frac{1}{2}$, I could still envision an operator that takes this spin value to $\frac{3}{2}$ or $\frac{-1}{2}$, ie, an operator that excites by +1 and annihiliates by -1. With your answer in mind, my confusion seems rooted in using the difference in the levels after excitation or relaxation to characterize the fundamental unit s. – Yajibromine Sep 4 '18 at 12:01
• @Glycoholic The point is that $a^\dagger$ is interpreted as creating one photon, and creating one photon changes the helicity by $\pm 1$. So that means each photon has helicity $\pm 1$. – knzhou Sep 4 '18 at 19:37
• @Glycoholic I suppose your interpretation might make sense if we were talking about, say, the electron in a hydrogen atom. In that case, the electron could carry very large orbital angular momentum, and we could define raising and lowering operators that go between these states in multiples of $\hbar$. But note how the interpretation is different. The raising/lowering operators there aren't creating or destroying electrons, they are just changing the electron state. Here, the raising/lowering operators change the photon quantum field state, by creating or destroying photons. – knzhou Sep 4 '18 at 19:39