I was reading Goldbring's article on the implications of $\text{MIP}^\ast=\text{RE}$ which contains a crash course-like overview of (classical, non-relativistic, non-QFT) quantum mechanics. This made me consider some questions. My question is ultimately concerned with a photon, i.e. a massless spin-$1$-particle, but I will start with what I understood about electrons, i.e. a massiv spin-$\frac12$-particle, and maybe there are already some mistakes or misconceptions there, of which I would be very happy if someone would point them out to me. My background: mathematician with some interest in computer science and physics, but not a professional.

Spin of electrons

Most of this is based on a wikipedia article about spin and I also found this PSE question and that one helpful, although the last two are about spin-$\frac32$.

For $n=1$ electron the spin state space is $\mathcal{H}_1=\mathbb{C}^2=\text{span}(e_1,e_2)$ or more precisely the unit sphere in $\mathcal{H}_1$ is the set of all pure states. Here $e_i$ are standard basis vectors of tuple vector spaces.
The three Pauli matrices $$ \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \sigma_y=\begin{pmatrix}0&i\\-i&0\end{pmatrix}, \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix} $$ are observables and any observable is a (real) linear combination of these three: $\sigma_u=u_x\sigma_x+u_y\sigma_y+u_z\sigma_z$ is up to normalization the spin measurement in direction $u=(u_x,u_y,u_z)^\top$. Up to multiplication by $i$, the imaginary unit, the 3-dimensional $\mathbb{R}$-vector space of observables is the Lie algebra $\mathfrak{su}(2)$ and thus the observables can be endowed with an algebra structure, i.e. a multiplication, which is useful to define the commutator of two observables.
Any (pure) state can be realized as an eigenstate of some observable, e.g. for $\sigma_z$ these are $\lvert1\rangle_z=\lvert\uparrow\rangle_z=\begin{pmatrix}1\\0\end{pmatrix}=e_1$ and $\lvert-1\rangle_z=\lvert\downarrow\rangle_z=\begin{pmatrix}0\\1\end{pmatrix}=e_2$ usually denoted as spin up/down in $z$-direction. However, one could equally well take $\widetilde{\lvert1\rangle}_z=\begin{pmatrix}-1\\0\end{pmatrix}=-e_1$ and $\widetilde{\lvert-1\rangle}_z=\begin{pmatrix}0\\\frac{1}{\sqrt{2}}(-1+i)\end{pmatrix}=e^{3/4\pi i}e_2$ as scaling by a unit norm complex scalar only changes the phase, which is not observable in this situation. Also in the spin-$\frac12$-case one might take the scaled Pauli matrix $\frac12\sigma_z$ (or $\frac\hbar2\sigma_z$) to get half-integral eigenvectors and rather write $\lvert\frac12\rangle_z$ and $\lvert-\frac12\rangle_z$ etc.

For $n=2$ electrons the (pure) spin state space is $\mathcal{H}_2=\mathcal{H}_1\otimes\mathcal{H}_1=\mathbb{C}^2\otimes\mathbb{C}^2\cong\mathbb{C}^4=\text{span}(e_i)_{i=1}^4$ where a possible isomorphism could take the form $e_i\otimes e_j\mapsto e_{2(i-1)+j}$. Since $(e_i)_{i=1}^2$ were seen to be eigenstates for the $z$-direction spin measurement this expresses a basis of $\mathcal{H}_2$ as a tensor product of componentwise eigenstates. However this is not a good description with respect to the observables it turns out. There is a decomposition into singlet and triplet states (for clarity we start with a cumbersome notation and introduce convenient abbriviations): $$ \mathcal{H}_2=\mathcal{H}_1\otimes\mathcal{H}_1= \text{span}\left(\frac{1}{\sqrt{2}}\left(\lvert\uparrow\rangle_z\otimes\lvert\downarrow\rangle_z-\lvert\downarrow\rangle_z\otimes\lvert\uparrow\rangle_z\right)\right) \oplus \text{span}\left(\lvert\uparrow\rangle_z\otimes\lvert\uparrow\rangle_z,\frac{1}{\sqrt{2}}\left(\lvert\uparrow\rangle_z\otimes\lvert\downarrow\rangle_z+\lvert\downarrow\rangle_z\otimes\lvert\uparrow\rangle_z\right),\lvert\downarrow\rangle_z\otimes\lvert\downarrow\rangle_z\right)= \text{span}\left(\frac{1}{\sqrt{2}}\left(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle\right)\right) \oplus \text{span}\left(\lvert\uparrow\uparrow\rangle,\frac{1}{\sqrt{2}}\left(\lvert\uparrow\downarrow\rangle+\lvert\downarrow\uparrow\rangle\right),\lvert\downarrow\downarrow\rangle\right)= \text{span}\left(\lvert0,0\rangle\right) \oplus \text{span}\left(\lvert1,1\rangle,\lvert1,0\rangle,\lvert1,-1\rangle\right). $$ In the last notation $\lvert s,m\rangle$ stands for the combined eigenstate for the two observables $S^2=S_x^2+S_y^2+S_z^2$ with eigenvalue $s(s+1)$ and $S_z$ with eigenvalue $m$. Here the $S_u$ are the directed spin observables on the tensor products induced from the Pauli matrices $\sigma_u$.
We focused on $\sigma_z$ but we could have instead employed any $\sigma_u$ for any direction $u$ and got the same up to isomorphism. Also note that we got basis elements that are not pure tensors (mathematically) but are still pure states (quantum mechanically).
Coming to the observables: Clearly $S_z=\sigma_z\otimes\sigma_z$ (Kronecker product) gives one and there are 8 further such observables (e.g. $\sigma_z\otimes\sigma_x$) spanning a 9-dimensional space of observables. On the other hand one can perform a measurement on one of the tensor factors without interfering with the other component: $\text{id}\otimes\sigma_u$ or $\sigma_u\otimes\text{id}$. This gives 6 more dimensions, however these measurements produce mixed states and not pure states. While it is clear to me, how one adds observables in the 9 dimensional space of Kronecker products of component observables, and within the two 3-dimensional spaces with one component the identity, the status of expressions like $\text{id}\otimes\sigma_x+\sigma_x\otimes\text{id}$ or $\text{id}\otimes\sigma_x+\sigma_y\otimes\sigma_z$ is not clear to me. In particular $\text{id}\otimes\sigma_x+\sigma_x\otimes\sigma_x=\left(\text{id}+\sigma_x\right)\otimes\sigma_x$ would suggest that $\text{id}+\sigma_x$ should be an observable for $n=1$ particle case and I am not sure about the status of this.
I assume that again any pure state can be realized as an eigenstate of some observable, although I am not sure how that observable would look like for a pure state like $\frac{1}{\sqrt{2}}\lvert0,0\rangle-\frac{1}{\sqrt{2}}\lvert1,-1\rangle$ which "mixes" singlet and triplet states.

Spin of a photon

Now lets consider a photon, a massless spin-$1$-"particle". Since the photon has no inner structure I would have assumed that singlet subspace is no longer relevant and the spin is governed by a triplet space $\cong\mathbb{C}^3$. However wikipedia on photons tells me that the state space is actually generated by the two circular polarizations (L,R) as a possible basis. However just in the next section they speak about three eigenvalues $-1,0,1$ and explain in a footnote that the 0-eigenstate corresponds to a linear combination of L and R (resp. $-1$- and $1$-states) - this does not make sense to me.

Question(s): How does one explicitly describe the space of (pure) states for the spin of a photon? Can this be done in isolation or does one always have to also consider the momentum and/or angular momentum of the photon? What are the (spin-)observables for the photon, how would one describe the set of all observables explicitly? Are there mixed states for (the spin of) a single photon and which observables produce them, if so?


1 Answer 1


I came across this answer to a similar question by ACuriousMind which gives a detailed background for an answer to my questions. I extract short answers to my questions, the detailed explanations to them can be found in the link. It seems that for a proper explanation you have to go to relativistic quantum mechanics, classical quantum mechanics will not do.

  • Massless particles do not have spin, but they have something analogous, namely helicity. The difference lies in the different algebra representations for massive and massless particles in relativistic QM.
  • Helicity is aligned with (linear) momentum. In contrast spin for massive particles can be measured in any direction $u$ via the spin operator $S_u$. To ensure gauge invariance only the spin operator in direction of the momentum is applicable and it measures helicity. There is another degree of freedom which is orbital angular momentum, which seems to be essentially independent of helicity (resp. spin angular momentum as it is also called).
  • The two eigenvalues for the helicity of a "spin 1 photon" are $\pm 1$ (resp. $\pm \hbar$) associated to the spin operator aligned with momentum as observable. The eigenvalues are also sometimes referred to as L (corresponding to $1$, i.e. helicity points in direction of momentum) and R (corresponding to $-1$, i.e. helicity points in opposite direction of momentum). L/R stand for left/right polarization.
  • There is no $0$ eigenstate, however superposition of pure L and pure R (to equal parts) gives you linear polarized photons. Measuring helicity of a linear polarized photon will give you either $1$ or $-1$, but $0$ on average for repeated measurement with several linear polarized photons.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.