# Is there a connection between $\hbar$ being in the path integral and being the unit of spin? [duplicate]

On the one hand, Planck constant $$\hbar$$ seems to fundamentally enter quantum theory via the path integral, in the factor $$e^{iS/\hbar}$$. Or via the Schrodinger or Dirac equations, in those versions of the theory. But we know that when you measure the spin of a fundamental particle you also get an integer or half-integer multiple of $$\hbar$$. Can the second fact be derived from the first? Or does $$\hbar$$ appear yet again in the specific form of the Lagrangian, in such a way as to ensure the spin measurement is $$\hbar$$ as well? Or is spin measurement being $$\hbar$$ an independent postulate?

I read this question whose answers seem to imply the derivation of spin from first principles is very complicated and esoteric, possibly involving Grassmann variables, so I guess I'm looking for something a bit more conceptual if possible, but anything is appreciated.

Yes. Briefly speaking, the $$\color{red}{i\hbar}$$ in the CCRs \begin{align} [\hat{q}^j,\hat{q}^k]~=~&0, \cr [\hat{q}^j,\hat{p}_k]~=~&+\color{red}{i\hbar}\delta^j_k\hat{\bf 1},\cr [\hat{p}_j,\hat{q}^k]~=~&-\color{red}{i\hbar}\delta^k_j\hat{\bf 1},\cr [\hat{p}_j,\hat{p}_k]~=~&0,\end{align}\tag{A} is directly correlated with:

1. The $$so(3)$$ Lie algebra $$[\hat{L}_j,\hat{L}_k]~=~\color{red}{i\hbar}\epsilon_{jk\ell}\hat{L}_{\ell}\tag{B}$$ and the eigenvalues $$\in\color{red}{\hbar}\mathbb{Z}$$ of the orbital angular momentum operator $$\hat{L}_i~=~\epsilon_{ijk} \hat{q}^j \hat{p}_k, \tag{C}$$ cf. e.g. my Phys.SE answer here. (It turns out that the eigenvalues $$\in\frac{\color{red}{\hbar}}{2}\mathbb{Z}$$ for spin angular momentum.)

2. The free propagator/2-point function \begin{align} \langle T[\hat{q}^j(t)\hat{q}^k(t^{\prime})]\rangle~=~& 0,\cr \langle T[\hat{q}^j(t)\hat{p}_k(t^{\prime})]\rangle ~=~&+\frac{\color{red}{i\hbar}}{2}{\rm sgn}(t\!-\!t^{\prime}) \delta^j_k,\cr \langle T[\hat{p}_j(t)\hat{q}^k(t^{\prime})]\rangle ~=~&-\frac{\color{red}{i\hbar}}{2}{\rm sgn}(t\!-\!t^{\prime}) \delta^k_j,\cr \langle T[\hat{p}_j(t)\hat{p}_k(t^{\prime})]\rangle~=~& 0,\end{align}\tag{D} if we treat the Hamiltonian as a perturbation, i.e. eq. (D) is the diabatic/short time approximation.

3. The argument $$\frac{i}{\color{red}{\hbar}}S$$ of the exponential in the path integral because of the $$\int \!dt ~p_j \dot{q}^j$$ term in the action $$S$$, cf. point 2 and my Phys.SE answer here.

I actually disagree with @Qmechanic: my answer is no, half integer spin cannot be derived from the path integral.

That massive particles come in discrete spins of multiples of half integer values of $$\hbar$$ is due to two postulates: a system is represented by a state in a Hilbert space (in particular, single particle states live in a Hilbert space) and, separately, relativity, that the universe is symmetric under the 4D Poincaré group IO(1,3). Neither of these postulates is related in any way to correlation functions and/or the time evolution of the system.

To expand slightly, Wigner proved that symmetry transformations are effected by linear, unitary operators on the Hilbert space. (Or anti-linear, anti-unitary op's.) Thus single particle states live in irreducible representations of the symmetry group. When one works out the irreps of the Poincaré group, one finds that massive particles have spins that are multiples of half integers times $$\hbar$$. (See, e.g., Ch. 2.5 of Weinberg's QFT book.)

That correlation functions are given by the path integral <=> time evolution is determined by the Schroedinger equation is a totally separate postulate. This postulate holds independent of the spacetime symmetries of the universe; in particular, this postulate is agnostic as to whether the universe is governed by non-relativistic physics (the Galilean group) or relativistic physics (the Poincaré group).

I do agree with @Qmechanic that there is a relationship between the fundamental unit of action, which normalizes and makes dimensionless the argument of the exponential in the path integral, and the fundamental unit of angular momentum: these two quantities have the same units, kg m$$^2$$/s, so we shouldn't be surprised that they're one and the same.

Edit: As was pointed out by @Ryder Rude, one can derive half integer spin non-relativistically from the rotation group SO(3); i.e. the Galilean group will still have half integer spin. But the point still stands that the half integer spin comes from the combination of quantum mechanics requiring a Hilbert space for states to live in and the states themselves living in an irrep of the symmetry group; i.e. there's nothing to do with time evolution and/or path integrals. It's worth remembering further that for 2 spatial dimensions one has anyons, so the half integer nature of spin is a quirk of the number of spatial dimensions in the problem.

• Could you elaborate on your last paragraph? It is not clear to me why, because two unrelated (as you posit) fundamental constants of physics happen to have the same units, they should also be expected to have the same value. What is the physical (as opposed to, esthetical or philosophical) rationale here? Dec 12, 2022 at 10:43
• You're probably right. If the two are exactly the same, there should be a good reason. The $\hbar$ in the spin will come from the postulate of quantum mechanics that the classical Poisson bracket goes over to the commutator with the RHS being $i\hbar$ times the classical result (in operator form). Presumably the normalization in the path integral phase is necessary to reproduce the classical commutation relation for, e.g., position and momentum in non-relativistic QM; i.e. @Qmechanic's answer. Dec 12, 2022 at 13:05

On the one hand, Planck constant $$\hbar$$ seems to fundamentally enter quantum theory via the path integral, in the factor $$e^{iS/\hbar}$$. Or via the Schrodinger or Dirac equations, in those versions of the theory. But we know that when you measure the spin of a fundamental particle you also get an integer or half-integer multiple of $$\hbar$$.

Planck's constant is the characteristic scale of quantum effects, which has the dimensionality of action, Energy$$\times$$Time - which also happens to be the dimensionality of angular momentum. This is the reason why $$\hbar$$ enters both the path integral and the expression for spin.

Taking $$\hbar$$ to be very small in path integrals leads to quasiclassical and eventually classical limit in path integrals, and makes spin (an essentially quantum phenomenon) to vanish.

Remark: It is worth also pointing out that in non-relativistic QM spin is introduced is an ad-hoc fashion, i.e., the corresponding term is simply added to the Hamiltonian to reflect the experimental facts (and our awareness of the rotational symmetry.) In relativistic QM it indeed arises naturally from the fundamental symmetries.

• Spinors can be derived as representations of the rotation group, or by taking the square root of the Non relativistic Hamiltonian. Relativity isn't needed and this isn't ad-hoc imo. Dec 12, 2022 at 12:32
• @RyderRude my remark says that much. However, existence of rotational symmetry is not a reason to add spin to the Hamiltonian. Dec 12, 2022 at 13:18
• @RyderRude One could probably build QM this way. However, standard canonical quantization procedure usually departs from classical theory - in QM this means taking classical Hamiltonian, replacing variables by operators... and adding spin, since this one does not have a classical analog. Dec 12, 2022 at 13:39
• I agree. Though there are attempts in deformation quantisation to arrive at spinors by quantising the ( phase space $\times S^2$) symplectic manifold. Dec 12, 2022 at 13:42
• @RyderRude Another possibility is crystal in two-band approximation - there we could, in principle, write a square root Hamiltonian that is non-relativistic - I have been wondering about this analogy for some time Dec 12, 2022 at 14:15

You do not need to look at spin to see a fundamental connection between angular momentum and $$\hbar$$. Just look at the $$L_z$$ operator $$i\hbar \frac{d}{d\theta}$$. Its eigenvectors are $$\psi (r,\theta)=R(r)e^{im\theta}$$, and these eigenstates have to obey the condition $$\psi (r, 0) =\psi (r, 2\pi)$$, which results in the condition:

$$2\pi m= 2\pi k$$

$$k\in z$$. You can see that the eigenvalhes have the spectrum $$\hbar k$$, which increments in jumps of $$\hbar$$.

All this isn't specific to angular momentum. In the particle in a box problem, a similar boundary conditions arises as $$\psi (L) =\psi (0) =0$$. This leads to quantisation of momentum in incremets proportional to $$\hbar$$.

Lastly, energy eigenvalues of bound states also have similar spectra. For example, the Harmonic oscillator has eigenvalues as $$n\hbar \omega$$. This is also a consequence of boundary conditions of requiring the wavefunction to exponentially fall off in the classically forbidden region.

• Exactly, and $L_z$ comes from the momentum operator (equivalent to the commutation relation $[x,p]$), whose $\hbar$ also goes into the Hamiltonian, whose $\hbar$ enters into the path integral through the time evolution operator. And so the source of all of the $\hbar$ can be seen to be the commutation relation or momentum operator. Mystery solved I believe. Dec 12, 2022 at 13:25
• @doublefelix I agree. It all starts with that commutation relation. It deforms the Poisson bracket which leads to $\hbar$ being present in the time evolution. It also sort of pixelates the phase space which leads to these discrete spectra in $\hbar$ increments. Dec 12, 2022 at 13:46