I'm attempting to understand how to motivate quantum mechanics through the use of path integrals. Because the path integral approach provides such a direct connection to classical mechanics (via the principle of least action), and because it provides such a direct physical understanding of the difference between classical and quantum mechanics, I think it's a really valuable lens through which to understand quantum theory. However, I'm having some conceptual difficulties regarding the introduction of spin.

The problem is as follows: suppose we know nothing about quantum mechanics except that the space of states is a complex Hilbert space whose inner product maps quantum states to probability amplitudes. Given that a system is in some quantum state $|\psi,t_0\rangle$, we can write the state of the system at some future time $t$ as $$|\psi,t\rangle = \hat{U}(t,t_0)|\psi,t_0\rangle.$$ We know that $\hat{U}$ must be a unitary operator due to the probabilistic interpretation of the inner product $\langle\psi|\psi\rangle$, and therefore that $$U(t_0,t) = U^\dagger(t,t_0).$$ Taking a time derivative, we have that $$\frac{\partial}{\partial t}|\psi,t\rangle = \frac{\partial\hat{U}(t,t_0)}{\partial t}|\psi,t_0\rangle = \frac{\partial\hat{U}(t,t_0)}{\partial t}\hat{U}^\dagger(t,t_0)|\psi,t\rangle.$$ If we define the operator $$\Lambda(t,t_0) = \frac{\partial\hat{U}(t,t_0)}{\partial t}\hat{U}^\dagger(t,t_0),$$ we can show (using only the unitarity of $\hat{U}$ that $\Lambda(t,t_0)$ is a.) independent of $t_0$ and b.) anti-Hermitian. We then define the operator $$H(t) = i \hbar \Lambda(t),$$ such that the time evolution of a quantum state is given by the Schrodinger equation as $$i\hbar\frac{\partial}{\partial t}|\psi,t\rangle = H(t)|\psi,t\rangle.$$ Now all that remains is to determine the operator $H(t)$. We could simply postulate the form of this operator, but instead we will use the path integral approach as a postulate and derive the form of $H(t)$ from there. We assume that the matrix elements of the unitary time evolution operator $\hat{U}(t,t_0$ are given by $$\langle\vec{r}|\hat{U}(t,t_0)|\vec{r}_0\rangle = \int\mathcal{D}\vec{r}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]},$$ where $S$ is the classical action and $\int\mathcal{D}\vec{r}$ denotes integration (however it may be defined) over the space of paths connecting the spacetime points $(t_0,\vec{r}_0)$ and $(t,\vec{r})$. Using this postulate, we can write the matrix elements of the operator $H(t)$ as \begin{align}\langle\vec{r}|H(t)|\vec{r}_0\rangle &= i\hbar\int \mathrm{d}\vec{r}'\frac{\partial}{\partial t}\left[\langle\vec{r}|\hat{U}(t,t_0)|\vec{r}'\rangle\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\frac{\partial}{\partial t}\left[\int\mathcal{D}\vec{r}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\frac{\partial}{\partial t}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\frac{i}{\hbar}\frac{\partial S}{\partial t}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= \int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\left(-\frac{\partial S}{\partial t}\right)\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle.\end{align}

The Hamilton-Jacobi equations tells us that $-\frac{\partial S}{\partial t}$ is simply equal to the Hamiltonian. Now I have two related questions:

1.) Generally speaking, functions on phase space which appear inside a path integral correspond to operators which appear outside the path integral. Can the same be done for the quantity $-\frac{\partial S}{\partial t}$ via the appropriate use of the resolution of the identity?

2.) If so, then this equation would seem to derive the functional form of the operator $H(t)$ as the classical Hamiltonian with the usual substitution $(\vec{r},\vec{p})\rightarrow(\hat{\vec{r}},-i\hbar\vec{\nabla})$. However, the quantum Hamiltonian often contains spin operators, which (to my knowledge) do not have classical analogues. How do spin operators enter into $H(t)$ in this formalism?


2 Answers 2


There is an extensive literature on path integrals for spin. They are usually derived by using the theory of generalized coherent states. I'm not sure what is a good entry point into the subject. I can advertise one of my own papers: arXiv:cond-mat/0111139, but perhaps the best starting point is the paper by Kaluder (J. R. Klauder, Ann. Phys. (N.Y.), 11, 123 (1960)) if you have access to it.

It is with pointing out that quantum spin do have classical analogues. See the sections on spin (near pages 54 254 and 259) in my online lecture notes at https://courses.physics.illinois.edu/phys509/sp2017/


As you probably know, there are two different subjects, often referred to as "quantum mechanics" and "quantum field theory." They are closely related (i.e., QFT is just what happens when you quantize a field) but are not the same.

The path integral is most useful in QFT where some simple tricks can be used to derive the Feynman rules. A path integral approach can be used derive regular quantum mechanics as well, although it's usually not particularly useful.

Within the context of single particle quantum mechanics, spin does need to be jammed into the set up in an ad-hoc manner. The notion of spin arises a bit more naturally in QFT.

Having said all of that, there certainly is a generalization of the path integral that will "yield" spin, even in single particle quantum mechanics. It's pretty weird, though.

Basically there's something you've taken for granted your entire life: variables commute. I don't mean operators, I mean variables. That is, if you have two numbers $a$ and $b$, then $ab = ba$. $a$ and $b$ can be locations in space, for example. In order to set up your path integral for spin you need to use anticommuting variables. That is, variables where ab = -ba. Mathematically speaking, this requires an "algebra" that is "generated" by such variables. They are often referred to as "Grassmann variables" and you should look them up.

I won't go into a whole course on Grassmann variables, but a (usually subpar) discussion of them is usually hidden away in many introductory QFT books.

Basically what happens is that you define "pseudo-classical mechanics" with these Grassmann variables using an anti-Poisson bracket instead of the usual Poisson bracket used in classical mechanics. (Look up "Hamiltonian Mechanics" if you don't know about this.) Upon quantization, or the path integral, you get your good old spin quantum system.

In other words, this is definitely an area where the path integral is a much more complicated way of doing things!

In your set up, you just have to cram these "pseudo-classical" mechanics into your path integral along with your regular classical mechanics.

(A side note: in the context of QFT, the path integral done the regular way will yield a "bosonic" field and the path integral done with Grassman variables will yield a "fermionic" field, so if you look into this subject you'll sometimes see it referred to as a "fermionic" path integral.)

I've made these Grassmann variables seem pretty confusing, but they're actually not so bad. Don't feel too intimidated by them.

Sorry my answer was long and rambling, but you asked a very tricky question!

  • $\begingroup$ " often referred to as "quantum mechanics" and "quantum field theory"" Maybe one should be clearer on this : QFT is a meta-level of Quantum Mechanics as its ground state is the QM free particle wavefunction for the given field. QFT is based on all the postulates of quantum mechanics. $\endgroup$
    – anna v
    Commented Jul 9, 2017 at 10:54
  • $\begingroup$ Thanks for your answer! I only have a passing familiarity with QFT, but I have heard of Grassmann variables and fermionic path integrals, and your answer gave me a good intuition about those concepts, which I really appreciate. $\endgroup$ Commented Jul 10, 2017 at 5:49

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