I ask this question because many of the books I'm familiar with assumes a familiarity with the operator formulation and then develops the path-integral formulation partly based on a mixture of operator formulation and path-integrals (as if it is necessary to know the operator formulation). But I think these two are two independent but equivalent approaches. To this point, I have the following questions.

  1. What are the minimal set of postulates in the path integral formulation of quantum mechanics in addition to the postulate that the amplitude of propagation $$G(\textbf{x}_2,t_2;\textbf{x}_1,t_1)=\sum\limits_{\textbf{x(t)}}\exp\Big[iS[\textbf{x}(t)]\Big]\tag{1}\hspace{0.9cm}?$$

  2. Is it enough to derive all of quantum mechanics from (1) if the operator formulation is not used?

  3. Is the Schrodinger equation taken as a postulate in the path-integral formulation? If not, what is the equivalent information that encodes the dynamics in the path-integral approach?

  4. Does/can one regard $$\psi(\textbf{x}_2,t_2)=\int d^3\textbf{x}_1G(\textbf{x}_2,t_2;\textbf{x}_1,t_1)\psi(\textbf{x}_1,t_1)\tag{2}$$ to be an additional posulate, independent of (1), to descrbe the dynamics instead of the Schrodinger equation (which $G$ given by Eqn.(1))?

  • 4
    $\begingroup$ One important postulate is "things will still somehow work out okay even if the math isn't rigorously well-defined." $\endgroup$
    – tparker
    Jun 16, 2017 at 23:02
  • 1
    $\begingroup$ @tparker In quantum mechanics, that postulate is unnecessary. $\endgroup$
    – user1504
    Jun 17, 2017 at 0:36
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    $\begingroup$ @user1504 In the operator formulation, true. In the path-integral formulation, the postulate is necessary. See Haag's theorem and en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure. $\endgroup$
    – tparker
    Jun 17, 2017 at 2:44
  • 1
    $\begingroup$ @tparker This is a question about quantum mechanics, so complete rigor is also available for the path integral. Haag's theorem is not an obstacle there. And you don't need infinite-dimensional Lebesgue measure. If you repeatedly insert states, and thereby chop up intervals finer and finer, what you are lead to is Wiener measure. Because Haag's theorem doesn't apply to QM, you can deform Wiener measure by turning on potentials via Feynman-Kac. The infinite-dimensional Lebesgue measure is just sloppy notation, not a conceptual challenge. $\endgroup$
    – user1504
    Jun 17, 2017 at 2:56
  • $\begingroup$ @user1504 Ah, I see, you are distinguishing non-relativistic quantum mechanics from quantum field theory. Some people define the term "quantum mechanics" to encompass both, others don't. $\endgroup$
    – tparker
    Jun 17, 2017 at 7:03

1 Answer 1


The path integral formalism doesn't really replace the operator formalism. You still need to know that the state space is a Hilbert space, that probabilities are computed by norm-squaring inner products in this space, that observables are operators on this space, that time evolution is implemented by a unitary operator, and so forth.

Basically, you have to start with the Dirac-von Neumann axioms (or something very similar):

  1. States are rays in a Hilbert space $\mathcal{H}$.
  2. Observables are self-adjoint operators on $\mathcal{H}$.
  3. The expectation value of an observable $A$ in a state represented by a unit vector $\psi$ is $E[A] = \langle \psi, A \psi\rangle$.

You need these postulates to make any connection to experiment. The path integral doesn't provide alternatives. (You're implicitly using these basic postulates when you talk about superpositions of states and when you label the endpoints of trajectories with observable quantities.) But these postulates are rather minimal. They don't even mention time evolution. So people usually add another axiom:

  1. Time evolution is represented on the state space by unitary operators $U(t)$, for $t \in \mathbb{R}$.

The path integral formalism and the canonical formalism aren't really competing sets of physical postulates. They're just different methods for setting up examples which satisfy the physical postulates above.

Let's look at an example, a particle moving in a potential $V$ in 3-space.

In the canonical formalism, you declare the state space is $L^2(\mathbb{R}^3)$, that the observables are functions of the standard multiplication and differentiation operators $Q$ and $P$, and that the time evolution is given by $U(t) = e^{-itH/\hbar}$ where $H$ is the Hamiltonian $P^2/2m + V(Q)$.

In the path integral formalism, you start with the same input data (the configuration space $\mathbb{R}^3$, the mass $m$, and the potential $V$), and you use them to define integrals on the space of paths. Then you use these path integrals to construct a Hilbert space, observables acting on the Hilbert space, and time evolution operators $U(t)$. The path integral directly constructs the matrix elements of $U(t)$, and you then derive Schrodinger's equation.

Modulo some normalization problems, this is what your equations 1 and 2 say: Your $G(x_2,t_2; x_1, t_1)$ is the matrix element $\langle x_2 | U(t_2 - t_1) | x_1 \rangle$, and equation 2 just says that $\psi(t_2) = U(t_2 - t_1) \psi(t_1)$.

Note also: The derivation of Schrodinger's equation is not deep. All Schrodinger's equation says is that time translations act unitarily on the Hilbert space. $i\hbar \partial_t \psi(t) = H \psi(t)$ is the equivalent infinitesimal form of $\psi(t) = e^{-itH/\hbar} \psi(0)$. It's interesting if you've chosen $H$ by some classical analogy and now need to find $U(t)$ / solve the eigenstate problem. In the path integral, however, you have $U(t)$ already, and you find $H$ via $H = i\hbar \partial_t U(t)|_{t=0}$.

Likewise, you prove that the Hilbert space you get from the path integral is actually $L^2(\mathbb{R}^3)$, that the observables are the usual position and momentum operators, and that the Hamiltonian is really the one you'd expect.

So, how do you actually get a Hilbert space from a path integral? Basically, you divide a time segment $[0,t]$ into $[0,s] \cup [s,t]$, and you look at how the path integral decomposes into integrals over the smaller intervals and over the field values on the boundary. The integral over the boundaries defines the Hilbert space inner product. Then observables are defined by using the path integral to compute their matrix elements. Appendix 1 of Polchinksi Vol 1 has a pretty nice explanation of this.

-- Footnote --

If you're being careful about the mathematical details, you construct Euclidean path integrals, and you get the Wiener measure $d\mu(\phi) = e^{-S(\phi)} d\phi$ associated to $V$ and $m$. Then you use the Osterwalder-Schrader Reconstruction theorem to define the Hilbert space and the operators. This is covered in some detail in Chapters 3 & 6 of Glimm & Jaffe's Quantum Physics book.

The basic idea is this:

  1. Wiener measures allow you to integrate certain functions on the space of paths, namely those which can be systematically approximated by functions of the form $F_{t_i}(\phi) = \phi(t_1) \phi(t_2) .. \phi(t_n)$.
  2. Wiener measures are reflection positive: Let $\mathcal{F}^+$ be the collection of functions of the form above for which the approximations only involve $t_i > 0$. Then the bilinear form defined by $B(F_{t_i},G_{t_j}) = \int \overline{F_{-t_i}(\phi)} G_{t_j}(\phi) d\mu(\phi)$ is non-negative.
  3. The Hilbert space is the quotient $\mathcal{H} =\mathcal{F}^+ / \operatorname{kernel}(B)$.
  4. Any function $F$ in $\mathcal{F}^+$ acts on $\mathcal{F}^+$ via multiplication. These multiplications descend to define observables on $\mathcal{H}$.
  5. Translations by $t > 0$ act on $\mathcal{F}^+$. This descends to the action of a semi-group $\mathbb{R}_{\geq 0}$, and this semigroup action analytically continues to a unitary action of $\mathbb{R}$.

There are axioms, due to Osterwalder & Schrader, which guarantee that a Euclidean path integral measure defines a Hilbert space and observables satisfying a stronger form of the axioms above (the Wightman axioms). I don't believe the OS axioms are physical postulates, however. They don't refer to observable quantities, and they make use of imaginary time in a way which doesn't generalize to all quantum theories.

  • $\begingroup$ Can you specifically address the questions (1-4) I've raised in the body of the question? @user1504 $\endgroup$
    – SRS
    Jun 18, 2017 at 12:37
  • $\begingroup$ I made some edits.. $\endgroup$
    – user1504
    Jun 22, 2017 at 1:42

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