We are all familiar with the version of Quantum Mechanics based on state space, operators, Schrodinger equation etc. This allows us to successfully compute relevant physical quantities such as expectation values of operators in certain states and then compare with experiment.

However, it is often claimed that the path integral is an "equivalent" way to do all of this. To me, the "equivalent" part is a bit vague. I understand that the Feynman path integral allows you compute the propagator $\langle x | e^{-iHt} |x' \rangle $ by just using the classical Lagrangian of the system. Then any expectation value in a state can be computed by two resolutions of the identity to get an integral over this propagator. This shows that the path integral is a way to compute a quantity that's very useful, but not much more than that, since we still need the concept of operators, and their representation in position space, as well as position space wave functions, all of these objects along with their usual interpretations.

Ultimately, regardless of how you compute things, QM will still be based on probabilities and thus wave functions, however my question is, is there anything analogous to the axioms of Quantum mechanics usually mentioned in textbooks that are based on the path integral?

The path integral if seen as an independent object gives us the propagator, correlation functions, and the partition function (and maybe other objects which I'm not aware of). Are all these sufficient to give us the same information that quantum mechanics based on Hilbert space and operators give us? I would really appreciate if someone can make these connections precise.

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    $\begingroup$ This is off-topic but what in the world makes you start a question with "We are all familiar with...". Opening like "We all know..." are cringeworthy in any case, but with this straight forward question in particular, it's probable that the users here are not on your level of knowledge, but rather above or below. $\endgroup$
    – Nikolaj-K
    Sep 18, 2013 at 7:16
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    $\begingroup$ Sorry for that. I meant "we all know..." in the sense that it's the way a person familiar with the subject will traditionally think about things. $\endgroup$ Sep 18, 2013 at 17:32

2 Answers 2


In the context of axiomatic quantum field theory, there is a theorem (see theorem 3-7 in PCT, Spin and Statistics, and All That by Streater and Wightman, who I will refer to as "SW"), which SW call the "reconstruction theorem," essentially stating that correlation functions serve to completely determine a corresponding field theory in the Hilbert Space formalism. Specifically, they show that (I paraphrase for brevity)

Given a sequence $\mathscr W^{(n)}$ of tempered distributions defined by $n$ spacetime points (correlation functions) that satisfy certain technical properties (cluster decomposition, relativistic transformation law, etc.) there exists a separable Hilbert space $\mathscr H$, a continuous unitary representation $U$ of $\mathbb R^{3,1}\rtimes \mathrm{SO}^+(3,1)$ (the proper, orthochronus Poincare group) on $\mathscr H$, a unique Poincare-invariant vacuum state $|0\rangle$, and a hermitian scalar field $\phi$ with appropriate domain such that \begin{align} \langle 0|\phi(x_1)\cdots \phi(x_n)|0\rangle = \mathscr W^{(n)}(x_1, \dots, x_n) \end{align} Furthermore, any other field theory with these vacuum expectation values (vevs) is unitary equivalent to this one.

In other words, vevs determine a field theory up to unitary equivalence, and a sequence of sufficiently well-behaved correlation functions completely determines a field theory with given vevs, so correlators determine a field theory up to unitary equivalence.

The Upshot. Since the path integral allows you to compute all correlators in principle through the (somewhat schematic) formula \begin{align} \langle \phi(x_1)\cdots \phi(x_n)\rangle = \frac{\int [d\phi] \phi(x_1)\cdots \phi(x_n)e^{i S[\phi]}}{\int [d\phi] e^{i S[\phi]} } \end{align} the path integral gives a complete characterization of a given field theory.

Note. I am by no means an expert on axiomatic quantum field theory, so if I have said anything here that isn't strictly, mathematically correct, I apologize ahead of time. Also, I'm not certain how general SW's characterization of field theory is, so my remarks are not completely general, but I would think that the spirit of these remarks is thought to hold for all (or most) physical quantum field theories.

Also, this is certainly not a particularly physical answer. I'd be curious to hear from another user about the physical intuition behind why one might expect correlators to be so fundamental and all-encompassing.

  • $\begingroup$ By Unitary equivalence, do they mean that if we take one set of fields, and apply Unitary transformations to the set(This is just like a change of basis), we will get the other set of fields, and that is why the correlation function would also be the same. Essentially, are the two field theories just different repackaging of each other? $\endgroup$ Jan 27, 2019 at 17:24
  • $\begingroup$ @TusharGopalka Yes that's essentially what it means, but you should really take a look at Streater and Wightman and read the definitions/theorems yourself. It's actually pretty concise and readable imo. $\endgroup$ Jan 27, 2019 at 19:08

Ultimately, the quantities that you have to calculate, and compare to the reality are probabilities or transition probabilities, which are the square of amplitudes or transition amplitudes. The path integral formalism represents directly transition amplitudes. In quantum mechanics, you may recover the Schrodinger equation from the expression of the path integral representing the transition amplitude $\langle x',t'|xt\rangle$. You have, for "wavefunctions", the integral equation $\langle x',t'|\psi \rangle = \int dx \langle x',t'|xt\rangle \langle x,t|\psi \rangle$,- with the expression $\langle x',t'|xt\rangle = \int [dC] e^{iS(C)}$, where $C$ is a path from $x$ to $x'$, $t$ to $t'$, and $S(c)$ is the action for this path - and taking the limit $t' \to t$, will give you the Schrodinger equation.

The quantum formalism that you may use (path integral formalism, operator formalism , Schrodinger/Heinsenberg representation, etc..), is secondary, in the sense that these formalisms are equivalent. It is very interesting to look at different formalisms, but, practically, depending on your problem, you will choose the simplest one.

Let' take the example of the quantum harmonic oscillator. You have different eigenstates $\psi_n(x,t)$ corresponding to energies $(n+ \dfrac{1}{2})\hbar \omega$. Suppose you want to calculate the transition probability : $|\langle\psi_{n}|X| \psi_{n+1}\rangle|^2$. You may imagine doing the integral on $x$ with the expressions of $\psi_n(x,t),\psi_{n+1}(x,t)$. You will work with the "wavefunctions" and Schrodinger representation. Now it's far more interesting, in this particular case, to work with Heinseberg representation, with an operator $X(t)$. The equation for the operator $X(t)$ is simply $\ddot X(t) + \omega^2 X(t)=0$, with the Heinsenberg constraints : $[X(t), P(t')]_{t=t'} = i \hbar$. A solution is $X(t) = \sqrt{\dfrac{\hbar}{m \omega}}(a e^{i \omega t} + a^+e^{-i \omega t})$, and, in the energy basis, the non- null terms of $a$ are $a_{n,n+1} = \sqrt{n+1}$. Now, the probability that we are looking for is simply $|X_{n,n+1}|^2 = \dfrac{\hbar}{m \omega}(n+1)$.

[In fact, to correctly understand Quantum mechanics, it is better to think in terms of Heinseberg representation, than Schrodinger representation. For instance, in Quantum Field Theory, you are working in a " Heinsenberg representation", that is: you work with operators $\Phi(x,t)$ depending on space and time, you are not usually working with wavefunctions $\psi(\Phi,x,t)$ - even if this formalism is possible].

Now, turning back to path integrals, it is the same logic. If you consider for instance QFT, depending on your problem, it could be more interesting to use the path integral formalism, or use the operator (canonical) formalism. For instance, you may want to calculate the vaccum energy for a bosonic field or a fermionic field. It is simpler to use the operator formalism, but you may use too the integral path formalism (see for instance Zee, QFT in a nutshell Chapter II.5 p 121/126, first edition), and you will find the same result.

If you want to calculate Green Functions and propagators, it is totally natural to use the integral path formalism, for instance in a perturbative field theory, and this leads naturally to Feynmann diagrams, and Feynmann diagrams is that you need to calculate transition amplitudes and transition probabilities, in collision particles process.


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