The definition of the propagator $\Delta(x)$ in the path integral formalism (PI) is different from the definition in the operator formalism (OF). In general the definitions agree, but it is easy to write down theories where they do not. In those cases, are the PI and OF actually inequivalent, or is it reasonable to expect that the $S$ matrices of both theories agree?

For definiteness I'll consider a real scalar field $\phi$, with action $$ S_0=\int\mathrm dx\ \frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2 \tag{1} $$

Path integral formalism

In PI we insert a source term into the action, $$ S_J=S_0+\int\mathrm dx\ \phi(x)\ J(x) \tag{2} $$

The mixed term $\phi J$ can be simplified with the usual trick: by a suitable change of variables, the action can be written as two independent terms

$$ S_J=S_0+\int \mathrm dx\;\mathrm dy\ J(x)\Delta_\mathrm{PI}(x-y)J(y) \tag{3} $$ and this relation defines $\Delta_\mathrm{PI}(x)$: to get the action into this form we have to solve $(\partial^2+m^2)\Delta_{\mathrm{PI}}=\delta(x)$, i.e., in PI the propagator is defined as the Green function of the Euler-Lagrange equations of $S_0$. This definition is motivated by the the fact that when $S_J$ is written as $(3)$ the partition function can be factored as $$ Z[J]=Z[0]\exp\left[i\int \mathrm dx\;\mathrm dy\ J(x)\Delta_\mathrm{PI}(x-y)J(y) \right] \tag{4} $$ which makes functional derivatives trivial to compute. For example, if we differente $Z[J]$ two times we get $$ \langle0|\mathrm T\ \phi(x)\phi(y)|0\rangle=\Delta_\mathrm{PI}(x-y) \tag{5} $$

In this formalism, the propagator is always the Green function of the differential operator of the theory.

Operator formalism

In OF the propagator is defined as the contraction of two fields: $$ \Delta_\mathrm{OF}(x-y)\equiv \overline{\phi(x)\phi}(y)\equiv \begin{cases}[\phi^+(x),\phi^-(y)]&x^0>y^0\\ [\phi^+(y),\phi^-(x)]&x^0<y^0\end{cases} \tag{6} $$ where $\phi^\pm$ are the positive and negative frequency parts of $\phi$.

In general, $\Delta_\mathrm{OF}$ is an operator, but if it commutes with everything (or, more precisely, if the propagator is in the centre of the operator algebra) we can prove Wick's theorem, which in turns means that $$ \langle 0|\mathrm T\ \phi(x)\phi(y)|0\rangle=\Delta_\mathrm{OF}(x-y) \tag{7} $$ i.e., the propagator coincides with the two-point function. This makes it very easy to see, for example, that $$ \Delta_\mathrm{OF}=\Delta_\mathrm{PI} \tag{8} $$

In this theory, the fact that the propagator is a Green function is a corollary and not a definition. The theorem may fail if the assumptions are not satisfied.

The discrepancy

The positive/negative frequency parts of $\phi$ are the creation and annihilation operators, which in OF usually satisfy $$ [\phi^+(x),\phi^-(y)]\propto\delta(x-y)\cdot1_\mathcal H \tag{9} $$ and therefore $\overline{\phi(x)\phi}(y)$ is a c-number. This means that the assumptions of Wick's theorem are satisfied and $(8)$ holds.

The relation $(9)$ can be derived from one of the basic assumptions of OF: the canonical commutation relations: $$[\phi(x),\pi(y)]=\delta(x-y)\cdot1_\mathcal H\tag{10} $$

But if we use any non-trivial operator in the r.h.s. of $(10)$ instead of a $1_\mathcal H$, Wick's theorem is violated and in general $\Delta_\mathrm{PI}\neq \Delta_\mathrm{OF}$. One could argue that the r.h.s. of $(10)$ is fixed by the Dirac prescription $\{\cdot,\cdot\}_\mathrm{D}\to\frac{1}{i\hbar}[\cdot,\cdot]$, where $\{\cdot,\cdot\}_\mathrm{D}$ is the Dirac bracket. In the Standard Model, it's easy to prove that $\{\cdot,\cdot\}_\mathrm{D}$ is always proportional to the identity, but in a more general theory we may have complex constraints which would make the Dirac bracket non-trivial - read, not proportional to the identity - and therefore $\Delta_\mathrm{OF}\neq\Delta_\mathrm{PI}$.

Scalar, spinor and vector QFT's always satisfy a relation similar to $(10)$, and therefore OF and PI formalisms agree. But in principle it is possible to study more general QFT's where we use a commutation relation more complex than $(10)$. I don't know of any practical use of this, but to me it seems to me that we can have a perfectly consistent theory where PI and OF formalisms predict different results. Is this correct? I hope someone can shed any light on this.


I think it can be useful to add some details to what I said about Dirac brackets, defined as $$ \{a,b\}_\mathrm{DB}=\{a,b\}_\mathrm{PB}-\{a,\ell^i\}_\mathrm{PB} M_{ij}\{\ell^j,b\}_\mathrm{PB}\tag{11} $$ where $\ell^i$ are the constraints and $M^{ij}=\{\ell^i,\ell^j\}_\mathrm{PB}$. As $\{q,p\}_\mathrm{PB}=\delta(x-y)$, the only way to get non-trivial Dirac brackets is through the second term. This may happen if we have non-linear constrains such that the second term is a function of $p,q$. If in some case we have non-linear constrains the matrix $M$ will depend on $p,q$ and $\{p,q\}_\mathrm{DB}$ will be a function of $p,q$. If we translate this into operators, we shall find $$ [\pi,\phi]=\delta(x-y)\cdot1_\mathcal H+f(\pi,\phi)\tag{12} $$ and so $[\pi,\phi]$ won't commute with neither $\pi$ nor $\phi$ as required by Wick's theorem. (This term $f$ is perhaps related to the $\hbar^2$ terms in QMechanic's answer, and the higher order terms in, e.g., the Moyal bracket).

  • 1
    $\begingroup$ I know that the title is not very descriptive but its the best I could think of. If someone comes up with a better one feel free to edit the post. $\endgroup$ Commented Apr 26, 2016 at 15:43
  • $\begingroup$ Related: physics.stackexchange.com/q/14481/2451 , physics.stackexchange.com/q/71987/2451 , physics.stackexchange.com/q/74795/2451 $\endgroup$
    – Qmechanic
    Commented Apr 26, 2016 at 15:48
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    $\begingroup$ The fields $\phi$ that enter the path integral formalism are required to be the canonical coordinates. This you can see by matching the way the path integral formalism is "derived" from the canonical formalism. For all such fields, $[\phi,\pi]$ will always be a $c$-number. $\endgroup$
    – Prahar
    Commented Apr 26, 2016 at 15:55
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    $\begingroup$ To complement Prahar comment, if you did not have the canonical commutation relation, the path integral would not be the one you have written (assuming you could write one, since you would not have the standard creation/annihilation operators and the coherent states and so on). $\endgroup$
    – Adam
    Commented Apr 26, 2016 at 16:49
  • $\begingroup$ @Prahar thank you for your comment. AFAIK, canonical variables can in principle have non-trivial Dirac brackets. If we follow Dirac's prescription "Dirac bracket $\to$ commutator", we might end up with a non-trivial commutator if the original Driac bracket was non-trivial. $\endgroup$ Commented Apr 27, 2016 at 20:36

3 Answers 3


General comments to the question (v1):

  1. Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post.

  2. Rather than claiming complete understanding and existence of the correspondence (1), it is probably more fair to say that we have a long list of theories (such as e.g. Yang-mills, Cherns-Simons, etc.), where both sides of the correspondence (1) have been worked out.

  3. The correspondence (1) is mired with subtleties. Example: Consider a non-relativistic point particle on a curved target manifold $(M,g)$ with classical Hamiltonian $$\tag{2} H_{\rm cl} ~=~\frac{1}{2} p_i p_jg^{ij}(x), $$ which we use in the Hamiltonian action of the phase space path integral. Then one may show that the corresponding Hamiltonian operator is $$\tag{3}\hat{H}~=~ \frac{1}{2\sqrt[4]{g}} \hat{p}_i\sqrt{g}~ g^{ij} ~\hat{p}_j\frac{1}{\sqrt[4]{g}}+ \frac{\hbar^2R}{8} +{\cal O}(\hbar^3),$$ cf. Refs. 1 & 2. The first term in eq. (3) is the naive guess, cf. my Phys.SE answer here. The two-loop correction proportional to the scalar curvature $R$ is a surprise, which foretells that a full understanding of the correspondence (1) is going to be complicated.


  1. F. Bastianelli and P. van Nieuwenhuizen, Path Integrals and Anomalies in Curved Space, 2006.

  2. B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.

  • $\begingroup$ Then which one is the correct theory if they both differ? Can we say that the path integral cannot be taken to define the theory, and that it must be derived from the operator formalism? $\endgroup$
    – Ryder Rude
    Commented Sep 3, 2022 at 8:16
  • $\begingroup$ Hi Ryder Rude: That is a popular opinion among physicists. $\endgroup$
    – Qmechanic
    Commented Sep 3, 2022 at 9:53

Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent.

A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard to understand. Lagrangian are only defined through their classical action (the limit $\hbar \to 0$,) which is nothing but the weak coupling limit (I've realized that many QFT books do not mention this fact.) If you have a theory that only lives at a strongly coupled fixed point, then you cannot construct a Lagrangian. As an example of this, just google "non Lagrangian theories." In my case, I get this article http://arxiv.org/abs/1505.05834 with a very nice introduction.

This is just to answer the title.

Regarding the main body, I think you are confused. First, you have to understand that the equivalence of the path integral and the operator formalism that you see in your QFT textbook is not by chance, it is by construction. So no, you are never going to spam a different Hilbert space by working with two formalism of the "same theory." Point (2) of Qmechanic is accurate in this sense. Since you know how to formulate the PI formalism for those theories, you are going to get the same answer.

Second, you argued that the LFH of (10) might be different. Truth is, it can't. The reason is simple: the propagator just move a state in space-time, it does not change it, it is by definition. Therefore, you can only get $1_{\mathcal H}$. If you ever get something else, then you are looking at the wrong object as the propagator.

Finally, you must always be careful when talking about the Path Integral of anything: it is full of subtle points. Actually, as of today, it is not know what is QFT, so you should also be careful when talking about the Operator Formalism of anything!


The real equivalence should be searched in a Euclidean formulation, both for the operator approach (now an algebra of comutative self adjoint field operators and the (now well defined as a mathematical object in a cuttoff scheme) Nelson-Feynman-Kac-Schwinger Euclidean QFT Path Integral. The real Minkowskian QFT is expected to be obtained through analytic continuation on the euclidean time of the (non time ordered) n point euclidean QFT Green functions.


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