If one solves the Schrodinger equation

$$i\hbar\partial_tU(t,0) = H U(t,0)$$

for time evolution operator $U(t,0)$, one can get the following Dyson series

$$U(t,0) = \sum_n(\dfrac{-i}{\hbar})^n\int_0^t dt_1 \int_0^{t_1}dt_2 \cdots \int_0^{t_{n-1}} d t_n H(t_1)H(t_2) \cdots H(t_n) .$$

So my question is: is there any relationship between every term in the Dyson series and the possible path in Feynman's path integral method for quantum mechanics?

  • 4
    $\begingroup$ Yes, the nth order term in the Dyson series corresponds to a Feynman diagram with n vertices. But I'm wondering, do you want a more detailed answer than that? $\endgroup$
    – DanielSank
    Commented Nov 30, 2016 at 5:47
  • $\begingroup$ Related: physics.stackexchange.com/q/103503/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Nov 30, 2016 at 5:59
  • $\begingroup$ Yes,I want to a more detailed answer.Could you give me some advice ? $\endgroup$
    – Jack
    Commented Nov 30, 2016 at 7:02

1 Answer 1


Here we will discuss the relation between the Dyson series and Feynman's functional integral. We will content ourselves with obtaining a non-perturbative expression for the partition function $Z[J]$ in both formalisms; once this is done, one may expand this object in terms of Feynman diagrams as is described in any introductory text.

In general, the partition function is defined as the exponential generating function of correlation functions: $$ Z[J]\equiv \sum_{n=0}^\infty \frac{1}{n!} J_{I_1}J_{I_2}\cdots J_{I_n} G^{I_1I_2\cdots I_n}\tag{1} $$ where $G_n$ is the $n$-point function. In the operator formalism, this function is given by $$ G^{I_1I_2\cdots I_n}=\langle \Omega|\mathrm T\ \phi^{I_1}\phi^{I_2}\cdots \phi^{I_n}|\Omega\rangle\tag{2} $$ where $\phi^{I_i}$ are operators in the Heisenberg picture; $\Omega$ is the exact (interacting) vacuum; and $\mathrm T$ is the (covariant) time ordering symbol. Moreover, $J:\mathbb R^d\times \mathbb N^k\to \Lambda_\infty$ is a $c$-current (with the same Grassmann parity as $\phi$).

As usual, the $n$-point function can be recovered from the known value of $Z[J]$ through repeated functional differentiation: $$ G^{I_1I_2\cdots I_n}=\delta^{I_1}\delta^{I_2}\cdots\delta^{I_n}Z[J]\,\bigg|_{J=0}\tag{3} $$

Note that we are using DeWitt's condensed notation, where $I_i\in \mathbb R^d\times\mathbb N^k$ contains both the discrete (spin/flavour/colour) indices as well as the continuous (space-time) coordinates.

On the other hand, in the functional-integral approach, the $n$-point function is defined as $$ G^{I_1I_2\cdots I_n}=N^{-1}\int \varphi^{I_1}\varphi^{I_2}\cdots\varphi^{I_n}\mathrm e^{i\ S[\varphi]+J\cdot\varphi}\ \mathrm d\varphi\tag{4} $$ where $\varphi:\mathbb R^d\times \mathbb N^k\to \Lambda_\infty$ is a $c$-field (with the same Grassmann parity as $\phi$). Note that most books use the same symbol for $\varphi$ and $\phi$; but this is a clear abuse of notation, inasmuch $\varphi$ is a function and $\phi$ is an operator: these objects are different. Here, to keep the notation as clear as possible, we will use two different symbols (the alternative, to use a hat over the operators $\hat\phi$, is slightly more cumbersome). In any case, as $\varphi$ is an integration variable, one can in principle use an arbitrary symbol for it.

The constant $N\in\mathbb C$ is a normalisation constant that makes $G_0=1$: $$ N=\int \mathrm e^{i\ S[\varphi]}\ \mathrm d\varphi\tag{5} $$

We now claim the following: $$ N^{-1}\int \mathrm e^{i\ S[\varphi]+J\cdot\varphi}\ \mathrm d\varphi=Z[J]=\langle \Omega|\mathrm T\ \mathrm e^{i J\cdot\phi}|\Omega\rangle\tag{6} $$ where, in the l.h.s., $S[\varphi]$ is the classical action.

The proof that $(6)$ is correct is rather trivial: the l.h.s. clearly reproduces $(4)$ upon repeated functional differentiation, while the r.h.s clearly reproduces $(2)$, as per the definition of the time-ordered exponential.

In essence, the link between the operator formalism and the functional-integral formalism is provided by $(6)$; but, in order to make this link transparent, we need to invoke the Dyson series. The reason is that we only know how to compute things in perturbation theory; and this is immediate in the functional-integral formalism, but not quite in the operator formalism: we need to derive, first, the Gell-Mann and Low theorem which, together with the Dyson series, allows us to compute the $n$-point functions in a straightforward manner.

We thus want to introducing the Dirac (interaction) picture, $$ \phi(x)=U^\dagger(t)\Phi(x)U(t)\qquad |\Omega\rangle\propto U(\pm\infty)|0\rangle\tag{7} $$ where $\Phi(x)$ is a "free" field; $|0\rangle$ is the "free" vacuum; and $$ U(t,t_0)=\mathrm e^{iH_0(t-t_0)}\mathrm e^{-iH(t-t_0)}=\mathrm T\ \mathrm e^{iS_\mathrm{int}(\Phi)}\tag{8} $$ where $S_\mathrm{int}$ is the interacting part of the action, $$ S_\mathrm{int}\equiv\int_{t_0}^t\int_{\mathbb R^{d-1}} \mathcal L_\mathrm{int}(\Phi)\ \mathrm d\tau\int\mathrm d\boldsymbol x \tag{9} $$

With this, equation $(6)$ can also be written as $$ \int\mathrm d\varphi\ \mathrm e^{i\ S[\varphi]+J\cdot\varphi}\propto Z[J]\propto\langle 0|\mathrm T\ \mathrm e^{i\ S_\mathrm{int}(\Phi)+J\cdot\Phi}|0\rangle\tag{10} $$ where now the constants of proportionality are determined by imposing $Z[0]=1$ (diagrammatically, we omit vacuum diagram in perturbation theory; but note that the equation above is non-perturbative).

We can now expand $(10)$ in power series in $S_\mathrm{int}$; this leads to the standard set of Feynman diagrams. To a given order in perturbation theory, we obtain an equivalence between Feynman's functional-integral and the Dyson series, as required.

More succinctly, $$ Z[j]\propto \exp\left[iS_\mathrm{int}(\delta)\right]\exp\left[-\frac i2 J_{I_1}\Delta^{I_1I_2}J_{I_2}\right]\tag{11} $$

The fact that $(11)$ follows from $(10)$ is trivial to prove in the path integral formalism. On the other hand, in the operator formalism it is a theorem (which easily follows from the Hori formula); see here for a proof. The equation $(11)$ is one of the cornerstones of perturbation theory: the expansion of this expression in powers of $S_\mathrm{int}$ leads directly to Feynman diagrams.

I sketch the proof of the implication "operator formalism" $\to$ "functional-integral formalism" in this PSE post. The other direction is more subtle, but it can be done (this is essentially what is done in DeWitt's The Global Approach to Field Theory, where the author takes the functional integral as a primitive object, and defines operators through the former). This essentially proves the equivalence between these formalisms, but there are some subtleties: in the operator formalism, unitarity is explicit but covariance is not; and vice-versa in the functional integral formalism. This seems to preclude a completely satisfactory proof of equivalence, because we know by experience that radiative corrections may spoil the nice properties of a theory we may initially expect. If covariance is spoiled in the first case, or unitarity in the second one, it becomes clear that the equivalence cannot possibly hold. Fortunately, we can say the following

  • In the operator formalism, Weinberg's QFT argues that it is always possible to add (potentially non-covariant) terms to the Hamiltonian so as to maintain covariance safe. This can be made manifest by using an explicitly covariant propagator.

  • In the functional formalism, DeWitt's QFT argues that it is always possible to add terms to the functional-integral measure so as to maintain unitarity safe. Perhaps this could be made manifest by noting that the measure always guarantees that we can analytically continue to Euclidean space, where the Osterwalder-Schrader positivity property ensures the unitarity of the real time theory.

In this sense, we see that in either formulation both covariance and unitarity are in principle safe, so the equivalence of the formalisms becomes much more plausible.

Further reading:

  • The fundamental object of the perturbative expansion of $Z[J]$ is the propagator. In principle, this object can be defined either in the functional-integral formalism or in the operator formalism. One may wonder if $Z_\mathrm{FI}[J]=Z_\mathrm{OF}[J]$ may fail as a consequence of a possible $\Delta_\mathrm{FI}\neq \Delta_\mathrm{OF}$. This is discussed in Are the path integral formalism and the operator formalism inequivalent?.

  • In introductory textbooks, perturbation theory based on the interaction picture is usually done in the Hamiltonian framework. If the interactions do not include derivatives, then $H_{\mathrm{int}}=L_{\mathrm{int}}$ and the Hamiltonian and Lagrangian frameworks are clearly the same. If the interactions do contain derivatives, then there is a subtlety that is addressed in this PSE post.

  • $\begingroup$ As a side note, the question in the OP seems to be about standard QM instead of QFT; but, as usual, the former is easily seen to be a specific case of the latter for $d=1+0$ dimensions, so the discussion above contains standard QM as a corollary. $\endgroup$ Commented Jan 1, 2017 at 13:00
  • $\begingroup$ Note to self: remove \int in (9) (and $\mathbb R^d\times\mathbb N^k\to\Lambda_\infty$ should be replaced by $\mathbb R^d\to\Lambda_\infty^k$). $\endgroup$ Commented Mar 26, 2018 at 2:14

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