Given operator equation like: $$i{\frac d{dt}}U(t,t_{0}) =V_I(t)U(t,t_{0})\tag{1} $$

The Dyson series solution is \begin{array}{lcl}U(t,t_{0})&=&1-i\int _{{t_{0}}}^{{t}}{dt_{1}V_I(t_{1})}+(-i)^{2}\int _{{t_{0}}}^{t}{dt_{1}\int _{{t_{0}}}^{{t_{1}}}{dt_{2}V_I(t_{1})V_I(t_{2})}}+\cdots \\&&{}+(-i)^{n}\int _{{t_{0}}}^{t}{dt_{1}\int _{{t_{0}}}^{{t_{1}}}{dt_{2}\cdots \int _{{t_{0}}}^{{t_{{n-1}}}}{dt_{n}V_I(t_{1})V_I(t_{2})\cdots V_I(t_{n})}}}+\cdots .\tag{2}\end{array}

Is this solution convergent? I always confused about this when I learn qft.

We have known that we can use feynman diagram (Dyson series) to perturbatively calculate every order, and even though after renormalization the result of every order becomes finite, the series of finite results is still a divergent asymptotic series. See also the radius of convergence is $0$ for QED

The most famous example is $0+0$-dim $\phi^4$, the partition function is $$Z(g)=\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}\exp(-\frac{1}{2} x^2 -g x^4)=\sum_{n}g^n I_n$$ with $$g^nI_n=\frac{(-g)^n}{n!}\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}x^{4n}=\frac{(-g)^n}{n!}(4n-1)!!\propto(\frac{-gn}{e})^n$$ We see no matter how small of $g$ the sum is still divergent. But we konw the original integral must be convergent, $$Z(g)=\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}\exp(-\frac{1}{2} x^2 -g x^4)= \frac{1}{4\sqrt{g\pi}}e^{\frac{1}{32g}}K_{1/4}(\frac{1}{32g})$$ In this explicit case, it's easy to explain since we can't exchange the order of integral and infinite summation. Is this the root of divergence of Dyson series in general?

From the definition of $U(t,t_{0})$ $(1)$, it must be a finite quantity. But why in general we get a divergent series when we act $U(t,t_{0})$ between any state. The derivation seems exact when we get Dyson series $(2)$ from $(1)$. So where is the loophole or where are we cheated in the derivation of Dyson series?

Certainly there are some cases, we can get finite result from $(2)$, like $H_0=p^2/2m + 1/2 m w^2 x^2$ with linear external potential $V= - e x$ in quantum mechanics. So it seems there should exist some convergent criteria for $(2)$ which are not said in textbook.

Another related question:

We know given any $n\times n$ matrix $A$, the series $$\sum_{n=0}^\infty \frac{A^n}{n!}$$ is finite for any component. So for any $n\times n$ matrix $A$, $$e^A=\sum_{n=0}^\infty \frac{A^n}{n!}$$ is always well-defined.

In principle $U(t,t_{0})$ can also be written as a formalism of exponential of some operator: $$U(t,t_{0})=T \exp(-i\int _{{t_{0}}}^{{t}}{dt_{1}V_I(t_{1})})$$ What's about the convergence of $e^A=\sum_{n=0}^\infty \frac{A^n}{n!}$ when $A$ is a infinte dimensional matrix, i.e. operator? I never learnt functional analysis, so I don't the answer about this question.

  • 3
    $\begingroup$ Note that if $v\equiv\max_{\tau\in(t_0,t)}||V(\tau)||<\infty$, then \begin{equation} \begin{aligned} \left|\left|\int _{{t_{0}}}^{t}{dt_{1}\int _{{t_{0}}}^{{t_{1}}}{dt_{2}\cdots \int _{{t_{0}}}^{{t_{{n-1}}}}{dt_{n}V_I(t_{1})V_I(t_{2})\cdots V_I(t_{n})}}}\right|\right|&\le v^n\int _{{t_{0}}}^{t}{dt_{1}\int _{{t_{0}}}^{{t_{1}}}{dt_{2}\cdots \int _{{t_{0}}}^{{t_{{n-1}}}}{dt_{n}}}}\\ &=\frac{(t-t_0)^n}{n!}v^n \end{aligned} \end{equation}, which means that the Dyson series converges. Thus, $v<\infty$ is sufficient -- but I don't think it is necessary -- for convergence. $\endgroup$ Sep 15, 2017 at 19:50

1 Answer 1


Like you pointed out mathematically speaking the series is converging, so where are the divergences coming from? The derivation of the Dyson series is exact it is our methods of estimating it and the approximation we make that gives divergences.

There is a theorem that states that the partition function is analytical (at least in temperature) for a finite system. My understanding is that some of the divergences (not all) are coming from the thermodynamic limit. Rigorously speaking the particle number is finite, but it is so large that it is obviously a very good approximation to say that $N\to\infty$, the price to pay is then some artificial divergences.

This is just a part of the story, the Dyson expansion treats the interaction in its full glory so you can imagine that summing it is simply an impossible task since you really are trying to sum over every physical process possible. The result is that we can only do partial sums and there is really no reason to expect thees partial sums to be converging. If you somehow manage to sum the entire series then you will recover the convergence.

As an example. The bubble diagrams are divergent but strictly speaking they scale with $N$ which is finite. The expansion in the Coulomb interaction is divergent but if you take into account the screening (which is just summing a larger subset of diagrams) it is not. The susceptibilities are divergent at the phase transition this is again an artifact of the limit $N\to\infty$.

In conclusion, we can't sum the Dyson series so we try to guess what physical process is important and try to estimate it using physical intuition and approximations rather then rigorous mathematical statement. As a result of these approximations we sometimes encounter divergences.

  • $\begingroup$ Great answer, but I don't think it's due to taking the thermodynamic limit. Though the partition function is smooth (infinitely differentiable) in inverse temperature $\beta$ (or coupling strength $g$), it may not be analytic (written as a series) at $g=0$. An example would be the OP's partition function $Z(g)$. Therefore, what the Feynmann diagrams are calculating our the Taylor coefficients, yet whether the Taylor coefficients converge or not is another story. $\endgroup$ Feb 13, 2021 at 7:51

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