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In dealing with interacting Hamiltonians, it's common to do expansions of some sort to make the problem tractable. Two common methods are the Dyson and Magnus expansion. Of the two, the Magnus expansion is quite interesting in that is preserves unitary time evolution at every order. This is not true for the Dyson series however. But, since this is such a nice property, my question is when and why would you prefer using the Dyson series over the Magnus expansion?

KF Gauss
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    The Dyson series seems better for Lorentz invariance stuff, because it has a $d^4 x$ and all four parameters are integrated from $-\infty$ to $\infty$. Ultimately, our goal in QFT isn't the Dyson series. It's to get to Feynman rules. Dyson series is just a way. – Ryder Rude Nov 11 '22 at 06:43
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    If you're talking about integrating over the Hamiltonian density in position and time, then the Magnus expansion appears to do that as well. – KF Gauss Nov 11 '22 at 13:11
  • I am saying that we don't use the Dyson series in QFT anyway. It's only used as part of the derivation of Feynman rules. There are other derivations too like the Schwinger Dyson equation. You can also perhaps work out a derivation using the Magnus expansion, but in the end, we only use the Feynman rules. – Ryder Rude Nov 11 '22 at 14:02
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    @RyderRude I understand on a practical level why you say the goal is to get to the Feynman rules. But I would argue a more correct statement is that our goal is to understand the dynamics of the QFT. In most cases, the Feynman rules are the end product of a perturbative expansion to do that. But in other cases (instantons, sphalerons, ...), perturbative methods and Feynman diagrams just aren't enough. Sorry if that comment is overlay pedantic. – Andrew Nov 12 '22 at 13:38
  • Unfortunately I can’t comment yet, but I would like to point out that the time-ordered Dyson series is equivalent to the Magnus expansion for those interested (see eq (9) in https://arxiv.org/abs/1206.3990). But it’s clear that one would need to expand the exponential matrix and time-order each expansion term from the Magnus exponent to see the relationship. – Kristian Barajas Oct 01 '23 at 03:32
  • @KristianBarajas is that simply the statement that the two series aim to approximate the same unitary operator? I think that is expected – KF Gauss Oct 02 '23 at 03:51

1 Answers1

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  1. The Dyson and Magnus series are expanding $U(t_2,t_1)$ and $\ln U(t_2,t_1)$ wrt. the coupling constant $\lambda$, respectively. Here $$\begin{align} U(t_2,t_1)~&=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~\lambda V(t)\right] &\text{for}& t_1 ~<~t_2 \cr\cr AT\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~\lambda V(t)\right] &\text{for}& t_2 ~<~t_1 \end{array}\right. \end{align}\tag{1} $$ is the evolution operator, which satisfies two TDSEs $$\begin{align} i\hbar \frac{\partial }{\partial t_2}U(t_2,t_1) ~=~&\lambda V(t_2)U(t_2,t_1), \cr i\hbar \frac{\partial }{\partial t_1}U(t_2,t_1) ~=~&-U(t_2,t_1)V(t_1)\lambda,\end{align}\tag{2} $$ along with the boundary condition $$ U(t,t)~=~{\bf 1}.\tag{3}$$

  2. From the perspective of an all-order formal power series in $\lambda$, cf. perturbation theory, the Dyson series is much simpler than the Magnus series. (For starters, Wikipedia only provides a recursive formula for the latter.)

  3. We also note that the Dyson series is time-ordered (i.e. respects the time-ordering), while the Magnus series isn't. (The Magnus series contains commutators, which generically contain terms of opposite time-ordering.)

  4. However, as OP already mentions, in terms of a truncated expansion (in particular in numerical work), the Magnus expansion is often useful because of manifest unitarity in QM (which transcribes to a symplectic integrator in classical mechanics).

Qmechanic
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  • Thanks! So in terms of being "simpler" - I take that to mean that Magnus expansion terms are harder to analytically/numerically compute? – KF Gauss Nov 12 '22 at 13:12
  • And in regards to the comment by @RyderRude, I suppose manifest Lorentz invariance (and other symmetries) don't change between Dyson and Magnus? – KF Gauss Nov 12 '22 at 13:15
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    I'm not very familiar with the Magnus expansion, so this is just a naive question. From the wikipedia page, based on the statement "one wishes to solve the initial-value problem" and the appearance of the commutators in the expansion, it seems the Magnus expansion may be the correct expansion for the time evolution operator in the "in-in" formalism, whereas the Dyson series is the correct expression for the "in-out" formalism, in which case it seems there is a fundamental difference in what these series are expressing. Is that reading incorrect? – Andrew Nov 12 '22 at 13:34
  • The basic version concerns the in-out formalism. Presumably it can be adapted to the in-in formalism. – Qmechanic Nov 12 '22 at 14:25