Quantum Field Theory Course

Total: 440+ pages covering classical field theory through the Standard Model

Part IV, Chapter 1 | Page 2 of 3

The Dyson Series

Solving the time evolution operator order by order in the interaction strength

1.6 From Differential to Integral Equation

Recall from Page 1 that the time evolution operator satisfies:

$$i\frac{\partial}{\partial t}U(t,t_0) = H_{\text{int},I}(t)\,U(t,t_0), \qquad U(t_0,t_0) = \mathbb{1}$$

We convert this into an integral equation by integrating both sides from $t_0$ to $t$:

$$U(t,t_0) = \mathbb{1} - i\int_{t_0}^{t}dt_1\,H_{\text{int},I}(t_1)\,U(t_1,t_0)$$

This is exact but implicit — U appears on both sides. We solve it by iteration. Substituting the equation into itself for the U on the right-hand side:

$$U(t,t_0) = \mathbb{1} - i\int_{t_0}^{t}dt_1\,H_{\text{int},I}(t_1)\Bigg[\mathbb{1} - i\int_{t_0}^{t_1}dt_2\,H_{\text{int},I}(t_2)\,U(t_2,t_0)\Bigg]$$

Expanding:

$$U(t,t_0) = \mathbb{1} + (-i)\int_{t_0}^{t}dt_1\,H_{\text{int},I}(t_1) + (-i)^2\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\,H_{\text{int},I}(t_1)H_{\text{int},I}(t_2) + \cdots$$

Continuing to all orders, we obtain the Dyson series:

$$\boxed{U(t,t_0) = \sum_{n=0}^{\infty}(-i)^n\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_n\,H_{\text{int},I}(t_1)H_{\text{int},I}(t_2)\cdots H_{\text{int},I}(t_n)}$$

Note the crucial ordering constraint: $t \geq t_1 \geq t_2 \geq \cdots \geq t_n \geq t_0$. The Hamiltonians appear in time-ordered sequence from left to right, with later times on the left.

1.7 The Time-Ordering Operator

The nested integration region $t_1 \geq t_2 \geq \cdots \geq t_n$ is inconvenient. We introduce the time-ordering operator T to write the integrals over the full hypercube.

For two operators, T is defined as:

$$T\big[A(t_1)B(t_2)\big] = \begin{cases} A(t_1)B(t_2) & \text{if } t_1 > t_2 \\ B(t_2)A(t_1) & \text{if } t_2 > t_1 \end{cases}$$

For fermionic operators, each swap introduces a minus sign (required for Grassmann-valued fields). More generally, T rearranges any product of operators so that later times stand to the left.

The key identity that relates the time-ordered integral over the full region to the ordered region is:

$$\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\,H_I(t_1)H_I(t_2) = \frac{1}{2!}\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\,T\big[H_I(t_1)H_I(t_2)\big]$$

Proof: The double integral over the full square $[t_0,t]\times[t_0,t]$ can be split into two triangular regions: $t_1 > t_2$ and $t_2 > t_1$. In the region $t_1 > t_2$, the time-ordering does nothing. In the region $t_2 > t_1$, T swaps the operators, giving the same integrand as the first region (after relabeling dummy variables). So both halves contribute equally, giving the factor of 1/2!.

At nth order, the n! permutations of the time variables cover the full hypercube, giving:

$$\int_{t_0}^{t}dt_1\cdots\int_{t_0}^{t_{n-1}}dt_n\,H_I(t_1)\cdots H_I(t_n) = \frac{1}{n!}\int_{t_0}^{t}dt_1\cdots\int_{t_0}^{t}dt_n\,T\big[H_I(t_1)\cdots H_I(t_n)\big]$$

💡Why Time Ordering?

Time ordering is not just a mathematical convenience — it reflects causality. In quantum field theory, events at later times must account for what happened at earlier times. The T-product ensures that when we evolve a state forward, each interaction "knows" about all prior interactions. This is why the time-ordering prescription emerges naturally from the iterative solution of the Schrodinger equation in the interaction picture.

1.8 The Dyson Formula: S = T exp(-i$\int$ Hint dt)

Substituting the time-ordering identity into the Dyson series:

$$U(t,t_0) = \sum_{n=0}^{\infty}\frac{(-i)^n}{n!}\int_{t_0}^{t}dt_1\cdots\int_{t_0}^{t}dt_n\,T\big[H_{\text{int},I}(t_1)\cdots H_{\text{int},I}(t_n)\big]$$

This sum has exactly the structure of a Taylor expansion of an exponential, leading to the compact notation:

$$\boxed{U(t,t_0) = T\exp\!\left(-i\int_{t_0}^{t}dt'\,H_{\text{int},I}(t')\right)}$$

This is Dyson's formula. It is important to understand that the time-ordered exponential is defined by its series expansion — you cannot simply exponentiate and then time-order. The T acts on the entire series.

For the S-matrix, we take the limits $t_0 \to -\infty$ and $t \to +\infty$:

$$\boxed{S = T\exp\!\left(-i\int_{-\infty}^{+\infty}dt\,H_{\text{int},I}(t)\right) = T\exp\!\left(-i\int d^4x\,\mathcal{H}_{\text{int},I}(x)\right)}$$

where in the last step we wrote $H_{\text{int}} = \int d^3x\,\mathcal{H}_{\text{int}}$ and combined the spatial and temporal integrals into a Lorentz-invariant four-dimensional integral. For theories where $\mathcal{H}_{\text{int}} = -\mathcal{L}_{\text{int}}$, this becomes:

$$S = T\exp\!\left(i\int d^4x\,\mathcal{L}_{\text{int},I}(x)\right)$$

1.9 Perturbative Expansion in Practice

For $\phi^4$ theory with $\mathcal{L}_{\text{int}} = -\frac{\lambda}{4!}\phi^4$, the S-matrix expansion reads:

$$S = \mathbb{1} + \frac{(-i\lambda)}{4!}\int d^4x\,T[\phi_I^4(x)] + \frac{1}{2!}\left(\frac{-i\lambda}{4!}\right)^2\int d^4x\,d^4y\,T[\phi_I^4(x)\phi_I^4(y)] + \cdots$$

Each term involves time-ordered products of free fields. To evaluate these, we need a systematic method for reducing time-ordered products to normal-ordered products (which have vanishing vacuum expectation values) plus contractions (propagators). This method is Wick's theorem, the subject of the next page.

The nth-order term in the expansion is proportional to $\lambda^n$ (or $e^{2n}$ in QED). If the coupling is small, higher-order terms are suppressed, justifying truncation of the series. The first few orders typically suffice for excellent agreement with experiment.

💡Convergence of the Dyson Series

The Dyson series is an asymptotic series, not a convergent one (Dyson's argument, 1952): if it converged for coupling $\lambda > 0$, it would also converge for $\lambda < 0$, but negative $\lambda$ makes the potential unbounded below, so the theory is sick. Nevertheless, the first few terms give extraordinarily accurate predictions. In QED, perturbation theory with $\alpha \approx 1/137$ matches experiment to 12 decimal places.

Key Concepts

  • The integral equation for U(t, t0) is solved iteratively to produce the Dyson series
  • Time ordering T rearranges operators so later times stand to the left, introducing a factor of 1/n!
  • The Dyson formula $U = T\exp(-i\int H_{\text{int},I}\,dt)$ compactly encodes all orders of perturbation theory
  • The S-matrix is $S = T\exp(i\int d^4x\,\mathcal{L}_{\text{int},I})$ in Lorentz-invariant form
  • The perturbative expansion is an asymptotic series, but its first few terms are extraordinarily accurate
  • Evaluating T-products of free fields systematically requires Wick's theorem (next page)
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