6 Perturbation Theory

Dyson’s formula

We wish for a series expansion of \(e^{t(H+V)}\). Consider the following expansion: \[ f(s)\big|_{s=0}^{s=t} = e^{s(H+V)} e^{(t-s)H} \big|_{s=0}^t = e^{t(H+V)} - e^{tH} = \int_0^t e^{s(H+V)} (H+V-H) e^{(t-s)H}\, dx \] Repeating this, we can consider \[\begin{align} e^{t(H+V)} &= e^{tH} + \int_0^t e^{s_1(H+V)} V e^{(t-s)H}\, ds_1 \\ &= e^{tH} + \int_0^t e^{s_1H} V e^{(t-s_1)H}\, ds_1 + \int_0^t ds_1 \int_0^{s_1} ds_2\, e^{-s_1(H+V)}V e^{-(s-s_1)H} V e^{-sH} \\ &= \sum_{n=0}^\infty \int_{0\leq s_1\leq \dots \leq s_{n+1}=t} e^{-s_{n+1}H} V e^{-s_n H} V \dots e^{-s_2H} V e^{-s_1H}\, ds \end{align}\] In mathematics, this was developed by Hille Phillips, and in physics by Dyson. Dyson proposed to write this in terms of time-ordered products. \[ e^{-s_{n+1}H} V e^{-s_n H} V \dots e^{-s_2H} V e^{-s_1H} = \dots \left(e^{(s_2-s_1)H} V e^{-(s_2-s_1)H}\right) \left( e^{s_1 H} V e^{-s_1 H} \right) \] Define \(V(s) = e^{s H} V e^{-s H}\), we obtain \[\begin{align} e^{t(H+V)} &= \sum_n \int_{s_j\text{ simplex}} e^{s_{n+1}H} V(s_n)\dots V(S_1) \\ &= \sum_n \dfrac 1 {n!} \int ds\, e^{tH} \left[ V(s_1) \dots V(s_n) \right]_+ \end{align}\] Here, the time-ordered product is defined as \[\begin{align} \left[ V(s_1)\dots V(s_n) \right]_+ = V(s_{\pi_1})\dots V(s_{\pi_n}) \text{ such that } s_\pi \text{ is ordered.} \end{align}\] With this convention, we obtain Dyson’s formula \[\begin{align} e^{t(H+V)} &= \sum_{n=0}^\infty \dfrac 1 {n!} \int e^{tH} \left[ V(s_1)\dots V(s_n) \right]_+ \, ds = \exp \left[tH + \left( \int_0^t V(s)\, dx \right)_+\right] \end{align}\]

Example: Fourier transform

Consider the Fourier transform \(f(x) = \frac 1 {\sqrt{2\pi}} e^{ipx} \tilde f(p)\, dp\). \[\begin{align} \dfrac 1 2 \left(-i \partial_{x}\right)^2f(x) &= \dfrac 1 {\sqrt{2\pi}} \int e^{ipx} \dfrac{p^2}{2} \tilde f(p)\, dp \end{align}\] Note that \(p^2/2\) is the kinetic energy. Exponentiating this solution, we claim that \(e^{t \partial_{x}^2/2}\) acts on a function as multiplying its Fourier transform by \(e^{-t p^2/2}\), which is equivalent to convolving \(f\) with a Gaussian: \[\begin{align} e^{t\partial_{x}^2/2} f(x) = \dfrac 1 {\sqrt{2\pi}} \int e^{ipx + t\frac{p^2}{2}} \tilde f(p)\, dp = \dfrac 1 {\sqrt{2\pi t}} \int e^{-\frac{(x-y)^2}{2t}}f(y)\, dy \end{align}\] This is the famous solution to the heat diffusion equation \(\dot f = \dfrac 1 2 \partial_{x}^2 f\). The diffusion kernel \(\mathcal K_t f = e^{t\partial_{x}^2/2} f =\) Gaussian convolution of \(f\). This is a one-parameter exponential group. Consider the decomposition \(\mathcal K_t = K_{t/n}^n\). \[\begin{align} \mathcal K_t(x_0, x_n) &= \int \mathcal K_{t/n}(x_0, x_1)\dots \mathcal K_{t/n}(x_{n-1, x_n})\, dx_1 \dots dx_{n-1} \\ &= \dfrac 1 {(2\pi t)^{n/2}} \int dx_1\dots dx_{n-1} \exp \left[ -\sum \dfrac{(x_j-x_{j+1})^2}{2t/n} \right] \end{align}\] If we replace \(t\mapsto it\), this resembles the path integral from quantum theory. Treating the integrand as finite approximations, we can write \[\begin{align} \lim_{n\to \infty} \mathcal K_t(x_0, x_n) \approx \dfrac 1 {\sqrt{(2\pi t)^n}} \int \exp \left( \dfrac 1 2 \int v^2\, dt \right) \end{align}\]