From 42d409fa774efb8206ae5c701d5cbcc4ae1d9cad Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 14 Sep 2021 21:20:30 +0200 Subject: Expand knowledge base --- content/know/concept/interaction-picture/index.pdc | 214 +++++++++++++++++++++ 1 file changed, 214 insertions(+) create mode 100644 content/know/concept/interaction-picture/index.pdc (limited to 'content/know/concept/interaction-picture') diff --git a/content/know/concept/interaction-picture/index.pdc b/content/know/concept/interaction-picture/index.pdc new file mode 100644 index 0000000..1ce330d --- /dev/null +++ b/content/know/concept/interaction-picture/index.pdc @@ -0,0 +1,214 @@ +--- +title: "Interaction picture" +firstLetter: "I" +publishDate: 2021-09-13 +categories: +- Physics +- Quantum mechanics + +date: 2021-09-09T21:15:37+02:00 +draft: false +markup: pandoc +--- + +# Interaction picture + +The **interaction picture** or **Dirac picture** +is an alternative formulation of quantum mechanics, +equivalent to both the Schrödinger picture +and the [Heisenberg picture](/know/concept/heisenberg-picture/). + +Recall that Schrödinger lets states $\ket{\psi_S(t)}$ evolve in time, +but keeps operators $\hat{L}_S$ fixed (except for explicit time dependence). +Meanwhile, Heisenberg keeps states $\ket{\psi_H}$ fixed, +and puts all time dependence on the operators $\hat{L}_H(t)$. + +However, in the interaction picture, +both the states $\ket{\psi_I(t)}$ and the operators $\hat{L}_I(t)$ +evolve in $t$. +This might seem unnecessarily complicated, +but it turns out be convenient when considering +a time-dependent "perturbation" $\hat{H}_{1,S}$ +to a time-independent Hamiltonian $\hat{H}_{0,S}$: + +$$\begin{aligned} + \hat{H}_S(t) + = \hat{H}_{0,S} + \hat{H}_{1,S}(t) +\end{aligned}$$ + +With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian. +We define the unitary conversion operator: + +$$\begin{aligned} + \hat{U}(t) + \equiv \exp\!\Big( i \frac{\hat{H}_{0,S} t}{\hbar} \Big) +\end{aligned}$$ + +The interaction-picture states $\ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$ +are then defined to be: + +$$\begin{aligned} + \boxed{ + \ket{\psi_I(t)} + \equiv \hat{U}(t) \ket{\psi_S(t)} + \qquad + \hat{L}_I(t) + \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t) + } +\end{aligned}$$ + + +## Equations of motion + +To find the equation of motion for $\ket{\psi_I(t)}$, +we differentiate it and multiply by $i \hbar$: + +$$\begin{aligned} + i \hbar \dv{t} \ket{\psi_I} + &= i \hbar \Big( \dv{\hat{U}}{t} \ket{\psi_S} + \hat{U} \dv{t} \ket{\psi_S} \Big) + \\ + &= i \hbar \Big( i \frac{\hat{H}_{0,S}}{\hbar} \Big) \hat{U} \ket{\psi_S} + \hat{U} \Big( i \hbar \dv{t} \ket{\psi_S} \Big) +\end{aligned}$$ + +We insert the Schrödinger equation into the second term, +and use $\comm*{\hat{U}}{\hat{H}_{0,S}} = 0$: + +$$\begin{aligned} + i \hbar \dv{t} \ket{\psi_I} + &= - \hat{H}_{0,S} \hat{U} \ket{\psi_S} + \hat{U} \hat{H}_S \ket{\psi_S} + \\ + &= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \ket{\psi_S} + \\ + &= \hat{U} \big( \hat{H}_{1,S} \big) \hat{U}{}^\dagger \hat{U} \ket{\psi_S} +\end{aligned}$$ + +Which leads to an analogue of the Schrödinger equation, +with $\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$: + +$$\begin{aligned} + \boxed{ + i \hbar \dv{t} \ket{\psi_I(t)} + = \hat{H}_{1,I}(t) \ket{\psi_I(t)} + } +\end{aligned}$$ + +Next, we do the same with an operator $\hat{L}_I$ +to find a description of its evolution in time: + +$$\begin{aligned} + \dv{t} \hat{L}_I + &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger + \\ + &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger + - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger + + \Big( \dv{\hat{L}_S}{t} \Big)_I + \\ + &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I + - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I} + + \Big( \dv{\hat{L}_S}{t} \Big)_I + = \frac{i}{\hbar} \comm*{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I +\end{aligned}$$ + +The result is analogous to the equation of motion in the Heisenberg picture: + +$$\begin{aligned} + \boxed{ + \dv{t} \hat{L}_I(t) + = \frac{i}{\hbar} \comm*{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{t} \hat{L}_S(t) \Big)_I + } +\end{aligned}$$ + + +## Time evolution operator + +Recall that an alternative form of the Schrödinger equation is as follows, +where a **time evolution operator** or +**generator of translations in time** $K_S(t, t_0)$ +brings $\ket{\psi_S}$ from time $t_0$ to $t$: + +$$\begin{aligned} + \ket{\psi_S(t)} + = \hat{K}_S(t, t_0) \ket{\psi_S(t_0)} + \qquad \quad + \hat{K}_S(t, t_0) + \equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big) +\end{aligned}$$ + +We want to find an analogous operator in the interaction picture, satisfying: + +$$\begin{aligned} + \ket{\psi_I(t)} + \equiv \hat{K}_I(t, t_0) \ket{\psi_I(t_0)} +\end{aligned}$$ + +Inserting this definition into the equation of motion for $\ket{\psi_I}$ yields +an equation for $\hat{K}_I$, with the logical boundary condition $\hat{K}_I(t_0, t_0) = 1$: + +$$\begin{aligned} + i \hbar \dv{t} \Big( \hat{K}_I(t, t_0) \ket{\psi_I(t_0)} \Big) + &= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \ket{\psi_I(t_0)} \Big) + \\ + i \hbar \dv{t} \hat{K}_I(t, t_0) + &= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0) +\end{aligned}$$ + +We turn this into an integral equation +by integrating both sides from $t_0$ to $t$: + +$$\begin{aligned} + i \hbar \int_{t_0}^t \dv{t'} K_I(t', t_0) \dd{t'} + = \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'} +\end{aligned}$$ + +After evaluating the left integral, +we see an expression for $\hat{K}_I$ as a function of $\hat{K}_I$ itself: + +$$\begin{aligned} + K_I(t, t_0) + = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'} +\end{aligned}$$ + +By recursively inserting $\hat{K}_I$ once, we get a longer expression, +still with $\hat{K}_I$ on both sides: + +$$\begin{aligned} + K_I(t, t_0) + = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} + + \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'} +\end{aligned}$$ + +And so on. Note the ordering of the integrals and integrands: +upon closer inspection, we see that the $n$th term is +a [time-ordered product](/know/concept/time-ordered-product/) $\mathcal{T}$ +of $n$ factors $\hat{H}_{1,I}$: + +$$\begin{aligned} + \hat{K}_I(t, t_0) + &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1} + + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2} + + \: ... + \\ + &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} + \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n} + \\ + &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} + \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\} +\end{aligned}$$ + +Here, we recognize the Taylor expansion of $\exp$, +leading us to a final expression for $\hat{K}_I$: + +$$\begin{aligned} + \boxed{ + \hat{K}_I(t, t_0) + = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\} + } +\end{aligned}$$ + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. + -- cgit v1.2.3