From 8b8caf2467a738c0b0ccac32163d426ffab2cbd8 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 15 Oct 2024 18:08:29 +0200 Subject: Improve knowledge base --- source/know/concept/interaction-picture/index.md | 182 +++++++++++------------ 1 file changed, 86 insertions(+), 96 deletions(-) (limited to 'source/know/concept/interaction-picture') diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md index de469fa..8428bf3 100644 --- a/source/know/concept/interaction-picture/index.md +++ b/source/know/concept/interaction-picture/index.md @@ -13,17 +13,19 @@ 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, +Recall that in the Schrödinger picture, +the states $$\Ket{\psi_S(t)}$$ evolve in time, +but time-independent operators $$\hat{L}_S$$ are fixed. +Meanwhile in the Heisenberg picture, +the states $$\Ket{\psi_H}$$ are constant, +and all time dependence is on the operators $$\hat{L}_H(t)$$ instead. + +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" $$\hat{H}_{1,S}$$ +This may seem unnecessarily complicated, +but it turns out to be convenient when considering +a system with a time-dependent "perturbation" $$\hat{H}_{1,S}$$ to a time-independent Hamiltonian $$\hat{H}_{0,S}$$: $$\begin{aligned} @@ -31,29 +33,43 @@ $$\begin{aligned} = \hat{H}_{0,S} + \hat{H}_{1,S}(t) \end{aligned}$$ -With $$\hat{H}_S(t)$$ the full Schrödinger Hamiltonian. -We define the unitary conversion operator: +Despite being called a perturbation, +$$\hat{H}_{1, S}$$ need not be weak compared to $$\hat{H}_{0, S}$$. +Basically, any way of splitting $$\hat{H}_S$$ is valid +as long as $$\hat{H}_{0, S}$$ is time-independent, +but only a few ways are useful. + +We now define the unitary conversion operator $$\hat{U}(t)$$ as shown below. +Note its similarity to the [time-evolution operator](/know/concept/time-evolution-operator/) +$$\hat{K}_S(t)$$, but with the opposite sign in the exponent: $$\begin{aligned} \boxed{ \hat{U}(t) - \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg) + \equiv \exp\!\bigg( \frac{i}{\hbar} \hat{H}_{0,S} t \bigg) } \end{aligned}$$ -The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$ -are then defined to be: +The interaction-picture states $$\Ket{\psi_I(t)}$$ +and operators $$\hat{L}_I(t)$$ are then defined as follows: $$\begin{aligned} \boxed{ \Ket{\psi_I(t)} \equiv \hat{U}(t) \Ket{\psi_S(t)} - \qquad + } + \qquad\qquad + \boxed{ \hat{L}_I(t) \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t) } \end{aligned}$$ +Because $$\hat{H}_{0, S}$$ is time-independent, +it commutes with $$\hat{U}(t)$$, +so conveniently $$\hat{H}_{0, I} = \hat{H}_{0, S}$$. + + ## Equations of motion @@ -61,17 +77,17 @@ To find the equation of motion for $$\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) + i \hbar \dv{}{t} \Ket{\psi_I} + &= i \hbar \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg) \end{aligned}$$ -We insert the Schrödinger equation into the second term, -and use $$\comm{\hat{U}}{\hat{H}_{0,S}} = 0$$: +We insert the definition of $$\hat{U}$$ in the first term +and the Schrödinger equation into the second, +and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}} = 0$$ +thanks to the time-independence of $$\hat{H}_{0, S}$$: $$\begin{aligned} - i \hbar \dv{}{t}\Ket{\psi_I} + 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} @@ -84,129 +100,103 @@ with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$: $$\begin{aligned} \boxed{ - i \hbar \dv{}{t}\Ket{\psi_I(t)} + i \hbar \dv{}{t} \Ket{\psi_I(t)} = \hat{H}_{1,I}(t) \Ket{\psi_I(t)} } \end{aligned}$$ Next, we do the same with an operator $$\hat{L}_I$$ -to find a description of its evolution in time: +in order to describe its evolution in time: $$\begin{aligned} - \dv{}{t}\hat{L}_I - &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger + \dv{\hat{L}_I}{t} + &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger \\ &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger - + \Big( \dv{\hat{L}_S}{t} \Big)_I + + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \\ &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I} - + \Big( \dv{\hat{L}_S}{t} \Big)_I - = \frac{i}{\hbar} \comm{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I + + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \end{aligned}$$ The result is analogous to the equation of motion in the Heisenberg picture: $$\begin{aligned} \boxed{ - \dv{}{t}\hat{L}_I(t) - = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_I + \dv{}{t} \hat{L}_I(t) + = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_I } \end{aligned}$$ +In other words, in the interaction picture, +the "simple" time-dependence (from $$\hat{H}_{0, S}$$) is given to the operators, +and the "complicated" dependence (from $$\hat{H}_{1, S}$$) to the states. +This means that the difficult part of a problem +can be solved in isolation in a kind of Schrödinger picture. -## 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 $$t$$: -$$\begin{aligned} - \Ket{\psi_S(t)} - = \hat{K}_S(t, t_0) \Ket{\psi_S(t_0)} - \qquad \quad - \hat{K}_S(t, t_0) - \equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big) -\end{aligned}$$ +## Time evolution operator -We want to find an analogous operator in the interaction picture, satisfying: +What about the time evolution operator $$\hat{K}_S(t)$$? +Its interaction version $$\hat{K}_I(t)$$ +is unsurprisingly obtained by the standard transform +$$\hat{K}_I = \hat{U} \hat{K}_S \hat{U}^\dagger$$: $$\begin{aligned} \Ket{\psi_I(t)} - \equiv \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} -\end{aligned}$$ - -Inserting this definition into the equation of motion for $$\Ket{\psi_I}$$ yields -an equation for $$\hat{K}_I$$, with the logical boundary condition $$\hat{K}_I(t_0, t_0) = 1$$: - -$$\begin{aligned} - i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big) - &= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big) + &= \hat{U}(t) \Ket{\psi_S(t)} \\ - i \hbar \dv{}{t}\hat{K}_I(t, t_0) - &= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0) -\end{aligned}$$ - -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: - -$$\begin{aligned} - K_I(t, t_0) - = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'} + &= \hat{U}(t) \: \hat{K}_S(t) \: \hat{U}^\dagger(t) \Ket{\psi_I(0)} + \\ + &\equiv \hat{K}_I(t) \Ket{\psi_I(0)} \end{aligned}$$ -By recursively inserting $$\hat{K}_I$$ once, we get a longer expression, -still with $$\hat{K}_I$$ on both sides: +But we can do better. By inserting this definition of $$\hat{K}_I$$ +into the interaction picture's analogue of Schrödinger's equation, +we get the following relation for $$\hat{K}_I$$: $$\begin{aligned} - K_I(t, t_0) - = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} - + \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'} + i \hbar \dv{}{t} \hat{K}_I(t) + &= \hat{H}_{1,I}(t) \: \hat{K}_I(t) \end{aligned}$$ -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 $$\hat{H}_{1,I}$$: +In other words, $$\hat{K}_I$$ can be said to also obey +the standard equation of motion for states, despite being an operator. +We integrate both sides and use $$\hat{K}_I(0) = 1$$: $$\begin{aligned} - \hat{K}_I(t, t_0) - &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1} - + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2} - + \: ... - \\ - &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} - \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n} - \\ - &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} - \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\} + K_I(t) + = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \: \hat{K}_I(\tau) \dd{\tau} \end{aligned}$$ -This construction is occasionally called the **Dyson series**. -We recognize the well-known Taylor expansion of $$\exp(x)$$, -leading us to a final expression for $$\hat{K}_I$$: +This equation can be recursively inserted into itself forever. +We recognize the resulting so called *Dyson series* +from the derivation of $$\hat{K}_S(t)$$ +for time-dependent Hamiltonians in the Schrödinger picture +([given here](/know/concept/time-evolution-operator/)), +so we know that the result is given by: $$\begin{aligned} \boxed{ - \hat{K}_I(t, t_0) - = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\} + \hat{K}_I(t) + = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \dd{\tau} \bigg) \bigg\} } \end{aligned}$$ +Where $$\mathcal{T}$$ is the +[time-ordering meta-operator](/know/concept/time-ordered-product/), +which is conventionally written in this way +to say that it applies to the terms of a Taylor expansion of $$\exp(x)$$. +This means that the evolution of a quantum state in the interaction picture +is determined by the perturbation $$\hat{H}_{1, I}$$. + ## References 1. H. Bruus, K. Flensberg, *Many-body quantum theory in condensed matter physics*, 2016, Oxford. - -- cgit v1.2.3