Categories: Physics, Quantum mechanics.

# 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.

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 $$\mathcal{T}$$ 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.