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} \boxed{ \hat{U}(t) \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg) } \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}

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 $\hat{K}_I$:

\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.