Interaction picture

Suppose for a moment that we assumed our Hamiltonian \(H\) came in two pieces:

\[ H = H^{(0)} + V \]

Where \(H^{(0)}\) is a Hamiltonian for which you know all the dynamics and solutions for, and the addition of \(V\) ruins your knowledge of it. Is there any way we can split the solutions so that we use what we know about \(H^{(0)}\) as best we can? Yes.

Time propagator

Sometimes in QM you can just make shit up. And as long as it ends up being well-defined, you can use that. Let's do that now. I want a magic operator \(U\) that encodes all the time evolution of my states:

\[ \ket{\psi(t)} = U(t, t_{0}) \ket{\psi(t_{0})} \]

\(U\) depends on the start and end times of the time-evolution if the Hamiltonian of the system has time-dependence \(H(t)\). If \(H\) is time-independent \(U\) only depends on the duration of the propagation.

The Schrodinger equation is

\[ \partial _{t} \ket{\psi(t)} = \frac{i 2\pi }{h} H \ket{\psi(t)} \]

But now use our definition of \(U\) to get

\[ \partial _{t} U(t)\ket{\psi(0)} = \frac{i 2\pi }{h} H U(t)\ket{\psi(0)} \]

But now \(\ket{\psi(0)}\) has no time-dependence, so we can just factor it out of both sides to get

\[ \partial_{t}U(t) = \frac{i 2\pi}{h} H U(t) \]

Now we have an equation of motion for \(U\). This is still the Schrodinger equation. Actually, this equation has a 'formal solution' of

\[ U(t) \equiv \exp \left(\frac{i2\pi}{h} Ht \right)\]

Plug in this definition of \(U\) into the Schrodinger equation and you will see it is a solution. However, the definition of \(U\) involves exponentiating the operator \(H\). Turns out, that's quite annoying to do. To do it, you need to know the eigenstates of \(H\). Which is just the problem of solving the Schrodinger equation again. So we don't gain any calculational power here. But we have gained a lot of conceptual simplicity, as we will see.

By the way, if \(U\) depends on the start and end times it actually has the form

\[ U(t, t_{0}) \equiv \exp \left( \frac{i 2\pi}{h} \int_{t_{0}}^{t} |dt_{1}|\ H(t_{1}) \right)\]

But this more complicated form does not change any of the results, so let's just leave it in the simpler form.

\(U\) is our new best friend. This is a super useful way to think about QM. In fact, when I think about QM in other ways, you can easily get confused between the Schrodinger picture, Heisenberg picture, etc. etc. But if you think in terms of \(U\), everything stays easy.

Sometimes I get confused about what is my system's state vector, and what is a basis vector. The difference is simple I think. Your system is the thing being hit with the \(U\) operator.

Finally, note that \(U\) is unitary, meaning \(U^{\dagger}U = 1\). This is a very useful property, and it also means that \(U^{\dagger}\) is the anti-propagator that propagates states backwards in time. After all, if \(U(t)\) sends a state \(t\) time into the future, \(U(t)^{\dagger}\) brings it back to where it started only if it sends it \(t\) time into the past.

Time propagation due to \(V\)

But now think in terms of our split from earlier. \(H = H^{(0)} + V\), so we can say

\[ U(t) = \exp\left( \frac{i 2\pi}{h} (H^{(0)} + V)t \right) \]

But wait, usually, \(e^{ A + B } = e^{ A }e^{ B }\). So could we factor this into

\[ U(t) =^{?} \exp\left( \frac{i 2\pi}{h} H^{(0)}t \right) \exp\left( \frac{i 2\pi}{h} V t \right)= U_{0}U_{V} \]

Ah, unfortunately no. The identity \(e^{ A + B } = e^{ A }e^{ B }\). Is only true if \(A, B\) commute. And \(H^{(0)}, V\) might not commute, that depends on their specific forms.

However...

We can play an extremely devilish trick. Instead of deriving this factorization, just DEFINE

\[ \underline{U} \equiv U_{0}^{\dagger}U \]

Now if \(H^{(0)}, V\) commute, then you can show that this is equivalent to

\[ \underline{U} = \exp\left( \frac{i 2\pi}{h} V t \right) \]

But if \(H^{(0)}, V\) don't commute (and I won't assume that they do!), it doesn't matter! We can still use \(U_{V} \equiv U_{0}^{\dagger} U\) as our definition of \(U_{V}\). And \(U_{V}\) will still have all the properties that you expect of an operator which causes time evolution just do the the presence of \(V\).

Defining the interaction picture

Wow, this is a very good idea. Could we write down a Schrodinger equation just for \(U_{V}\)? Actually, yes.

Let's introduce the interaction picture. For all operators \(\mathcal{O}\) that are not \(U\), their interaction picture version is marked with an underline:

\[ \underline{\mathcal{O}} \equiv U^{\dagger}_{0} \mathcal{O} U_{0} \]

And this applies to \(V\) also. The only operator this definition doesn't apply to is \(\underline{U} \equiv U_{0}^{\dagger}U\). Let's see if we can write down a Schrodinger equation for it:

\[ \begin{align*} \partial_{t} \underline{U} &= \partial_{t}(U^{\dagger}_{0} U) \\ &= (\partial_{t} U_{0}^{\dagger})U + U_{0}^{\dagger}( \partial_{t} U) \\ &= -\frac{i 2\pi}{h}H^{0} U_{0}^{\dagger} U + U_{0}^{\dagger}\left( \frac{i 2\pi}{h}(H^{(0)} + V) U \right) \\ &= \frac{i 2\pi}{h}[-H^{0} U_{0}^{\dagger} U + U_{0}^{\dagger} H^{0}U + U_{0}^{\dagger}V U] \end{align*} \]

But \(U_{0} = \exp(i 2\pi H^{(0)} t/h)\), which means it commutes with \(H^{(0)}\) and so cancels the other term that looks like this. Now insert \(1 = U_{0}U_{0}^{\dagger}\) and find

\[ \begin{align*} \partial_{t} \underline{U} &= \frac{i 2\pi}{h}U_{0}^{\dagger}V(U_{0}U_{0}^{\dagger}) U \\ \partial_{t} \underline{U} &= \frac{i 2\pi}{h}(U_{0}^{\dagger}VU_{0})\ (U_{0}^{\dagger} U) \\ \partial_{t} \underline{U} &= \frac{i 2\pi}{h}\underline{V} \underline{U} \end{align*} \]

Amazing! Everything about \(H^{(0)}\) has seemingly dropped out of our interaction-picture Schrodinger equation. It's called the interaction picture because our interaction Hamiltonian \(V\) is the only Hamiltonian term that explicitly appears here.

Review

Our system's state is whatever we hit with \(U\). In the interaction picture, we instead hit it with

\[ \underline{U} \equiv U_{0}^{\dagger}U \]

which causes evolution only due to \(V\). \(\underline{U}\) is simpler to work with because we find its form by only solving

\[ \partial_{t} \underline{U} = \frac{i 2\pi}{h}\underline{V} \underline{U} \]

which is simpler than the full Schrodinger equation. In the interaction picture, every non-\(U\) operator, \(\mathcal{O}\), is related to its Schrodinger (regular picture) operator by

\[ \underline{\mathcal{O}} \equiv U^{\dagger}_{0} \mathcal{O} U_{0} \]