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1 files changed, 86 insertions, 96 deletions

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.
-