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1 files changed, 64 insertions, 44 deletions

diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 359ecfe..54bf397 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -8,99 +8,117 @@ categories:
 layout: "concept"
 ---
 
-The **Heisenberg picture** is an alternative formulation of quantum
-mechanics, and is equivalent to the traditionally-taught Schrödinger equation.
+The **Heisenberg picture** is an alternative formulation of quantum mechanics,
+and is equivalent to the traditional Schrödinger equation.
 
-In the Schrödinger picture, the operators (observables) are fixed
-(as long as they do not depend on time), while the state
-$$\Ket{\psi_S(t)}$$ changes according to the Schrödinger equation,
-which can be written using the generator of translations $$\hat{U}(t)$$ like so,
-for a time-independent $$\hat{H}_S$$:
+In the Schrödinger picture,
+time-independent operators are constant by definition,
+and the state $$\Ket{\psi_S(t)}$$ varies as follows, where $$\hat{U}(t)$$
+is the [time evolution operator](/know/concept/time-evolution-operator/):
 
 $$\begin{aligned}
-    \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
-    \qquad \quad
-    \boxed{
-        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
-    }
+    \Ket{\psi_S(t)}
+    = \hat{U}(t) \Ket{\psi_S(0)}
 \end{aligned}$$
 
-In contrast, the Heisenberg picture reverses the roles:
-the states $$\Ket{\psi_H}$$ are invariant,
-and instead the operators vary with time.
-An advantage of this is that the basis states remain the same.
+In the Heisenberg picture, the roles are reversed:
+the states $$\Ket{\psi_H}$$ are constants,
+and instead the operators vary in time.
+In some situations this approach can be more convenient,
+and since we usually care about the evolution of observable quantities,
+studying the corresponding operators directly
+may make more sense than finding abstract quantum states.
+Another advantage is that basis states remain fixed,
+which can simplify calculations.
 
-Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$, and operator
-$$\hat{L}_S(t)$$ which may or may not depend on time, they can be
-converted to the Heisenberg picture by the following change of basis:
+Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$
+and an operator $$\hat{L}_S(t)$$ that may or may not depend on time,
+they can be converted to the Heisenberg picture by the following transformation:
 
 $$\begin{aligned}
     \boxed{
-        \Ket{\psi_H} \equiv \Ket{\psi_S(0)}
-        \qquad
-        \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
+        \Ket{\psi_H}
+        \equiv \Ket{\psi_S(0)}
+    }
+    \qquad\qquad
+    \boxed{
+        \hat{L}_H(t)
+        \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
     }
 \end{aligned}$$
 
-Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged:
+Note that if $$\hat{H}_S$$ is time-independent,
+then it commutes with $$\hat{U}(t)$$,
+meaning $$\hat{H}_H = \hat{H}_S$$,
+so it can simply be labelled $$\hat{H}$$.
+This is not true for time-dependent Hamiltonians.
+
+Since $$\hat{U}(t)$$ is unitary,
+the expectation value of a given operator is unchanged:
 
 $$\begin{aligned}
     \expval{\hat{L}_H}
     &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
-    = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
+    \\
+    &= \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
     \\
     &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
-    = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
-    = \expval{\hat{L}_S}
+    \\
+    &= \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
+    \\
+    &= \expval{\hat{L}_S}
 \end{aligned}$$
 
 The Schrödinger and Heisenberg pictures therefore respectively
 correspond to active and passive transformations by $$\hat{U}(t)$$
 in [Hilbert space](/know/concept/hilbert-space/).
-The two formulations are thus entirely equivalent,
+The two formulations are entirely equivalent,
 and can be derived from one another,
-as will be shown shortly.
+as we will show shortly.
 
 In the Heisenberg picture, the states are constant,
 so the time-dependent Schrödinger equation is not directly useful.
-Instead, we will use it derive a new equation for $$\hat{L}_H(t)$$.
-The key is that the generator $$\hat{U}(t)$$ is defined from the Schrödinger equation:
+Instead, we use it derive a new equation for $$\hat{L}_H(t)$$,
+with the key being that $$\hat{U}(t)$$ itself
+satisfies the Schrödinger equation by definition:
 
 $$\begin{aligned}
-    \dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
 \end{aligned}$$
 
 Where $$\hat{H}_S(t)$$ may depend on time. We differentiate the definition of
-$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation
-when necessary:
+$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation when necessary:
 
 $$\begin{aligned}
-    \dv{}{\hat{L}H}{t}
+    \dv{\hat{L}_H}{t}
     &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
     + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
     + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
     \\
     &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
     - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
-    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
     \\
     &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
     - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
-    + \Big( \dv{\hat{L}_S}{t} \Big)_H
-    = \frac{i}{\hbar} \comm{\hat{H}_H}{\hat{L}_H} + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
 \end{aligned}$$
 
-We thus get the equation of motion for operators in the Heisenberg picture:
+We thus get the following equation of motion for operators in the Heisenberg picture:
 
 $$\begin{aligned}
     \boxed{
-        \dv{}{t}\hat{L}_H(t) = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_H
+        \dv{}{t}\hat{L}_H(t)
+        = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_H
     }
 \end{aligned}$$
 
-This equation is closer to classical mechanics than the Schrödinger picture:
-inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
-gives the following Newton-style equations:
+This result is arguably more intuitive than the Schrödinger picture,
+because it allows us to think about observables (i.e. operators) in a more classical way.
+For example, inserting the position $$\hat{X}$$
+and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
+gives the following Newton-style relations:
 
 $$\begin{aligned}
     \dv{\hat{X}}{t}
@@ -112,5 +130,7 @@ $$\begin{aligned}
     = - \dv{V(\hat{X})}{\hat{X}}
 \end{aligned}$$
 
-For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
-which is closely related to the Heisenberg picture.
+Where the commutators have been treated as known.
+These equations would not be valid in the Schrödinger picture,
+unless we took their expectation value
+to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/).