From c157ad913aa9f975ea8c137e24175d134486f462 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 24 Feb 2021 21:27:41 +0100
Subject: Expand knowledge base

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+---
+title: "Heisenberg picture"
+firstLetter: "H"
+publishDate: 2021-02-24
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-24T16:46:26+01:00
+draft: false
+markup: pandoc
+---
+
+# Heisenberg picture
+
+The **Heisenberg picture** is an alternative formulation of quantum
+mechanics, and is equivalent to the traditionally-taught 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
+$\hat{U}(t) = \exp{} (- i t \hat{H} / \hbar)$ like so:
+
+$$\begin{aligned}
+    \ket{\psi_S(t)} = \hat{U}(t) \ket{\psi_S(0)}
+\end{aligned}$$
+
+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.
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \ket{\psi_H} = \ket{\psi_S(0)}
+        \qquad
+        \hat{L}_H(t) = \hat{U}^\dagger(t) \hat{L}_S(t) \hat{U}(t)
+    }
+\end{aligned}$$
+
+Since $\hat{U}(t)$ is unitary, the expectation value of a given operator is unchanged:
+
+$$\begin{aligned}
+    \expval*{\hat{L}_H}
+    &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
+    = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
+    \\
+    &= \matrixel*{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
+    = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
+    = \expval*{\hat{L}_S}
+\end{aligned}$$
+
+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,
+and can be derived from one another,
+as will be shown 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:
+
+$$\begin{aligned}
+    \dv{t} \hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \hat{U}(t)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \dv{\hat{L}_H}{t}
+    &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
+    + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
+    + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
+    \\
+    &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
+    - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
+    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    \\
+    &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
+    - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
+    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    = \frac{i}{\hbar} [\hat{H}_H, \hat{L}_H] + \Big( \dv{\hat{L}_S}{t} \Big)_H
+\end{aligned}$$
+
+We thus get the equation of motion for operators in the Heisenberg picture:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{t} \hat{L}_H(t) = \frac{i}{\hbar} [\hat{H}_H(t), \hat{L}_H(t)] + \Big( \dv{t} \hat{L}_S(t) \Big)_H
+    }
+\end{aligned}$$
+
+This equation is closer to classical mechanics than the Schrödinger picture:
+inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: d/d\hat{X}$
+gives the following Newton-style equations:
+
+$$\begin{aligned}
+    \dv{\hat{X}}{t}
+    &= \frac{i}{\hbar} [\hat{H}, \hat{X}]
+    = \frac{\hat{P}}{m}
+    \\
+    \dv{\hat{P}}{t}
+    &= \frac{i}{\hbar} [\hat{H}, \hat{P}]
+    = - \dv{V(\hat{X})}{\hat{X}}
+\end{aligned}$$
+
+For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
+which is closely related to the Heisenberg picture.
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