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+---
+title: "Ehrenfest's theorem"
+date: 2021-02-24
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
+time evolution of an observable's expectation value $\expval{\hat{L}}$.
+
+The time-dependent Schrödinger equation is as follows,
+where prime denotes differentiation with respect to time $t$:
+
+$$\begin{aligned}
+    \Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
+    \qquad
+    \Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
+\end{aligned}$$
+
+Given an observable operator $\hat{L}$ and a state $\Ket{\psi}$,
+the time-derivative of the expectation value $\expval{\hat{L}}$ is as follows
+(due to the product rule of differentiation):
+
+$$\begin{aligned}
+    \dv{\expval{\hat{L}}}{t}
+    &= \matrixel{\psi}{\hat{L}}{\psi'} + \matrixel{\psi'}{\hat{L}}{\psi} + \matrixel{\psi}{\hat{L}'}{\psi}
+    \\
+    &= \frac{1}{i \hbar} \matrixel{\psi}{\hat{L}\hat{H}}{\psi}
+    - \frac{1}{i \hbar} \matrixel{\psi}{\hat{H}\hat{L}}{\psi}
+    + \Expval{\dv{\hat{L}}{t}}
+\end{aligned}$$
+
+The first two terms on the right can be rewritten using a commutator,
+yielding the general form of Ehrenfest's theorem:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{\expval{\hat{L}}}{t}
+        = \frac{1}{i \hbar} \Expval{[\hat{L}, \hat{H}]} + \Expval{\dv{\hat{L}}{t}}
+    }
+\end{aligned}$$
+
+In practice, since most operators are time-independent,
+the last term often vanishes.
+
+As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+this relation proves itself,
+when one simply wraps all terms in $\Bra{\psi}$ and $\Ket{\psi}$.
+
+Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$.
+Applying the above theorem to $\hat{X}$ yields the following,
+which we reduce using the fact that $\hat{X}$ commutes
+with the potential $V(\hat{X})$,
+because one is a function of the other:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{X}}}{t}
+    &= \frac{1}{i \hbar} \Expval{[\hat{X}, \hat{H}]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2]}
+    \\
+    &= \frac{1}{2 i \hbar m} \Expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}}
+    = \frac{2 i \hbar}{2 i \hbar m} \expval{\hat{P}}
+    = \frac{\expval{\hat{P}}}{m}
+\end{aligned}$$
+
+This is the first part of the "original" form of Ehrenfest's theorem,
+which is reminiscent of classical Newtonian mechanics:
+
+$$\begin{gathered}
+    \boxed{
+        \dv{\expval{\hat{X}}}{t} = \frac{\expval{\hat{P}}}{m}
+    }
+\end{gathered}$$
+
+Next, applying the general formula to the expected momentum $\expval{\hat{P}}$
+gives us:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{P}}}{t}
+    &= \frac{1}{i \hbar} \Expval{[\hat{P}, \hat{H}]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]}
+    = \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
+\end{aligned}$$
+
+To find the commutator, we go to the $\hat{X}$-basis and use a test
+function $f(x)$:
+
+$$\begin{aligned}
+    \Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
+    &= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx}
+    = - i \hbar \frac{dV}{dx} f(x)
+\end{aligned}$$
+
+By inserting this result back into the previous equation, we find the following:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{P}}}{t}
+    &= - \frac{i \hbar}{i \hbar} \Expval{\frac{d V}{d \hat{X}}}
+    = - \Expval{\frac{d V}{d \hat{X}}}
+\end{aligned}$$
+
+This is the second part of Ehrenfest's theorem,
+which is also similar to Newtonian mechanics:
+
+$$\begin{gathered}
+    \boxed{
+        \dv{\expval{\hat{P}}}{t} = - \Expval{\pdv{V}{\hat{X}}}
+    }
+\end{gathered}$$
+
+There is an important consequence of Ehrenfest's original theorems
+for the symbolic derivatives of the Hamiltonian $\hat{H}$
+with respect to $\hat{X}$ and $\hat{P}$:
+
+$$\begin{gathered}
+    \boxed{
+        \Expval{\pdv{\hat{H}}{\hat{P}}}
+        = \dv{\expval{\hat{X}}}{t}
+    }
+    \qquad \quad
+    \boxed{
+        - \Expval{\pdv{\hat{H}}{\hat{X}}}
+        = \dv{\expval{\hat{P}}}{t}
+    }
+\end{gathered}$$
+
+These are easy to prove yourself,
+and are analogous to Hamilton's canonical equations.
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