1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
|
---
title: "Ehrenfest's theorem"
sort_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 $$\expval{\hat{L}}$$.
Recall the time-dependent Schrödinger equation,
where prime denotes differentiation with respect to time $$t$$:
$$\begin{aligned}
\Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
\qquad \qquad
\Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
\end{aligned}$$
Given an observable operator $$\hat{L}$$ and a state $$\Ket{\psi}$$,
the $$t$$-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.
Note that this relation is trivial to prove
in the [Heisenberg picture](/know/concept/heisenberg-picture/),
by wrapping all terms in $$\Bra{\psi}$$ and $$\Ket{\psi}$$.
Two observables of particular interest
are position $$\hat{X}$$ and momentum $$\hat{P}$$.
Applying the theorem to $$\hat{X}$$ yields the following,
using $$\hat{H} = \hat{P}^2 / (2 m) + V(\hat{X})$$
and a few basic properties of commutators:
$$\begin{aligned}
\dv{\expval{\hat{X}}}{t}
&= \frac{1}{i \hbar} \Expval{[\hat{X}, \hat{H}]}
\\
&= \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
\\
&= \frac{1}{2 i \hbar m} \Expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}}
\\
&= \frac{2 i \hbar}{2 i \hbar m} \expval{\hat{P}}
\end{aligned}$$
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 evaluate the commutator,
we go to the $$\hat{X}$$-basis and use a test function $$f(x)$$:
$$\begin{aligned}
\Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
&= - i \hbar \dv{}{x} \Big( V(x) \: f(x) \Big) - V(x) \Big( \!-\! i \hbar \dv{}{x} \Big) f(x)
\\
&= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx}
\\
&= - i \hbar \frac{dV}{dx} f(x)
\end{aligned}$$
By inserting this result back into the previous equation,
we find the second part of Ehrenfest's original theorem,
which is again reminiscent 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 $$\hat{P}$$:
$$\begin{gathered}
\boxed{
\Expval{\pdv{\hat{H}}{\hat{P}}}
= \dv{\expval{\hat{X}}}{t}
}
\qquad \qquad
\boxed{
- \Expval{\pdv{\hat{H}}{\hat{X}}}
= \dv{\expval{\hat{P}}}{t}
}
\end{gathered}$$
These are easy to prove yourself,
and are analogous to Hamilton's canonical equations.
|
