summaryrefslogtreecommitdiff
path: root/source/know/concept/heisenberg-picture/index.md
blob: 359ecfeb5fb4efc2809e9973b885ff1f995b498e (plain)
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
---
title: "Heisenberg picture"
sort_title: "Heisenberg picture"
date: 2021-02-24
categories:
- Quantum mechanics
- Physics
layout: "concept"
---

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 $$\hat{U}(t)$$ like so,
for a time-independent $$\hat{H}_S$$:

$$\begin{aligned}
    \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
    \qquad \quad
    \boxed{
        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
    }
\end{aligned}$$

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:

$$\begin{aligned}
    \boxed{
        \Ket{\psi_H} \equiv \Ket{\psi_S(0)}
        \qquad
        \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
    }
\end{aligned}$$

Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged:

$$\begin{aligned}
    \expval{\hat{L}_H}
    &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
    = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
    \\
    &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
    = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
    = \expval{\hat{L}_S}
\end{aligned}$$

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:

$$\begin{aligned}
    \dv{}{\hat{L}H}{t}
    &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
    + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
    + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
    \\
    &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
    - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
    + \Big( \dv{\hat{L}_S}{t} \Big)_H
    \\
    &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
    - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
    + \Big( \dv{\hat{L}_S}{t} \Big)_H
    = \frac{i}{\hbar} \comm{\hat{H}_H}{\hat{L}_H} + \Big( \dv{\hat{L}_S}{t} \Big)_H
\end{aligned}$$

We thus get the equation of motion for operators in the Heisenberg picture:

$$\begin{aligned}
    \boxed{
        \dv{}{t}\hat{L}_H(t) = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_H
    }
\end{aligned}$$

This equation is closer to classical mechanics than the Schrödinger picture:
inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
gives the following Newton-style equations:

$$\begin{aligned}
    \dv{\hat{X}}{t}
    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{X}}
    = \frac{\hat{P}}{m}
    \\
    \dv{\hat{P}}{t}
    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{P}}
    = - \dv{V(\hat{X})}{\hat{X}}
\end{aligned}$$

For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
which is closely related to the Heisenberg picture.