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
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
|
---
title: "Imaginary time"
firstLetter: "I"
publishDate: 2021-11-11
categories:
- Physics
- Quantum mechanics
date: 2021-11-05T15:19:29+01:00
draft: false
markup: pandoc
---
# Imaginary time
Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture.
Then, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
consider the following expectation value
with respect to thermodynamic equilibium
(as found in [Green's functions](/know/concept/greens-functions/) for example):
$$\begin{aligned}
\expval*{\hat{A}_H(t) \hat{B}_H(t')}
&= \frac{1}{Z} \Tr\!\Big( \exp\!(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big)
\end{aligned}$$
Where the "simple" Hamiltonian $\hat{H}_{0,S}$ is time-independent.
Suppose a (maybe time-dependent) "difficult" $\hat{H}_{1,S}$ is added,
so that the total Hamiltonian is $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$.
Then it is easier to consider the expectation value
in the [interaction picture](/know/concept/interaction-picture/):
$$\begin{aligned}
\expval*{\hat{A}_H(t) \hat{B}_H(t')}
&= \frac{1}{Z} \Tr\!\Big( \exp\!(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big)
\end{aligned}$$
Where $\hat{K}_I(t, t_0)$ is the time evolution operator of $\hat{H}_{1,S}$.
In front, we have $\exp\!(-\beta \hat{H}_S(t))$,
while $\hat{K}_I$ is an exponential of an integral of $\hat{H}_{1,I}$, so we are stuck.
Keep in mind that exponentials of operators
cannot just be factorized, i.e. in general
$\exp\!(\hat{A} \!+\! \hat{B}) \neq \exp\!(\hat{A}) \exp\!(\hat{B})$
To get around this, a useful mathematical trick is
to use an **imaginary time** variable $\tau$ instead of the real time $t$.
Fixing a $t$, we "redefine" the interaction picture along the imaginary axis:
$$\begin{aligned}
\boxed{
\hat{A}_I(\tau)
\equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \: \hat{A}_S \: \exp\!\bigg( \!-\! \frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
}
\end{aligned}$$
Ironically, $\tau$ is real; the point is that this formula
comes from the real-time definition by replacing $t \to -i \tau$.
The Heisenberg and Schrödinger pictures can be redefined in the same way.
In fact, by substituting $t \to -i \tau$,
all the key results of the interaction picture can be updated,
for example the Schrödinger equation for $\ket{\psi_S(\tau)}$ becomes:
$$\begin{aligned}
\hbar \dv{t} \ket{\psi_S(\tau)}
= - \hat{H}_S \ket{\psi_S(\tau)}
\quad \implies \quad
\ket{\psi_S(\tau)}
= \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \ket{\psi_H}
\end{aligned}$$
And the interaction picture's time evolution operator $\hat{K}_I$
turns out to be given by:
$$\begin{aligned}
\boxed{
\hat{K}_I(\tau, \tau_0)
= \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{1}{\hbar} \int_{\tau_0}^\tau \hat{H}_{1,I}(\tau') \dd{\tau'} \bigg) \bigg\}
}
\end{aligned}$$
Where $\mathcal{T}$ is the
[time-ordered product](/know/concept/time-ordered-product/)
with respect to $\tau$.
This operator works as expected:
$$\begin{aligned}
\ket{\psi_I(\tau)}
= \hat{K}_I(\tau, \tau_0) \ket{\psi_I(\tau_0)}
\end{aligned}$$
Where $\ket{\psi_I(\tau)}$ is related to
the Schrödinger and Heisenberg pictures as follows:
$$\begin{aligned}
\ket{\psi_I(\tau)}
\equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \ket{\psi_S(\tau)}
= \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
\end{aligned}$$
It is interesting to combine this definition
with the action of time evolution $\hat{K}_I(\tau, \tau_0)$:
$$\begin{aligned}
\ket{\psi_I(\tau)}
&= \hat{K}_I(\tau, \tau_0) \ket{\psi_I(\tau_0)}
\\
\exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
&= \hat{K}_I(\tau, \tau_0) \exp\!\bigg(\frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau_0 \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
\end{aligned}$$
Rearranging this leads to the following useful
alternative expression for $\hat{K}_I(\tau, \tau_0)$:
$$\begin{aligned}
\boxed{
\hat{K}_I(\tau, \tau_0)
= \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
\exp\!\bigg(\!-\! \frac{(\tau \!-\! \tau_0) \hat{H}_{S}}{\hbar}\bigg)
\exp\!\bigg(\!-\! \frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg)
}
\end{aligned}$$
Returning to our initial example,
we can set $\tau = \hbar \beta$ and $\tau_0 = 0$,
so $\hat{K}_I(\tau, \tau_0)$ becomes:
$$\begin{aligned}
\hat{K}_I(\hbar \beta, 0)
&= \exp\!\big(\beta \hat{H}_{0,S}\big) \exp\!\big(\!-\! \beta \hat{H}_{S}\big)
\\
\implies \quad
\exp\!\big(\!-\! \beta \hat{H}_{S}\big)
&= \exp\!\big(\!-\! \beta \hat{H}_{0,S}\big) \hat{K}_I(\hbar \beta, 0)
\end{aligned}$$
Using the easily-shown fact that
$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$,
we can therefore rewrite the thermodynamic expectation value like so:
$$\begin{aligned}
\expval*{\hat{A}_H(\tau) \hat{B}_H(\tau')}
&= \frac{1}{Z} \Tr\!\Big(\! \exp\!(-\beta \hat{H}_{0,S}) \hat{K}_I(\hbar \beta, \tau)
\hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
\end{aligned}$$
We now introduce a time-ordering $\mathcal{T}$,
letting us reorder the (bosonic) $\hat{K}_I$-operators inside,
and thereby reduce the expression considerably:
$$\begin{aligned}
\expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
&= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, \tau) \hat{K}_I(\tau, \tau') \hat{K}_I(\tau', 0)
\hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
\\
&= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
\end{aligned}$$
Where $Z = \Tr\!\big(\exp\!(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp\!(-\beta \hat{H}_{0,S})\big)$.
If we now define $\expval{}_0$ as the expectation value with respect
to the unperturbed equilibrium involving only $\hat{H}_{0,S}$,
we arrive at the following way of writing this time-ordered expectation:
$$\begin{aligned}
\boxed{
\expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
= \frac{\expval{\mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\}}_0}{\expval{\hat{K}_I(\hbar \beta, 0)}_0}
}
\end{aligned}$$
For another application of imaginary time,
see e.g. the [Matsubara Green's function](/know/concept/matsubara-greens-function/).
## References
1. H. Bruus, K. Flensberg,
*Many-body quantum theory in condensed matter physics*,
2016, Oxford.
|
