---
title: "Imaginary time"
sort_title: "Imaginary time"
date: 2021-11-11
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

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.