Categories: Physics, Quantum mechanics.

# Imaginary time

Let $$\hat{A}_S$$ and $$\hat{B}_S$$ be time-independent in the Schrödinger picture. Then, in the Heisenberg picture, consider the following expectation value with respect to thermodynamic equilibium (as found in Green’s 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:

\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 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.

1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.