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.