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+---
+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.