Categories: Physics, Quantum mechanics.

Imaginary time

Let A^S\hat{A}_S and B^S\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):

A^H(t)B^H(t)=1ZTr ⁣(exp(βH^0,S(t))A^H(t)B^H(t))\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 H^0,S\hat{H}_{0,S} is time-independent. Suppose a (maybe time-dependent) “difficult” H^1,S\hat{H}_{1,S} is added, so that the total Hamiltonian is H^S=H^0,S+H^1,S\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}. Then it is easier to consider the expectation value in the interaction picture:

A^H(t)B^H(t)=1ZTr ⁣(exp(βH^S(t))K^I(0,t)A^I(t)K^I(t,t)B^I(t)K^I(t,0))\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 K^I(t,t0)\hat{K}_I(t, t_0) is the time evolution operator of H^1,S\hat{H}_{1,S}. In front, we have exp(βH^S(t))\exp(-\beta \hat{H}_S(t)), while K^I\hat{K}_I is an exponential of an integral of H^1,I\hat{H}_{1,I}, so we are stuck. Keep in mind that exponentials of operators cannot just be factorized, i.e. in general exp(A^ ⁣+ ⁣B^)exp(A^)exp(B^)\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 τ\tau instead of the real time tt. Fixing a tt, we “redefine” the interaction picture along the imaginary axis:

A^I(τ)exp ⁣(τH^0,S)A^Sexp ⁣( ⁣ ⁣τH^0,S)\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, τ\tau is real; the point is that this formula comes from the real-time definition by replacing tiτt \to -i \tau. The Heisenberg and Schrödinger pictures can be redefined in the same way.

In fact, by substituting tiτt \to -i \tau, all the key results of the interaction picture can be updated, for example the Schrödinger equation for ψS(τ)\Ket{\psi_S(\tau)} becomes:

ddtψS(τ)=H^SψS(τ)    ψS(τ)=exp ⁣( ⁣ ⁣τH^S)ψH\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 K^I\hat{K}_I turns out to be given by:

K^I(τ,τ0)=T{exp ⁣( ⁣ ⁣1τ0τH^1,I(τ)dτ)}\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 T\mathcal{T} is the time-ordered product with respect to τ\tau. This operator works as expected:

ψI(τ)=K^I(τ,τ0)ψI(τ0)\begin{aligned} \Ket{\psi_I(\tau)} = \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)} \end{aligned}

Where ψI(τ)\Ket{\psi_I(\tau)} is related to the Schrödinger and Heisenberg pictures as follows:

ψI(τ)exp ⁣(τH^0,S)ψS(τ)=exp ⁣(τH^0,S)exp ⁣( ⁣ ⁣τH^S)ψH\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 K^I(τ,τ0)\hat{K}_I(\tau, \tau_0):

ψI(τ)=K^I(τ,τ0)ψI(τ0)exp ⁣(τH^0,S)exp ⁣( ⁣ ⁣τH^S)ψH=K^I(τ,τ0)exp ⁣(τ0H^0,S)exp ⁣( ⁣ ⁣τ0H^S)ψH\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 K^I(τ,τ0)\hat{K}_I(\tau, \tau_0):

K^I(τ,τ0)=exp ⁣(τH^0,S)exp ⁣( ⁣ ⁣(τ ⁣ ⁣τ0)H^S)exp ⁣( ⁣ ⁣τ0H^0,S)\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 τ=β\tau = \hbar \beta and τ0=0\tau_0 = 0, so K^I(τ,τ0)\hat{K}_I(\tau, \tau_0) becomes:

K^I(β,0)=exp ⁣(βH^0,S)exp ⁣( ⁣ ⁣βH^S)    exp ⁣( ⁣ ⁣βH^S)=exp ⁣( ⁣ ⁣βH^0,S)K^I(β,0)\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 K^I(β,0)K^I(0,τ)=K^I(β,τ)\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:

A^H(τ)B^H(τ)=1ZTr ⁣( ⁣exp(βH^0,S)K^I(β,τ)A^I(τ)K^I(τ,τ)B^I(τ)K^I(τ,0) ⁣)\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 T\mathcal{T}, letting us reorder the (bosonic) K^I\hat{K}_I-operators inside, and thereby reduce the expression considerably:

T{A^HB^H}=1ZTr ⁣(T{K^I(β,τ)K^I(τ,τ)K^I(τ,0)A^I(τ)B^I(τ)}exp(βH^0,S))=1ZTr ⁣(T{K^I(β,0)A^I(τ)B^I(τ)}exp(βH^0,S))\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 ⁣(exp(βH^S))=Tr ⁣(K^I(β,0)exp(βH^0,S))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 0\Expval{}_0 as the expectation value with respect to the unperturbed equilibrium involving only H^0,S\hat{H}_{0,S}, we arrive at the following way of writing this time-ordered expectation:

T{A^HB^H}=T{K^I(β,0)A^I(τ)B^I(τ)}0K^I(β,0)0\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.