diff options
Diffstat (limited to 'source/know/concept/imaginary-time/index.md')
| -rw-r--r-- | source/know/concept/imaginary-time/index.md | 173 |
1 files changed, 173 insertions, 0 deletions
diff --git a/source/know/concept/imaginary-time/index.md b/source/know/concept/imaginary-time/index.md new file mode 100644 index 0000000..5dc9264 --- /dev/null +++ b/source/know/concept/imaginary-time/index.md @@ -0,0 +1,173 @@ +--- +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. |