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Diffstat (limited to 'source/know/concept/imaginary-time')
| -rw-r--r-- | source/know/concept/imaginary-time/index.md | 56 |
1 files changed, 28 insertions, 28 deletions
diff --git a/source/know/concept/imaginary-time/index.md b/source/know/concept/imaginary-time/index.md index 1dbdf11..245c3c1 100644 --- a/source/know/concept/imaginary-time/index.md +++ b/source/know/concept/imaginary-time/index.md @@ -8,7 +8,7 @@ categories: layout: "concept" --- -Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture. +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 @@ -19,9 +19,9 @@ $$\begin{aligned} &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big) \end{aligned}$$ -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 $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$. +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/): @@ -30,16 +30,16 @@ $$\begin{aligned} &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big) \end{aligned}$$ -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. +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 -$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$ +$$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$$ 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: +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: $$\begin{aligned} \boxed{ @@ -48,13 +48,13 @@ $$\begin{aligned} } \end{aligned}$$ -Ironically, $\tau$ is real; the point is that this formula -comes from the real-time definition by replacing $t \to -i \tau$. +Ironically, $$\tau$$ is real; the point is that this formula +comes from the real-time definition by replacing $$t \to -i \tau$$. The Heisenberg and Schrödinger pictures can be redefined in the same way. -In fact, by substituting $t \to -i \tau$, +In fact, by substituting $$t \to -i \tau$$, all the key results of the interaction picture can be updated, -for example the Schrödinger equation for $\Ket{\psi_S(\tau)}$ becomes: +for example the Schrödinger equation for $$\Ket{\psi_S(\tau)}$$ becomes: $$\begin{aligned} \hbar \dv{}{t}\Ket{\psi_S(\tau)} @@ -64,7 +64,7 @@ $$\begin{aligned} = \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \Ket{\psi_H} \end{aligned}$$ -And the interaction picture's time evolution operator $\hat{K}_I$ +And the interaction picture's time evolution operator $$\hat{K}_I$$ turns out to be given by: $$\begin{aligned} @@ -74,9 +74,9 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\mathcal{T}$ is the +Where $$\mathcal{T}$$ is the [time-ordered product](/know/concept/time-ordered-product/) -with respect to $\tau$. +with respect to $$\tau$$. This operator works as expected: $$\begin{aligned} @@ -84,7 +84,7 @@ $$\begin{aligned} = \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)} \end{aligned}$$ -Where $\Ket{\psi_I(\tau)}$ is related to +Where $$\Ket{\psi_I(\tau)}$$ is related to the Schrödinger and Heisenberg pictures as follows: $$\begin{aligned} @@ -94,7 +94,7 @@ $$\begin{aligned} \end{aligned}$$ It is interesting to combine this definition -with the action of time evolution $\hat{K}_I(\tau, \tau_0)$: +with the action of time evolution $$\hat{K}_I(\tau, \tau_0)$$: $$\begin{aligned} \Ket{\psi_I(\tau)} @@ -105,7 +105,7 @@ $$\begin{aligned} \end{aligned}$$ Rearranging this leads to the following useful -alternative expression for $\hat{K}_I(\tau, \tau_0)$: +alternative expression for $$\hat{K}_I(\tau, \tau_0)$$: $$\begin{aligned} \boxed{ @@ -117,8 +117,8 @@ $$\begin{aligned} \end{aligned}$$ Returning to our initial example, -we can set $\tau = \hbar \beta$ and $\tau_0 = 0$, -so $\hat{K}_I(\tau, \tau_0)$ becomes: +we can set $$\tau = \hbar \beta$$ and $$\tau_0 = 0$$, +so $$\hat{K}_I(\tau, \tau_0)$$ becomes: $$\begin{aligned} \hat{K}_I(\hbar \beta, 0) @@ -130,7 +130,7 @@ $$\begin{aligned} \end{aligned}$$ Using the easily-shown fact that -$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$, +$$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$$, we can therefore rewrite the thermodynamic expectation value like so: $$\begin{aligned} @@ -139,8 +139,8 @@ $$\begin{aligned} \hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big) \end{aligned}$$ -We now introduce a time-ordering $\mathcal{T}$, -letting us reorder the (bosonic) $\hat{K}_I$-operators inside, +We now introduce a time-ordering $$\mathcal{T}$$, +letting us reorder the (bosonic) $$\hat{K}_I$$-operators inside, and thereby reduce the expression considerably: $$\begin{aligned} @@ -151,9 +151,9 @@ $$\begin{aligned} &= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp(-\beta \hat{H}_{0,S}) \Big) \end{aligned}$$ -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 $\hat{H}_{0,S}$, +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} |