diff options
Diffstat (limited to 'source/know/concept/matsubara-greens-function')
| -rw-r--r-- | source/know/concept/matsubara-greens-function/index.md | 390 |
1 files changed, 390 insertions, 0 deletions
diff --git a/source/know/concept/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md new file mode 100644 index 0000000..d54bbf2 --- /dev/null +++ b/source/know/concept/matsubara-greens-function/index.md @@ -0,0 +1,390 @@ +--- +title: "Matsubara Green's function" +date: 2021-11-12 +categories: +- Physics +- Quantum mechanics +layout: "concept" +--- + +The **Matsubara Green's function** is an +[imaginary-time](/know/concept/imaginary-time/) version +of the real-time [Green's functions](/know/concept/greens-functions/). +We define as follows in the imaginary-time +[Heisenberg picture](/know/concept/heisenberg-picture/): + +$$\begin{aligned} + \boxed{ + C_{AB}(\tau, \tau') + \equiv -\frac{1}{\hbar} \Expval{\mathcal{T} \big\{ \hat{A}(\tau) \hat{B}(\tau') \big\}} + } +\end{aligned}$$ + +Where the expectation value $\Expval{}$ is with respect to thermodynamic equilibrium, +and $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator. +Because the Hamiltonian $\hat{H}$ cannot depend on the imaginary time, +$C_{AB}$ is a function of the difference $\tau \!-\! \tau'$ only: + +$$\begin{aligned} + C_{AB}(\tau, \tau') + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big) + \\ + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar} + e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big) + \\ + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} \Big) +\end{aligned}$$ + +For $\tau > \tau'$, we see by expanding in the many-particle eigenstates $\Ket{n}$ +that we need to demand $\hbar \beta > \tau \!-\! \tau'$ to prevent +$C_{AB}$ from diverging for increasing temperatures: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau') + &= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} + \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} + \\ + &= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} e^{-\beta E_n} e^{(\tau - \tau') E_n / \hbar} +\end{aligned}$$ + +And likewise, for $\tau < \tau'$, +we must demand that $\tau \!-\! \tau' > -\hbar \beta$ +for the same reason: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau') + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big) + \\ + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} \Big) + \\ + &= \mp \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} e^{-\beta E_n} e^{- (\tau - \tau') E_n / \hbar} +\end{aligned}$$ + +With $-$ for bosons, and $+$ for fermions, +due to the time-ordered product for $\tau > \tau'$. + +On this domain $[-\hbar \beta, \hbar \beta]$, +the Matsubara Green's function $C_{AB}$ +obeys a useful shift relation: +it is $\hbar \beta$-periodic for bosons, +and $\hbar \beta$-antiperiodic for fermions: + +$$\begin{aligned} + \boxed{ + C_{AB}(\tau \!-\! \tau') = + \begin{cases} + \pm C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta) + & \mathrm{if\;} \tau \!-\! \tau' < 0 + \\ + \pm C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta) + & \mathrm{if\;} \tau \!-\! \tau' > 0 + \end{cases} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-period"/> +<label for="proof-period">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-period">Proof.</label> +First $\tau \!-\! \tau' < 0$. +We insert the argument $\tau \!-\! \tau' \!+\! \hbar \beta$, +and use the cyclic property: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta) + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau' + \hbar \beta) \hat{H} / \hbar} + \hat{A} e^{-(\tau - \tau' + \hbar \beta) \hat{H} / \hbar} \hat{B} \Big) + \\ + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{B} \Big) + \\ + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} + e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar} \Big) + \\ + &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big) +\end{aligned}$$ + +Since $\tau < \tau'$ by assumption, +we can bring back the time-ordered product $\mathcal{T}$: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta) + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big) + \\ + &= \pm C_{AB}(\tau \!-\! \tau') +\end{aligned}$$ + +Moving on to $\tau \!-\! \tau' > 0$, the proof is perfectly analogous: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta) + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau' - \hbar \beta) \hat{H} / \hbar} + \hat{B} e^{(\tau - \tau' - \hbar \beta) \hat{H} / \hbar} \hat{A} \Big) + \\ + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{A} \Big) + \\ + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar} + e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big) + \\ + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big) + \\ + &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big) + \\ + &= \pm C_{AB}(\tau \!-\! \tau') +\end{aligned}$$ +</div> +</div> + +Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$, +the [Fourier transform](/know/concept/fourier-transform/) +of $C_{AB}(\tau)$ consists of discrete frequencies +$k_n \equiv n \pi / (\hbar \beta)$. +The forward and inverse Fourier transforms +are therefore defined as given below (with $\tau' = 0$). +It is convention to write $C_{AB}(i k_n)$ instead of $C_{AB}(k_n)$: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + C_{AB}(i k_n) + &\equiv \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} + \\ + C_{AB}(\tau) + &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty C_{AB}(i k_n) e^{-i k_n \tau} + \end{aligned} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-FT-def"/> +<label for="proof-FT-def">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-FT-def">Proof.</label> +We will prove that one is indeed the inverse of the other. +We demand that the inverse FT of the forward FT of $C_{AB}(\tau)$ +is simply $C_{AB}(\tau)$ again: + +$$\begin{aligned} + C_{AB}(\tau) + &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty + \bigg( \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: e^{i k_n \tau'} \dd{\tau'} \bigg) e^{-i k_n \tau} + \\ + &= \frac{1}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') + \bigg( \frac{1}{2} \sum_{n = -\infty}^\infty e^{i k_n (\tau' - \tau)} \bigg) \dd{\tau'} + \\ + &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') + \bigg( \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i \pi n (\tau' - \tau) / \hbar \beta} \bigg) \dd{\tau'} +\end{aligned}$$ + +Here, the inner expression turns out to be +a [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i n x} + = \delta(x) +\end{aligned}$$ + +From which the rest of the proof follows straightforwardly: + +$$\begin{aligned} + C_{AB}(\tau) + &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta\big( (\tau' \!-\! \tau) \pi / \hbar \beta \big) \dd{\tau'} + \\ + &= \frac{\pi \hbar \beta}{\pi \hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'} + \\ + &= \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'} + \\ + &= C_{AB}(\tau) +\end{aligned}$$ +</div> +</div> + +Let us now define the **Matsubara frequencies** $\omega_n$ +as a species-dependent subset of $k_n$: + +$$\begin{aligned} + \boxed{ + \omega_n \equiv + \begin{cases} + \displaystyle\frac{2 n \pi}{\hbar \beta} + & \mathrm{bosons} + \\ + \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta} + & \mathrm{fermions} + \end{cases} + } +\end{aligned}$$ + +With this, we can rewrite the definition of the forward Fourier transform as follows: + +$$\begin{aligned} + \boxed{ + C_{AB}(i \omega_n) + = \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau} + = \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-FT-alt"/> +<label for="proof-FT-alt">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-FT-alt">Proof.</label> +We split the integral, shift its limits, +and use the (anti)periodicity of $C_{AB}$: + +$$\begin{aligned} + C_{AB}(i k_n) + &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} + + \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} + \\ + &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} + + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau \!-\! \hbar \beta) \: e^{i k_n (\tau - \hbar \beta)} \dd{\tau} + \\ + &= \frac{1}{2} \int_0^{\hbar \beta} \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{-i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau} + \\ + &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} +\end{aligned}$$ + +With $+$ for bosons, and $-$ for fermions. Since $k_n \equiv n \pi / (\hbar \beta)$, +we know $e^{-i k_n \hbar \beta} \in \{-1, 1\}$, +so for bosons all odd $n$ vanish, and for fermions all even $n$, +yielding the desired result. + +For the other case, we simply shift the first integral's limits instead of the seconds': + +$$\begin{aligned} + C_{AB}(i k_n) + &= \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau \!+\! \hbar \beta) \: e^{i k_n (\tau + \hbar \beta)} \dd{\tau} + + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} + \\ + &= \frac{1}{2} \int_{-\hbar \beta}^0 \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau} + \\ + &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} +\end{aligned}$$ +</div> +</div> + +If we actually evaluate this, +we obtain the following form of $C_{AB}$, +which is almost identical to the +[Lehmann representation](/know/concept/lehmann-representation/) +of the "ordinary" retarded and advanced Green's functions: + +$$\begin{aligned} + \boxed{ + C_{AB}(i \omega_m) + = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}} + \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-Lehmann"/> +<label for="proof-Lehmann">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-Lehmann">Proof.</label> +For $\tau \!-\! \tau' > 0$, we start by expanding +in the many-particle eigenstates $\Ket{n}$: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau') + &= - \frac{1}{\hbar Z} \sum_{n} + \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} + \\ + &= - \frac{1}{\hbar Z} \sum_{n n'} \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n'} + \Matrixel{n'}{e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} + \\ + &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'} + \matrixel{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar} +\end{aligned}$$ + +We take the Fourier transform by integrating over $[0, \hbar \beta]$: + +$$\begin{aligned} + C_{AB}(i \omega_m) + &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'} + \matrixel{n'}{\hat{B}}{n} \int_0^{\hbar \beta} e^{(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau} + \\ + &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n} + \bigg[ \frac{\hbar e^{(i \hbar \omega_m + E_n - E_{n'}) \tau / \hbar}}{i \hbar \omega_m + E_n - E_{n'}} \bigg]_0^{\hbar \beta} + \\ + &= - \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}} + \Big( e^{(i \hbar \omega_m + E_n - E_{n'}) \beta} - 1 \Big) + \\ + &= - \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}} + \Big( e^{i \hbar \omega_m \beta} e^{-\beta E_{n'}} - e^{-\beta E_n} \Big) + \\ + &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}} + \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) +\end{aligned}$$ + +Moving on to $\tau \!-\! \tau' < 0$, +we again expand in the many-particle eigenstates $\Ket{n}$: + +$$\begin{aligned} + C_{AB}(\tau \!-\! \tau') + &= \mp \frac{1}{\hbar Z} \sum_{n} + \Matrixel{n}{e^{-\beta \hat{H}} e^{- (\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} + \\ + &= \mp \frac{1}{\hbar Z} \sum_{n n'} \Matrixel{n}{e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n'} + \Matrixel{n'}{e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} + \\ + &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'} + \matrixel{n'}{\hat{A}}{n} e^{-(E_n - E_{n'})(\tau - \tau') / \hbar} +\end{aligned}$$ + +Since $\tau \!-\! \tau' < 0$ this time, +we take the Fourier transform over $[-\hbar \beta, 0]$: + +$$\begin{aligned} + C_{AB}(i \omega_m) + &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'} + \matrixel{n'}{\hat{A}}{n} \int_{-\hbar \beta}^0 e^{-(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau} + \\ + &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n} + \bigg[ \frac{\hbar e^{(i \hbar \omega_m - E_n + E_{n'}) \tau / \hbar}}{i \hbar \omega_m - E_n + E_{n'}} \bigg]_{-\hbar \beta}^0 + \\ + &= \mp \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}} + \Big( 1 - e^{(-i \hbar \omega_m + E_n - E_{n'}) \beta} \Big) + \\ + &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}} + \Big( e^{-\beta E_n} - e^{-i \hbar \omega_m \beta} e^{-\beta E_{n'}} \Big) + \\ + &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}} + \Big( e^{- \beta E_n} \pm e^{-\beta E_{n'}} \Big) + \\ + &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}} + \Big( e^{- \beta E_{n'}} \mp e^{-\beta E_n} \Big) +\end{aligned}$$ + +Where swapping $n$ and $n'$ gives the desired result. +</div> +</div> + +This gives us the primary use of the Matsubara Green's function $C_{AB}$: +calculating the retarded $C_{AB}^R$ and advanced $C_{AB}^A$. +Once we have an expression for Matsubara's $C_{AB}$, +we can recover $C_{AB}^R$ and $C_{AB}^A$ by substituting +$i \omega_m \to \omega \!+\! i \eta$ and $i \omega_m \to \omega \!-\! i \eta$ respectively. + +In general, we can define the **canonical Green's function** $C_{AB}(z)$ +on the complex plane: + +$$\begin{aligned} + C_{AB}(z) + = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{z + E_n - E_{n'}} + \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) +\end{aligned}$$ + +This is a [holomorphic function](/know/concept/holomorphic-function/), +except for poles on the real axis. +It turns out that $C_{AB}(z)$ must have these properties +for the substitution $i \omega_n \to \omega \!\pm\! i \eta$ to be valid. + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. |