Categories: Physics, Quantum mechanics.

# Matsubara Green’s function

The Matsubara Green’s function is an imaginary-time version of the real-time Green’s functions. We define it as follows in the imaginary-time 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 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}

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}

Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$, the 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}

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:

\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}

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}

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}

If we actually evaluate this, we obtain the following form of $C_{AB}$, which is almost identical to the 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}

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.

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, 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.