---
title: "Matsubara Green's function"
sort_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.