path: root/content/know/concept/greens-functions/index.pdc

---
title: "Green's functions"
firstLetter: "G"
publishDate: 2021-11-03
categories:
- Physics
- Quantum mechanics

date: 2021-11-01T09:46:27+01:00
draft: false
markup: pandoc
---

# Green's functions

In many-body quantum theory, a **Green's function**
can be any correlation function between two given operators,
although it is usually used to refer to the special case
where the operators are particle creation/annihilation operators
from the [second quantization](/know/concept/second-quantization/).

They are somewhat related to
[fundamental solutions](/know/concept/fundamental-solution/),
which are also called *Green's functions*,
but in general they are not the same,
except in a special case, see below.


## Single-particle functions

If the two operators are single-particle creation/annihilation operators,
then we get the **single-particle Green's functions**,
for which the symbol $G$ is used.

The **retarded Green's function** $G_{\nu \nu'}^R$
and the **advanced Green's function** $G_{\nu \nu'}^A$
are defined like so,
where the expectation value $\expval{}$ is
with respect to thermodynamic equilibrium,
$\nu$ and $\nu'$ are labels of single-particle states,
and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            G_{\nu \nu'}^R(t, t')
            &\equiv -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
            \\
            G_{\nu \nu'}^A(t, t')
            &\equiv \frac{i}{\hbar} \Theta(t' - t) \expval{\comm*{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
        \end{aligned}
    }
\end{aligned}$$

Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
and $[,]_{\mp}$ is a commutator for bosons,
and an anticommutator for fermions.
We are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent,
but keep in mind that time-dependent Hamiltonians are allowed,
so it might not be trivial.

Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
and **lesser Green's function** $G_{\nu \nu'}^<$ are:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            G_{\nu \nu'}^>(t, t')
            &\equiv -\frac{i}{\hbar} \expval{\hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')}
            \\
            G_{\nu \nu'}^<(t, t')
            &\equiv \mp \frac{i}{\hbar} \expval{\hat{c}_{\nu'}^\dagger(t') \: \hat{c}_{\nu}(t)}
        \end{aligned}
    }
\end{aligned}$$

Where $-$ is for bosons, and $+$ is for fermions.
The retarded and advanced Green's functions can thus be expressed as follows:

$$\begin{aligned}
    G_{\nu \nu'}^R(t, t')
    &= \Theta(t - t') \Big( G_{\nu \nu'}^>(t, t') - G_{\nu \nu'}^<(t, t') \Big)
    \\
    G_{\nu \nu'}^A(t, t')
    &= \Theta(t' - t) \Big( G_{\nu \nu'}^<(t, t') - G_{\nu \nu'}^>(t, t') \Big)
\end{aligned}$$

If the Hamiltonian involves interactions,
it might be more natural to use quantum field operators $\hat{\Psi}(\vb{r}, t)$
instead of choosing a basis of single-particle states $\psi_\nu$.
In that case, instead of a label $\nu$,
we use the spin $s$ and position $\vb{r}$, leading to:

$$\begin{aligned}
    G_{ss'}^R(\vb{r}, t; \vb{r}', t')
    &= -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{\Psi}_{s}(\vb{r}, t)}{\hat{\Psi}_{s'}^\dagger(\vb{r}', t')}_{\mp}}
    \\
    &= \sum_{\nu \nu'} \psi_\nu(\vb{r}) \: \psi^*_{\nu'}(\vb{r}') \: G_{\nu \nu'}^R(t, t')
\end{aligned}$$

And analogously for $G_{ss'}^A$, $G_{ss'}^>$ and $G_{ss'}^<$.
Note that the time-dependence is given to the old $G_{\nu \nu'}^R$,
i.e. to $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$,
because we are in the Heisenberg picture.

If the Hamiltonian is time-independent,
then it can be shown that all the Green's functions
only depend on the time-difference $t - t'$:

$$\begin{aligned}
    G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t')
    \qquad \quad
    G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t')
    \\
    G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
    \qquad \quad
    G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-time-diff"/>
<label for="proof-time-diff">Proof</label>
<div class="hidden">
<label for="proof-time-diff">Proof.</label>
We will prove that the thermal expectation value
$\expval*{\hat{A}(t) \hat{B}(t')}$ only depends on $t - t'$
for arbitrary $\hat{A}$ and $\hat{B}$,
and it trivially follows that the Green's functions do too.

Suppose that the system started in thermodynamic equilibrium.
This could sometimes be in the [canonical ensemble](/know/concept/canonical-ensemble/)
(for two-particle Green's functions, see below),
but usually it will be in the
[grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
since we are adding/removing particles.
In the latter case, we assume that the chemical potential $\mu$
is already included in the Hamiltonian.

In any case, at equilibrium, we know that the
[density operator](/know/concept/density-operator/)
$\hat{\rho}$ is as follows:

$$\begin{aligned}
    \hat{\rho} = \frac{1}{Z} \exp\!(- \beta \hat{H})
\end{aligned}$$

Where $Z \equiv \Tr\!(\exp\!(- \beta \hat{H}))$ is the partition function.
In that case, the expected value of the product
of the time-independent operators $\hat{A}$ and $\hat{B}$ is calculated like so:

$$\begin{aligned}
    \expval*{\hat{A}(t) \hat{B}(t')}
    &= \frac{1}{Z} \Tr\!\big( \hat{\rho} \hat{A}(t) \hat{B}(t') \big)
    \\
    &= \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i t \hat{H} / \hbar} \hat{A} e^{-i t \hat{H} / \hbar}
    e^{i t' \hat{H} / \hbar} \hat{B} e^{-i t' \hat{H} / \hbar} \Big)
\end{aligned}$$

Using that the trace $\Tr$ is invariant
under cyclic permutations of its argument,
and that all functions of $\hat{H}$ commute, we find:

$$\begin{aligned}
    \expval*{\hat{A}(t) \hat{B}(t')}
    = \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i (t - t') \hat{H} / \hbar} \hat{A} e^{-i (t - t') \hat{H} / \hbar} \hat{B} \Big)
\end{aligned}$$

As expected, this only depends on the time difference $t - t'$,
because $\hat{H}$ is time-independent by assumption.
Note that thermodynamic equilibrium is crucial:
intuitively, if the system is not in equilibrium,
then it evolves in some transient time-dependent way.
</div>
</div>

If the Hamiltonian is both time-independent and non-interacting,
then the time-dependence of $\hat{c}_\nu$
can simply be factored out as
$\hat{c}_\nu(t) = \hat{c}_\nu \exp\!(- i \varepsilon_\nu t / \hbar)$.
Then the diagonal ($\nu = \nu'$) greater and lesser Green's functions
can be written in the form below, where $f_\nu$ is either
the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/)
or the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/).

$$\begin{aligned}
    G_{\nu \nu}^>(t, t')
    &= -\frac{i}{\hbar} \expval{\hat{c}_{\nu} \hat{c}_{\nu}^\dagger} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
    \\
    &= -\frac{i}{\hbar} (1 - f_\nu) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
    \\
    G_{\nu \nu}^<(t, t')
    &= \mp \frac{i}{\hbar} \expval{\hat{c}_{\nu}^\dagger \hat{c}_{\nu}} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
    \\
    &= \mp \frac{i}{\hbar} f_\nu \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
\end{aligned}$$


## As fundamental solutions

In the absence of interactions,
we know from the derivation of
[equation-of-motion theory](/know/concept/equation-of-motion-theory/)
that the equation of motion of $G^R(\vb{r}, t; \vb{r}', t')$
is as follows (neglecting spin):

$$\begin{aligned}
    i \hbar \pdv{G^R}{t}
    = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
    + \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\comm*{\hat{H}_0}{\hat{\Psi}(\vb{r}, t)}}{\hat{\Psi}^\dagger(\vb{r}', t')}}
\end{aligned}$$

If $\hat{H}_0$ only contains kinetic energy,
i.e. there is no external potential,
it can be shown that:

$$\begin{aligned}
    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-commH0"/>
<label for="proof-commH0">Proof</label>
<div class="hidden">
<label for="proof-commH0">Proof.</label>
In the second quantization,
the Hamiltonian $\hat{H}_0$ is written like so:

$$\begin{aligned}
    \hat{H}_0
    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \braket{\psi_\nu}{\nabla^2 \psi_{\nu'}}
    \\
    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
    \\
    &= - \frac{\hbar^2}{2 m}
    \int \Big( \sum_{\nu} \psi_\nu^*(\vb{r}') \hat{c}_\nu^\dagger \Big) \Big( \nabla^2 \sum_{\nu'} \psi_{\nu'}(\vb{r}') \hat{c}_{\nu'} \Big) \dd{\vb{r}'}
    \\
    &= - \frac{\hbar^2}{2 m}
    \int \hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$

We then insert this into the commutator that we want to prove, yielding:

$$\begin{aligned}
    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
    &= - \frac{\hbar^2}{2 m} \int \comm{\hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})} \dd{\vb{r}'}
    \\
    &= - \frac{\hbar^2}{2 m} \int \hat{\Psi}^\dagger(\vb{r}') \comm{\nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})}
    + \comm{\hat{\Psi}^\dagger(\vb{r}')}{\hat{\Psi}(\vb{r})} \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
    \\
    &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu' \nu''}
    \Big( \hat{c}_\nu^\dagger \comm*{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm*{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''} \Big)
    \psi_{\nu'}(\vb{r}) \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu''}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$

When deriving equation-of-motion theory,
we already showed that the following identity
holds for both bosons and fermions:

$$\begin{aligned}
    \hat{c}_\nu^\dagger \comm*{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm*{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''}
    = - \delta_{\nu \nu'} \hat{c}_{\nu''}
\end{aligned}$$

Such that the commutator can be significantly simplified to:

$$\begin{aligned}
    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
    &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
    \int \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r}) \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$

We know that the $\psi_\nu$ form a *complete* basis,
which implies (see [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)):

$$\begin{aligned}
    \sum_{\nu} \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r})
    = \delta(\vb{r} - \vb{r}')
\end{aligned}$$

With this, the commutator can be reduced even further as follows:

$$\begin{aligned}
    \comm*{\hat{H}_0}{\hat{\Psi}(\vb{r})}
    &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
    \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
    \\
    &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
</div>
</div>

After substituting this into the equation of motion,
we recognize $G^R(\vb{r}, t; \vb{r}', t')$ itself:

$$\begin{aligned}
    i \hbar \pdv{G^R}{t}
    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
    + \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}}
    \\
    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2
    \Big( \!-\! \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}} \Big)
    \\
    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
    - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 G^R(\vb{r}, t; \vb{r}', t')
\end{aligned}$$

Rearranging this leads to the following,
which is the definition of a fundamental solution:

$$\begin{aligned}
    \Big( i \hbar \pdv{t} + \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 \Big) G^R(\vb{r}, t; \vb{r}', t')
    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
\end{aligned}$$

Therefore, the retarded Green's function
(and, it turns out, the advanced Green's function too)
is a fundamental solution of the Schrödinger equation
if there is no potential,
i.e. the Hamiltonian only contains kinetic energy.


## Two-particle functions

The above can be generalized to two arbitrary operators $\hat{A}$ and $\hat{B}$,
giving us the **two-particle Green's functions**,
or just **correlation functions**.
The **retarded correlation function** $C_{AB}^R$
and the **advanced correlation function** $C_{AB}^A$ are defined as
(in the Heisenberg picture):

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            C_{AB}^R(t, t')
            &\equiv -\frac{i}{\hbar} \Theta(t - t') \expval{\comm*{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
            \\
            C_{AB}^A(t, t')
            &\equiv \frac{i}{\hbar} \Theta(t' - t) \expval{\comm*{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
        \end{aligned}
    }
\end{aligned}$$

Where the expectation value $\expval{}$ is taken of thermodynamic equilibrium.
The name *two-particle* comes from the fact that $\hat{A}$ and $\hat{B}$
will often consist of a sum of products
of two single-particle creation/annihilation operators.

Like for the single-particle Green's functions,
if the Hamiltonian is time-independent,
then it can be shown that $C_{AB}^R$ and $C_{AB}^A$
only depend on the time-difference $t - t'$:

$$\begin{aligned}
    G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
    \qquad \quad
    G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
\end{aligned}$$



## References
1.  H. Bruus, K. Flensberg,
    *Many-body quantum theory in condensed matter physics*,
    2016, Oxford.