path: root/source/know/concept/greens-functions/index.md

---
title: "Green's functions"
sort_title: "Green's functions"
date: 2021-11-03
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

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 **time-ordered** or **causal Green's function** $$G_{\nu \nu'}$$ is as follows,
where $$\mathcal{T}$$ is the [time-ordered product](/know/concept/time-ordered-product/),
$$\nu$$ and $$\nu'$$ are single-particle states,
and $$\hat{c}_\nu$$ annihilates a particle from $$\nu$$, etc.:

$$\begin{aligned}
    \boxed{
        G_{\nu \nu'}(t, t')
        \equiv -\frac{i}{\hbar} \Expval{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t') \Big\}}
    }
\end{aligned}$$

The expectation value $$\Expval{}$$ is
with respect to thermodynamic equilibrium.
This is sometimes in the [canonical ensemble](/know/concept/canonical-ensemble/)
(for some two-particle Green's functions, see below),
but usually 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 $$\hat{H}$$.
Explicitly, for a complete set of many-particle states $$\Ket{\Psi_n}$$, we have:

$$\begin{aligned}
    G_{\nu \nu'}(t, t')
    &= -\frac{i}{\hbar Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}} \Big)
    \\
    &= -\frac{i}{\hbar Z} \sum_{n}
    \Matrixel{\Psi_n}{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}}}{\Psi_n}
\end{aligned}$$

Arguably more prevalent are
the **retarded Green's function** $$G_{\nu \nu'}^R$$
and the **advanced Green's function** $$G_{\nu \nu'}^A$$
which are defined like so:

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

Where $$\Theta$$ is a [Heaviside function](/know/concept/heaviside-step-function/),
and $$[,]_{\mp}$$ is a commutator for bosons,
and an anticommutator for fermions.
Depending on the context,
we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/)
or in the [interaction picture](/know/concept/interaction-picture/),
hence $$\hat{c}_\nu$$ and $$\hat{c}_{\nu'}^\dagger$$ are time-dependent.

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 $$+$$ for fermions.
With this, the causal, retarded and advanced Green's functions
can thus be expressed as follows:

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

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:

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

And analogously for $$G_{ss'}^R$$, $$G_{ss'}^A$$, $$G_{ss'}^>$$ and $$G_{ss'}^<$$.
Note that the time-dependence is given to the old $$G_{\nu \nu'}$$,
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 $$t - t'$$:

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

<div class="accordion">
<input type="checkbox" id="proof-time-diff"/>
<label for="proof-time-diff">Proof</label>
<div class="hidden" markdown="1">
<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.

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

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

The expected value of the product
of the time-independent operators $$\hat{A}$$ and $$\hat{B}$$ is then:

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

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" markdown="1">
<label for="proof-commH0">Proof.</label>
In the second quantization,
the Hamiltonian $$\hat{H}_0$$ is written like so:

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

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

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

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

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 the two-particle functions
only depend on the time-difference $$t - t'$$:

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



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