Categories: Physics, Quantum mechanics.

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.

They are somewhat related to fundamental solutions, 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, $\nu$ 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 (for some two-particle Green’s functions, see below), but usually in the 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, and $[,]_{\mp}$ is a commutator for bosons, and an anticommutator for fermions. Depending on the context, we could either be in the Heisenberg picture or in the 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}

Proof

Proof. 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 $\hat{\rho}$ 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.

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 or the 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 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}

Proof

Proof. 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):

\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}

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

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