# Nonvanishing expectation value lesser Green's function

Consider bosonic field operators in the Heisenberg picture: \begin{align} \Psi(x)=\int \frac{d^{3}p}{(2\pi)^{3}}e^{-ip\cdot x}a_{\bf{p}}\\ \Psi^{\dagger}(x)=\int \frac{d^{3}p}{(2\pi)^{3}}e^{+ip\cdot x}a^{\dagger}_{\bf{p}}, \end{align}

where the dot product $$p\cdot x$$ is Minkowskian with the signature $$(+,-,-,-)$$. The retarded Green's function is defined (see for example: Introduction to Many-body quantum theory in condensed matter physics by Henrik Bruus and Karsten Flensberg, Eq. 8.28) as: $$$$G_{R}(x,x^{\prime})=\theta(t-t^{\prime})\langle 0|[\Psi(x),\Psi^{\dagger}(x^{\prime})]|0 \rangle.$$$$ The first term in the commutator gives the greater Green's function and the second gives the lesser: \begin{align} G^{>}(x,x^{\prime})=\langle 0|\Psi(x)\Psi^{\dagger}(x^{\prime})|0 \rangle\\ G^{<}(x,x^{\prime})=\langle 0|\Psi^{\dagger}(x^{\prime})\Psi(x)|0 \rangle. \end{align} The first one is nonzero since the contribution $$a_{\bf{k}}a^{\dagger}_{\bf{p}}$$ gives a delta function. The second term has the order $$a^{\dagger}_{\bf{k}}a_{\bf{p}}$$ and gives zero. Shouldn't it vanish?

In condensed matter many-body physics vacuum usually means a state filled up to the Fermi energy (such as the ground state of conduction band in a metal), i.e., $$a_\mathbf{p}|0\rangle = 0 \text{ if } \epsilon_\mathbf{p}>\epsilon_F,\\ a_\mathbf{p}^\dagger|0\rangle = 0 \text{ if } \epsilon_\mathbf{p}<\epsilon_F$$ Therefore $$\langle 0|a_\mathbf{p}^\dagger a_\mathbf{p}|0\rangle=\theta(\epsilon_F-\epsilon_\mathbf{p})=n_F(\epsilon_\mathbf{p}),\\ \langle 0|a_\mathbf{p}a_\mathbf{p}^\dagger|0\rangle=1-\theta(\epsilon_F-\epsilon_\mathbf{p})=1-n_F(\epsilon_\mathbf{p})=\theta(\epsilon_\mathbf{p}-\epsilon_F)$$

More general interpretation (at finite temperatures and/or when interaction is taken into account) is that the lesser Green's function gives the distribution of the occupied states, whereas the greater one gives the distribution of the empty states (or holes, depending on the context).

• I see. but then, is there any point for normal ordering when evaluating Wick's theorem? Thanks btw
– M91
Commented May 24, 2022 at 14:25
• @M91 Normal ordering is usually defined in such a way that the operators on the right annihilate the ground state - that is one has to separate the parts of the operators acting above and below the Fermi sea - this saves WIck's theorem (at zero temperature, at finite temperatures it is the volume factor). If you are interested in equilibrium techniques I would suggest using Fetter&Walecka or AGD - these may seem out of date and pedestrian in comparison to modern books, such as Bruus&Flensberg, but they are more thorough on the details. Commented May 24, 2022 at 14:35
• The reason for this sub-question is indeed due to AGD (Abrikosov, Gorlcov , Dyaloslinski). They use both your explanation and Wick's theorem. If creation operators are to the right and annihilate the vacuum then its expt. value vanishes. If it doesn't ( Fermi see argument), then what is the point of Wick's theorem? I'm not sure if you understand me. my point is: either this method or that
– M91
Commented May 24, 2022 at 14:45
• @M91 I think we are comparing apples and oranges: the physical states are filled up to the Fermi energy, and the creation operator kills the ground state, if it acts below the Fermi level. Now, knowing this, I can separate this operator in electron and hole parts, e.g., by defining $h_\mathbf{p}=a_\mathbf{p}^\dagger$ for $\epsilon_\mathbf{p}<\epsilon_F$, and then all the new operators kill the ground state, so I can use the usual proof of the Wick's theorem... which will mean that it holds for the original operators as well. Commented May 24, 2022 at 14:53
• I worked out the time-ordered product: $\langle 0|T(\Psi(x)\Psi^{\dagger}(x^{\prime}))|0 \rangle$. for $t>t^{\prime}$ it gives $D(x-x^{\prime})$. For $t<t^{\prime}$, I evaluate Wick's theorem: $\Psi^{\dagger}(x^{\prime})\Psi(x) \sim N(a_{p^{\prime}}^{\dagger}a_{p}-a_{p}a_{p^{\prime}}^{\dagger} +a_{p}a_{p^{\prime}}^{\dagger}) = -D(x-x^{\prime})=D(x^{\prime}-x)$. Thus for negative energies, I get propagation backwards in time. Seems the same result as for the Feynmann propagator
– M91
Commented May 30, 2022 at 11:24