Positive polarity of a retarded Green's function's imaginary part

[gonadas91]

Hi all! I am currently working on a problem that involves calculation of Green functions. I am having serious doubts on some part of the calculation, as I am getting a GF that has positive values of its imaginary part. According to Lehman's rule, the retarded GF can be decomposed by the spectrum and this decomposition is positive definite. Moreover, it represents a density of states. I know this for LOCAL green functions, but is this also true if I calculate G_ret( 0, 1 ) that is, non-diagonal elements of the retarded GF matrix?

In summary: Terms G_ret(i,i) have Im( G_ret(i,i) ) < 0, but I am having terms with:
$Im[G_{ij}(\omega)] > 0$
for $$i\neq j$$, is that possible? Thanks!EDIT: I found out that in general we can write the retarded Green's function as:
$G_{ij}^{ret}(\omega) = \sum_{n}\frac{\langle i |A |n\rangle\langle n| B^{\dagger}|j\rangle}{\omega - \varepsilon_{n} + i\eta}$

for any basis of states satisfying the closure relation. I guess the positivity of the (-) imaginary part is only satisfied when $i=j$, but if I am wrong here please correct me!

 

[vanhees71]

The only thing that must be positive is the $\eta$ in the above equation, because that ensures that in the time domain it's $\propto \Theta(t)$, which is the defining property of this particular Green's function being "retarded".

 

[gonadas91]

The only thing that must be positive is the $\eta$ in the above equation, because that ensures that in the time domain it's $\propto \Theta(t)$, which is the defining property of this particular Green's function being "retarded".

Many thanks for the reply. The question was more oriented towards the matrix elements of G outside the diagonal, and I forgot to mention the operators are supposed to be fermionic. In that case, the (-) imaginary part of the local retarded GF has to be definite positive if we have c_i*c_j^dagger for i=j, but if I and j are not the same, then the imaginary part of the retarded GF does not need to be negative?

 

[vanhees71]

Why do you think the sign of the imaginary part is important? From the retardation condition you only get that there must not be any singularities in the upper complex energy-half plane. The reason is that
$$G(t)=\int_{\mathbb{R}} \frac{\mathrm{d}E}{2 \pi} \tilde{G}(E) \exp(-\mathrm{i} E t).$$
Now to evaluate the integral you close the integration path in the upper (lower) half-plane for $t<0$ ($t>0$). In order that $G(t) \propto \Theta(t)$ (retardation condition) the only condition is that $\tilde{G}(E)$ has no singularities in the upper half-plane.

 

[gonadas91]

Why do you think the sign of the imaginary part is important? From the retardation condition you only get that there must not be any singularities in the upper complex energy-half plane. The reason is that
$$G(t)=\int_{\mathbb{R}} \frac{\mathrm{d}E}{2 \pi} \tilde{G}(E) \exp(-\mathrm{i} E t).$$
Now to evaluate the integral you close the integration path in the upper (lower) half-plane for $t<0$ ($t>0$). In order that $G(t) \propto \Theta(t)$ (retardation condition) the only condition is that $\tilde{G}(E)$ has no singularities in the upper half-plane.

Yes, but the imaginary part of the retarded Green function physical is connected to the density of states, and therefore, to probability conservation and sum rule. If your imaginary part is positive, then you would have negative density of states, which does not make any sense physically speaking.

 

[vanhees71]

I see, so you want the specific Green's function, solving the Schrödinger equation, not a general two-point function characterized by arbitrary operators $A$ and $B$.

This Green's function is defined by (using natural units with $\hbar=1$)
$$(\mathrm{i} \partial_t-\hat{H}_x) G(t,\vec{x},\vec{x}')=\delta t \delta^{(3)}(\vec{x}-\vec{x}').$$
Now suppose you have a complete set of energy eigenstates, $u_E(\vec{x})$. For simplicity we also assume that there's no degeneracy of the energy eigenvalues. If there is you must add more observables to diagonalize simultaneously with $\hat{H}$ to get an unambigous complete set of energy eigenvalues.

Then you can expand the Green's function in terms of these in the form
$$G(t,\vec{x},\vec{x}')=\int_{\mathbb{R}} \mathrm{d} E A_E(t,\vec{x}') u_E(\vec{x}).$$
Plugging this into the equation for the Green's function, gives
$$\int(\mathrm{i} \partial_t A_E - E A_E) u_E(\vec{x})=\delta(t) \delta^{(3)}(\vec{x}-\vec{x}')=\delta(t) \int_{\mathbb{R}} \mathrm{d} E u_E^*(\vec{x}') u_E(\vec{x}), \qquad (1)$$
where the completeness relation for the energy eigenstates has been used.

Obviously the $\delta$-distribution in time must come from the time derivative on the left-hand side. Thus $A_E$ must have a jump of height 1 at $t=0$. For $t>0$ it must satisfy
$$\mathrm{i} \partial_t A_E=E A_E \; \Rightarrow \; A_E \propto \exp(-\mathrm{i} E t).$$
The coefficient is then determined by the right-hand side to be $u_E^*(\vec{x}')$.

Thus you get the ansatz (together with the retardation condition $G(t,\vec{x},\vec{x}')=0$ for $t<0$)
$$A_E=\Theta(t) u_E^*(\vec{x}') \exp(-\mathrm{i} E t).$$
Indeed, this solves the equation of motion for the Green's function, because
$$\mathrm{i} \partial_t A_E - E A_E=\delta(t) u_E^*(\vec{x}')$$
and plugging this into (1) shows that this indeed solves the equation.

Now let's consider the Fourier transformation, which is unique thanks to the $\Theta$ function from the retardation condition (using a regulator in terms of an infinitesimal positive imaginary part added to the frequency):
$$\tilde{G}(\omega,\vec{x},\vec{x}')=\int_{\mathbb{R}} \mathrm{d} t G(t,\vec{x},\vec{x}') \exp(\mathrm{i} (\omega+\mathrm{i} 0^+)= \int_{\mathbb{R}} \mathrm{d} E u_E^*(\vec{x}') u_E(\vec{x}) \frac{1}{\omega-E+\mathrm{i} 0^+}.$$
Now using the identity
$$\frac{1}{\omega-E+\mathrm{i} 0^+}=\mathcal{P} \frac{1}{\omega-E} - \mathrm{i} \pi \delta(\omega-E).$$
Then setting $\vec{x}'=\vec{x}$ gives
$$\mathrm{Im} \tilde{G}(\omega,\vec{x},\vec{x})=-\pi |u_{\omega}(\vec{x})|^2.$$
So in this case indeed you have a negative imaginary part.

 

[gonadas91]

I see, so you want the specific Green's function, solving the Schrödinger equation, not a general two-point function characterized by arbitrary operators $A$ and $B$.

This Green's function is defined by (using natural units with $\hbar=1$)
$$(\mathrm{i} \partial_t-\hat{H}_x) G(t,\vec{x},\vec{x}')=\delta t \delta^{(3)}(\vec{x}-\vec{x}').$$
Now suppose you have a complete set of energy eigenstates, $u_E(\vec{x})$. For simplicity we also assume that there's no degeneracy of the energy eigenvalues. If there is you must add more observables to diagonalize simultaneously with $\hat{H}$ to get an unambigous complete set of energy eigenvalues.

Then you can expand the Green's function in terms of these in the form
$$G(t,\vec{x},\vec{x}')=\int_{\mathbb{R}} \mathrm{d} E A_E(t,\vec{x}') u_E(\vec{x}).$$
Plugging this into the equation for the Green's function, gives
$$\int(\mathrm{i} \partial_t A_E - E A_E) u_E(\vec{x})=\delta(t) \delta^{(3)}(\vec{x}-\vec{x}')=\delta(t) \int_{\mathbb{R}} \mathrm{d} E u_E^*(\vec{x}') u_E(\vec{x}), \qquad (1)$$
where the completeness relation for the energy eigenstates has been used.

Obviously the $\delta$-distribution in time must come from the time derivative on the left-hand side. Thus $A_E$ must have a jump of height 1 at $t=0$. For $t>0$ it must satisfy
$$\mathrm{i} \partial_t A_E=E A_E \; \Rightarrow \; A_E \propto \exp(-\mathrm{i} E t).$$
The coefficient is then determined by the right-hand side to be $u_E^*(\vec{x}')$.

Thus you get the ansatz (together with the retardation condition $G(t,\vec{x},\vec{x}')=0$ for $t<0$)
$$A_E=\Theta(t) u_E^*(\vec{x}') \exp(-\mathrm{i} E t).$$
Indeed, this solves the equation of motion for the Green's function, because
$$\mathrm{i} \partial_t A_E - E A_E=\delta(t) u_E^*(\vec{x}')$$
and plugging this into (1) shows that this indeed solves the equation.

Now let's consider the Fourier transformation, which is unique thanks to the $\Theta$ function from the retardation condition (using a regulator in terms of an infinitesimal positive imaginary part added to the frequency):
$$\tilde{G}(\omega,\vec{x},\vec{x}')=\int_{\mathbb{R}} \mathrm{d} t G(t,\vec{x},\vec{x}') \exp(\mathrm{i} (\omega+\mathrm{i} 0^+)= \int_{\mathbb{R}} \mathrm{d} E u_E^*(\vec{x}') u_E(\vec{x}) \frac{1}{\omega-E+\mathrm{i} 0^+}.$$
Now using the identity
$$\frac{1}{\omega-E+\mathrm{i} 0^+}=\mathcal{P} \frac{1}{\omega-E} - \mathrm{i} \pi \delta(\omega-E).$$
Then setting $\vec{x}'=\vec{x}$ gives
$$\mathrm{Im} \tilde{G}(\omega,\vec{x},\vec{x})=-\pi |u_{\omega}(\vec{x})|^2.$$
So in this case indeed you have a negative imaginary part.

Thanks very much for the detailed reply! My GF in the diagonal (x=x') do follow this rule, but out of the diagonal they don't, which is what puzzled me at first!

 

[vanhees71]

Thanks very much for the detailed reply! My GF in the diagonal (x=x') do follow this rule, but out of the diagonal they don't, which is what puzzled me at first!

But for $\vec{x}' \neq \vec{x}$ the numerator in this Callen-representation of the propagator is not real, because in general $u_E^*(\vec{x}') u_E(\vec{x}) \notin \mathbb{R}$.