# Spectral function with negative value

How does one understand a negative value in the spectral function

$$\chi=-\mathrm{Im(G)}$$

where $G$ is the Green function and $\chi$ is a spectral function?

• Isn't it just the phase? – Floris May 31 '15 at 14:18
• Sorry, i don't get your point, could you give more details? thanks – Xinloong Han May 31 '15 at 14:50
• Isn't negative amplitude just a signal with a 180 degree phase shift? – Floris May 31 '15 at 14:51
• Maybe you could give some details of the specific problem? For a typical spectral function I think this should never be the case, because it should look like the product of a density of states and an occupation distribution. – Rococo May 31 '15 at 16:53

Actually, I guess details aren't necessary. Working pretty much directly from Xiao-Gang Wen's many-body textbook:

A general (zero-temperature) spectral function of a two-operator correllator can be defined as: $$A_+(\omega)=\sum_m \delta(\omega-(\epsilon_m-\epsilon_0)) \langle 0|O_2 |\psi_m \rangle \langle \psi_m|O_1 |0 \rangle$$

So for each energy eigenstate $m$ it is the product of two amplitudes, representing the probability amplitude to go from the vacuum to that state using the operator $O_1$ or $O_2$. If these operators are different, this can certainly take a negative or complex value that is just the product of two phases. However, we normally use the spectral function for a creation operator of a momentum eigenstate, $\psi_k^\dagger$, in which case this reduces to:

$$A_{k,+}(\omega)=\sum_m \delta(\omega-(\epsilon_m-\epsilon_0)) |\langle \psi_m| \psi_k^\dagger |0 \rangle|^2$$

This expression is always positive or zero. If you integrate over $k$, it becomes the density of states.

The Green's function and the spectral function can be defined as $$G(\mathbf r,\mathbf r',\omega) = \sum_n \frac{f_n(\mathbf r)f_n^*(\mathbf r')}{\omega - \epsilon_n + i\eta},$$ where $\eta \rightarrow 0^+$, and $$A(\mathbf r,\mathbf r',\omega) = \sum_n f_n(\mathbf r)f_n^*(\mathbf r') \delta(\omega - \epsilon_n).$$ In order to get from one to the other we make use of the identity $$\mathrm{Im} \frac{1}{x+i\eta} = -\frac{\eta}{x^2+\eta^2} = -\pi\delta(x),$$ in which case it is easy to see that $$A(\mathbf r,\mathbf r',\omega) = -\frac{1}{\pi} \mathrm{Im} G(\mathbf r,\mathbf r',\omega).$$ If you forgot the minus sign, then the spectral function would have negative values, which would be unphysical.