How to unify the electrons and holes in Feynman Diagram? As we know, following diagram describes the electron-hole pairs excitation. Namely, the right-going line describes creating electrons propagating with energy $(\omega_n+\nu_n)$, and the left-going line describes the creating  holes propagating with energy $\nu_n$. Thus, the total energy of such "virtual excitation" in system is $(\omega_n+2\nu_n)$.

However,we can also alter the arrow of the lower line, which gives:

I think now the physical meaning for the lower lines is: creating electrons with energy $-\nu_n$ propagates, but then the total energy of such virtual excitation will just be $\omega_n$, or, may be it even cannot be considered as virtual excitation.
I am confused of the discrepancy here, there must be some mistakes I have taken.
Addition: Comments says if I just switch the arrow like the second figure, the quantum number conservation is not consistent. But does exists some standard procedure if I must want to change the "label arrow" (for standardized automated calculations )?
 A: The energy in the first diagram is conserved, it means that the total energy (or Matsubara frequency) is $i\omega$ in all parts of the diagram. I the middle it is $(i\omega+i\nu)-i\nu$, the second term is subtracted because the corresponding arrow is directed to the left. So I don't think the quantity $i\nu$ is somehow connected with the virtual excitation energy.
To look at virtual excitations we need to take into account the exact stationary states $\{\psi_a\}$ available in the system, where the system can spend some time. In order to do it, we can use the spectral representation for the Green function
$$
G(\mathbf{r}_1,\mathbf{r}_2,i\nu)=\sum_a\frac{\psi_a(\mathbf{r}_1)\psi^*_a(\mathbf{r}_2)}{i\nu-E_a}.
$$
Thus you diagram can be written as (if the points on the left and on the right are $\mathbf{r}_1$ and $\mathbf{r}_2$)
$$
T\sum_\nu G(\mathbf{r}_1,\mathbf{r}_2,i\omega+i\nu)G(\mathbf{r}_2,\mathbf{r}_1,i\nu)=\\
T\sum_{ab\nu}
\frac{\psi_a(\mathbf{r}_1)\psi^*_a(\mathbf{r}_2)}{i\omega+i\nu-E_a}\frac{\psi_b(\mathbf{r}_2)\psi^*_b(\mathbf{r}_1)}{i\nu-E_b}.
$$
After frequency summation we get
$$
\sum_{ab}\psi_a(\mathbf{r}_1)\psi^*_a(\mathbf{r}_2)
\psi_b(\mathbf{r}_2)\psi^*_b(\mathbf{r}_1)
\frac{f_b-f_a}{i\omega+E_b-E_a},
$$
where $f_a$, $f_b$ are occupation numbers.
Here we can see the virtual excitations $b\rightarrow a$, where the particle in the $a$ state and the hole in the $b$ state are created with the total energy $E_a-E_b$. The denominator $i\omega-(E_a-E_b)$ has the typical resonant form of the difference between the actual and virtual energies.
If you want to transform one of the particles into the hole, you may introduce the hole energy $\tilde{E}_b=-E_b$, wave function $\tilde\psi_b=\psi^*_b$, and the occupation number $\tilde{f}_b=1-f_b$ to get the diagram in the explicit form of particle-hole pair propagator:
$$
\sum_{ab\nu}\psi_a(\mathbf{r}_1)\tilde\psi_b(\mathbf{r}_1)\psi^*_a(\mathbf{r}_2)
\tilde\psi^*_b(\mathbf{r}_2)
\frac{1-f_a-\tilde{f}_b}{i\omega-(E_a+\tilde{E_b})}.
$$
