# Trace of the Operator

I want to ask a question about the fundamental knowledge of trace of the an operator. The operator $$A$$ is $$A = v (G_r-G_a)$$ where v is the velocity operator of the Hamiltonian ($$v=dH/dk$$); $$G_r$$ and $$G_a$$ are retarded and advanced green functions $$G_r=\frac{1}{E_f-H+i \gamma},\;\; G_a=\frac{1}{E_f-H-i \gamma}$$ $$E_f$$ is the Fermi energy of the system, $$H$$ is the Hamiltonian matrix, $$i$$ is the complex number $$(0.0, 1.0)$$ and $$\gamma$$ is a real number. I want to calculate the trace of operator A and I have the following equation $$\rm{Tr}(A)=\sum_i \langle i|v(G_r-G_a)|i\rangle= \sum_{i,i} \langle i|v|j\rangle \langle j|(G_r-G_a)|i\rangle =\;\;\sum_{i,j}\langle i|v|j\rangle \langle j|(G_{r_i}-G_{a_i})|i\rangle$$ where, $$G_{r_i}=\frac{1}{E_f-e_i+i\gamma}$$ and $$G_a=\frac{1}{E_f-e_i-i\gamma}$$. In other words, the Hamiltonian matrix in $$G_r$$ and $$G_a$$ is converted into eigenvalue.

I want to ask whether $$|i\rangle$$ and $$j\rangle$$ must be the eigenvectors of the operator $$A$$? Can $$|i\rangle$$ and $$|j\rangle$$ be the eigenvectors of $$H$$ matrix; not of the $$A$$ operator?

My second question is that suppose $$A$$ is a 2 by 2 matrix and the eigen vector matrix $$|i\rangle$$ or $$|j\rangle$$ of $$H$$ is 2 by 1 matrix. In order to compute $$\sum_{i,j}\langle i|v|j\rangle \langle j|(G_{r_i}-G_{a_i})|i\rangle$$, I should use the following combination. $$\sum_{i,j}\langle i|v|j\rangle \langle j|(G_{r_i}-G_{a_i})|i\rangle=\langle 1|v|1\rangle \langle 1|(G_{r_i}-G_{a_i})|1\rangle+\langle 1|v|2\rangle \langle 2|(G_{r_i}-G_{a_i})|1\rangle+\langle 2|v|1\rangle \langle 1|(G_{r_i}-G_{a_i})|2\rangle+\langle 2|v|2\rangle \langle 2|(G_{r_i}-G_{a_i})|2\rangle$$ Is my understanding correct or not? Thank you very much.

In order for your formula to be valid, the states $$|i\rangle$$ must be eigenvectors of $$H$$ with the energy $$e_i$$. Otherwise you will not get $$\langle j | G_r | i \rangle = \langle j| \frac{1}{E_f-e_i+i\gamma} |i\rangle$$ as what you did was explicitly act with $$H$$ on the state to the right. While $$|j\rangle$$ might be any basis of states whatsoever, it will be convenient to also be the eigenstates of the Hamiltonian, as this will render the double-sum over $$i$$ and $$j$$ a single sum, since for such states $$\langle j | G_r | i \rangle = \langle j| \frac{1}{E_f-e_i+i\gamma} |i\rangle = \delta_{i,j} \frac{1}{E_f-e_i+i\gamma}$$

• Thank you for the answer. Would you please give me some more suggestions on my second question that I just added to my post? Commented Mar 7, 2020 at 3:44

The trace of any matrix/operator will be same regardless of what basis you use, provided they are complete.

So it doesn't matter whether you choose eigenvectors of $$A$$ or $$H$$, but you must be consistent and use the complete basis.

If you decide to use the eigenbasis of $$A$$, then you can't simply substitute the scalar energy $$E_i$$ for $$H$$, you must keep $$H$$ an operator. The only exception is if you have a simultaneously diagonalizable basis for $$A$$ and $$H$$, which rarely happens.

• while answer is true in general, the OP's formula where one replaces $H$ with its eigenvalues holds only for the eigenvectors of $H$.
– user245141
Commented Mar 6, 2020 at 12:25
• @yu-v you're right. Let me add that point. Commented Mar 6, 2020 at 12:27
• @KFGauss Thank you for the answer. Would you please give me some more suggestions on my second question that I just added to my post? Commented Mar 7, 2020 at 3:45