How can I show that $\mathrm{Tr}\left( f(G^\dagger G)\right)=\mathrm{Tr}\left( f(G G^\dagger)\right)$?

Question

I'm slightly stuck on the following question:

Prove that: $\mathrm{Tr}\left( f(G^\dagger G)\right)=\mathrm{Tr}\left( f(G G^\dagger)\right)$ where $G$ is any operator.

Using the definition of the trace and definition $B = G^{\dagger} G$ and $C = GG^{\dagger}$, I have the expressions:

$$\mathrm{Tr}\left( f(G^\dagger G)\right)=\mathrm{Tr}\left( f(B)\right)=\sum_jf(b_j)$$ and $$\mathrm{Tr}\left( f(GG^\dagger)\right)=\mathrm{Tr}\left( f(C)\right)=\sum_jf(c_j),$$ where $b_j$ and $c_j$ are the eignevalues of $B$ and $C$, which are hermitian. I have taken the trace w.r.t to the same basis which spans the Hilbert Space: $\sum_{j} \cdots = 1 $.

I'm unable to show how and why must the traces be equal.

Answer 1

This follows from the cyclicity of the trace, i.e. the property that $$\mathrm{Tr}(AB)=\mathrm{Tr}(BA),$$ which extends to the cyclic permutation $\mathrm{Tr}(ABC\cdots XYZ)=\mathrm{Tr}(BC\cdots XYZA)$ for larger products. Thus, if you expand $f$ in its Taylor series, you get \begin{align} \mathrm{Tr}\left(f(G^\dagger G)\right) & = \mathrm{Tr}\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(G^\dagger G)^n\right) \\& = \mathrm{Tr}\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}G^\dagger G\cdots G^\dagger G\right) \\& = \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\mathrm{Tr}\left(G^\dagger G\cdots G^\dagger G\right) \\& = \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\mathrm{Tr}\left(GG^\dagger \cdots G^\dagger GG^\dagger \right) \\& = \mathrm{Tr}\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}GG^\dagger \cdots G^\dagger GG^\dagger \right) \\& = \mathrm{Tr}\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(G G^\dagger)^n\right) \\& = \mathrm{Tr}\left(f(GG^\dagger )\right) . \end{align}

Alternatively, you can calculate the trace via the eigenvalues of $B=G^\dagger G$ and $C=GG^\dagger$, in which case you need to relate the two sets of eigenvalues; this can be done by noting that if $B\psi=b\psi$ gives an eigenpair of $B$, then $$C(G\psi)=GG^\dagger G\psi=GB\psi=Gb\psi=b(G\psi)$$ also gives an eigenpair with the same eigenvalue. (In infinite dimensions, of course, this argument needs to be handled carefully, in case $\psi$ is normalized but $G\psi$ is not as well-behaved, but you're assuming loads in just getting to the point-spectrum sum you use in the first place.)

These are probably the two main ideas behind most proofs of that fact - the rest of the details revolve around what properties $G$ is known to have, and how you're defining $f$ at an operator in the first place.

Answer 2

$G^\dagger G$ and $G G^\dagger$ are unitary equivalent in the finite dimensional case :

There exists the polar decomposition $G = R U$ where $R \geq 0$ and $U$ unitary. Then $G^\dagger G = U^\dagger R^2 U = U^\dagger (G G^\dagger) U$.

But from this it follows that $G^\dagger G$ and $G G^\dagger$ have the same eigenvalues.

In the infinite dimensional case we know only that U is a partial isometry .