The Functional Determinants in Peskin and Schroeder (Eq.9.77)

Question

I'm working on the Eq.9.77 in Peskin (page 304):

To demonstrate this, we need only apply standard identities from linear algebra.  First notice that, if a matrix $B$ has eigenvalues $b_i ,$ we can write its determinant as
$$
det{B}
~=~ prod_{i}{b_i}
~~= exp{left[ sum_{i}{log{b_i}} right]}
~=~ exp{left[ operatorname{Tr}{left(log{B}right)} right]}
,,
tag{9.77}
$$
where the logarithm of a matrix is defined by its power series.

However, I test the identity and find it doesn't work.

For $B={{3, -2, 4}, {-2, 6, 2}, {4, 2, 3}} ,$ the left-hand side of the Eq.(9.77) is -98, and the right is 54.
Why does this happen? What are the conditions for this identity to be true? How to prove it?

JulianDeV · Answer

Obviously, since we use the $ln$ operation, none of the eigenvalues may be negative or 0.
It's fairly trivial to prove that this must hold:
begin{align}
det(B) &= b_1...b_n 
&= e^{ln(b_1)}...e^{ln(b_n)} 
&= exp(sum_i ln(b_i)) 
&= exp[Tr(ln B)]
end{align}
Since we can find the eigenvalues of $B$, we can also diagonalise it with the property that
begin{equation}
ln[diag(b_1,...,b_n)] = diag(ln b_1,..., ln b_n)
end{equation}
Also remember that diagonalisation doesn't alter the trace due to the cyclic property: let $S$ be the matrix that diagonalises $B$ then
begin{equation}
Tr(B') = Tr(S^{-1}BS) = Tr(B S S^{-1}) = Tr(B)
end{equation}
The same holds for the $det$.
I am not familiar with your code, but I hope your $log$ is nog the standard log with base 10, it has to be the natural one.

Qmechanic · Answer

The problem seems to be that Mathematica here evaluates ${rm Log}[text{matrix}]$ by taking ${rm Log}$ of each matrix element, which is not the correct mathematical definition of the complex natural logarithm of a matrix.

Answered by Qmechanic on February 11, 2021

The Functional Determinants in Peskin and Schroeder (Eq.9.77)

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