# Making sense of the formula $\operatorname{Det} (I+M )= e^{\operatorname{Tr} \ln (I+M)}$, especially in the infinite dimensional cases

$$\DeclareMathOperator\Det{Det}\DeclareMathOperator\Tr{Tr}$$In physics literature dealing with quantum field theory, the formula $$$$\Det(I+M) = e^{\Tr \ln(I+M)}$$$$ appears frequently. Moreover, in the so-called infinitesimal cases, $$$$\ln(I+M) \simeq M$$$$ is used in analogy with the Taylor series for the usual logarithm.

What is worse, $$M$$ can be not only unbounded but singular as well, containing $$\delta(x-y)$$ for example and yielding $$$$\Tr\delta(x-y)=\delta(x-x).$$$$

I searched everywhere in order to find relevant materials and make sense of the outrageous formulae, especially the "Mathematical Methods of Modern Physics" series by Reed & Simon. However, I cannot find reference to start with.

• For trace class $M$ the standard definition in use is that of the Fredholm determinant: en.wikipedia.org/wiki/Fredholm_determinant . Beyond trace class, one should proceed with caution if one wants to be fully rigorous, although often the formal manipulations favored by physicists can still lead to accurate answers in such regimes. Commented Nov 10, 2021 at 15:49
• Yes, but one does not have as broadly applicable a theory as that of the Fredholm determinant once one leaves the trace class and each individual manipulation of a non-Fredholm determinant often has to be treated by its own arguments. One common technique is to somehow approximate or discretise or analytically continue the operator to be trace class or even finite-dimensional, take determinants there, and pass to a limit (or perform continuation), possibly after performing one or more renormalisations to ensure convergence of the limit or existence of meromorphic continuation. Commented Nov 10, 2021 at 15:54
• For instance zeta function methods are popular en.wikipedia.org/wiki/Functional_determinant . The story here is basically a noncommutative version of the story of divergent summation en.wikipedia.org/wiki/Divergent_series Commented Nov 10, 2021 at 15:56
• TeX note: Use $\operatorname{Det} M$ \operatorname{Det} M, which takes care of the spacing automatically, rather than doing it manually with $\text{Det }M$ \text{Det }M. I have edited accordingly. More importantly, I think it should be $\operatorname{Det}(I + M)$ (as in the body of your post), not $\operatorname{Det} M$, in your title. Commented Nov 10, 2021 at 17:16
• @TerryTao Is the story you refer to ("a noncommutative version of the story of divergent summation") written out in detail somewhere? Thank you. Commented Nov 10, 2021 at 19:23

rather than continuing the long sequence of comments, I switch to the answer box

Q: How to make sense of the outrageous formula $$\ln\operatorname{Det}M=\operatorname{Tr}\ln M$$ ?

A: You might first consider a diagonalizable $$M$$, and then since both determinant and trace do not change if you evaluate them in a basis where $$M$$ is diagonal, the identity follows immediately.

If $$M$$ is not diagonalizable, decompose it into Jordan blocks and use the identity $$\ln{{1\;1}\choose{0\;1}}={{0\;1}\choose{0\;0}}.$$

• I think the essence of the asker's worry is about the infinite-dimensional case, where the story is much less tidy. Commented Nov 10, 2021 at 23:38
• Right. I am worrying about infinite dimensional cases with 'singular' terms like $\delta(x-y)$.. Commented Nov 10, 2021 at 23:59