***Introduction:***


As a quick reminder, the Gel'fand Yaglom theorem uses the generalized zeta-function approach to compute functional determinants of differential operators. Given a differential operator on $x\in [0,1]$ 


$$O\_1=\frac{d^2}{dx^2}+V(x).$$


The functional determinant can be obtained from a solution $y(x)$ of the following differential equation with initial conditions $y(0)=0$ and $y'(0)=1$:


$$\frac{d^2}{dx^2}y(x)+V(x)y(x)=0,$$


so that the determinant is given by


$$\text{det}O\_1 = y(1).$$


***Question:***


Lets now assume that a more complicated differential operator is given on $x \in [0,1]$ by:


$$O\_2=\left(\frac{d^2}{dx^2}+V\_1(x)\right) \binom{1~~~0}{0~~~1}+V\_2 (x)\binom{0~~~1}{1~~~0}+V\_3 (x)\binom{0~-i}{i~~~~0}$$
$$=\left(\frac{d^2}{dx^2}+V\_1(x)\right) 1\_{2 \times 2}+V\_2 (x)\sigma\_1+V\_3 (x)\sigma\_2$$


The Pauli matrices $\sigma\_1$ and $\sigma\_2$ are orthogonal, and lets take $V\_2 (x)$ and $V\_3 (x)$ such that the operator indeed cannot be diagonalized by any transformation. Is the Gel'fand Yaglom theorem still aplicable here, in the sense of solving


$$O\_2~ {\vec y}(x) = {\vec 0} $$


with initial conditions ${\vec y}(0)={\vec 0}$ and ${\vec y}'(0)={\vec 1}$? And obtaining the determinant from:


$$\text{det}O\_2 = \prod\_i y\_i(1) = y\_1 (1) y\_2 (1)~?$$


Formally, I do not see any problem. Considering the generalized zeta-function origin of the technique I also do not see that any eigenvalues get lost. But maybe I miss some crucial point? A confirmation or falsification would be nice.



Gel'fand Yaglom functional determinant of non-diagonal operator?
Not an expert, but you caught my interest. Looks like you may need to be a little careful in higher dimensions, but there are versions of the theorem which work. See, for example:


<http://library.msri.org/books/Book57/files/70kirsten.pdf>


