# Gel'fand Yaglom functional determinant of non-diagonal operator?

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.

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