Spectral integral form of tensor products

Question

$T_{1}\otimes T_{2}$ is the operator acting on $\mathcal{H}\otimes \mathcal{H}$ then where $T_{1},T_{2}$ are self adjoint operators. What is the spectral resolution of identity for $T_{1}\otimes T_{2}$, further what is the spectral measure of $T_{1}\otimes T_{2}$?

Answer 1

Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 \otimes E^2$ be the spectral measure on $\mathbb{R}\times \mathbb{R}$ defined by $$ E^1 \otimes E^2 (B_1 \times B_2)=E^1 (B_1) \otimes E^2 (B_2), $$ where $B_1, B_2$ are Borel subsets of $\mathbb{R}$.

Remark: The above projection-valtued set function $E^1 \otimes E^2$ defined on rectangles $B_1 \times B_2$ has an unique extension to a spectral measure.

Let us denote by $E$ the spectral measure on $\mathbb{R}$ given by $$ E(B)=(E^1 \otimes E^2)(\{ (x,y):xy \in B \}). $$ Then $E$ is the spectral measure of $T_1 \otimes T_2$.

Sketch of the proof:

First, we prove the equality $T_1 \otimes T_2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy\ d(E^1 \otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).

Second, we use the above equality to prove $T_1 \otimes T_2 = \int_{-\infty}^{\infty} x\ d E (x)$.