$(S\otimes T)^{it}= S^{it}\otimes T^{it}$ for unbounded operators

Question

Let $S,T$ be unbounded, closed operators in Hilbert spaces $H,K$. In that case, we can form the tensor product operator $S\otimes T$ on the Hilbert space $H\otimes K$ which is the closure of the algebraic tensor product map $S\odot T$. In other words, the algebraic tensor product $H\odot K$ is a core for $S\otimes T$.

If necessary, we can assume that $S,T$ are strictly positive.

I want to prove that for $t\in \mathbb{R}$, we have $$(S\otimes T)^{it}= S^{it}\otimes T^{it}$$ where we make sense of both sides using functional calculus for unbounded self-adjoint operators.

How can I rigorously show this? Both sides are bounded operators defined on $H\otimes K$, so it suffices to check the equality on elementary tensors $\xi \otimes \eta$.

Thanks in advance!

Answer 1

If we ignore possible spectral multiplicity, then we can realize $S,T$ as multiplication by the variable in $H=L^2([0,\infty), \mu)$ and $K=L^2([0,\infty),\nu)$, respectively. That makes $S\otimes T$ the operator of multiplication by $st$ in $H\otimes K = L^2([0,\infty)\times[0,\infty), \mu\otimes\nu)$.

In general, if $A$ is the operator of multiplication by $a(x)$ on its natural domain, then $f(A)$ is multiplication by $f\circ a$ (on its natural domain). Thus $f(S\otimes T)$ is multiplication by $f(st)$, and for the function you consider, this equals $f(s)f(t)$, so $f(S\otimes T)=f(S)\otimes f(T)$.

In the general case, this argument still works, but we need to consider sums of $L^2$ spaces.