To understand a crucial example in representation theory, I need the explicit spectral decomposition of the differential operator $$ Df(x)=(1+x^2)f''(x)+2xf'(x) $$ on $L^2({\mathbb R})$. I'm not an expert, but at first glance, theory tells me the existence of a spectral measure, but not what it looks like. Is the spectral measure absolutely continuous with respect to the Lebesgue measure? What are the multiplicities? Has it been explicitly computed somewhere?
1 Answer
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This is a Sturm-Liouville operator $(Df)(x)=(pf')'$, with $p=1+x^2$. These can be rewritten as Schrodinger equations, by using what I would call a Kummer-Liouville transformation. By some conspiracy, all reference to these on the internet seems to have disappeared, but see perhaps my answer here.
We introduce the new variables $t=\int_0^x p^{-1/2}(s)\, ds$, $u=p^{1/4}f$, and then $-D$ becomes the Schrodinger operator $$ Lu = - \frac{d^2u}{dt^2} + V(t)u , \quad V= \frac{p''}{4} - \frac{p'^2}{16p} . $$ That gives $$ V = \frac{1}{4} - \frac{1}{4(1+x^2)} = \frac{1}{4} + O( e^{-|t|}) . $$ Since this is almost constant, with a rapidly decaying error term, standard results show that we have purely absolutely continuous spectrum of multiplicity $2$ on $\sigma_{ac}(-D)=[1/4, \infty)$, and some (non-empty, finite) discrete spectrum below $1/4$.
answered Jul 19, 2018 at 16:55
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$\begingroup$ Sorry, but I don't get these results. Can you check for errors or typos? $\endgroup$– user1688Jul 20, 2018 at 9:02
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$\begingroup$ I tried to reproduce your computations and I get different results. $\endgroup$– user1688Jul 20, 2018 at 18:01
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$\begingroup$ @Corbennick: The formulae look correct to me. You did notice that I'm discussing $-D$, not $D$ ? In particular, when working out $d^2 u/dt^2$, the term $(pf')'$ (with $'=d/dx$) must be set equal to $-\lambda f$. $\endgroup$ Jul 20, 2018 at 23:26
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$\begingroup$ I am still not getting it. Let's split it into two steps. First let $f(x)=h(t(x))$, then the operator becomes $h''(t)+\frac{p'(x(t))}{p(x)^{1/2}}h'(t)$. The second step would by multiplying $h$ by $p(x)^{1/4}$. BUt, as you already have coefficient 1 with $h''$, you won't get that coefficient with $u''$. Also, I don't get cancellation of the $u'$ term. It seems I am misinterpreting your solution completely. $\endgroup$– user1688Jul 23, 2018 at 7:14
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$\begingroup$ Sorry, I thought the problem was in my calculations. However, it turned out that my calculations were alright, only I interpreted them wrongly. $\endgroup$– user1688Jul 23, 2018 at 18:05