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Let $T: \operatorname{dom}(T) \subset H \rightarrow H$ be a positive self-adjoint unbounded operator, then I want to define a UNIQUE(!) operator $A$ such that $A^{*}A = T$. Actually, this construction is nothing new, but I am uncertain about the DOMAIN(!) of $A$ in the case of unbounded operators. Therefore, I was wondering if anybody here knows a good reference that treats this "polar decomposition" also in the case of unbounded operators. Alternatively, if somebody wants to comment on this problem, I would highly appreciate this.

I mean, one is somehow tempted to say $\operatorname{dom}(A) = \operatorname{dom}(T)$ and then it is necessary that $\{Ax:x \in \operatorname{dom}(T)\} \subseteq\operatorname{dom}(A^*)$, but how can I see this or is this completely wrong?

2 Answers2

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$\text{dom}(A) = \text{dom}(T)$ is wrong.

By the Spectral Theorem, you can essentially assume $T$ is multiplication by the variable $x$ on $L^2(\mu)$ for some positive measure $\mu$ on $[0,\infty)$, with $\text{dom}(T) = \{f \in L^2(\mu): x f \in L^2(\mu)\}$. Then you want $A$ to be multiplication by $\sqrt{x}$, with $\text{dom}(A) = \{f \in L^2(\mu): \sqrt{x} f \in L^2(\mu)\}$.

Robert Israel
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  • thank you. do you have a reference, cause I would like to read a little about this by myself? –  Dec 11 '14 at 00:38
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    You might try Reed and Simon, "Methods of Modern Mathematical Physics I: Functional Analysis". – Robert Israel Dec 11 '14 at 05:07
  • Hi Robert, I wonder if your answer can be adapted to say something about the following situation. Suppose $T = D^2+f$, where $D$ is a self-adjoint first-order differential operator and $f$ is multiplication by a non-negative function with compact support. Then $T:L^2\rightarrow L^2$ is an unbounded operator with unique positive square root. Then is it true that dom$(D)$ = dom$(\sqrt{D^2+f})$? – geometricK Dec 12 '17 at 23:58
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The spectral theorem for an unbounded selfadjoint operator allows you to represent $T$ as $$ Tx = \int_{0}^{\infty}\lambda dE(\lambda)x,\\ \mathcal{D}(T) = \left\{ x \in H : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty \right\}. $$ The unique positive square root of $T$ is $$ \sqrt{T}y = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)y,\\ \mathcal{D}(\sqrt{T})=\left\{ y \in H : \int_{0}^{\infty}\lambda d\|E(\lambda)y\|^{2} < \infty\right\}. $$ This operator $\sqrt{T}$ is selfadjoint on its domain. In terms of domains $$ x \in \mathcal{D}(T) \iff x \in \mathcal{D}(\sqrt{T}) \mbox{and} \sqrt{T}x\in\mathcal{D}(\sqrt{T}). $$ Note: I apologize for rolling back another user's edit. I understand that some do not like the notation I am using for the Spectral Theorem, but it has been standard since John von Neumann first stated and proved the Spectral Theorem for unbounded selfadjoint operators on a Hilbert Space. John von Neumann was the first to characterize the domain of $T$ as I stated it here; one can view the integral definition for $Tx$ as an improper integral, as one would for the Riemann integral. This notation has been standard for roughly 80 years, including von Neumann's use of $E$ for the spectral measure, motivated by the Schrodinger equation $E\psi=H\psi$.

Disintegrating By Parts
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  • actually, there is something strange about this. Let's take $T(f)=-f''$ then you can write $A(f)=f'$ and $A^(f)= -f'$ , so $T(f)=A^A(f)$, but $A$ is not self-adjoint, where does that come from? –  Dec 11 '14 at 10:53
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    @TobiasHurth : Or, $A(f) = -if'$ is selfadjoint and $T=A^{2}$. There are lots of roots but only one positive selfadjoint root. The positive one is great for Math, but not so great for Physics. A 'causal' root works better for Quantum, where support of the function is preserved under the action of the operator. – Disintegrating By Parts Dec 11 '14 at 12:26
  • very right, this is actually what I am looking for, in particular: If there is a positive solution $\psi>0$ to $H \psi= -\psi''+V\psi= \lambda \psi$, then we can define the operators $A = - f' + \phi$ and $A^* = f'+\phi$ for $\phi = \frac{\psi'}{\psi}$ such that $H = AA^$. Now my question would be: What do we know about the domains of $A$ and $A^$? –  Dec 11 '14 at 13:06
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    @TobiasHurth : I'm a little confused by your operator. At first I thought you were defining a linear $A$, but you're not, are you? – Disintegrating By Parts Dec 11 '14 at 13:44
  • http://mathoverflow.net/questions/190478/domains-of-raising-and-lowering-operators-in-qm –  Dec 11 '14 at 14:18
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    @TobiasHurth : Your previous comment here didn't make sense to me. Now, with more info, I see how you are defining the operators. I'm curious to see what will pop up at MathOverflow. The question is getting +'d which is a good sign. Nice question. – Disintegrating By Parts Dec 11 '14 at 19:01
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    @TobiasHurth : I see that you added on MathOverflow that your operator is defined on a bounded domain and $V$ is smooth on that domain. Then you need to specify endpoint conditions to make your operator selfadjoint. Part of choosing the domain of $A$ has to involve those conditions. – Disintegrating By Parts Dec 11 '14 at 21:04
  • yes, $H$ should definitely have boundary conditions, but they can be very different ranging from periodic conditions, as for $-f''$ on $[0,2\pi]$, to whatever. Despite, I don't think that we necessarily always need boundary conditions, as we still might be in the limit point case, don't you think so?- I mean for example $\frac{1}{\sin(x)}$ is $C^{\infty}$ on $(0,\pi)$, still this potential would not be regular. –  Dec 11 '14 at 21:44
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    @TobiasHurth : When you say $\mathcal{C}^{\infty}$, you should make it clear that this is over the interior of an interval. Without conditions, you can't know the domain of $T$, which means that you can know the domain of $A$. When you write $T=A^{\star}A$, that's an operator equation where $A^{\star}$ is an adjoint. Some conditions may rule out such a relation because $A^{\star}A \ge 0$, and $T \ge 0$ may not be true. – Disintegrating By Parts Dec 11 '14 at 22:39
  • somehow you are right. this question is probably to broad: I mean there are not so many conditions that admit a positive solution. The only prominent example that I am aware of are periodic boundary conditions. Do you see in that case for example which boundary conditions, we would need to take for $A$? So assume $T$ has periodic boundary conditions on a finite interval. What would this mean for $A$? I guess $\operatorname{dom}(A) = {f \in H^1: f(a)=f(b)}$ but I don't see what this would mean for $\operatorname{dom}(A^*)$. –  Dec 11 '14 at 22:45
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    @TobiasHurth : $((\partial+W)f,g)-(f,(-\partial+W)g)=(f',g)+(f,g')=\int_{a}^{b}(fg)',dx = fg|_{a}^{b}$. Looks like periodic for the domain of both $\partial+W$ and $-\partial +W$ works out just fine in terms of adjoints. The condition for $(-\partial+W)(\partial+W)f$ would be that $f$ and $(f'+Wf)$ are periodic. Ichh. That would work I suppose if $W$ is periodic because that then reduces to $f$, $f'$ being periodic, which is classical. But check these details for yourself. – Disintegrating By Parts Dec 12 '14 at 01:41
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    @T.A.E.: That's fine, also it looks cleaner this way. But are you missing an $E$ in $\int\sqrt{\lambda}d(\lambda)y$? – C-star-W-star Dec 12 '14 at 10:30
  • @T.A.E. $W$ is periodic, cause $W = \frac{\psi'}{\psi}$ and they are both periodic due to the boundary conditions of the Hamiltonian. But how do you get the domain of $(-\partial+W)$ if you take periodic boundary conditions for $(\partial+W)$? –  Dec 12 '14 at 10:41
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    @TobiasHurth : Look back up a couple of comments. – Disintegrating By Parts Dec 12 '14 at 16:07
  • @T.A.E. what do you mean, I see that you show that for $f$ or $g$ being periodic, both scalar products are the same, but I don't see how this implies that given $\operatorname{dom}(\partial+W)= { f \in AC(a,b): f(a)=f(b)}$, we have $\operatorname{dom}((\partial+W)^*) = \operatorname{dom}(-\partial+W)$ –  Dec 12 '14 at 17:10
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    @TobiasHurth : I'm assuming $\psi'/\psi$ is smooth on $[a,b]$, which seems warranted if $\psi > 0$ on $[a,b]$ (i.e., has no nodes.) Then the boundary functionals associated with $\pm\partial+W$ are endpoint evaluations only. So everything reduces to a 2d linear algebra in that case. – Disintegrating By Parts Dec 12 '14 at 19:03
  • @T.A.E. not sure if we talk about the same thing. From your calculation I could see that $(- \partial +W)$ is the adjoint to $( \partial + W)$. I also see that $\operatorname{dom}(\partial+W) \subset \operatorname{dom(-\partial + W)}$ holds. Now my question is: Why do we have equality here? Concerning your last comment: I don't know how functionals that evaluate functions at end-points are related to the domain issue? –  Dec 12 '14 at 19:18
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    @TobiasHurth: Imposing endpoint conditions on the unconstrained domain is how operator domains are defined. – Disintegrating By Parts Dec 13 '14 at 01:10
  • @T.A.E. sorry, I think you misunderstand me. If you choose a domain for $(\partial+W)$, then you automatically get a domain for $(-\partial+W)$. Now, you claimed that this domain for $(-\partial+W)$ is the same as the domain for $(\partial+W)$ and I am asking you, why is that the case? –  Dec 13 '14 at 12:30
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    @TobiasHurth : $\frac{1}{i}\partial$ is selfadjoint on absolutely continuous periodic functions on $[a,b]$. So $(\partial)^{\star}=-\partial$ on this same domain. It is easy to check that adding a bounded operator $W$ leaves $\pm \partial +W$ closed on the same domain with adjoints $\mp\partial + W$. The domain of $(-\partial+W)(\partial+W)$ consists of periodic absolutely continuous $f$ for which $f'+Wf$ is periodic and absolutely continuous. If $Wf$ leaves $f$ periodic, then $f'+Wf$ is absolutely continuous and periodic iff $f$ is twice a.c. with $f$, $f'$ periodic. – Disintegrating By Parts Dec 13 '14 at 15:49
  • @T.A.E. thank you very much, that was exactly what I wanted to know. –  Dec 13 '14 at 17:20
  • @TobiasHurth : Thanks for this problem; your problems are always interesting. I'm a little shocked that such a clean treatment is possible for the problem. – Disintegrating By Parts Dec 13 '14 at 18:52
  • @T.A.E. thank you, well they come from physics applications, where I try to apply spectral theory in order to understand this. Actually, I just asked a new question that is somehow related to: Can I make arbitrary substitutions in a differential equation and preserve the self-adjointness of the operator domain? http://math.stackexchange.com/questions/1066918/preserve-self-adjoint-properties –  Dec 13 '14 at 23:01
  • I would like to know how tricky is it to address the question of uniqueness here? Let's say $A : D(A) \to H$ is a densely defined, positive, self-adjoint operator. If $T : D(T) \to H$ is a densely defined, positive, self-adjoint operator with $T^2 = A$, is there only one such $T$? – JZS Jan 18 '23 at 21:48