I'm following some of the lectures of professor Schuller. I'm watching this video specifically

At about minute 23 he defines the PVM (Projected Valued Measure) as a map

$$ P_A : \sigma(\Theta_\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H}) $$

with the specific expression

$$ \langle\psi,P_A(\Omega)\phi\rangle=\int \chi_\Omega d\mu_{\psi,\phi}^A \;\;\forall\psi, \phi \in \mathcal{H} $$

Here $\mathcal{H}$ is an hilbert space, $A$ is a self-adjoint operator in $\mathcal{H}$, while $\mu_{\psi,\phi}^A$ is a complex valued measure constructed by polarization of a real valued measure $\mu_{\phi}^A$ (these concepts are expressed a bit earlier in the same lecture, but all the constructions are given later).

My question is why the PVM defined above does represent an operator when evaluated in a measurable set? suppose indeed a self-adjoint operator $A$ is given, and we also fix a measurable set $\Omega$, the PVM seems to me to take two arguments and not just one.

Schuller also says that the map is somehow given by that expression, so there's probably something I'm missing.

Can you help me in interpreting that expression?


I've watched the whole video, and I guess my question may be explained as follows. Suppose we are in a vector space of dimension $n$ embedded with a dot product. The dot product is characterized by a matrix $A \in \mathbb{R}^{n\times n}$. We can recover such matrix by knowing how the dot product acts on a basis. By taking the canonical base $V = \left\{e_1,\ldots,e_n \right\}$ (columns vectors), we can recover the element $a_{ij}$ as

$$ \langle e_i,e_j \rangle = e_i^T A e_j = a_{ij} $$

If the base is not the canonical one we can perform a transformation and still retrieving the matrix (which again is actually an operator). Hence for finite dimension the dot product is fully characterized once we now how it acts on a basis. (I haven't been very precise, but you've got my point I hope).

The expression

$$ \langle\psi,P_A(\Omega)\phi\rangle=\int \chi_\Omega d\mu_{\psi,\phi}^A \;\;\forall\psi, \phi \in \mathcal{H} $$

it seems to me be somehow related, therefore I was wondering if there's any theorem that characterize self-adjoints operator in a similar fashion as matrices and dot products in finite dimensions. Name of a theorem, any reference, or if you could prove it for me would be really useful.

  • $\begingroup$ The notation are unclear to me. Maybe you can relate it to this and this ? $\endgroup$
    – reuns
    Commented Aug 9, 2017 at 10:55
  • $\begingroup$ @reuns What is unclear specifically? $\endgroup$ Commented Aug 9, 2017 at 11:01
  • $\begingroup$ The RHS. You need to make it clear how it is a (bi)linear operator in $\psi,\phi$. Did you look at this explaining everything ? $\endgroup$
    – reuns
    Commented Aug 9, 2017 at 11:11
  • $\begingroup$ @reuns I still don't understand. $\psi,\phi$ are taken from the same Hilbert space, hence that is the dot product in that space I suppose. Regarding the link it seems to me the same stuff, namely it seems to me that Schuller in that video proves the theorem in your link. $\endgroup$ Commented Aug 9, 2017 at 11:56
  • 1
    $\begingroup$ Just out of curiosity, do these theorems actually help in practice? Looking at those lectures, which again I'm able both to follow and understand, I haven't actually seen it in pratice to decompose a self adjoint operator, I can understand now what it means. Unfortunately I can find lecture notes or problems of this professor, but I would help to find some examples where those notions are actually used (I'm not a physicist so I don't I'd be able to follow the applications he provides later). $\endgroup$ Commented Aug 9, 2017 at 13:14

1 Answer 1


Theorem [Bounded Form Representation]: Let $b(x,y)$ be complex function on a complex Hilbert space $\mathcal{H}$. Suppose $b$ is linear in $x$ and conjugate linear in $y$, and further suppose there is a constant $K$ such that $|b(x,y)| \le K\|x\|\|y\|$ for all $x,y\in\mathcal{H}$. Then there is a unique bounded linear operator $B : \mathcal{H}\rightarrow\mathcal{H}$ such that $b(x,y)=\langle Bx,y\rangle_{\mathcal{H}}$ for all $x,y\in\mathcal{H}$; furthermore, $B$ is bounded by the same $K$, which is to say that $\|Bx\| \le K\|x\|$ for all $x\in\mathcal{H}$.

Proof: Assuming $b$ is as stated, then, for every fixed $y$, the function $\Phi_{x}(y)=\overline{b(x,y)}$ is a bounded linear functional on $\mathcal{H}$. By the Riesz Representation Theorem, there exists a unique $Bx\in\mathcal{H}$ such that $$ \overline{b(x,y)} = \langle y,Bx\rangle_{\mathcal{H}},\;\;\; y\in\mathcal{H}. $$ Therefore, $$ b(x,y) = \langle Bx,y\rangle_{\mathcal{H}},\;\;\; x,y\in\mathcal{H}. $$ The function $B : \mathcal{H}\rightarrow\mathcal{H}$ is a linear operator on $\mathcal{H}$. To see why, first consider that the following holds for all $x,x',y\in\mathcal{H}$: \begin{align} \langle Bx+Bx',y\rangle_{\mathcal{H}}&=b(x,y)+b(x',y) \\ &=b(x+x',y) \\ &=\langle B(x+x'),y\rangle_{\mathcal{H}}. \end{align} Hence $B(x+x')=Bx+Bx'$ for all $x,x'\in\mathcal{H}$. Similarly, for all $x\in\mathcal{H}$, $y\in\mathcal{H}$, $\alpha\in\mathbb{C}$, \begin{align} \langle B(\alpha x),y\rangle_{\mathcal{H}}&=b(\alpha x,y) \\ &=\alpha b(x,y) \\ &=\alpha \langle Bx,y\rangle_{\mathcal{H}} \\ &=\langle \alpha Bx,y\rangle_{\mathcal{H}}. \end{align} Hence $B(\alpha x)=\alpha Bx$ holds for all $\alpha\in\mathbb{C}$, $x\in\mathcal{H}$. Therefore, $B$ is a linear operator on $\mathcal{H}$. And $B$ is bounded because of the boundedness of $b$: \begin{align} \|Bx\| & =\sup_{\|y\|=1}|\langle Bx,y\rangle| \\ & = \sup_{\|y\|=1}|b(x,y)| \\ & \le \sup_{\|y\|=1} K\|x\|\|y\| \\ & = K\|x\|.\;\; \blacksquare \end{align}


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