# Computing a Projection Valued Measure

I've recently begun learning about Projection Valued Measure and I'm a little confused. I understand that a Projection Valued Measure is a family of orthogonal projections $P(\Lambda)$ indexed by the Borel measurable sets on a Hilbert Space with some caveats ($P(0)=0$, $P(\mathbb{R})=1$, etc.) I also understand that for a self-adjoint operator, we have $P(\Lambda)=\chi_{\Lambda}(H)$ where $\chi$ is the characteristic function on the Borel set $\Lambda$.
In one of my questions, I was given a matrix that was just a linear combination of Pauli matrices and asked to find the projection valued measure of it. There was no mention of Borel sets or anything, so I became a little confused.
I found that for Hermitian matrices, we can use functional calculus to find $$f(H)=\sum\limits_{i} f(E_i)P_i$$ Where $f$ is a Borel function, $E_i$ the eigenvalues of $H$ and $P_i$ the corresponding projections into the eigenspaces. Is this the same as the Projection Valued Measure for a $n\times n$ Hermitian matrix?

EDIT: I think I might have figured it out. Wouldn't the projection valued measure just be $$P(\Lambda)=\sum\limits_{\{i|E_i\in\Lambda\}} P_i$$ Where $E_i$ are the eigenvalues and $P_i$ the projections into the eigenspaces? Again, no Borel sets are defined, but we are working in $\mathbb{C}^2$.

Typically, one has a locally compact or compact topological Hausdorff space $\Omega$, and the projection measure is defined on the Borel subsets of $\Omega$. In the context of the Spectral Theorem, $\Omega$ is the closed subset of $\mathbb{C}$ which is the spectrum of a normal operator $N$, but the abstraction to a more general space can be useful, especially when dealing with operator algebras.
In your case, the spectrum is discrete, and you'll use the subspace topology inherited from $\mathbb{C}$, which is the discrete topology. This is a compact topology. The projection measure is as you stated for this case: $E\{ \lambda\}$ is the orthogonal projection onto the eigenspace with eigenvalue $\lambda$ for the hermitian matrix $H$. Then $\int \lambda dE = H$.
• Thank you, I'm starting to get it. I'm finding this material kind of hard to follow. I'm having trouble visualizing what is going on in my head. For example, what does $$\int_{\mathbb{R}} \lambda dE$$ look like if you were to write it out explicitly? Also, is there a way to express the PVM in terms of eigenfunctions since the spectrum is discrete? – TinaBelcher Nov 19 '14 at 2:00
• For a Hermitian matrix with distinct eigenvalues $\lambda_{j}$, you get $\int_{\mathbb{R}}\lambda dE = \sum_{j}\lambda_{j}E\{\lambda_{j}\}$ because $E$ is a discrete measure. $E\{\lambda\}$ is $0$ if $\lambda$ is not one of the $\lambda_{j}$ and $E\{\lambda_{j}\}$ is the orthogonal projection onto the eigenspace associated with eigenvalue $\lambda_{j}$, which is the null space $\mathcal{N}(H-\lambda_{j} I)$. If you have an orthonormal basis of eigenvectors $e_{\lambda_{j},k}$ where $1 \le k \le n_{j}$, then $(E\{\lambda_{j}\})x=\sum_{k=1}^{n_{j}}(x,e_{\lambda_{j},k})e_{\lambda_{j},k}$. – DisintegratingByParts Nov 19 '14 at 2:52
• Hi again, thanks very much for the explanation for the discrete case. I'm now having some trouble with the continuous case though. For example, how would one calculate the PVM for a self adjoint operator such as $-i\partial_x$ on $H^1(\mathbb{R})$? The spectrum is the whole real line, so I'm confused as to how to calculate the measure. Would it just be the projection of the Hilbert space onto the range of the operator? – TinaBelcher Nov 26 '14 at 4:18
• @TinaBelcher : This is a little long-winded, but I show how to derive the $L^{2}$ theory of the Fourier transform using the Spectral Theorem for $-i\partial_{x}$, without assuming anything about the existence or properties of the Fourier transform. math.stackexchange.com/questions/1007071/… – DisintegratingByParts Nov 26 '14 at 4:25
• @TinaBelcher : Thank you. I really wanted that basic problem of $-i\partial_{x}$ to be out there where people could see how to constructively obtain Fourier transform theory from the Spectral Theorem. So much of this classical theory is being lost, and that's a shame. These techniques work for a lot of classical differential operators of Math/Physics, especially when there's a Green function solution. The spectral objects are constructively recovered from the jump discontinuity of the resolvent across the spectrum, both in the case of discrete residues and in the case of continuous spectrum. – DisintegratingByParts Nov 26 '14 at 5:46