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I'm trying to construct a simple explicit example of a 'spectral measure'. Here is the definition I have.

Let $A$ be a bounded self-adjoint opreator. Let $\psi \in H$ where $H$ is a Hilbert space. Then $f\to (\psi,f(A)\psi)$ is a positive linear functional on $C(\sigma(A))$. Thus by the Riesz-Markov theorem there exists a unique measure $\mu_\psi$ on the compact set $\sigma(A)$ such that $$(\psi,f(A)\psi) = \int_{\sigma(A)} f(\lambda)d\mu_\psi.$$ $\mu_\psi$ is the spectral measure associated with $\psi$.

So take $H = \mathbb{R}$. Let $f(x) = x^2$ and let $$A=\begin{pmatrix}3 & 0 \\ 0 & 1\end{pmatrix}\quad \text{and} \quad \psi = \begin{pmatrix}1 \\ 1\end{pmatrix}, $$ which means $\sigma(A) = \{3,1\}$. Then

$$ \begin{align} (\psi,f(A)\psi) & = (\begin{pmatrix}1 \\ 1\end{pmatrix},\begin{pmatrix}3 & 0 \\ 0 & 1\end{pmatrix}^2\begin{pmatrix}1 \\ 1\end{pmatrix}) \\ & = (\begin{pmatrix}1 \\ 1\end{pmatrix},\begin{pmatrix}9 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}) \\ & = 9 + 1 = 10. \end{align} $$

So we should have $$ \int_{\sigma(A)} f(\lambda)d\mu_\psi(\lambda) = \int_{\{3,1\}} \lambda^2 d\mu_\psi(\lambda) = \text{something...} = 10. $$

But what is this 'something'? To be precise, what is an explicit expression for $\mu_\psi$ and then how do I interpret an integral over the discrete set $\{3,1\}$ with respect to this measure $\mu_\psi$?

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The integral is (by definition of Lebesgue integral) $$\tag1 \int_{\{3,1\}}f(\lambda)\,d\mu(\lambda)=f(3)\,\mu_\psi(\{3\})+f(1)\,\mu_\psi(\{1\}). $$ This is supposed to agree with $$\tag2 \langle \psi,f(A)\psi\rangle=f(3)+f(1). $$ As $(1)$ and $(2)$ are agree for all $f$, we get $\mu_\psi(\{3\})=\mu_\psi(\{1\})=1$.

In this example, because the form of $A$ is so simple, one can calculate the spectral measure and thus obtain $d\mu_\psi$ for any $\psi$. Because $$ f(A)=f(3)E_{11}+f(1)E_{22}=\int_{\sigma(A)} f(\lambda)\,d\mu, $$ where $\mu$ is the spectral measure $\mu(\{3\})=E_{11}$, $\mu(\{1\})=E_{22}$. And then $$\tag3 \mu_\psi(R)=\langle \psi,\mu(R)\psi\rangle $$ for any Borel set $R\subset \sigma(A)$ (in this case the only possibilities are $\varnothing$, $\{1\}$, $\{3\}$, $\{1,3\}$).

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  • $\begingroup$ I thought $\mu_\psi$ was the spectral measure, but here you say $\mu_\psi$ is a function of the spectral measure $\mu$ and the vector $\psi$. So what is this $\mu_\psi$ as you have defined it? $\endgroup$ – eurocoder May 23 '18 at 6:13
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    $\begingroup$ I just noticed that I missed several $\psi$ at the beginning. My $\mu_\psi$ is your $\mu_\psi$; it is a (usual) measure, that depends on $\psi$. "The" spectral measure is projection valued; that's my $\mu$. Try to read carefully and distinguish which equations are made up of numbers, and which are made up of operators (matrices, in this case). $\endgroup$ – Martin Argerami May 23 '18 at 6:28
  • $\begingroup$ The reference I'm using is Reed and Simon Vol I and they call the measure $\mu_\psi$ the 'spectral measure associated with the vector $\psi$ (page 225). I just checked the book again now and it seems they introduce 'projection valued measures' on page 235 (which I haven't reached yet) - I assume that is what you are referring to. So it seems there are two notions of 'spectral measures' which is confusing..do you know if there is a connection between 'spectral measure' and 'projection valued measure' as they are defined in Reed and Simon? $\endgroup$ – eurocoder May 23 '18 at 6:38
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    $\begingroup$ Yes, that's equality $(3)$ in my answer. It is true that theoretically one constructs the $\mu_\psi$ first, and then use them to define $\mu$ via $(3)$. But I would say that the projection valued spectral measure is "the" spectral measure. $\endgroup$ – Martin Argerami May 23 '18 at 6:46
  • $\begingroup$ Ok cheers, I will come to review this answer once I get to spectral projection valued measures! $\endgroup$ – eurocoder May 23 '18 at 8:38
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In this case, the spectral measure is discrete, with atoms at $3$ and at $1$. $\mu\{3\}$ is the orthogonal projection onto the eigenspace associated with eigenvalue $3$, and similarly for $\mu\{1\}$. So $$ \mu\{1\}x = \mu\{1\}\left[\begin{array}{c}x_1\\x_2\end{array}\right]=x_2\left[\begin{array}{c}0 \\ 1\end{array}\right], \\ \mu\{3\}x = \mu\{3\}\left[\begin{array}{c}x_1\\x_2\end{array}\right]= x_1\left[\begin{array}{c}1 \\ 0\end{array}\right]. $$ Then, $$ \int f(\lambda)d\mu(\lambda)x=f(1)\mu\{1\}x+f(3)\mu\{3\}x. $$ For example, $$ \int \lambda^2 d\mu(\lambda)x=1^2\mu\{1\}x+3^2\mu\{3\}x = \left[\begin{array}{cc}3^2 & 0 \\ 0 & 1^2\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right] $$

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  • $\begingroup$ I think there is a typo in the first part - $x_1$ and $x_2$ - should be switched as '3' is associated with the $x_1$ element? Anyway, its a very interesting answer, it seems you are saying that the spectral measure $\mu$ is a projection matrix in this case? I.e. for eigenvalue $3$, $$\mu\{3\} = \begin{pmatrix}1&0\\ 0 & 0\end{pmatrix}.$$ But a measure should be a real number? In Martin's answer he has $\mu=1$. Also, $\int_{\sigma(A)}f(\lambda)d\mu_\psi x=(\psi,f(A)\psi) x$ and $(\psi,f(A)\psi)x$ is simply a scalar times a vector, so I don't see how it can act differently on each element of $x$? $\endgroup$ – eurocoder May 23 '18 at 5:59
  • $\begingroup$ @eurocoder : Yes, the $x_1,x_2$ were swapped and that is fixed now. The spectral measure is a an operator valued measure. The quadratic form of that measure is the measure you are using: $d\mu_{\psi}=\langle d\mu \psi,\psi\rangle$. $\endgroup$ – DisintegratingByParts May 23 '18 at 10:49

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