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This is the exercise 6 of chapter 8 of Hall's book: Quantum Theory for Mathematicians.

Suppose $A\in\mathcal{B}(H)$ is self-adjoint and $v$ is an eigenvector for $A$ with $Av = cv$ for some $c\in\mathbb{R}$. Then for any bounded and measurable function $f$ on the spectrum $\sigma(A)$ we have $f(A)v = f(c)v$

where $H$ is a Hilbert space over $\mathbb{C}$ and $f(A)$ is an operator defined as

$$f(A) = \int_{\sigma(A)}f(c)d\mu^A(c)$$

with $\mu^A$ being such a unique projection-valued measure on the Borel $\sigma$-algebra in $\sigma(A)$ with values in projections on $H$ such that

$$\int_{\sigma(A)}cd\mu^A(c) = A$$

A given hint is to use the prior exercise 5:

Suppose $A\in \mathcal{B}(H)$ is self-adjoint operator and $V$ is a closed subspace of $H$ that is invariant under $A$. Then a.) the spectrum of the restriction to $V$ of $A$ is contained in the spectrum of $A$ b.) if $f$ is a bounded measurable function on $\sigma(A)$ then $V$ is invariant under $f(A)$ and $f(A)\mid_V = f(A\mid_V)$.

Thoughts/problem: My problem is that I don't really know how to apply $f(A)$ to $v$ or any element $w\in H$ for that matter. Connection to exercise 5 is probably by the span of $v$. Namely, define $S:=\{\alpha v\mid \alpha\in\mathbb{C}\}$. It is not hard to show that the span of $v$, $S$, is a closed subspace of $H$ invariant under $A$. Then if we can determine what $f(A)w$ for $w\in H$ is in terms of $w$, the result may follow quite easily. But I am lost even on what $\chi_E(A)\psi := \int_{\sigma(A)}\chi_E(c)d\mu^A(c) = \mu^A(E)\psi$ should be for a measurable $E\subset\sigma(A)$, let alone how to move the limit outside the integral when transition from simple functions to measurable functions.

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    $\begingroup$ The quantity $f(c)$ is not well defined. $\endgroup$ Commented May 25, 2023 at 12:18
  • $\begingroup$ @RyszardSzwarc How so? $\endgroup$ Commented May 25, 2023 at 12:43
  • $\begingroup$ The function $f$ is measurable. If it was continuuos or monotonic left continuous then the point values would be well defined. $\endgroup$ Commented May 25, 2023 at 15:28
  • $\begingroup$ The evaluation is well defined! Here, $f$ is an actual function and not just an element of $L^\infty.$ $\endgroup$
    – Stewan
    Commented May 30, 2023 at 23:56

2 Answers 2

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Another approach:

Let $V$ be the span of the eigenvector $v.$ To apply the other exercise, note that $\sigma(A\vert_V)=\{c\},$ then: $$f(A)\vert_V=f(A\vert_V)=\int_{\sigma(A\vert_V)}f(\lambda)d\mu^{A\vert_V}(\lambda)=f(c)\int_{\sigma(A\vert_V)}1d\mu^{A\vert_V}(\lambda)=f(c)\mu^{A\vert_V}(\mathbb{R})=f(c)\operatorname{id}_V.$$

The first equality is the other exercise; the second is the spectral theorem for $A\vert_V;$ the third follows since the spectrum consists only of the single point $c,$ so $f$ is constant with value $f(c)$ on $\sigma(A\vert_V)$; the last equality is a property of the projection-valued measure $\mu^{A\vert_V}.$

Now apply both sides of $f(A)\vert_V=f(c)\operatorname{id}_V$ to $v$ to conclude $f(A)v=f(c)v.$

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Based on my limited understanding, here is my answer,

I presume or think that: Let $V_{E_j} = <\{v:Av = qv, q \in E_{j}\}>.$ Since $ f(A \restriction_V ) = f(A)\restriction_V \implies \int_{E_{j}^c} f(c) d\mu_A(c) w = 0$ for all $w \in V_{E_j}.$

Now, $$f(A) = \int f(c) d\mu_A(c)$$ and let $$\int_{E_{j}^c} f(c) d\mu_A(c) + f(u_j) \mu_A(E_j) \rightarrow \int f(c) d\mu_A(c)$$ as $j \rightarrow \infty.$

Now by continuity of measure and for now assuming continuity of $f$, $$ f(u_j) \mu_A(E_{j}) v \rightarrow f(c) \mu_A(c) v.$$

as $j \rightarrow \infty.$

where $E_j = (c-\epsilon_j,c+\epsilon_j)$, $\epsilon_j \rightarrow 0$, $u_j = c \in E_j$ for all $j.$

For $e \notin E_j$ for all $j \geq J$ with $Av_e = e v_e$, since $\int_{E_{j}^c} f(c) d\mu_A(c)v_e = \int f(c) d\mu_A(c)v_e = f(A) v_e$, we have that, $$ f(u_j) \mu_A(E_{j}) v_e \rightarrow 0.$$

But $f(u_j) \mu_A(E_{j}) v_e \rightarrow f(c) \mu_A(c) v_e.$ Hence $f(c) \mu_A(c) v_e = 0$

By choosing $f(x) = 1$ for all $x$, we have that $$\mu_A(c) v_q = 0$$ for all $q \neq c$ and $Av_q = qv_q.$

Now by choosing $f(x) = x$ for all $x$, we have that, $c\mu_A(c)v = Av = cv \implies$ $$\mu_A(c)v = v.$$ Here in the above i assumed $c \neq 0$ because otherwise $\mu_A(c)$ may not be well defined by the integral definition $\int c d\mu_A = A.$

Now if $f$ is measurable and not continuous, we now choose $u_j \rightarrow f(c)$ and $E_j = f^{-1}((f(c)-\epsilon_j,f(c)+\epsilon_j))$, $\epsilon_j \rightarrow 0.$ Now by continuity of measure, by same argument as above with above choice of $E_j$, $$ u_j \mu_A(E_{j}) v \rightarrow f(c)\mu_A(f^{-1}(f(c)))v = f(c) \sum_{\{c_i: f(c_i) = f(c)\}} \mu_A(c_i) v = f(A)v.$$

But then $\mu_A(c_i)v = 0$ for all $c_i \neq c.$ Hence, we have

$$ u_j \mu_A(E_{j}) v \rightarrow f(c)\mu_A(f^{-1}(f(c)))v = f(c) \sum_{\{c_i: f(c_i) = f(c)\}} \mu_A(c_i) v = f(c) \mu_A(c)v = f(c)v= f(A)v.$$

Now if $\{c_i: f(c_i) = f(c)\}$ is not a discrete set, you need to use continuity of $f(x) = 1$ to conclude:

$$f(c) \mu_A(f^{-1}(f(c)))v = f(c) \int_{\{c_i: f(c_i) = f(c)\}} d\mu_A v = f(c) \mu_A(c)v = f(c)v = f(A)v.$$

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  • $\begingroup$ Sorry for the late response. I have few questions: 1.) Why does $f(u_j)\mu_A(E_j)\to 0,j\to\infty$?, 2.) What is $u_j$? $\endgroup$ Commented May 29, 2023 at 13:18
  • $\begingroup$ $u_j = c$ (written in the answer) and $f(u_j) \mu_A(E_j) \rightarrow f(x) \mu_A(c)$ by continuity of measure. $\endgroup$
    – Balaji sb
    Commented May 30, 2023 at 2:19
  • $\begingroup$ Ah, so you have taken $\left(u_j\right)_{j}$ to be a constant sequence with $u_j = c$ for all $j$? $\endgroup$ Commented May 30, 2023 at 7:01
  • $\begingroup$ yes. $u_j$ is constant for all $j$. $\endgroup$
    – Balaji sb
    Commented May 30, 2023 at 10:46
  • $\begingroup$ Similar question: math.stackexchange.com/questions/4709333/… $\endgroup$
    – Balaji sb
    Commented May 30, 2023 at 13:19

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