I've read the following proposition in a book on quantum theory.

Proposition. If a quantum system is in a state described by a unit vector $\psi$ and for some quantum observable $\hat{f}$ we have $\hat{f}\psi=\lambda \psi$ for some $\lambda\in \mathbb{R}$, then

(1)$$E(f^m)=\langle \psi, (\hat{f})^m\psi\rangle=\lambda^m$$ for all positive integers $m$.

(2) The unique probability measure consistent with this condition is the one in which $f$ has the definite value $\lambda$, with probability one.

Part (1) is clear to me. To explain and prove (2) the author writes

"...we want to find a probability measure $\mu$ on $\mathbb{R}$ such that $$\int_\mathbb{R} x^m\, d\mu=\lambda^m,$$ for all non-negative integers $m$. The proposition is claiming that there is one and only one such measure, namely the $\delta$-measure at the point $\lambda$.''

I'm not sure how the integral is formed and why it does not explicitly depend on $f$. Prior to reading this paragraph I was thinking that for a function $g$, the expectation value is computed by $E(g)=\int_{\mathbb{R}} g\, d\sigma$ where $\sigma$ is a suitable probability measure.

[Edit: According to the spectral theorem, there exists a unique spectral measure $\mu_{\hat{f}}$ such that $$\hat{f}=\int_{\sigma(\hat{f})}x\, d\mu_{\hat{f}}(x),$$ and $$\langle \phi, \hat{f}\phi\rangle=\int_{\sigma(\hat{f})}x\, d\mu(x),$$ where $\mu(E)=\langle \phi, \mu_{\hat{f}}(E)\phi\rangle$. Thus, I guess the author's explanation of part 2 is implicitly based on the spectral theorem (and functional calculus), although the spectral theorem appears much later in the book. Q: Is there any way to get the integral in question without resorting to the spectral theorem?]

Could someone, please, clarify this confusion about the definition/computation of the expectation value via integration against a probability measure?

I would like to edit my question as follows:

a) Isn't the proposition claiming the uniqueness of a measure $\sigma$ satisfying $E(f)=\int_\mathbb{R} f\, d\sigma$ rather than the uniqueness of $\mu$? (It's not a 100 percent clear to me that these two are equivalent, although I've not thought about it carefully yet.)

b) Since the author proves the uniqueness of $\mu$ (by referring to a theorem about moment problems in probability theory) I wonder if the uniqueness of $\mu$ can be derived from the uniqueness of the spectral measure $\mu_{\hat{f}}$ associated to $\hat{f}$.

  • 1
    $\begingroup$ You have a typo $\lambda > \psi$ in the first quote. This part of the book is here. I will try to explain this tomorrow. $\endgroup$ Aug 19, 2017 at 4:54
  • $\begingroup$ I spent a few minutes googling axioms of QM and they were similar to what I remember. His Axiom 3 is strange to me. I guess he has his reasons for this approach, but I can't guess why from the sample chapters. I can't see where he defines "the probability distribution for the measurement of some observable" - it seems like Axiom 3 is the definition. But given his definitions (as I understand them), the application of the moment theorem seems straightforward. $\endgroup$ Aug 20, 2017 at 3:18
  • $\begingroup$ There are unique spectral measures associated with some symmetric operators (those that are self-adjoint) and unique measures associated with some, not all moment sequences. (Spectral theory was built on concepts from the moment problem.) If $A$ is self-adjoint I'm not sure whether the moment sequence $(\psi, A^n \psi)$ has a unique measure for all $\psi \in C^\infty (A)$. $\endgroup$ Aug 20, 2017 at 3:48
  • $\begingroup$ The answer to the last statement is no. Krein showed that there is a positive $w$ such that sequence $\int x^n w(x)dx$ does not have a unique measure. Thus, if $A$ is multiplication by $x$ and $\vert\psi(x) \vert^2 = w(x)$ the statement fails. So the uniqueness of the spectral measure does not imply the uniqueness of the probability measure. $\endgroup$ Aug 21, 2017 at 3:49


You must log in to answer this question.

Browse other questions tagged .