I'm a math student and a totally newcomer to quantum mechanics and I'm trying to teach myself this subject by studying Faddeev and Yakubovski's Lectures on Quantum Mechanics for Mathematics Students.

The following paragraph appears on pp. 49-50 of the book.

For any self-adjoint operator there is a spectral function $P_A(\lambda)$, that is, a family of projections with the following properties:

  1. $P_A(\lambda)\le P_A(\mu)$ for $\lambda<\mu$, that is, $P_A(\lambda)P_A(\mu)=P_A(\lambda)$;
  2. $P_A$ is right-continuous;
  3. $P_A(-\infty)=0$ and $P_A(\infty)=I$;
  4. $P_A(\lambda)B=BP_A(\lambda)$ if $B$ is any bounded operator commuting with $A$.

A vector $\phi$ belongs to the domain of the operator $A$ if $$\int_{-\infty}^\infty \lambda^2\, d(P_A(\lambda)\phi, \phi)<\infty,$$ and then $$A\phi=\int_{-\infty}^\infty \lambda\, dP_A(\lambda)\phi.$$

I've seen the construction of $P_A(\lambda)$ for the finite-dimensional case as $\theta(\lambda-A)$ where $\theta$ is the Heaviside function. However,

  1. I wonder whether the reason that the authors say "there is a spectral function $P_A(\lambda)$" without introducing it explicitly in the infinite-dimensional case is because of the complexity of the derivation/formula or the fact that the existence of $P_A$ with the above 4 properties is just an "axiom?"
  2. Are the integrals merely consequences of items 1--4? (The book does not explain the meaning of $d(P_A(\lambda)\phi, \phi)$ and $dP_A(\lambda)\phi$ in integrations, and I'm not even sure if I get them correctly.)

Any help with clarifications is greatly appreciated!

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    $\begingroup$ Probably you would appreciate a somewhat more mathematically oriented approach (by the way there are a number of ways to treat the spectral decomposition problem). I would suggest to have a look at the four volumes "Methods of modern mathematical physics" by Read and Simon and also "Perturbation btheory of linear operators" by Kato. $\endgroup$ – Urgje Mar 14 '16 at 20:32
  • $\begingroup$ Just to make this absolutely clear... what you are learning there is not quantum mechanics but operator theory. The math above is of almost zero consequence to the actual physics. $\endgroup$ – CuriousOne Mar 14 '16 at 20:44
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    $\begingroup$ @Urgje Good recommendations, but can I also add that there's a nice book called "Introduction to Hilbert Space and the Theory of Spectral Multiplicity" by Halmos, which talks about the idea of spectral measures. $\endgroup$ – snulty Mar 15 '16 at 13:56
  • $\begingroup$ Yes, I actually have it. It is rather high level, though. $\endgroup$ – Urgje Mar 15 '16 at 20:32

This has nothing to do with physics, but it's pure functional analysis.

What you have here is nothing but a slightly convoluted version of the spectral theorem, which is formulated for all self-adjoint operators (bounded or unbounded). It's slightly convoluted, because it includes the necessary definitions of spectral measure in the theorem itself - I'd usually see them defined beforehand.

To understand how this theorem comes about is nontrivial, because the spectrum of a (possibly un)bounded operator in infinite dimensions contains parts that are not eigenvalues (continuous spectrum). The way you construct $P$ and define $\int dP$ at least for bounded operators is the same as you do it for arbitrary measures in measure theory and/or probability theory. For unbounded operators, you need to take heed with domain issues.

Remember probability theory? If you have a discrete measure $p_i$, an expectation value of a random variable $X$ is nothing else but $\sum_i X_ip_i$. If you have a continuous measure $\mu$, the expectation value will be given by something like $\int Xd\mu$. It's the same here: Given a self-adjoint matrix (or a compact operator), the spectral theorem reads $Av=\sum_i \lambda_i \langle v_i,v\rangle v_i$ where $v_i$ are the eigenvectors and $\lambda_i$ the eigenvalues. Your last expression for $A\phi$ is the continuous counterpart. Another complication arising here is that your measures are not real or complex-valued, but operator-valued in contrast to the usual measures.

If you want, I can try to add a rough sketch of how to prove the theorem.

  • $\begingroup$ Thanks for the clarification, Martin. I think now I can see the connection between the finite-dimensional version $Av=\sum_i \lambda_i\langle v_i, v\rangle v_i$ and the infinite-dimensional (from wikipedia) $A=\int_{\sigma(A)} \lambda\, dE_\lambda$. $\endgroup$ – EPS Mar 15 '16 at 18:31
  • $\begingroup$ As an aside. The subject of spectral measures was investigated for operators on more general Banach spaces by Dunford and Schwartz in the third volume of their "Linear Operators". It turned out that now generators of norm-preserving continuous groups (self-adjoint operators in Hilbert space) and operators defined by means of a spectral measure usually do not coincide. This is regrettable since it excludes a powerful technique in dealing with density operators which are elements of the trace class, a Banach space but not a Hilbert space. $\endgroup$ – Urgje Mar 15 '16 at 20:45

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