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the following is a short question regarding a theorem from a quantum mechanics book I am working through but the question is a mathematical one.

There is a theorem which states:

Theorem: The eigenstates of a Hermitian operator defines a complete set of mutually orthonormal basis states. The operator is diagonal in this eigenbasis with its diagonal elements equal to the eigenvalues.

Question:

  • For a finite dimensional space we can describe the Hermitian operator as a Hermitian matrix, it is then clear that a diagonal operator is then a diagonal matrix. When considering an operator on an infinite dimensional Hilbert space, what is a diagonal operator defined as? Could it be described as an infinite diagonal matrix. If so, is it required that this is a special case where the spectrum is discrete?

Thanks.

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  • $\begingroup$ In the infinite dimensional case, this usually means the operator, say $A$, is unitarily equivalent to a multiplication operator on some $L^{2}$-space. In general, $A$ cannot be described by an $\infty\times\infty$ matrix. $\endgroup$ – James May 14 '16 at 19:41
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If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space.

For example, start with the position operator $M$ on an interval $[0,1]$. This operator is multiplication by $x$, meaning $(Mf)(x)=xf(x)$. This is a perfectly legitimate selfadjoint linear operator on the Hilbert space $L^2[0,1]$, but it has no eigenvectors, meaning that there are no non-trivial functions $f \in L^2[0,1]$ for which $Mf = \lambda f$, regardless of the choice of $\lambda$. The operator $M$ has "continuous spectrum" but no eigenvectors or eigenvalues. There just are no such "eigenstate" objects.

Physicists typically introduce the $\delta$ function into such a space (which is logically inconsistent) and claim that $(M \delta_{y})(x) = x\delta(x-y)=y\delta_{y}$. For many reasons, on many levels, that is Mathematical non-sense, and cannot be salvaged within $L^2[0,1]$ in a logically consistent way. The axioms of Quantum use Hilbert spaces, but $\delta$ functions cannot live in spaces such as $L^2[0,1]$ because two functions that are equal a.e. in $L^2[0,1]$ are identical, which means that pointwise values have no meaning for elements of $L^2[0,1]$. Dirac knew this, but he liked the intuition.

John von Neumann, who was a contemporary of P.A.M. Dirac, created consistent ways of dealing with the selfadjoint operators of Physics through the Spectral Theorem, and this was available to Dirac at the time; but Dirac chose an intuitive presentation over logical consistency. And it wasn't because rigorous Mathematics was unavailable at the time to handle everything; the correct Math was there.

All of the Hilbert spaces of Quantum must be separable, meaning that the space contains a countable dense subset. The reason for this has to do with constructibility, and being able to bootstrap to general answers through finite approximations. $L^2[0,1]$ is a separable space. One of the consequences of having a separable space, is that an orthonormal basis of such a space is always finite or countably infinite. You cannot have mutually orthogonal objects that are indexed by an interval of the real line, for example. That's impossible in separable spaces. And, if you have a selfadjoint operator $M$ on a Hilbert space, and you have $Mf=\lambda f$ and $Mg = \mu g$ for $\lambda\ne \mu$, then $(f,g)=0$ must hold. So, a selfadjoint operator on a Hilbert space that is allowed in Quantum Mechanics cannot have more than a countable number of actual eigenvalues. That's a consequence of correct axiomatic systems for Quantum. You can have continuous spectrum, but that's different than having a continuum of eigenvalues. If you enlarge the space to allow for such things, then you lose the ability to approximate in a finite way, which disconnects the theory from a setting where finite approximation makes sense.

Sorry to disappoint you. The theorem you have--as stated--is not true. Correct spectral theory gives you essentially the same thing, but not exactly that.

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  • $\begingroup$ is there a way to recognize that a normal operator on a separable Hilbert space is (or is not) of the form $T x = \sum_{n=1}^\infty \lambda_n \langle e_n,x\rangle e_n$ for some orthonormal basis $(e_n)$ ? if say $T$ is compact, it automatically has a SVD, and hence if it is normal it has this form, right ? $\endgroup$ – reuns May 14 '16 at 21:26
  • $\begingroup$ @user1952009 : Operators with a continuous spectrum would not be compact. You can have a bounded position observable on a bounded interval, but it is not compact. Compact resolvents make more sense, such as for a Hamiltonian. $\endgroup$ – DisintegratingByParts May 14 '16 at 21:31
  • $\begingroup$ "and you say $\delta_x$ is not so consistent". the operator $P_x : f \to \langle \delta_x, f \rangle$ with $\|f\|^2 = \int_{X} |P_x f|^2 d\mu$ is a definition of a Hilbert space space of functions. if all the $P_x$ are bounded then we are probably in $l^2(X)$, but if they are not then $P_x$ can be defined as some sort of weak-* limit of a sequence of operators of $H^*$ ? is there a problem with those considerations ? $\endgroup$ – reuns May 14 '16 at 21:44
  • $\begingroup$ @user1952009 : There are all kinds of extended structures designed to salvage the Dirac idea in some way for cases of continuous spectrum, but you can't do this in a separable Hilbert space. Sometimes it's just better to start doing correct Math, live without the "ideals," but have a consistent system that really isn't any more difficult to deal with than the kludged-up alternative. $\endgroup$ – DisintegratingByParts May 14 '16 at 21:50

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