Let us assume that we know only some basic QM notions which are part of the Heisenberg picture of quantum mechanics and Dirac quantization

  • Physical observables are represented by Hermitian operators $\hat{O}$, which generally do not commute
  • $[\hat{x},\hat{p}]=i$ , the commutation relation between momentum and postion
  • The existence of a Hamiltonian operator $\hat{H}$ such that $$\frac{d}{dt}<\hat{O}>=\left<i[\hat{O},\hat{H}]\right>$$ where $<...>$ is a probabilistic expectation.

Note that we do not know how to obtain what $<...>$ is, only its equation of mootion. But its enough to completely solve, fo instance, the simple harmonic oscillator, and from this obtain both the notions of intrinsic uncertainity of position and momentum and quantization of energy level.

From this and also notions of representation theory, hermiticity, and other strictly mathematical properties of operators as well as usual properties of probabilistic averages, can one prove that a "density" observable exists, $\hat{\rho}$ such that for any other observable $\hat{O}$, $$<\hat{O}>=Tr \left( \hat{O}\hat{\rho}\right)$$

The reason I am asking is, the above setup is, at least to me, very physically intuitive (measurements do not commute so you represent them by operators, and so on) and free of the usual apparently arbitrary and paradoxical notions associated with QM (wavefunction collapse and so on). If the existence of the density matrix can be understood this way, it would be easy to continue with usual Eigenvalue and EigenVector notions to construct the wavefunction as a generic basic for the density matrix, and derive the rest of quantum mechanics from this. It would be, in my opinion, the way to introduce quantum mechanics with the least amount of philosophical mumbo-jumbo. However, it relies on the density matrix following from the above assumptions, and I was not able to prove it or find such a proof (a kind of reverse Gleason's theorem). Do you think this is doable?

  • 2
    $\begingroup$ Use '\langle' and '\rangle' instead of '<' and '>'. $\langle\hat O\rangle$ vs. $<\hat O>$ $\endgroup$ – BioPhysicist Oct 5 '19 at 13:17
  • $\begingroup$ We can formulate QM starting with an abstract algebra $\mathcal{A}$ of observables (a C*-algebra) and define a state to be a map $X\to\rho(X)$ from operators $X$ to complex numbers $\rho(X)$ satisfying certain properties, as described in my answer here. Then the GNS theorem implies the existence of a traditional representation of $\mathcal{A}$ as linear operators on a Hilbert space. (The measurement problem is still there, but this might put it in a different perspective.) Does the GNS theorem sound like what you're looking for? $\endgroup$ – Chiral Anomaly Oct 5 '19 at 17:04
  • $\begingroup$ Thank you! I actually thought about this a bit more, and I think I found something that partially answers my own question and makes alink to Chiral Anomaly's answer. The expectation value, from the axioms of probability theory, is a sum of the type $$\langle \hat{O} \rangle = \sum O_i P(i)$$ where $P(i)$ are probabilities and $O_i$ are outcomes. The only way to generalize such an expression so that one recovers the expression above to operators is via traces $$ \langle O \rangle = Tr\left( \hat{\rho} \hat{O} \right) $$ $\endgroup$ – Giorgio Torrieri Oct 5 '19 at 17:58
  • $\begingroup$ All that would be left is to prove the uniqueness of $\hat{\rho}$ for all operators, all operators have the same $\rho$. Which I think can be done using the property of hermiticity of operators, I think Chiral Anomaly's reasoning is a rigorous way to arrive at this. $\endgroup$ – Giorgio Torrieri Oct 5 '19 at 18:05
  • $\begingroup$ The measurement problem is I think considerably lessened, since in this approach what is fundamental is that all physical quantities are probabilistic, and this probability is a result of non-commutation of observables. This is a much more "physical" way to understand the measurement problem than "collapse". It relies on pretty sophisticated mathematics, but in my experience it is a lot more useful to come up with intuitive postulates and say "look, and then theorem X says Y, if you want to understand why take this representation theory class" than to have seemingly ad hoc assumptions. $\endgroup$ – Giorgio Torrieri Oct 5 '19 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.