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I am familiar with the notion of infinite-dimensional linear operators from quantum mechanics, such as the Hamiltonian $\hat H = -\frac{1}{2m} \frac{\partial^2}{\partial x^2}$ which has eigenstates $e^{ipx}$ and eigenvalues $\frac{p^2}{2m}$ , for $-\infty < p < \infty$ .

Now for Hermitian matrices knowing the eigenvalues determines the matrix up to a unitary similarity transformation, or in other words two finite-dimensional Hamiltonians with exactly the same eigenvalues will describe the same physical system, since we can map the eigenstates of one to the other with a unitary transformation, and the time evolution will be the same because the eigenvalues are the same.

How we can determine if two infinite-dimensional Hermitian operators with a continuous spectrum are similar ?
e.g. if we know that some other Hamiltonian has eigenvalues $0 < E < \infty$ , how can we tell if it describes the same physical system as the one above ?

More precisely is there a way to describe the spectrum such that we can tell if two operators are similar just by comparing their spectra (without having to know what the eigenvectors are) ?

  • Due diligence. Formally diagonalize each and equate the diagonal answers, then transport diagonalizing operators to blend with the other's ones. – Cosmas Zachos Jan 26 '22 at 15:15
  • " two finite-dimensional Hamiltonians with exactly the same eigenvalues will describe the same physical system" . That is of course not true. In fact, the systems will be mathematically the same but could be physically completely different, v.g a 2-level atom vs a spin-1/2 particle. – ZeroTheHero Jan 26 '22 at 15:23
  • related to https://physics.stackexchange.com/a/15285/36194 – ZeroTheHero Jan 26 '22 at 15:26
  • @ZeroTheHero thanks for the reference. To your point yes I guess you can call it "mathematical" equivalence. what I meant in this context was that there is a one-to-one correspondence between the physical observables, so in that sense they are the same – user341440 Jan 26 '22 at 18:21
  • well... not really but it seems semantic. With proper scaling the motion of a harmonic oscillator and of a pendulum (in the small angle limit) satisfy $\ddot{q}=- q$ but the physical systems are not the same. Of course the strict mathematical equivalence is actually very powerful and hugely interesting as it reveals ways distinct physical system nevertheless share some similarities. – ZeroTheHero Jan 26 '22 at 19:39
  • @ZeroTheHero you say that the two systems are not the same because there is some physical observable (e.g. the height of the pendulum) that exist in one and not the other, but that's not the case in question. A better analogy is for example two water molecules, you can still argue that they are "not the same system", but that's really purely semantic – user341440 Jan 26 '22 at 20:11
  • interesting. No I really mean like the math. of a two-level system can be applied to physically different systems, not two copies of the same. The celebrated qubit can be realized as polarization or spin degrees of freedoms in obviously distinct systems, i.e. polarization with light and spin with a spin-1/2 particle. The observables of one (i.e. the Stokes parameter) are NOT the same as the observables of the others (i.e. the components of spin). So it’s a bit of a leap to suggest that because two Hamiltonians have the same spectrum they represent the same physical system. – ZeroTheHero Jan 26 '22 at 21:39
  • Don’t get me wrong. The interesting thing is often precisely the mapping you refer to, i.e. if we “discover” that something is possible with polarization is very natural to ask of the same effect will also appear with spin, but they still are different systems. – ZeroTheHero Jan 26 '22 at 21:41

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