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I have two quantum mechanical Hamiltonians such that \begin{equation} [\hat{H}_1,\hat{H}_2] = 0, \end{equation} where $\hat{H}_1$ and $\hat{H}_2$ act on the same set of states. What is there to infer physically about these two Hamiltonians? Are there further mathematical subtleties that were not brought out in "What is the Physical Meaning of Commutation of Two Operators?" for the case of two Hamiltonians?

From looking around I have these properties:

  • Treating them as observables I can measure them simultaneously.
  • They share the same set of eigenstates and thus any state can be expanded as a sum of these.
  • They can both be simultaneously diagonalised.

Are there further properties or subtleties to this relationship?

EDIT : Edited after comments pointing out that there is little to be said if they individually act on different subsystems, apart from the fact that they share eigenstates when acting on both systems together.

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Matta
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    Hmmm... can you explain why they should share the same set of eigenstates? If $H_1$ acts on one subspace and $H_2$ acts on an independent one, they will commute, but structurally they can be completely different and have a completely different set of eigenstates. That's not what you mean, right? – CuriousOne Apr 30 '15 at 09:14
  • @CuriousOne If the eigenvalues of $H_1$ are $|n>$ and those of $H_2$ are $|m>$, then in the complete system they will share the set of eigenstates $|n,m>$. – Noiralef Apr 30 '15 at 09:23
  • If $H_1$ and $H_2$ act on independent subspaces $V_1$ and $V_2$ respectively, we can choose any basis of $V_1$ for a basis of eigenstates of $H_2$, so in particular a basis of eigenstates associated to $H_1$. That means there exists a shared set of eigenstates. – Hrodelbert Apr 30 '15 at 09:23
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    Two "Hamiltonians" generically act on different spaces of states, i.e. there is a Hilbert space associated to each one. It's not obvious what you mean when you speak of the commutator of operators on different spaces. – ACuriousMind Apr 30 '15 at 09:55
  • @Noiralef: That's kind of what I thought was meant... it's just not a very strong property. – CuriousOne Apr 30 '15 at 09:55
  • @Hrodelbert: $H_1$ and $H_2$ don't even have to have the same dimensionality, so there is absolutely no reason to assume that one can express one set of eigenstates in the bases of another. – CuriousOne Apr 30 '15 at 09:57
  • You are all being incredibly helpful, thank you. What if they act on the same space of states? Is there something to be said then? – Matta Apr 30 '15 at 10:01
  • If it turns out to be more interesting then I can make the question more specific so that it is possible to answer. – Matta Apr 30 '15 at 10:09
  • Question updated - hopefully correctly. – Matta Apr 30 '15 at 10:23
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    @CuriousOne I am not sure why we are disagreeing on this: the two hamiltonians should be defined on the same Hilbert space $\mathcal{H}$ (otherwise such a commutator identity is meaningless from the start). Therefore "acting on different subspaces" can only mean that we can write $\mathcal{H} = V_1 \oplus V_2$ and write $H_1= \tilde{H_1} \oplus 1$ and $H_2= 1\oplus \tilde{H_2}$, in line with my previous comment – Hrodelbert Apr 30 '15 at 10:26
  • @Hrodelbert: The non-trivial parts of both Hamiltonians do not have to have the same dimensionality, so solving one doesn't tell you anything about the other. – CuriousOne Apr 30 '15 at 10:33
  • @CuriousOne Your are absolutely right, but that was never the claim: we only want a basis of eigenstates, which is possible for these hamiltonians. – Hrodelbert Apr 30 '15 at 10:35
  • Trivial separation of variables doesn't give you much, though. I (and probably the OP, too) was hoping for something more "juicy". – CuriousOne Apr 30 '15 at 10:39
  • @CuriousOne Understandable. I was only trying to convey the point that commuting hamiltonians really share a basis of eigenstates since your were questioning this in your first comment. – Hrodelbert Apr 30 '15 at 10:42
  • @Hrodelbert: granted... I am still hoping for a little more. Any ideas? – CuriousOne Apr 30 '15 at 10:43
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    In analyzing evolution of a dynamical system - if the Hamiltonian of the system commutes with itself in different times - there is a "simple" solution to the Schrodinger equation ($i\hbar\frac{\partial}{\partial t}U=HU$) , $|\psi(t)\rangle=U(t)|\psi(t_0)\rangle$ - $U(t)=e^{\frac{-i}{\hbar}\int_{t_0}^tH(t)\text{d}t}$. If it is not commuting - $U$ is given by kind of "taylor expansion" called "Dyson series", which is not very pretty https://en.wikipedia.org/wiki/Dyson_series – Alexander Jul 12 '15 at 20:42

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Comments to the question (v2):

  1. It seems that the question does not explain how a 'Hamiltonian' $H$ differs from a self-adjoint operator $A$ (presumably bounded from below). This would make OP's question a duplicate of the linked Phys.SE post.

  2. Perhaps a 'Hamiltonian' $H$ is also supposed to generate 'time'-evolution for some distinguished parameter $t$, which may or may not be actual time? Then consider a universe with two 'time' directions $t_1$ and $t_2$, cf. e.g. this Phys.SE post. The two commuting Hamiltonians $[H_1,H_2]=0$ means that one gets the same result if one first 'time'-evolve $e^{iH_1t_1/\hbar}$ wrt. $t_1$ and then 'time'-evolve $e^{iH_2t_2/\hbar}$ wrt. $t_2$, as one would get if one does it the other way around. In other words, $t_1$ and $t_2$ represent commuting flows, and it makes sense to specify a state with two 'time'-coordinates $(t_1,t_2)$.

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Qmechanic
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At least a partial answer to your question is that commuting hamiltonians help you to solve the physical system described by one of them: in particular, if your system has $N$ degrees of freedom and you have $N$ commuting hamiltonians, there is good hope that you can trivialize the problem and solve it exactly. In classical mechanics, this is known as Liouville integrability (where there the commutativity is associated to a Poisson bracket). In quantum mechanics, the notion is not completely well-defined, although searching for the term quantum integrability will provide you with ample reading material. Due to the equation of motion in the Heisenberg picture $$ \frac{dA}{dt} = [A,H], $$ where $A$ is an operator, we see that if $H_1$ is the hamiltonian governing time evolution, $H_2$ is conserved in time. But since many known Hilbert spaces are infinite dimensional, using the above in practice is harder.

Note that for quantum mechanical models such as spin chains ($N$ fixed particles interacting via spin degrees of freedom), the space of states is finite dimensional and all of the above does help.

Hrodelbert
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  • I am not aware that even the one-dimensional Schroedinger equation can be integrated in general for arbitrary potentials, so even the complete independence of all variables is no guarantee that the problem has a simple solution. – CuriousOne Apr 30 '15 at 10:31
  • No, that is true. But the one-dimensional Schroedinger equation is also not represented by a one-dimensional Hamiltonian: it acts on a space of functions that is usually infinite-dimensional – Hrodelbert Apr 30 '15 at 10:32
  • Yep, QM is a beast, even for the most simple case. We can be glad, historically, that the hydrogen problem has such a nice solution, or Schroedinger would have been in a lot of trouble to make a case for his equation... – CuriousOne Apr 30 '15 at 10:37
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    Yes, hurray hydrogen:) – Hrodelbert Apr 30 '15 at 10:39