Clearly it is possible to derive a set of commutation relations from some Hamiltonian, and certainly they give useful and interesting invariants when investigating the behavior of quantum systems. But my question is can we go backwards?

In other words, can one set up a collection of (anti)commutation relations which determine a particular C* algebra, all of whose representations are solutions to some Schrodinger equation?

And if not, how much extra data do we need to reconstruct the exact Schrodinger equation?

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    $\begingroup$ Can you say exactly what is the given data? Do you know that there is an x operator, and you are given [x,p], and [p,H]? In this case the reconstruction is simple. Are you given an abstract algebra of operators with a commutator map? It is impossible to answer this question as posed. $\endgroup$
    – Ron Maimon
    Nov 2, 2011 at 7:53
  • $\begingroup$ What I am asking is if the set of all solutions to the Schrödinger equation is equivalent to what you would get if you just pick a set of appropriate commutator relations and then look at the set of all operators which satisfy those. In an algebraic sense, this is a more restricted version of looking at the representations of the c-star algebra generated by a finite set of operators (ie p, x, H, etc.) mod some commutator relations. $\endgroup$
    – Mikola
    Nov 2, 2011 at 8:31
  • $\begingroup$ You haven't answered my question: do you know the names of all the operators ahead of time? Do you know which one is x? Or do you have to figure it out from the structure of the algebra? The algebra is intrinsic--- it only gives you a structure on itself. The physics include extrinsic information, like what is an "x measurement". Are you given the extrinsic information, in the form of knowning which operator is the Hamiltonian? $\endgroup$
    – Ron Maimon
    Nov 2, 2011 at 17:08
  • $\begingroup$ I am beginning to think I should have asked this on math overflow instead. At a formal level, what I am asking is: Does the ideal cut out by the Schrodinger equation have a presentation as a collection of generators and commutator relations? (It is true for example that any system of differential equations cuts out an ideal in a C* algebra, the irreducible representations of which correspond to its solutions. However, in quantum mechanics special emphasis is placed on these commutator relations. What I am wondering is if those commutators are enough to determine the ideal of the SEqn.) $\endgroup$
    – Mikola
    Nov 2, 2011 at 18:06
  • $\begingroup$ you made it clearer now, but I am still confused. The ideal of a differential equation is the one which, when you quotient by it, makes the differential operators appearing on opposite sides algebraically equal to each other. This could be (formally) used for the time-dependent Schrodinger equation to get H - p^2 - V(x) as the generator of the ideal (one generator, no new relations beyond [p,x]=i) But does the C* algebra itself care about the form of V? Isn't it always the algebra of operators on L_2? For good V, is the operator algebra changed by SE? Or does SE only specify one operator as H. $\endgroup$
    – Ron Maimon
    Nov 3, 2011 at 9:38

1 Answer 1


I believe this question is suffering from a confusion regarding the C* algebra of operators in the quantum mechanics. To verify that I am thinking the same as the OP, I will quickly recapitulate the question in my own words:

When you have a linear differential equation, like

$$ (\partial_y^2 - \partial_x^2 - x\partial_x)\psi = 0 $$

You can think of the operator algebra containing $p_x=-i\partial_x$,$p_y = -i\partial_y$ as well as functions of $x,y$ as having an algebraic relation imposed on it, namely;

$$ p_x^2 - p_y^2 + i x p_x = 0 $$

This algebraic relation implies other relations, together with $[x,p]=i$, and whatever else follows. The collection of all things in the old C* algebra that are made equal to zero by the relation of the equation is called the ideal corresponding to the differential equation. The question is what is this ideal for the Schrodinger equation, and can it be expressed using commutators?

The problem is that the Schrodinger equation is not naturally living in the space of functions on x,t and operators on them. The space of operators is for functions only of x. So the ideal corresponding to the Schrodinger equation is all due to the relation defining the Hamiltonian

$$ i\partial_t = {p^2\over 2m} + V(x)$$

And this relation doesn't alter the algebraic relations between p and x, it just serves to define H. So there is no alteration in the C* algebra of operators by defining the Hamiltonian, and the question doesn't make sense.

Perhaps you mean to define the space of all operators on functions of space and time together, and then quotient that by a nontrivial ideal. But I couldn't figure out how to to do this. A reference would be nice, because the question is very terse.


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