Construction of Hamiltonian from Casimir operators

metroplex021

In Greiner & Muller's 'Quantum Mechanics: Symmetries' (section 3.5) they explain that where a system possesses a symmetry, the corresponding Hamiltonian must be 'built up' from the Casimir operators of the corresponding symmetry group.

Does anyone know of a reference where this is gone into in any detail?

Does anyone know what happens when the system possesses a product of symmetries (such as, say, Poincare x SU(2))?

Any help / references appreciated!

Bill_K

That statement seems a bit of a stretch. A weaker form of it might be, "the group operators can only appear in the Hamiltonian in combinations corresponding to the Casimir operators." But even this seems too strong. I don't see any Casimir operators in a Hamiltonian with a gauge symmetry.

strangerep

Are you sure you quoted the correct book and section?

In my 1989 (1st edition) copy of G+M sect 3.5 is indeed about Casimir operators,
but I don't see anything in that (small) section about constructing Hamiltonians.
I also checked the 2004 edition on Amazon, but it's much the same. Maybe you
were thinking of a different book or section?

For a completely integrable system, one can express the Hamiltonian in terms
of conserved quantities, but they're different from Casimirs in general.

I guess that you're talking about the meaning of "symmetry" as a conserved
(time-invariant) symmetry. In that case the Hamiltonian commutes with all
the generators of the symmetry and therefore must be a function of the
Casimirs of the symmetry group (else the Hamiltonian would simply be
another Casimir). G+M go into more detail about why this is so in their
later section 3.8.

Oh, here's the passage you probably meant. In my edition it's at
the end of section 3.11 (Completeness Relation for Casimir Operators).
They're talking in the context of multiplets (i.e., a subspace of the
Hilbert space which transforms into itself under the action of the
symmetry operators).

I'm not sure what precisely you find unclear about their presentation
so if you're still not happy you'll need to ask a more specific question...

Bill_K

strangerep, Since you seem to agree with the original poster, maybe you can clear up my doubts on this. For every example I've been able to think of, the statement is not true.

1) Schrodinger particle in a spherically symmetric potential. Hψ = h²/2m [- (1/r²)∂/∂r(r²∂ψ/∂r) + L²/r²] ψ + V(r) ψ. I see the Casimir operator L² in there, but H is certainly not a function of L².

2) Spin-orbit coupling, in which a L·S term appears in the Hamiltonian. This is not a function of the Casimir operator J² for the rotation group. While you can write J = L + S and therefore L·S = (J² - L² - L²)/2, you now have terms L² and S² which are not separately conserved and moreover not the Casimir operators of anything.

3) The Dirac Equation, Hψ = β mc² ψ - ihc α·∇ ψ. This is invariant under the Poincare group, whose Casimir operators P² and W² do not appear.

4) Electromagnetism, L = - A^μ,ν A_μ,ν, invariant under gauge transformations A_μ → A_μ + λ_,μ. How does the statement about Casimir operators apply to this case?

Quoting what you said,
I don't see that this statement is true, and I think the examples above illustrate what's wrong with it. In each case, H commutes with all the generators of the symmetry group, but is not a function of the Casimir operators.

Polyrhythmic

That the Hamiltonian commutes with all the generators of the symmetry group doesn't mean that it is actually part of the group and therefore somehow related to the Casimir operators.

strangerep

Actually, I was trying to encourage the OP to re-study the context in G&M in
which their statement was made.

G&M are talking strictly in the context of multiplets on which the Hamiltonian
is degenerate.

But more general cases are more complicated. E.g., for the spherically-symmetric
Coulomb potential (non-rel hydrogen atom, let's say) the full dynamical group is SO(4,2)
not merely SO(3), though this is a subgroup of SO(4,2). So in that case, one must
work with the 3 Casimirs of SO(4,2) to get the full picture.

In other words, the statement that "H itself has to be built up from invariant
operators of the symmetry group" should be understood to mean the largest
dynamical group applicable for that Hamiltonian.

In a specific representation, the Casimirs take on numerical values appropriate to
that representation. In the Dirac case, these are mass=m and spin=1/2, and
there's obviously an "m" in the Dirac Hamiltonian.

To say anything on the spin-orbit case, you'd need to specify the particular
Hamiltonian. Generically, the dynamical group gets bigger, hence (probably)
more Casimirs.

[Gotta run, sorry. I hope Arnold Neumaier stops by. He could answer this
much better than me.]

A. Neumaier

Probably you meant ''may be''. The technique assumes that the Hamiltonian can be written in terms of generators from a representation of a finite-dimensional so-called ''dynamical group'', of which the symmetry group is a subgroup. In the simplest case of a chain of only two Lie algebras,, one writes H=H_0+V, where H_0 is invariant under the whole group, and the perturbation V only under the subgroup.

The values of the possible Casimirs define the irreducible representations involved, and the generators of a Cartan subalgebra together with Casimirs of the chain of subalgebras define a commutative algebra in a concrete diagonal representation in which H_0 is diagonal. The problem then becomes a small matrix diagonalization problem.

The details are outlined in Section 23.6 of my book

Arnold Neumaier and Dennis Westra,
Classical and Quantum Mechanics via Lie algebras,
2008, 2011. http://lanl.arxiv.org/abs/0810.1019

Much more detailed treatments, complete with examples, can be found for example in papers and books by Jachello.

Bill_K

Ok, this has gradually evolved to a statement that I can agree with. IF the Hamiltonian can be written in terms of the generators of a finite-dimensional symmetry group, then those terms can be manifestly expressed as a combination of the Casimir operators.

But I do not think it's tenable to claim that the Dirac equation is "built from" the values m and 1/2.

samalkhaiat

[QUOTE=metroplex021;3369556]This is not a theorem that one can prove. However, in the absence of a Lagrangian ( usually the case in nuclear physics) and if experiments indicate that the system possesses certain symmetries, then one tries one best to construct a (bound-state-) “interaction Hamiltonian” out of the set of all “Good” quantum numbers (or invariants) available to him. Take (for example) the Deuteron (2-fermion bound state). Experiments indicate rotational invariance, iso-spin invariance, translational invariance and some other discrete symmetries. So one starts with the most general interaction Hamiltonian,
[tex]H(1,2) = F(r_{1}, s_{1},t_{1};r_{2},s_{2},t_{2}),[/tex]
where [itex](r_{i}, s_{i}, t_{i}), i=1,2[/itex] are the position, spin and isospin vectors,
then one uses all known symmetries to restrict its functional form:
1) translational invariance implies
[tex]H(1,2) = F(r,s_{1},t_{1},s_{2},t_{2})[/tex]
where [itex]r = |r_{1}-r_{2}|[/itex].
2) invariance under parity implies even powers of r;
[tex]H(1,2) = F(r^{2n}, s_{1},t_{1}, s_{2},t_{2})[/tex]
for simplicity one takes n = 1.
3) SO(3) invariance implies
[tex]
H(1,2) = F\left(r^{2}, (s_{1}.s_{2}), \frac{(s_{1}.\vec{r})(s_{2}.\vec{r})}{r^{2}}, t_{1},t_{2}\right)
[/tex]
these are the only SO(3) invariant objects available; [itex](s_{1}.s_{2})^{n}[/itex] can be reduced to [itex]a + b (s_{1}.s_{2})[/itex]
4) and finally invariance under the isospin group SU(2) implies
[tex]
H(1,2) = F\left(r^{2}, (s_{1}.s_{2}), \frac{(s_{1}.\vec{r})(s_{2}.\vec{r})}{r^{2}}, (t_{1}.t_{2})\right)
[/tex]
So now you can write
[tex]
H=f_{1}(r^{2})+f_{2}(r^{2})s_{1}.s_{2}+f_{3}(r^{2})\frac{(s_{1}.\vec{r} )(s_{2}.\vec{r})}{r^{2}}+f_{4}(r^{2})(s_{1}.s_{2})(t_{1}.t_{2})+f_{5}(r ^{2})(t_{1}.t_{2})
[/tex]
and ask your experimental physicist friend to find the functions
[tex]f_{i}= f_{i}(r^{2}).[/tex]

Noether theorem gives you a unique Hamiltonian in this case, so there is no need for the garbage I described above.

Sam

samalkhaiat

[QUOTE=Bill_K;3373073]
Actually, it is! Provided one knows what one is talking about! Equations of motion (field equations) and more stuff follow from certain constraints on the (conformal) algebra. For example, for the massless representations, the constraints on “wave functions” are
[tex]P^{2} = 0 \ \ (B)[/tex]
and
[tex](1/2)[iK^{a}, P^{2}] = \{2 \Sigma^{ac}P_{c} + 2s P^{a}\} = 0\ \ \ (A)[/tex]
where s is (in n-dimensional space-time ) the real number
[tex]s = d - \frac{n-2}{2},[/tex]
d is the scaling dimension of [itex]\Psi (x)[/itex],
and
[tex]P_{a}\Psi_{r} (x) = -i\partial_{a}\Psi_{r} (x).[/tex]

Now, for [itex]\Sigma = 0[/itex], we get from B and A
[tex]\partial_{c}\partial^{c}\phi (x)=0, \ \ s = 0.[/tex]
If we take [itex]\Sigma[/itex] to be the Clifford number
[tex]\Sigma^{ac}= \frac{1}{4}[\gamma^{a},\gamma^{c}],[/tex]
we find
[tex]i\gamma^{a}\partial_{a}\psi (x) = 0, \ \ \ s = \frac{1}{2}[/tex]
Take
[tex](\Sigma^{ac})^{b}_{d} = \delta^{a}_{d}\eta^{cb}-\delta^{c}_{d}\eta^{ab}[/tex]
you find
[tex]\partial_{a}V_{c}(x)-\partial_{c}V_{a}(x) = 0, \ \partial^{a}V_{a}=0 \mbox{and} \ \ s = 1[/tex]
This shows that massless spin-1 vector field is equivalent to the scalar field
[tex]V_{a}(x) = \partial_{a}\phi (x)[/tex]
But if you take the appropriate [itex]\Sigma[/itex] for atisymmetric tensor field, you end up with the whole set of Maxwell equations with s =1. And this way you can carry on generating all possible field equations.

Sam