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Construction of Hamiltonian from Casimir operatorsby metroplex021
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#1
Jun2211, 04:43 AM

P: 119

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! 


#2
Jun2211, 05:12 AM

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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.



#3
Jun2311, 01:55 AM

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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 (timeinvariant) 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). so if you're still not happy you'll need to ask a more specific question... 


#4
Jun2311, 09:49 AM

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Construction of Hamiltonian from Casimir operators
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 = h^{2}/2m [ (1/r^{2})∂/∂r(r^{2}∂ψ/∂r) + L^{2}/r^{2}] ψ + V(r) ψ. I see the Casimir operator L^{2} in there, but H is certainly not a function of L^{2}. 2) Spinorbit coupling, in which a L·S term appears in the Hamiltonian. This is not a function of the Casimir operator J^{2} for the rotation group. While you can write J = L + S and therefore L·S = (J^{2}  L^{2}  L^{2})/2, you now have terms L^{2} and S^{2} which are not separately conserved and moreover not the Casimir operators of anything. 3) The Dirac Equation, Hψ = β mc^{2} ψ  ihc α·∇ ψ. This is invariant under the Poincare group, whose Casimir operators P^{2} and W^{2} 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, 


#5
Jun2311, 10:18 AM

P: 342

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.



#6
Jun2311, 11:19 PM

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which their statement was made. is degenerate. But more general cases are more complicated. E.g., for the sphericallysymmetric Coulomb potential (nonrel 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. 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 spinorbit 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.] 


#7
Jun2411, 08:00 AM

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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. 


#8
Jun2411, 11:46 AM

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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 finitedimensional 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. 


#9
Jun2511, 07:29 PM

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[QUOTE=metroplex021;3369556]
[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] Sam 


#10
Jun2511, 10:41 PM

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P: 892

[QUOTE=Bill_K;3373073]
[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 ndimensional spacetime ) the real number [tex]s = d  \frac{n2}{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 spin1 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 


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