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Here is my Hamiltonian: $$H = \frac{1}{2m} p_1^2 + \frac{1}{4m} p_2^2 + \frac{1}{6m} p_3^2 + \frac{1}{2} k^2 (q_1^2 + 2q_2^2 + 3q_3^2),$$ where $q_i$ and $p_i$are canonical variables, in the sense that $[q_a,p_b] = \delta_{a,b} i$.

How can I identify the Lie Group of the continuous symmetries of $H$ acting on $(q_a,p_b)$?

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  • $\begingroup$ The WP section on SU(3) is not clear? $\endgroup$ Commented Feb 8, 2020 at 23:07
  • $\begingroup$ Very clear, but I have a few more questions: 1) How can I show that at each energy level the states form an irriducible representation of G? I think that is sufficient to say that the elements of SU(3) map a state into another state with the same energy, because the Hamiltonian is invariant under SU(3). Therefore is sufficient to calculate the number of states at each energy level and show that this number corresponds to the dimension of an irriducible representation of SU(3). $\endgroup$
    – dfgoe55
    Commented Feb 8, 2020 at 23:15
  • $\begingroup$ 2) If I perturbate the Hamiltonian: $H_\lambda = H + \lambda(q_1^2p_2^2 + 4q_2^2p_1^2 -2q_1p_2q_2p_1 -2q_2p_1q_1p_2)$, it is correct to say that the SU(2) subgroup of SU(3) is the group of symmetry of hamiltonian? $\endgroup$
    – dfgoe55
    Commented Feb 8, 2020 at 23:17
  • $\begingroup$ The degeneracy of the 3D isotropic oscillator is here. The dimensionality of the irreducible reps of SU(3) is, of course (p+1)(q+1)(p+q+2)/2. $\endgroup$ Commented Feb 8, 2020 at 23:27

1 Answer 1

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Through the obvious canonical rescaling, $$ q_1=Q_1, \qquad q_2=Q_2/\sqrt{2}, \qquad q_3=Q_3/\sqrt{3},\\ p_1=P_1, \qquad p_2=P_2 ~ \sqrt{2}, \qquad p_3=P_3~\sqrt{3}, $$ which preserves the commutation relation, $[Q_a,P_b] = \delta_{a,b} ~i I$.

It transforms the hamiltonian into the ("unmodified") isotropic 3D SHO, whose symmetry is well-known to be U(3). Since, however, an over-all phase change is immaterial/unphysical in QM, the focus of your attention, the symmetry group is pruned down to merely SU(3).

The coincidence of the state degeneracy (n+1)(n+2)/2 with the irreducible representations of SU(3) is to be found in this answer; hint: take zero antiquarks!

Note added on comment 2): I suspect you really are having me do your homework for you. In the "nice" variables above, your perturbation is but $$ 2\lambda (Q_1^2 P_2^2 + Q_2^2P_1^2 - Q_1P_2Q_2P_1- Q_2P_1Q_1P_2). $$ Proceed to collect to a square!

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  • $\begingroup$ Ok, now my Hamiltonian is: $H_\lambda = H_{SHO} + 2 \lambda (Q_1 P_2 - Q_2P_1)^2$. Now the variables $Q_1$, $Q_2$, $P_1$, $P_2$, are mixed together with a term in $\lambda$. I suspect that the SU(3) invariance is broken. I am sure that there is a U(1) residual symmetry because the perturbation term does not include $(Q_3,P_3)$. Now I have to figure out if there is a subgroup of SU(3) that acts on $(Q_1,Q_2,P_1,P_2)$ which leaves the Hamiltonian invariant. I do not want the answer immediately, tell me if I am on the right way. $\endgroup$
    – dfgoe55
    Commented Feb 9, 2020 at 11:20
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    $\begingroup$ Yes, you are. . . $\endgroup$ Commented Feb 9, 2020 at 11:31
  • $\begingroup$ I don’t figure out, it seems that the only symmetry that remains is a U(1) symmetry...Is that correct? I need a hint... $\endgroup$
    – dfgoe55
    Commented Feb 9, 2020 at 11:47
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    $\begingroup$ Of the 8 Gell-Mann matrices, $\lambda_2$ commutes with itself and $\lambda_8$. $\endgroup$ Commented Feb 9, 2020 at 13:04
  • $\begingroup$ I have no idea how to continue. Please can you explain? $\endgroup$
    – dfgoe55
    Commented Feb 9, 2020 at 15:48

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