Dynamical Symmetry

A dynamical symmetry uses an algebra to organize a spectrum or its time evolution. The term has several conventions: sometimes it means a Hamiltonian built from Casimir operators, and sometimes a larger algebra containing operators that connect different energies. Stating the operators and their commutators is more informative than the name alone. The oscillator makes the distinction concrete: creation operators change energy, while number-conserving operators can organize degeneracies.

Helpful background. Exact symmetry explains energy-preserving transformations, hidden symmetry introduces less visible conserved operators, and the oscillator ladder solution supplies creation and annihilation operators.

Spectrum-generating operators

Let $H$ be time independent, and suppose an operator $A$ satisfies

[H,A]=\Delta A

on an appropriate common domain, with constant energy increment $\Delta$ . For an eigenvector $H|E\rangle=E|E\rangle$ ,

H(A|E\rangle)=(E+\Delta)A|E\rangle.

If $A|E\rangle$ is nonzero and belongs to the Hamiltonian domain, it is an eigenvector at shifted energy. When $\Delta\ne0$ , $A$ is not a conserved time-independent generator; its Heisenberg evolution is $A_H(t)=e^{i\Delta t/\hbar}A$ . Such an operator need not be Hermitian or unitary. It need not itself represent a physical symmetry transformation.

This differs from a conserved operator $G$ satisfying $[H,G]=0$ : $G$ maps an energy eigenspace into itself, subject to the same domain qualifications. Conservation alone does not guarantee degeneracy, since $G$ may act only by a scalar in that eigenspace. See degeneracy and multiplets.

Oscillator ladders and shell symmetries

For one oscillator,

H=\hbar\omega(N+1/2),\qquad N=a^\dagger a, \qquad [a,a^\dagger]=I.

The commutators $[H,a^\dagger]=\hbar\omega a^\dagger$ and $[H,a]=-\hbar\omega a$ explain the equally spaced spectrum. Positivity and normalization then give

a|n\rangle=\sqrt n|n-1\rangle,\qquad a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle.

The lowest state is annihilated by $a$ ; repeated lowering therefore does not produce an unbounded negative-energy ladder. The full ladder derivation explains this boundary condition and the zero-point energy.

For $d$ oscillators of the same frequency,

H=\hbar\omega\left(\sum_{i=1}^{d}N_i+\frac d2\right), \qquad [H,a_i^\dagger a_j]=0.

The bilinear $a_i^\dagger a_j$ transfers one quantum between modes without changing the total number. The $d^2$ bilinears close under commutation,

[a_i^\dagger a_j,a_k^\dagger a_l] =\delta_{jk}a_i^\dagger a_l-\delta_{il}a_k^\dagger a_j,

giving the $\mathfrak u(d)$ algebra. Their Hermitian combinations generate transformations within a fixed shell. A single $a_i^\dagger$ , by contrast, changes the shell. For two modes at total number two, the states $|2,0\rangle,|1,1\rangle,|0,2\rangle$ are degenerate and connected by the transfer operators. These are distinct roles for operators built from the same oscillator variables.

Rotor energies from a Casimir

A spin-free linear rigid rotor has

H=\frac{J^2}{2I}=B\frac{J^2}{\hbar^2}, \qquad B=\frac{\hbar^2}{2I},

where $I$ is the moment of inertia and $B$ has energy units. Since $J^2$ is the quadratic Casimir of the rotation algebra,

E_j=Bj(j+1),\qquad j=0,1,2,\ldots .

The generators $J_i$ commute with this Hamiltonian and relate projections within a fixed $j$ ; they do not raise $j$ to the next rotational level. The special Hamiltonian form turns representation labels into energy values. Rotational invariance by itself would allow a more general function of $J^2$ , and additional radial or internal quantum numbers in a less restricted system. See rotational spectra.

More elaborate algebraic models use mutually commuting Casimirs along a nested chain of algebras. Their representation labels diagonalize the Hamiltonian when it has the required form. This is a useful solvable limit, not a consequence of merely being able to expand a Hamiltonian in some operator basis; Iachello and Arima develop this approach for nuclear collective models.

Anharmonicity changes the ladder relation

Adding $\lambda X^4$ with $\lambda>0$ to the oscillator provides a useful test while keeping the potential confining. Since $X=\sqrt{\hbar/(2m\omega)}(a+a^\dagger)$ , the perturbation is still expressible in oscillator operators. What fails is the simple closure $[H,a^\dagger]=\hbar\omega a^\dagger$ , not the ability to use those operators at all.

At first order, the energy correction is

\Delta E_n=\lambda\left(\frac{\hbar}{2m\omega}\right)^2 (6n^2+6n+3).

Consequently the spacing between levels $n$ and $n+1$ changes by $12\lambda(\hbar/2m\omega)^2(n+1)$ . The spacing is no longer constant. Perturbation theory requires these corrections and the induced mixing to be small in the levels studied; small $\lambda$ does not give uniform control at arbitrarily large $n$ .

Exercises

Take two oscillators with unequal frequencies, $H=\hbar\omega_1(N_1+1/2)+\hbar\omega_2(N_2+1/2)$ . Compute $[H,a_1^\dagger a_2]$ and identify the energy change.

Solution

The first number operator counts one added quantum and the second counts one removed quantum, so

[H,a_1^\dagger a_2]=\hbar(\omega_1-\omega_2)a_1^\dagger a_2.

The transferred state, when nonzero, changes energy by $\hbar(\omega_1-\omega_2)$ . The transfer is conserved at equal frequencies. Unequal frequencies remove this particular degeneracy symmetry; rational frequency ratios can still create other degeneracies.

Verify the quartic diagonal matrix element without expanding every product in $(a+a^\dagger)^4$ .

Solution

Put $x_0=\sqrt{\hbar/(2m\omega)}$ . Acting twice gives

X^2|n\rangle=x_0^2\left[ \sqrt{n(n-1)}|n-2\rangle+(2n+1)|n\rangle +\sqrt{(n+1)(n+2)}|n+2\rangle\right].

References

F. Iachello and A. Arima, The Interacting Boson Model, Cambridge University Press, 1987.
M. Moshinsky and Y. F. Smirnov, The Harmonic Oscillator in Modern Physics, Harwood Academic Publishers, 1996.
J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020. See angular momentum and perturbation theory.