9
$\begingroup$

I have several questions about symmetry in quantum mechanics.

  1. It is often said that the degeneracy is the dimension of irreducible representation. I can understand that if the Hamiltonian has a symmetric group $G$, then the state space with the same energy eigenvalue will carry a representation of $G$. However, why this representation is usually irreducible?
  2. Is it true that the representation of continuous symmetric group must be unitary and cannot be anti-unitary?
  3. What is the difference between geometric symmetry and dynamical symmetry? By dynamical symmetry I mean for example the $\mathrm{SO}(4)$ symmetry of hydrogen. Some text refers dynamical symmetry to "internal" symmetry. How to state the definition of dynamical symmetry strictly?
$\endgroup$
4
$\begingroup$

Before going into the details, let me describe pictorially how the Hamiltonian, the Symmetry group, and the Dynamical group look in a basis in which the Hamiltonian is diagonal.

Hamiltonian

$$ H = \begin{bmatrix} \begin{bmatrix} \lambda_1 \mathbf{1} \end{bmatrix} & & & \\ & \begin{bmatrix} \lambda_2 \mathbf{1} \end{bmatrix} & & \\ & &\begin{bmatrix} \lambda_2 \mathbf{1} \end{bmatrix} & \\ & & &\begin{bmatrix} \lambda_3 \mathbf{1} \end{bmatrix} \\ \end{bmatrix}$$

Symmetry group

$$ g = \begin{bmatrix} \begin{bmatrix} \blacksquare \end{bmatrix}& & & \\ & \begin{bmatrix} \blacksquare \end{bmatrix}& & \\ & & \begin{bmatrix} \blacksquare \end{bmatrix} & \\ & & &\begin{bmatrix} \blacksquare \end{bmatrix} \\ \end{bmatrix}$$

A dynamical group

$$ g = \begin{bmatrix} & \blacksquare & & \\ & & \begin{bmatrix} \blacksquare & \blacksquare\\ \blacksquare & \blacksquare \\ \end{bmatrix} & \\ & & &\blacksquare \\ \end{bmatrix} $$

Full dynamical group

$$ g= \begin{bmatrix} \begin{bmatrix} \blacksquare & \blacksquare &\blacksquare & \blacksquare \\ \blacksquare & \blacksquare &\blacksquare & \blacksquare \\ \blacksquare & \blacksquare &\blacksquare & \blacksquare \\ \blacksquare & \blacksquare &\blacksquare & \blacksquare \\ \end{bmatrix} \end{bmatrix}$$ Where the black squares mean full submatrices, and square parentheses mean irreducible representations.

Of course, the full Hilbert space can be infinite dimensional; the figures capture only the conceptual framework.

The symmetry group is defined to be the group generated by all observables (Essentially self adjoint operators) commuting with the Hamiltonian. The generators of the symmetry group are constants of motion. The symmetry group cannot mix spaces with different eigenvalues of the Hamiltonian. Thus it must be block diagonal with respect to the Hamiltonian eigenspaces.

Please observe that in the illustrated example, there are two subspaces of eigenvalue $\lambda_2$, such that on each subspace the symmetry group acts irreducibly. This is a case of accidental degeneracy. In summary the answer to the first question is positive in the case where no accidental and negative if an accidental degeneracy is present.

The answer to the second question is positive in the case of a finite or compact group $G$ acting on the Hilbert space. In these cases due to the existence of a Haar measure $d\mu(g) ($, we can take any inner product $(.,.)_0$ on the Hilbert space and average it to obtain an invariant inner product $(.,.)$. This is called the Weyl unitarity trick. Explicitly:

$$ (u,v)^G = \int_G d\mu(g) (g \small{\circ} u,g \small{\circ} v) $$

due to the translation invariance of the Haar measure. Thus

$$ (g_1 \small{\circ} u,g_1 \small{\circ} v)^G = (u,v)^G, \; g_1 \in G$$

As a consequence the $G$ action can be either unitary or anti-unitary. When $G$ is a connected continuous group then there is a path connecting every element to the identity (which must be represented by a unitary operator), and a discrete property cannot change continuously, then the group element must act by a unitary operator.

A Dynamical group is a group generated by the observables of the system and has all the Hamiltonian spectrum under a single unitary irreducible representation. In particular a dynamical group resolves accidental degeneracies.

A dynamical group as illustrated in the third illustration. It has non-vanishing matrix elements between two degenerate representations of the symmetry group. In this way, we will be able to classify the degenerate vectors of the Hamiltonian by quantum numbers of a single irreducible representations of the dynamical group. Thus a dynamical group acting as in the third illustration can be used to label the degenerate subspaces of an accidental degeneracy. Of course there may be more accidental degeneracies in other sectors of the Hilbert space, thus a full dynamical group is defined as illustrated in figure (4).

Actually a more precise definition of the dynamical group requires a treatment of the system at the classical level. Please see the following thesis by S.G. Bartlett for details. The spectrum generating algebra is an algebra generated by functions separating points on the classical phase space. The dynamical group is a corresponding Lie group. The classical phase space becomes a coadjoint orbit of the dynamical group. An example is the 2-sphere

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.