21.2 so(4) symmetry and the Coulomb potential

21.2 so(4) symmetry and the Coulomb potential

Concept links · terms present in this machine draft; source roles are unverified: eigenvalue · Lie algebra · irreducible representation

The Coulomb potential problem is very special in that it has an additional symmetry, of a non-obvious kind. This symmetry appears even in the classical problem, where it is responsible for the relatively simple solution one can find to the essentially identical Kepler problem. This is the problem of finding the classical trajectories for bodies orbiting around a central object exerting a gravitational force, which also has $\mathrm{a}\frac{1}{r}$ potential.

Kepler’s second law for such motion comes from conservation of angular momentum, which corresponds to the Poisson bracket relation

$$\{l_j,h\}=0$$

Here we’ll take the Coulomb version of the Hamiltonian that we need for the hydrogen atom problem

$$h=\frac{1}{2m}|\mathbf{p}{|}^2-\frac{e^2}{r}$$

The relation $\{l_j,h\}=0$ can be read in two ways:

• The Hamiltonian $h$ is invariant under the action of the group $\left(SO(3)\right)$ whose infinitesimal generators are $l_j$

• The components of the angular momentum $(l_j)$ are invariant under the action of the group (R of time translations) whose infinitesimal generator is $h,$ so the angular momentum is a conserved quantity.

For this special choice of Hamiltonian, there is a diferent sort of conserved quantity. This quantity is, like the angular momentum, a vector, often called the Lenz (or sometimes Runge-Lenz, or even Laplace-Runge-Lenz) vector:

Definition (Lenz vector)

The Lenz vector is the vector-valued function on the phase space ${\dot{\mathbf{R}}}^6$ given by

$$\mathbf{w}=\frac{1}{m}(\mathbf{l}\times \mathbf{p})+e^2\frac{\mathbf{q}}{|\mathbf{q}|}$$

Simple manipulations of the cross-product show that one has

$$\mathbf{l}\cdot \mathbf{w}=0$$

We won’t here explicitly calculate the various Poisson brackets involving the components $w_j$ of $\mathbf{w}.$ since this is a long and unilluminating calculation, but will just quote the results, which are

$$\{w_j,h\}=0$$

This says that, like the angular momentum, the vector with components $w_j$ is a conserved quantity under time evolution of the system, and its components generate symmetries of the classical system.

$$\{l_j,w_k\}={ϵ}_{jkl}{w}_l$$

These relations say that the generators of the $SO(3)$ symmetry act on $w_j$ in the way one would expect for the components $w_j$ of a vector in ${\mathbf{R}}^3$

$$\left\{w_j,w_k\right\}={ϵ}_{jkl}{l}_l\left(\frac{-2h}{m}\right)$$

This is the most surprising relation, and it has no simple geometrical explanation (although one can change variables in the problem to try and give it one). It expresses a highly non-trivial relationship between the Hamiltonian h and the two sets of symmetries generated by the vectors $l$, $w$.

The $w_j$ are cubic in the $q$ and $p$ variables, so the Groenewold-van Hove no-go theorem implies that there is no consistent way to quantize this system by finding operators $W_j,L_j,H$ providing a representation of the Lie algebra generated by the functions $w_j,l_j,h$ (taking Poisson brackets). Away from the locus $h=0$ in phase space, the function $h$ can be used to rescale the $w_j.$ , defining

$$k_j=\sqrt{\frac{-m}{2h}}{w}_j$$

and the functions $l_j,k_j$ then do generate a finite dimensional Lie algebra. Quantization of the system can be performed by finding appropriate operators $W_j$ ， then rescaling them using the energy eigenvalue, giving operators $L_j,K_j$ that provide a representation of a finite dimensional Lie algebra on energy eigenspaces.

A choice of operators $W_j$ that will work is

$$\mathbf{W}=\frac{1}{2m}(\mathbf{L}\times \mathbf{P}-\mathbf{P}\times \mathbf{L})+e^2\frac{\mathbf{Q}}{|\mathbf{Q}|}$$

where the last term is the operator of multiplication by $e^2{q}_j/|\mathbf{q}|$ . By elaborate and unenlightening computations the $W_j$ can be shown to satisfy the commutation relations corresponding to the Poisson bracket relations of the $w_j:$

$$\begin{array}{rl}[W_j,H]& =0\\ & =i\hslash {ϵ}_{jkl}{W}_l\\ & =i\hslash {ϵ}_{jkl}{L}_l\left(-\frac{2}{m}H\right)\end{array}$$

as well as

$$\mathbf{L}\cdot \mathbf{W}=\mathbf{W}\cdot \mathbf{L}=0$$

The first of these shows that energy eigenstates will be preserved not just by the angular momentum operators $L_j,$ but by a new set of non-trivial operators, the $W_j$ , so will be representations of a larger Lie algebra than so(3). In addition, one has the following relation between $W^2$ , $H$ and the Casimir operator $L^2$

$$W^2=e^{4}1+\frac{2}{m}H(L^2+{\hslash }^{2}1)\text{\hspace{1em}(21.2)}$$

If we now restrict attention to the subspace ${\mathcal{H}}_E\subset \mathcal{H}$ of energy eigenstates of energy $E,$ on this space we can define rescaled operators

$$\mathbf{K}=\sqrt{\frac{-m}{2E}}\mathbf{W}$$

On this subspace, equation 21.2 becomes the relation

$$2H(K^2+L^2+{\hslash }^{2}1)=-m{e}^{4}1$$

and we will be able to use this to find the eigenvalues of $H$ in terms of those of $L^2$ and $K^2$

We will assume that $E<0.$ , in which case we have the following commutation relations

$$\begin{array}{rl}[L_j,L_k]& =i\hslash {ϵ}_{jkl}{L}_l\\ & =i\hslash {ϵ}_{jkl}{K}_l\\ & =i\hslash {ϵ}_{jkl}{L}_l\end{array}$$

Defining

$$\mathbf{M}=\frac{1}{2}(\mathbf{L}+\mathbf{K}),\mathbf{N}=\frac{1}{2}(\mathbf{L}-\mathbf{K})$$

one has

$$\begin{array}{rl}[M_j,M_k]& =i\hslash {ϵ}_{jkl}{M}_l\\ [N_j,N_k]& =i\hslash {ϵ}_{jkl}{N}_l\\ [M_j,N_k]& =0\end{array}$$

This shows that we have two commuting copies of $\mathfrak{so}(3)$ acting on states, spanned respectively by the $M_j$ and $N_j$ , with two corresponding Casimir operators $M^2$ and $N^2$

Using the fact that

$$\mathbf{L}\cdot \mathbf{K}=\mathbf{K}\cdot \mathbf{L}=0$$

one finds that

$$M^2=N^2$$

Recall from our discussion of rotations in three dimensions that representations of $\mathfrak{so}(3)=\mathfrak{su}(2)$ correspond to representations of $Spin(3)=SU(2)$ , the double cover of $SO(3)$ and the irreducible ones have dimension $2l+1$ , with $l$ half-integral. Only for l integral does one get representations of $SO(3)$ , and it is these that occur in the $SO(3)$ representation on functions on ${\mathbf{R}}^3$ . For four dimensions, we found that $Spin(4)$ , the double cover of $SO(4)$ , is $SU(2)\times SU(2)$ and one thus has spin $(4)=\mathfrak{so}(4)=\mathfrak{su}(2)\times \mathfrak{su}(2)=\mathfrak{so}(3)\times \mathfrak{so}(3)$ . This is exactly the Lie algebra we have found here, so one can think of the Coulomb problem at a fixed negative value of $E$ as having an $\mathfrak{so}(4)$ symmetry. The representations that will occur can include the half-integral ones, since neither of the two so $(3)$ factors is the $\mathfrak{so}(3)$ of physical rotations in 3-space (the physical angular momentum operators are the $\mathbf{L}=\mathbf{M}+\mathbf{N})$

The relation between the Hamiltonian and the Casimir operators $M^2$ and $N^2$ is

$$2H(K^2+L^2+{\hslash }^{2}1)=2H(2M^2+2N^2+{\hslash }^{2}1)=2H(4M^2+{\hslash }^{2}1)=-m{e}^{4}1$$

On irreducible representations of $\mathfrak{so}(3)$ of spin $\mu ,$ , we will have

$$M^2=\mu (\mu +1){\hslash }^{2}1$$

for some half-integral $\mu ,$ so we get the following equation for the energy eigenvalues

$$E=-\frac{-m{e}^4}{2{\hslash }^2(4\mu (\mu +1)+1)}=-\frac{-m{e}^4}{2{\hslash }^2(2\mu +1{)}^2}$$

Letting $n=2\mu +1$ , for $\mu =0,\frac{1}{2},1,...$ . we get $n=1,2,3,...$ . and precisely the same equation for the eigenvalues described earlier

$$E_n=-\frac{m{e}^4}{2{\hslash }^2{n}^2}$$

One can show that the irreducible representations of the product Lie algebra $\mathfrak{so}(3)\times \mathfrak{so}(3)$ are tensor products of irreducible representations of the factors, and in this case the two factors in the tensor product are identical due to the equality of the Casimirs $M^2=N^2$ . The dimension of the $\mathfrak{so}(3)\times \mathfrak{so}(3)$ irreducibles is thus $(2\mu +1{)}^2=n^2$ , explaining the multiplicity of states one finds at energy eigenvalue $E_n$

The states with $E<0$ are called “bound states” and correspond physically to quantized particles that remain localized near the origin. If we had chosen $E>0$ , our operators would have satisfied the relations for a diferent real Lie algebra, called so(3, 1), with quite diferent properties. Such states are called “scattering states”, corresponding to quantized particles that behave as free particles far from the origin in the distant past, but have their momentum direction changed by the Coulomb potential (the Hamiltonian is non-translation invariant, so momentum is not conserved) as they propagate in time.

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