Natural Science Forum / Physics / Research / July 2008

For any vector operator (A_1, A_2, A_3), one has
   [ L_r, A_s ] = -i hbar eps_{rst} A_t,
where eps_{rst} is the Levi-Civita symbol.  See, eg, Merzbacher,
"Quantum Mechanics".

This relation arises because L is the generator of rotations, i.e.,
the unitary operator
  U(n) = exp[ i L.n/hbar ]
has to rotate any given observable vector A, i.e.
 U(n)* A U(n) = R(n) A,
where R(n) denotes rotation in the direction n/|n| by an angle |n|.
In particular, for infinitesimal n, one can expand both sides of this
relation, to first order in n, to give
  (-i/hbar) [ L.n, A] = n x A,
and thence obtain the above commutation relation.

The rules are as follows:

[L_i, x_j] = i hbar eps(ijk) x_k

[L_i, p_j] = i hbar eps(ijk) p_k

where eps(ijk) is the Levi Civita symbol.

[L_i, V_j] = V_k for (i,j,k) = (1,2,3), (2,3,1), (3,1,2) for any 3-
vector V formed out of P, L, K, and the energy H by use of cross
products, dot products, multiplication by scalars and addition. [L_i,
S] = 0 for any scalar S formed out of P, L, K and H by the same set of
operations.

If you use vectors-valued dummy arguments to index the generators
with; i.e. L_u = L_1 u^1 + L_2 u^2 + L_3 u^3; similarly for K_u, P_u,
and V_u above, then [L_u, V_v] = V_{uxv}; [L, S] = (vector 0).

For ordinary QM, one has [K_u, K_v] = 0 for the boost generators; and
[K_u, P_v] = m (u.v). H is interpreted a kinetic energy, and one has
[K, H] = P.

For relativistic QM, one has [K_u, K_v] = -(1/c)^2 L_{uxv}, [K_u, P_v]
= M (u.v), where M = E/c^2 = m + H/c^2. H is the kinetic energy and E
is the total energy, here. The distinction between H and E, in this
context, is superfluous, but helps show what's going on in the non-
relativistic limit (1/c^2) -> 0.

The invariants in both cases are: m = M - H/c^2 (or just m = M for the
non-relativistic case); P^2 - 2MH + (1/c)^2 H^2 = P^2 - E^2/c^2 +
(mc)^2 (or just P^2 - 2MH for the non-relativistic case). Plus you
have the invariant W^2 - (1/c)^2 W_0^2 (or just W^2 if (1/c)^2 -> 0),
where W = ML + PxK and W_0 = P.L gives you mass x spin. This is
present both relativistically AND non-relativistically.

A consequence of the points raised above:
  [L_i, W_j] = W_k where (i,j,k) = (1,2,3), (2,3,1) and (3,1,2); or
[L_u, W_v] = W_{uxv};
and
  [L_i, W_0] = 0.

It's a remarkable, but little-known, fact that the same process used
to convert the invariant P^2 - E^2/c^2 + (mc)^2 into the Dirac
equation can be used for P^2 - 2MH + k H^2 for any k. The resulting
algebra is independent of what k is! (In fact, it's just the
complexified Dirac algebra gamma_0, gamma_1, gamma_2, gamma_3,
gamma_5). The "Dirac" equation for k = 0 is equivalent to the
Schroedinger equation. Though quadratic, it "reduces" to linear in H
because M reduces to a constant, m. So you more commonly see it as H =
P^2/2m, when really you ought to be writing P^2 - 2mH = 0 to get a
better picture of what's going on here.

So, that should answer your question and more.