Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

and [J 2 1 , J z] = [J 2 2 , J z] = 0:
[J1 2, J 2 ] = [J1 2, J xJ x + J y J y + J z J z ]
= J x [J1 2, J x] + [J1 2, J x]J x + (. . .)
= J x [J1 2, J 1x + J 2x ] + [J1 2, J 1x + J 2x ]J x + (. . .)
= J x [J1 2, J 1x] + [J1 2, J 1x]J x + (. . .)
i = 0
(C.3)
and
[J 2 1 , J z ] = [J 2 1 , J 1z + J 2z ] = [J 2 1 , J 1z ] = 0 (C.4)
Let us label with j 1 , j 2 , j and m the quantum numbers characterizing the
eigenvalues of our operators. For fixed j 1 and j 2 , j will assume values between
j min and j max . By definition, m = m 1 + m 2 , where m is the projection of
J z = J 1z + J 2z . Thus, we deduce that j max = j 1 + j 2 . j min can be obtained
by the condition that the total number of states is the same as the one for the
original representation, i.e.
j 1
∑+j 2
j=j min
(2j + 1) = (2j 1 + 1)(2j 2 + 1)
(C.5)
It can be verified that this conditions yields j min = |j 1 − j 2 |. o In summary:
j = |j 1 −j 2 |, . . . , j 1 +j 2 . One can use a ”vector” picture: for j = |j 1 −j 2 | the two
vectors have same direction and opposite sign, for j = j 1 +j 2 same direction and
same sign, while intermediate cases correspond to angular momenta pointing
in different directions.
As an important example, let us consider the case j 1 = 1/2 and j 2 = 1/2.
There are four independent states, m 1 = ±1/2 and m 2 = ±1/2. Let us move
to the coupled representation. The allowed values for j are j = 0 and j = 1.
For j = 0 only m = 0 is possible (“singlet state”). For j = 1 one has m = 0, ±1
(“triplet states”). In total, there are always four states.
The reason to introduce the coupled representation, apparently equivalent
to the “ordinary” one, is that in many cases there are terms in the Hamiltonian
that couple angular momenta. Quite common for instance is the case of a
“torque” that tends to align the two vectors:
H = . . . − AJ 1 · J 2
(C.6)
If such term is present, J 1z and J 2z are no longer conserved and their eigenvalues
are no longer good quantum numbers. In fact
[J 1z , −AJ 1 · J 2 ] = −A[J 1z , J 1x J 2x + J 1y J 2y + J 1z J 2z ] (C.7)
= −A[J 1z , J 1x ]J 2x − A[J 1z , J 1y ]J 2y (C.8)
= −i¯hA(J 1y J 2x − J 1x J 2y ) (C.9)
and this operator is in general not zero.
It is however immediate to verify that J 1z +J 2z is conserved, since [J 2z , −AJ 1·
J 2 ] is equal to the term calculated above, but with opposite sign.
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