( j )
( j )
1) argue that the eigenvalues of J x , y are the same as that of J z (namely: j , ( j 1) , ..., ( j ) . Blah diddly blah blah — symmetry. symmetry. ( j) J . Generalize the result to ˆ ( j) J be related to J z by a unitary similarity transformation, In order to preserve normalization, let ˆ
ˆ J
( j)
†
U J z U
[ .1] 1
†
Note that we we know the eigenvalue eigenvalue relation, relation, which we can front-multiply front-multiply by U, and insert insert I U U U U , and having [ .1] at the ready,
J z m m m
UJ z U †U m m U m
( ˆ J ( j ) ) †U m m U m
J Note that this implies implies U m is an eigenket of the (Hermitian!) operator ( ˆ
[ .2]
( j ) †
) , with the same eigenvalues m .
2) Show that the zero operator can be written as, O ( J
?????
m j
j )( J ( j 1) )( J ( j 2) )...( J j ) m j ( J m ) 0 J ˆ J ( j )
[ .3] [ .4]
Do a unitary transform to the operator O to make the J into J z, as in [ .1], which can be done by inserting the identiy †
operator I U U at ―strategic‖ points in [ .3] to get, O z
†
U OU U
†
m j m j
I( J
m )I U U
†
m j
m j
UU † ( J m )UU † U I m j ( J z I m )I m j
[ .5]
†
Let the operator O z U OU act on an arbitrary ket, and let us expand that – ket ket in the (complete) eigenket basis jm ,
jm | O z jm jm |
O z
m j
m
j
( J z m ) jm
[ .6]
In [ .6], .6], we notice the appearance of a sort of ―complimentary Kronecker delta‖ . Since j m j , we can write the operator eigenrelation,
( J z m ) jm (m m ) jm (m m)(1 mm ) jm Putting [ .7] into [ .6], and noting
m j
m
j
[ .7]
(1 mm ) 0 for j m j , we immediately get, O z
jm | 0 jm 0
[ .8]
It is then trivial to undo the similarity transformation effected in [ .5],
UO z U †
U 0U
†
0 UU †OUU † IOI O
O
the zero operator
[ .9]
2 j 1
3) it follows from the result above that J
0
1
2j
is a linear combination of J , J ,..., J . Argue the same goes for
J 2 j k , k 1, 2,... 2 j 1
huh? This seems useless. It’s obvious that since J combination of all the previous J’s.
Do this by mathematical induction. Assume way of sum-index-fiddling, Answer:
is block-diagonal, you of course have a J being a linear
is true, and this implies the following, by
[ .10]
The proof for
tricky — what is it meant when
follows from just repeating the same induction [ .10]. One of the steps looks a little
????
