Physics 506 : Solutions for Problem set 1
Due : September 5, 2014
1
Problem 1
a) Express ijk lmn in terms of Kronecker delta symbols.
Sol:
δil δim δin
ijk lmn = δjl δjm δjn
δ
δkm δkn
kl
.
Thus ijk lmn = (δil δjm δkn + δjl δkm δin + δkl δim δjn ) − (δin δjm δkl + δjn δkm δil + δkn δim δjl ) .
P
b) Evaluate ij ijk ijm .
Sol: From a) we have
X
X δii δij δim δji δjj δjm .
ijk ijm =
ij
ij δki
δkj δkm X
ijk ijm =
ij
X
(δii δjj δkm + δji δkj δim + δki δij δjm ) −
X
(δim δjj δki + δjm δkj δii + δkm δij δji )
ij
ij
= (9δkm + δkm + δkm ) − (3δkm + 3δkm + 3δkm ) = 2δkm .
P
P
Sol2: From i ijk ilm = δjl δkm − δjm δkl , we can get j (δjj δkm − δjm δkj ) = 2δkm by multiplying
P
δlj and summing over the index j. Where j δjj = 3.
P P
c) Evaluate ijk lmn ijk lmn Mil Mjm Mkn , where M is an arbitrary matrix.
Sol:
X
X
XX
ijk lmn Mil Mjm Mkn =
ijk ijk det M =
2δkk det M = 6 det M.
ijk lmn
ijk
k
d) Show explicitly that
Sol:
~ × B)
~ · (C
~ × D)
~ = ijk ilm Aj Bk Cl Dm = (δjl δkm − δjm δkl )Aj Bk Cl Dm
(A
= (Aj Cj )(Bk Dk ) − (Aj Dj )(Bk Ck )
~ · C)(
~ B
~ · D)
~ − (A
~ · D)(
~ B
~ · C)
~ .
= (A
2
Problem 2
~ given by
An electric charge q1 moving with velocity ~v1 produces a magnetic induction B
~ = µ0 q1 ~v1 × r̂ ,
B
4π
r2
~ is measured.
where r̂ is a unit vector that points from q1 to the point at which B
a) Evaluate the magnetic force F~2 exerted by q1 on the second charge q2 , with velocity ~v2 , in terms
of vector triple product.
~ 1 = 0)
Sol: From the Lorentz force equation, we get (for E
~ 1 + ~v2 × B
~ 1 = µ0 q1 q2 ~v2 × (~v1 × r̂) .
F~2 = q2 E
4π r2
b) What is the corresponding magnetic force F~1 that q2 exerts on q1 . Be careful about the unit
radial vector. Comments on these two forces.
Sol: The solution is the same as a) except the role of 1 and 2 are changed. One of the important
distinction is the radial vector ~r0 that is from q2 to q1 . Thus ~r0 = −~r as they are vector quantity.
~ 2 + ~v1 × B
~ 2 = µ0 q1 q2 ~v1 × (~v2 × (r̂0 )) = − µ0 q1 q2 ~v1 × (~v2 × r̂) .
F~1 = q1 E
4π (r0 )2
4π r2
In general there is no simple relation between F~1 and F~2 . Specifically, Newton’s third law, F~1 = −F~2
does not hold.
c) Calculate F~1 and F~2 for the case of q1 and q2 moving along parallel trajectories side by side. Are
there simple relation?
Sol: Parallel trajectory means ~v = ~v2 = ~v1 , and side by side means r̂ · ~v = 0. We already developed
the identity ~v × (~v × r̂) = ~v (~v · r̂) − r̂(~v · ~v ) = −v 2 r̂. Thus
µ0 q1 q2 2
F~2 = −
v r̂ = −F~1 .
4π r2
There are mutual attractions.
3
Problem 3
One description of spin 1

0 1
1 
Mx = √
1 0
2
0 1
particle uses the matrices


0
0 −i
1
1  , My = √  i 0
2
0
0 i

0
−i  ,
0

1
Mz =  0
0
0
0
0

0
0  .
−1
a) Derive commutation relations (with the commutator [A, B] = AB − BA) for the matrices, and
express them in a single formula using the Levi-Civita symbol.
Sol: One can check [Mx , My ] = iMz , [My , Mz ] = iMx , [Mz , Mx ] = iMy by doing explicit computations. They can be written in a simple formula as
[Mi , Mj ] = iijk Mk .
b) Evaluate M 2 = Mx2 + My2 + Mz2 .
Sol:
M 2 = Mx2 + My2 + Mz2 = 2I3×3 .
c) Construct the raising and lowering operators L+ = Mx + iMy and L− = Mx − iMy . Then show
[M 2 , Mi ] = 0 ,
[Mz , L+ ] = L+ ,
[L+ , L− ] = 2Mz .
Sol: Using the following raising and lowering operators



0 2 0
0
1
1
L− = √  2
L+ = √  0 0 2  ,
2
2
0 0 0
0
0
0
2

0
0  .
0
one can check the relations explicitly. These will be very useful later on when we are studying
angular momentum.
4
Problem 4
Repeat the Problem 3 using Mathematica for the following matrices of spin 3/2
√
√





0
3
0
0
0
−
3 0
0
3
√
√





i
1 0
1
3 0
2 √0 
3
0
−2
0 
√
, My = 
Mx = 
 , Mz = 2  0
2 √0
3 
2
0
−
3
2 0
2 0
√
0
0
0
0
0
3 0
3
0
Please print out your Mathematica file and hand in along the other problems.
Sol: The solution will be presented later ...
0
1
0
0
0
0
−1
0

0
0 
 .
0 
−3