Introduction to Quantum Mechanics, Fall 2014
Problem Set 3
Due Tuesday, October 7
Problem 1: On the Lie algebras g = su(2) and g = so(3) one can define the
Killing form K(·, ·) by
(X, Y ) ∈ g × g → K(X, Y ) = tr(XY )
1. For both Lie algebras, show that this gives a bilinear, symmetric form,
negative definite, with the basis vectors Xj in one case and lj in the other
providing an orthogonal basis if one uses −K(·, ·) as an inner product.
2. Another possible way to define the Killing form is as
K 0 (X, Y ) = tr(ad(X) ◦ ad(Y ))
Here the Lie algebra adjoint representation (ad, g) gives for each X ∈ g a
linear map
ad(X) : R3 → R3
and thus a 3 by 3 real matrix. This K 0 is determined by taking the trace
of the product of two such matrices. How are K and K 0 related?
Problem 2: Under the homomorphism
Φ : Sp(1) → SO(3)
what elements of SO(3) do the quaternions i, j, k (unit length, so elements
of Sp(1)) correspond to? Note that this is not the same question as that of
evaluating Φ0 on i, j, k.
Problem 3: In special relativity, we consider space and time together as
R4 , with an inner product such that hv, vi = v02 − v12 − v22 − v32 , where v =
(v0 , v1 , v2 , v3 ) ∈ R4 . The group of linear transformations of determinant one
preserving this inner product is written SO(1, 3) and known as the Lorentz
group. Show that, just as SO(4) has a double-cover Spin(4) = Sp(1) × Sp(1),
the Lorentz group has a double cover SL(2, C), with action on vectors given by
identifying R4 with 2 by 2 Hermitian matrices according to
v0 + v3 v1 − iv2
(v0 , v1 , v2 , v3 ) ↔
v1 + iv2 v0 − v3
and using the conjugation action of SL(2, C) on these matrices. (Hint: use
determinants).
1
Note that the Lorentz group has a spinor representation, but it is not unitary.
Problem 4: Consider a spin- 12 particle, with a state |ψ(t)i evolving in time
under the influence of a magnetic field of strength B = |B| in the 3-direction.
If the state is an eigenvector for S1 at t = 0, what are the expectation values
hψ(t)|Sj |ψ(t)i
at later times for the observables Sj (recall that Sj =
2
σj
2 )?