An eigenvalue ofℓ(ℓ+ 1) ̄h^2 = 2 ̄h^2 impliesℓ= 1. We will need a linear combination of theY 1 m
to get the eigenstate ofLx=L++ 2 L−.
L++L−
2
(AY 11 +BY 10 +CY 1 − 1 ) = 0
̄h
√
2
(AY 10 +BY 11 +BY 1 − 1 +CY 10 ) = 0
(BY 11 + (A+C)Y 10 +BY 1 − 1 ) = 0
Since this is true for allθandφ, each term must be zero.
B= 0
A=−C
The state is
1
√
2
(Y 11 −Y 1 − 1 )
The trivial solution thatA=B=C= 0 is just a zero state, not normalizable to 1.
- Write the (normalized) state which is an eigenstate ofL^2 with eigenvalueℓ(ℓ+1) ̄h^2 = 2 ̄h^2 and
also aneigenstate ofLywith eigenvalue 1 ̄hin terms of the usualYℓm.
- Calculate thecommutators[pz,Lx] and [L^2 x,Lz].
5.Derivethe relationL+Yℓm= ̄h
√
ℓ(ℓ+ 1)−m(m+ 1)Yℓ(m+1).
- A particle is in al= 1 state and is known to have angular momentum in thexdirection equal
to + ̄h. That isLxψ= ̄hψ. Since we knowl= 1,ψmust have the formψ=R(r)(aY 1 , 1 +
bY 1 , 0 +cY 1 ,− 1 ). Find the coefficientsa,b,andcforψnormalized.
- Calculate the following commutators: [x,Lz], [L+,L^2 ], [^12 mω^2 r^2 ,px].
- Prove that, if the Hamiltonian is symmetric under rotations, then[H,Lz] = 0.
- In 3 dimensions, a particle is in the state:
ψ(r) =C(iY 33 − 2 Y 30 + 3Y 31 )R(r)
whereR(r) is some arbitrary radial wave function normalized such that
∫∞
0
R∗(r)R(r)r^2 dr= 1.
a) Find the value of C that will normalize this wave function.
b) If a measurement ofLzis made, what are the possible measured values and what are
probabilities for each.
c) Find the expected value of< Lx>in the above state.
- Two (different) atoms of massesM 1 andM 2 are bound together into the ground state of a
diatomic molecule. The binding is such that radial excitations can be neglected at low energy
and that the atoms can be assumed to be a constant distancea= 3 ̊Aapart. (We will ignore
the small spread aroundr=a.)
a) What is the energy spectrum due to rotations of the molecule?
