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CEM 882, Problem Set 5 – Due Monday, April 13 in class
1.
a. (20 points) For the hydrogen atom, determine the m selection rule for the μ y dipole moment
operator.
For the μy component of a m 2 ← m 1 transition:
2 1 2 1
2 1 1 2 1 2
(^2) * 1 2 0 0 (^1 2 ) ( 1) ( 1) 0 0
(sin ) (2 ) (sin )
(2 ) (2 ) ( ) { 2 2
im im m m
im i i im i m m i m m
d e e d
e e e e d e d e d i i
π π (^) φ φ
π φ φ φ φ π φ
0
π φ
− −
− − − − − +
− −
d
Each integral is only non-zero if the argument of the exponential is 0 and results in a selection rule.
Δ m = ± 1
b. (20 points) For the hydrogen atom, determine the m selection rule for the μ z dipole moment
operator.
For the μz component of a m 2 ← m 1 transition:
2 1 1 2 2 1
(^2) * 1 2 1 2 ( ) 0 0 0 m m d^ (2^ )^ e^ im^ e^ im^ d^ (2^ ) ei m^ m
π π φ φ π φ
∫ Φ^ Φ^ φ^ =^ π−^ ∫ −^ φ^ = π −∫ −^ φ
The integral is only non-zero if the argument of the exponential is 0 and results in a selection rule.
Δ m = 0
a. (20 points) For magnetic resonance, determine the m selection rule for the μ y dipole moment
operator.
For the μy component of a S m , 2 ← S m , 1 transition:
2 1 2 1 2 1
1/2 1/ 2 1 2 1
( ){[ ( 1) ( 1)] , , 1 [ ( 1) ( 1)] , , 1 }
S m (^) y S m (^) i S m S S m S m S S m
S S m m S m S m S S m m S m S m i
h
h −
The first integral is non-zero for m 2 (^) = m 1 + 1 and the second integral is non-zero for m 2 (^) = m 1 − 1
and result in a Δ m = ± 1 selection rule.
b. (20 points) For magnetic resonance, determine the m selection rule for the μ z dipole moment
operator.
For the μz component of a S m , 2 ← S m , 1 transition:
S m , 2 μ z S m , 1 = ( γh) S m , 2 S z S m , 1 =( γh) S m , 2 S m , 1
The integral is non-zero for m 2 = m 1 and results in aΔ m = 0 selection rule.
c. (20 points) Use results from the class notes and from problem 2a,b to define the relative orientation of the uniaxial and the radiative magnetic fields in a magnetic resonance spectrometer. Provide a clear explanation of your reasoning using equations as necessary.
In a magnetic resonance spectrometer, the uniaxial magnetic field is along z direction and the
radiative magnetic field is in the xy plane. The radiative magnetic field will interact with the μ x
or μ y components of the magnetic dipole moment. According to the selection rule , the
interaction will induce a change in the spin states of the spin system. The associated energy difference will then be detected by the spectrometer.
Δ m = ± 1
3. In the simplest pulsed magnetic resonance experiment of a spin ½ particle, a “90o^ pulse” is applied followed by detection. After the 90o^ pulse, the wavefunction at time t = 0 is:
ψ (0) = (2)–½^ { | ½ , + ½〉 + | ½ , – ½〉 }
a. (20 points) Determine the wavefunction ψ ( t ) at some later time t. Your wavefunction should
contain ω 0 = γ B B 0.
The wavefunction at time 0 has the form (^1 )
ψ = c + + c − , where
1 2
c = c =. The time evolution of the “ c ” parameters is determined by the relation
i Ekt ck t ck e
− = ⋅ h^ where k = 1 or 2. The term Ek is determined by time-independent
Schrodinger Equation H Ψ = E Ψ where H = −h ω 0 Sz for the external magnetic field. Thus, for
the spin states
0 0 1
0 0 2
z
z
S E
S E
0
0
ω ω ω
ω ω
h h
h h ω
h
h
The expression of the wavefunction at time t is 0 0 ( ) 2 2 1 , 1 2 2 1 ,^1 2 2 2 2 2
i t i t t e e
ω ω
2
ψ
− = + + −
S
b. (30 points) The observable property in the magnetic resonance experiment is μ (^) + = μ (^) x + i μ (^) y where i is the imaginary number. Determine an expression for the average value of μ (^) + as a function of time. This calculation is most straightforward if you first express μ (^) + in terms of S (^) + and/or S (^) –. Also note that S (^) + | ½ , + ½〉 = 0 and S (^) – | ½ , – ½〉 = 0.
μ (^) + = μ x (^) + i μ (^) y = − (γ h S (^) x + i γh S (^) y ) = −γ h( Sx + iSy )= −γh +
The average value of μ + can be calculated by
