### Orbital angular momentum

#### Problem:

Consider a spinless particle of mass m.  What values can the magnitude of the angular momentum J take in quantum mechanics when it is measured?

Solution:

• Concepts:
Orbital angular momentum in QM
• Reasoning:
The only type of angular momentum for this particle is orbital angular momentum.
• Details of the calculation:
When measuring L2 we only can obtain an eigenvalue l(l +1)ħ2, with l a non-negative integer.
The magnitude of the angular momentum J therefore can only be measured as
J = (l(l +1))½ħ, l = 0, 1, 2, ... .

#### Problem:

The magnitude of the orbital angular momentum L of a hydrogen atom is found to be 30½ħ.  Lz is measured and found to be 3ħ.
(a)  Which values of the principle quantum number n are consistent with these measurements?
(b)  What is the value of Lx2 + Ly2?

Solution:

• Concepts:
Angular momentum, the hydrogen atom
• Reasoning:
For the hydrogen atom H, L2 and Lz are commuting observables.
• Details of the calculation:
(a)  L2 = l(l+1)ħ2 = 30ħ2,  l = 5.  For the hydrogen atom the allowed value for l are all non-negative integers equal to or less than n - 1.  So n must be a positive integer greater or equal to 6.
(b)  Lx2 + Ly2 = L2 - Lz2 = (30 - 9)ħ2  = 21ħ2

#### Problem:

A given normalized wave function is ψ(θ,φ) = N sinθ cosφ.
(a)  Write this wave function in terms of spherical harmonics.
(b)  Find the normalization constant N.
(c)  What is the mean value of L2 for a state with this wave function?
(d)  Lz is measured.  What are the possible results of this measurement and what are the respective probabilities?

Solution:

• Concepts:
The spherical harmonics, the postulates of quantum mechanics
• Reasoning:
If we want to find the probability of measuring the eigenvalue a of an observable A we must express the wave function of the system as a linear combination of eigenfunctions of A.  Then  P(a) = Σi|<ai|ψ>|2.
• Details of the calculation:
Express the wave function in terms of spherical harmonics.
(a)  cosφ = ½(exp(iφ) + exp(-iφ)).
ψ(θ,φ)  = -(N/2)(8π/3)½Y11 + (N/2)(8π/3)½Y1-1.
(b)  ∫0πsinθ dθ∫0dφ Y*l'm'(θ,φ)Ylm(θ,φ) = δl'lδm'm.
1 = (N2/2)(8π/3).  N2 = 2/(8π/3)½.
ψ(θ,φ)  = 2½Y1-1 - 2/Y11.
(c)  The given wave function is an eigenfuction of L2 with eigenvalue 2ħ2.
(d)  The possible results of measuring Lz are  ħ and -ħ.  The probability of obtaining each of these results is ½.

#### Problem:

Show that a state with position wave function (y - iz)k is an eigenstate of the orbital angular momentum operator Lx and find its eigenvalue.  Are there conditions on k?

Solution:

• Concepts:
Eigenfunctions, orbital angular momentum operators
• Reasoning:
The orbital angular momentum operator in Cartesian coordinate has the form L = r × p
We check if (y - iz)k is an eigenfunction of Lx = ypz - zpy = (ħ/i)(y∂/∂z - z∂/∂y).
• Details of the calculation:
∂(y - iz)k/∂z = -k*(y - iz)k-1*i, ∂(y - iz)k/∂y = k*(y - iz)k-1.
Lx(y - iz)k = (ħ/i)(-iky - kz) (y - iz)k-1 = -ħk(y - iz)k.  The eigenvalue is -ħk.
k must be an integer.

#### Problem:

For a simple particle moving in space, show that the wave function ψlm(r) = x2 + y2 - 2z2 represents a simultaneous eigenstate of L2 and Lz with eigenvalues l(l + 1)ħ2 and mħ.  Determine l and m.  Find a function with the same eigenvalue for L2 and the maximum possible eigenvalue for Lz.

Solution:

• Concepts:
The eigenfunctions of the orbital angular momentum operator, the spherical harmonics
• Reasoning:
The common eigenfunctions of L2 and Lz are the spherical harmonics.  We have to write the given wave functions in terms of the spherical harmonics.
• Details of the calculation:
ψlm(r) = x2 + y2 - 2z2 = r2sin2θcos2φ + r2sin2θsin2φ - 2r2cos2θ = r2(sin2θ - 2cos2θ)
in spherical coordinates.
Express ψlm(r) in terms of the spherical harmonics.
ψlm(r) =  r2(1 - cos2θ -  2cos2θ)  = r2(1 - 3cos2θ)
= -r2(16π/5)½Y20(θ,φ).
[Formulas for some of the spherical harmonics:
Y00 = (4π),  Y1±1 = ∓(3/8π)½sinθ exp(±iφ),  Y10 = (¾π)½cosθ,
Y2±2 = (15/32π)½sin2θ exp(±i2φ),  Y2±1 = ∓(15/8π)½sinθ cosθ exp(±iφ),
Y20 = (5/16π)½(3cos2θ - 1).]
ψlm(r) is an eigenfunction of Lz with eigenvalue 0 and an eigenfunction of L2 with eigenvalue  6ħ2 (l = 2).  For l = 2 the maximum possible eigenvalue of Lz is mħ = 2ħ.  The corresponding eigenfunction is
Cr2Y22(θ,φ) = Cr2(15/32π)½sin2θ exp(±i2φ).

#### Problem:

The wave function of a particle subjected to a spherically symmetric potential U(r) is given by ψ(r) = (x + y - 3z)f(r).
(a)  Is ψ an eigenfunction of L2?  If so, what is its l value?  If not, what are the possible values of l we may obtain when L2 is measured?
(b)  What are the probabilities for the particle to be found in various ml states?
(c)  Suppose it is known somehow that ψ(r) is an energy eigenfunction with eigenvalue E.  Indicate how we may find U(r).

Solution:

• Concepts:
The eigenfunctions of the orbital angular momentum operator, the spherical harmonics
• Reasoning:
The common eigenfunctions of L2 and Lz are the spherical harmonics.  We have to write the given wave functions in terms of the spherical harmonics.
• Details of the calculation:
ψ(r) = (x + y - 3z)f(r) = (rsinθcosφ + rsinθsinφ - 3rcosθ)f(r)
= rf(r)(sinθcosφ + sinθsinφ - 3cosθ)
= rf(r)(sinθ ½(e + e-iφ) - i sinθ ½(e - e-iφ) - 3cosθ)
= rf(r)(sinθ ½e (1 - i) + sinθ ½ e-iφ (1 + i) - 3cosθ)
= rf(r)(2exp(-iπ/2) sinθe  + 2exp(iπ/2) sinθ e-iφ - 3cosθ).

Y1±1 = ∓(3/8π)½sinθ exp(±iφ),  Y10 = (¾π)½cosθ.
ψ(r) = rf(r)(-exp(-iπ/2)(4π/3)½Y11(θ,φ) + exp(iπ/2)(4π/3)½Y1-1(θ,φ) - (12π)½Y10(θ,φ))
= rf(r)2√π(-exp(-iπ/2)(1/3)½Y11(θ,φ) + exp(iπ/2)(1/3)½Y1-1(θ,φ) - (3π)½Y10(θ,φ))
=  (44π/3)½rf(r)(-exp(-iπ/2)/11½Y11(θ,φ) + exp(iπ/2)/11½Y1-1(θ,φ) - 3/11½Y10(θ,φ))
= R(r) (-exp(-iπ/2)/11½Y11(θ,φ) + exp(iπ/2)/11½Y1-1(θ,φ) - 3/11½Y10(θ,φ)).
The function of angle is normalized.
ψ(r) is  an eigenfunction of L2 with eigenvalue 2ħ2, l = 1.

(b)  The possible values of ml are 1, 0, -1.
P(ml = 1) = 1/11,  P(ml = -1) = 1/11,  P(ml = 0) = 9/11.

(c)  The wave function ψklm(r,θ,φ) = Rkl(r)Ylm(θ,φ) = [ukl(r)/r]Ylm(θ,φ) is a product of a radial function Rkl(r) and the spherical harmonic Ylm(θ,φ).  The differential equation for ukl(r) is
[-(ħ2/(2m))(∂2/∂r2) + ħ2l(l+1)/(2mr2) + U(r)]ukl(r) = Eklukl(r).
Since ukl(r), Ekl and l are known, we can solve this equation for U(r).

#### Problem:

The ladder operator L+ is given by
L+ = ħ exp(iφ) [∂/∂θ + i cotθ ∂/∂φ].
The spherical harmonics are given by Ylm(θ,φ) = Flm(θ)exp(imφ).
(a)  Show that Yll(θ,φ) = cl sinl(θ)exp(ilφ), where cl is a constant.
(b)  Indicate how you would proceed to find the θ-dependence of Ylm(θ,φ).

Solution:

• Concepts:
The raising and lowering operators L+ = Lx + iLy  and L- = Lx - iLy, the general properties of angular momentum operators
• Reasoning:
L±|k,l,m> = [l(l+1) - m(m±1)]½ħ|k,l,m±1>.  Since
LzYlm(θ,φ) = mħYlm(θ,φ),  (ħ/i)(∂/∂φ)Ylm(θ,φ) = mħYlm(θ,φ),
we separate variables and look for solutions of the form Ylm(θ,φ) = Flm(θ)exp(imφ).
Wave functions must be continuous at all points in space.
We therefore need Ylm(θ,0) = Ylm(θ,2π) or exp(im2π) = 1 or m = integer.
Since L+Yll(θ,φ) = 0, we can find a differential equation satisfied Fll(θ) and solve it.
• Details of the calculation:
(a)  L+Yll(θ,φ) = ħ exp(iφ) [∂/∂θ + i cotθ ∂/∂φ]Yll(θ,φ) = 0
from the general theory.
[d/dθ + l cotθ]Fll(θ) = 0.
dFll(θ) = l cotθ dθ  Fll(θ), dFll(θ) = l Fll(θ) d(sinθ)/sinθ.
dFll(θ)/Fll(θ) = l d(sinθ)/sinθ = l d(ln(sinθ)) = d(ln(sinθ))l = d(sinθ)l/(sinθ)l
-->  Fll(θ) = cl(sinθ)l.
Yll(θ,φ) = cl sinl(θ)exp(ilφ).  Yll is uniquely defined up to the arbitrary constant cl.
0πsinθ dθ∫0dφ |Ylm(θ,φ)|2 = |cl|20πsinθ dθ∫0dφ sin2l(θ) = 1
determines |cl| = (1/(2ll!))((2l+1)!/(4π))½.
It is customary to choose the phase of cl such that
|cl| = [(-1)l/(2ll!)] ((2l+1)!/(4π))½.
Yll(θ,φ) is thus uniquely determined.

(b) L-|k,l,m> = [l(l+1) - m(m±1)]½ħ|k,l,m-1> from the general theory.
L-l-m|k,l,l> ∝ |k,l,m>.
L- is a differential operator.  Apply it (l - m) times to Yll(θ,φ) to obtain an expression for Ylm(θ,φ).  The Ylm(θ,φ) are thus uniquely defined.
To each pair (l,m) there corresponds one and only one eigenfunction Ylm(θ,φ).

#### Problem:

A particle in a spherical potential is known to be in an eigenstate of L2 and Lz with eigenvalues l(l + 1)ħ2 and mħ, respectively.  Prove that the expectation values between |l m> states  satisfy
<Lx> = <Ly> = 0,   <Lx2> = <Ly2> = (l(l + 1)ħ2 - m2ħ2)/2.

Solution:

• Concepts:
The raising and lowering operators L+ = Lx + iLy and L+ = Lx - iLy,
L ±|k,l,m> |k,l,m ± 1> follows from the commutation relations defining angular momentum.
• Reasoning:
We can express Lx and Ly in terms of L+ and L- and then find the matrix elements in the common eigenbasis of L2 and Lz.
• Details of the calculation:
Lx = ½(L+ + L-) and Ly = (-i/2)(L+ - L-).
L ±|k,l,m> |k,l,m ± 1>,  <k,l,m|L ±|k,l,m> = 0.
Therefore <Lx> = <Ly> = 0.
Eigenstates of L2 and Lz have azimuthal symmetry.
This symmetry requires that  <Lx2> = <Ly2> = ½<Lx2 + Ly2> = ½<L2 - Lz2>.
<Lx2> = <Ly2> = (l(l + 1)ħ2 - m2ħ2)/2.