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 L^{2}we only can obtain an eigenvalue l(l +1)ħ^{2}, with l a non-negative integer.

The magnitude of the angular momentum J therefore can only be measured as

J = (l(l +1))^{½}ħ, l = 0, 1, 2, ... .

The magnitude of** **the orbital** **angular momentum
**L** of a
hydrogen atom is found to be 30^{½}ħ. L_{z} 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 L_{x}^{2} + L_{y}^{2}?

Solution:

- Concepts:

Angular momentum, the hydrogen atom - Reasoning:

For the hydrogen atom H, L^{2}and L_{z}are commuting observables. - Details of the calculation:

(a) L^{2}= l(l+1)ħ^{2}= 30ħ^{2}, 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) L_{x}^{2 }+ L_{y}^{2}= L^{2}- L_{z}^{2}= (30 - 9)ħ^{2}= 21ħ^{2}.

Show that a state with position wave
function (y - iz)^{k} is an eigenstate of the orbital angular momentum
operator L_{x} 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 L_{x}= yp_{z}- zp_{y}= (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}.

L_{x}(y - iz)^{k}= (ħ/i)(-iky - kz) (y - iz)^{k-1}= -ħk(y - iz)^{k}. The eigenvalue is -ħk.

k must be an integer.

For a simple particle moving in space, show that the wave function ψ_{lm}(**r**) = x^{2} + y^{2}
- 2z^{2} represents a simultaneous eigenstate of L^{2} and L_{z}
with eigenvalues l(l + 1)ħ^{2}
and mħ. Determine l and m. Find a
function with the same eigenvalue for L^{2} and the maximum possible
eigenvalue for L_{z}.

Solution:

- Concepts:

The eigenfunctions of the orbital angular momentum operator, the spherical harmonics - Reasoning:

The common eigenfunctions of L^{2}and L_{z}are the spherical harmonics. We have to write the given wave functions in terms of the spherical harmonics. -
Details of the calculation:

ψ_{lm}(**r**) = x^{2}+ y^{2}- 2z^{2}=

r^{2}sin^{2}θcos^{2}φ + r^{2}sin^{2}θsin^{2}φ - 2r^{2}cos^{2}θ = r^{2}(sin^{2}θ - 2cos^{2}θ)

in spherical coordinates.

Express ψ_{lm}(**r**) in terms of the spherical harmonics.

ψ_{lm}(**r**) = r^{2}(1 - cos^{2}θ - 2cos^{2}θ) = r^{2}(1 - 3cos^{2}θ)

= -r^{2}(16π/5)^{½}Y_{20}(θ,φ).

[Formulas for some of the spherical harmonics:

Y_{00}= (4π)^{-½}, Y_{1±1}= ∓(3/8π)^{½}sinθ exp(±iφ), Y_{10}= (¾π)^{½}cosθ,

Y_{2±2}= (15/32π)^{½}sin^{2}θ exp(±i2φ), Y_{2±1}= ∓(15/8π)^{½}sinθ cosθ exp(±iφ),

Y_{20}= (5/16π)^{½}(3cos^{2}θ - 1).]

ψ_{lm}(**r**) is an eigenfunction of L_{z}with eigenvalue 0 and an eigenfunction of L^{2}with eigenvalue 6ħ^{2}(l = 2). For l = 2 the maximum possible eigenvalue of L_{z}is mħ = 2ħ. The corresponding eigenfunction is

Cr^{2}Y_{22}(θ,φ) = Cr^{2}(15/32π)^{½}sin^{2}θ exp(±i2φ).

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 L^{2}? If so, what is its l
value? If not, what are the possible values of l we may obtain when L^{2}
is measured?

(b) What are the probabilities for the particle to be found in various m_{l}
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 L^{2}and L_{z}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^{iφ}+ e^{-iφ}) - i sinθ ½(e^{iφ}- e^{-iφ}) - 3cosθ)

= rf(r)(sinθ ½e^{iφ}(1 - i) + sinθ ½ e^{-iφ}(1 + i) - 3cosθ)

= rf(r)(2^{-½}exp(-iπ/2)^{ }sinθe^{iφ}+ 2^{-½}exp(iπ/2) sinθ e^{-iφ}- 3cosθ).

Y_{1±1}= ∓(3/8π)^{½}sinθ exp(±iφ), Y_{10}= (¾π)^{½}cosθ.

ψ(**r**) = rf(r)(-exp(-iπ/2)(4π/3)^{½}Y_{11}(θ,φ) + exp(iπ/2)(4π/3)^{½}Y_{1-1}(θ,φ) - (12π)^{½}Y_{10}(θ,φ))

= rf(r)2√π(-exp(-iπ/2)(1/3)^{½}Y_{11}(θ,φ) + exp(iπ/2)(1/3)^{½}Y_{1-1}(θ,φ) - (3π)^{½}Y_{10}(θ,φ))

= (44π/3)^{½}rf(r)(-exp(-iπ/2)/11^{½}Y_{11}(θ,φ) + exp(iπ/2)/11^{½}Y_{1-1}(θ,φ) - 3/11^{½}Y_{10}(θ,φ))

= R(r) (-exp(-iπ/2)/11^{½}Y_{11}(θ,φ) + exp(iπ/2)/11^{½}Y_{1-1}(θ,φ) - 3/11^{½}Y_{10}(θ,φ)).

The function of angle is normalized.

ψ(**r**) is an eigenfunction of L^{2}with eigenvalue 2ħ^{2}, l = 1.

(b) The possible values of m_{l}are 1, 0, -1.

P(m_{l}= 1) = 1/11, P(m_{l}= -1) = 1/11, P(m_{l}= 0) = 9/11.

(c) The wave function ψ_{klm}(r,θ,φ) = R_{kl}(r)Y_{lm}(θ,φ) = [u_{kl}(r)/r]Y_{lm}(θ,φ) is a product of a radial function R_{kl}(r) and the spherical harmonic Y_{lm}(θ,φ). The differential equation for u_{kl}(r) is

[-(ħ^{2}/(2m))(∂^{2}/∂r^{2}) + ħ^{2}l(l+1)/(2mr^{2}) + U(r)]u_{kl}(r) = E_{kl}u_{kl}(r).

Since u_{kl}(r), E_{kl}and l are known, we can solve this equation for U(r).

The ladder operator L_{+} is given by

L_{+} = ħ exp(iφ) [∂/∂θ + i cotθ ∂/∂φ].

The spherical harmonics are given by
Y_{lm}(θ,φ)
= F_{lm}(θ)exp(imφ).

(a)
Show that
Y_{ll}(θ,φ)
= c_{l} sin^{l}(θ)exp(ilφ), where
c_{l} is a constant.

(b) Indicate how you would proceed to find the
θ-dependence of Y_{lm}(θ,φ).

Solution:

- Concepts:

The raising and lowering operators L_{+ }= L_{x }+ iL_{y}and L_{- }= L_{x }- iL_{y}, the general properties of angular momentum operators - Reasoning:

L_{±}|k,l,m> = [l(l+1) - m(m±1)]^{½}ħ|k,l,m±1>. Since

L_{z}Y_{lm}(θ,φ) = mħY_{lm}(θ,φ), (ħ/i)(∂/∂φ)Y_{lm}(θ,φ) = mħY_{lm}(θ,φ),

we separate variables and look for solutions of the form Y_{lm}(θ,φ) = F_{lm}(θ)exp(imφ).

Wave functions must be continuous at all points in space.

We therefore need Y_{lm}(θ,0) = Y_{lm}(θ,2π) or exp(im2π) = 1 or m = integer.

Since L_{+}Y_{ll}(θ,φ) = 0, we can find a differential equation satisfied F_{ll}(θ) and solve it. -
Details of the calculation:

(a) L_{+}Y_{ll}(θ,φ) = ħ exp(iφ) [∂/∂θ + i cotθ ∂/∂φ]Y_{ll}(θ,φ) = 0

from the general theory.

[d/dθ + l cotθ]F_{ll}(θ) = 0.

dF_{ll}(θ) = l cotθ dθ F_{ll}(θ), dF_{ll}(θ) = l F_{ll}(θ) d(sinθ)/sinθ.

dF_{ll}(θ)/F_{ll}(θ) = l d(sinθ)/sinθ = l d(ln(sinθ)) = d(ln(sinθ))^{l}= d(sinθ)^{l}/(sinθ)^{l }--> F_{ll}(θ) = c_{l}(sinθ)^{l}.

Y_{ll}(θ,φ) = c_{l }sin^{l}(θ)exp(ilφ). Y_{ll}is uniquely defined up to the arbitrary constant c_{l}.

∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ |Y_{lm}(θ,φ)|^{2}= |cl|^{2}∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ sin^{2l}(θ) = 1

determines |c_{l}| = (1/(2^{l}l!))((2l+1)!/(4π))^{½}.

It is customary to choose the phase of c_{l}such that

|c_{l}| = [(-1)^{l}/(2^{l}l!)] ((2l+1)!/(4π))^{½}.

Y_{ll}(θ,φ) is thus uniquely determined.(b) L

_{-}|k,l,m> = [l(l+1) - m(m±1)]^{½}ħ|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 Y_{ll}(θ,φ) to obtain an expression for Y_{lm}(θ,φ). The Y_{lm}(θ,φ) are thus uniquely defined.

To each pair (l,m) there corresponds one and only one eigenfunction Y_{lm}(θ,φ).

A particle is known to be in an eigenstate of L^{2} and L_{z}.
Prove that the expectation values satisfy

<L_{x}> = <L_{y}> = 0, <L_{x}^{2}> =
<L_{y}^{2}> = (l(l+1)ħ^{2}
- m^{2}ħ^{2})/2.

- Concepts:

The raising and lowering operators L_{+}= L_{x}+ iL_{y}and L_{+}= L_{x}- iL_{y},

L_{±}|k,l,m> = [l(l + 1) - m(m ± 1)]^{½}ħ|k,l,m ± 1>. - Reasoning:

We can express L_{x}and L_{y}in terms of L_{+}and L_{-}and then find the matrix elements in the common eigenbasis of L^{2}and L_{z}. -
Details of the calculation:

L_{x}= ½(L_{+}+ L_{-}) and L_{y}= (-i/2)(L_{+}- L_{-}).

L_{x}^{2}= ¼(L_{+}+ L_{-})(L_{+}+ L_{-}) = ¼(L_{+}^{2}+ L_{-}^{2}+ L_{-}L_{+}+ L_{+}L_{-}).

L_{y}^{2}= (-¼)(L_{+}- L_{-})(L_{+}- L_{-}) = ¼(-L_{+}^{2}- L_{-}^{2}+ L_{-}L_{+}+ L_{+}L_{-})

L_{±}|k,l,m> |k,l,m±1>.

The only matrix elements that can be nonzero are the matrix elements of L_{+}L_{-}and L_{-}L_{+}. Therefore <L_{x}> = <L_{y}> = 0, <L_{x}^{2}> = <L_{y}^{2}> = ½(L_{x}^{2}+ L_{y}^{2}) = ½)(L^{2}- L_{z}^{2}).

<L_{x}^{2}> = <L_{y}^{2}> = (l(l+1)ħ^{2}- m^{2}ħ^{2})/2.