Physics Problems & Solutions: Spin

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Nuclear & Particle Physics - Stern-Gerlach Experiment

A beam of neutral hydrogen atoms in their ground state is moving into the plane of this page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, a detector would find that it has been

A. deflected upward
B. deflected to the right
C. undeviated
D. split vertically into two beams
E. split horizontally into three beams

(GR9677 #11)

Solution:

The Stern–Gerlach experiment → spin discovery

A beam of neutral atom passes through inhomogeneous magnetic field will split vertically into 2 beams representing spin-up and spin-down electrons.

Answer: D

Nuclear & Particle Physics - Positronium

Positronium is the bound state of an electron and a positron. Consider only the states of zero orbital angular momentum l = 0. The most probable decay product of any such state of positronium with spin zero (singlet is)

A. 0 photons
B. 1 photons
C. 2 photons
D. 3 photons
E. 4 photons

(GR9677 #53)

Solution:

The singlet state, s = 0 of Positronium is known as para-Positronium, decays preferentially into two gamma rays.

It can decay into any even number of photons (2, 4, 6, ...), but the probability quickly decreases with the number.

Thus, the most probable decay product of singlet state of positronium is 2 photons.

Answer: C

Quantum Mechanics - Spin Angular Momentum

Two ions 1 and 2, at fixed separation, with spin angular momentum operators S₁ and S₂, have the interaction Hamiltonian H = −J S₁·S₂, where J > 0. The values of S₁² and S₂² are fixed at S₁(S₁+ 1) and S₂(S₂+ 1), respectively. Which of the following is the energy of the ground state of the system?

A. 0
B. –JS₁S₂
C. –J[S₁(S₁+ 1) – S₂(S₂+ 1)]
D. –(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) – S₁(S₁+1) – S₂(S₂+1)]
E. –(J/2)[(S₁(S₁+ 1) + S₂(S₂+ 1))/(S₁+ S₂)(S₁+ S₂+ 1)]

(GR9677 #77)

Solution:

S₁² ψ = S₁(S₁+ 1) ψ
S₂² ψ = S₂(S₂+ 1) ψ
S_i² ψ = S_i(S_i+ 1) ψ

H = −J S₁·S₂

Using general arithmetic equation: ab = ½ [(a + b)² − a² − b²]

H = −(J/2)[(S₁ + S₂)² − S₁² − S₂²]

Since S_i² ψ = S_i(S_i+ 1) ψ
For (S₁ + S₂)² → replace S_i with S₁ + S₂
→ (S₁ + S₂)² ψ = (S₁+ S₂)(S₁+ S₂+ 1) ψ

H = −(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) − S₁(S₁+ 1) − S₂(S₂+ 1)]

Answer: D

Nuclear & Particle Physics - Spin

The ground state of the helium atom is a spin

A. singlet
B. doublet
C. triplet
D. quartet
E. quintuplet

(GR9277 #59)

Solution:

Spectroscopic notation: N^2s+1 L_j

2s + 1 = multiplicity

Singlet: 2s + 1 = 1, s = 0
Doublet: 2s + 1 = 2, s = 1/2
Triplet: 2s + 1 = 3, s = 1

Electron configuration of Helium: 1s²

Pauli Exclusion Principle: no two electrons can have exactly the same quantum number.

m_s1 = ½ and m_s2 = −½
Total spin quantum number, s = ½ + (−½) = 0
Multiplicity, 2 · 0 + 1 = 1 → Singlet

Answer: A

Nuclear & Particle Physics - Gyromagnetic Ratio

Consider a heavy nucleus with spin 1/2. The magnitude of the ratio of the intrinsic magnetic moment of this nucleus to that of an electron is

A. Zero, because the nucleus has no intrinsic magnetic moment
B. Greater than 1, because the nucleus contains many protons
C. Greater than 1, because the nucleus is so much larger in diameter than the electron
D. Less than 1, because of the strong interactions among the nucleons in a nucleus
E. Less than 1, because the nucleus has a mass much larger than that of the electron

(GR9277 #77)

Solution:

The intrinsic magnetic moment is defined in terms of the gyromagnetic ratio and spin as μ_s = γS
where γ = eg/2m
g = the Lande g-factor

Thus, the magnetic moment is inversely related to mass.

Since the nucleus has the same spin as the electron, we can omit S.
And since m_e ≪ m_n, the ratio of the magnetic moment of a nucleus vs electron is μ_n/μ_e = m_e/m_n ≪1

Answer: E

Nuclear & Particle Physics - Spin

The hypothesis that an electron possesses spin is qualitatively significant for the explanation of all the following topics EXCEPT the

A. Structure of the periodic table
B. Specific heat of metals
C. Anomalous Zeeman effect
D. Deflection of moving electron by a uniform magnetic field
E. Fine structure of atomic spectra

(GR8677 #27)

Solution:

(A) TRUE
Spin → Pauli exclusion principle → electron configuration → Structure of the periodic table

(B) TRUE
Specific heat for Fermions (½-integer spin) is different from Bosons (integer spin).

(C) TRUE
Zeeman Effect: the splitting of spectral lines when an external magnetic field is applied.
"Normal" Zeeman effect → This type of splitting is observed for spin 0 states since the spin does not contribute to the angular momentum.
"Anomalous" Zeeman effect → When electron spin is included, there is a greater variety of splitting patterns.

(D) FALSE
Deflection of moving electron by a uniform magnetic field does not depend on spin

(E) TRUE
Fine structure = the splitting of the spectral lines of atoms due to quantum mechanical (electron spin) and relativistic corrections.

Answer: D

Quantum Mechanics - Wave Function

A system containing two identical particles is described by a wave function of the form

$\psi= \frac{1}{\sqrt{2}} \left[ \psi_\alpha(x_1)\psi_\beta(x_2)+ \psi_\beta(x_1)\psi_\alpha(x_2)\right ]$

Where x₁ and x₂ represent the spatial coordinates of the particles and α and β represent all the quantum numbers, including spin, of the states that they occupy. The particles might be

A. Electrons
B. Positrons
C. Protons
D. Neutrons
E. Deuterons

(GR8677 #89)

Solution:

Wave function: $\psi= \frac{1}{\sqrt{2}} \left[ \psi_\alpha(x_1)\psi_\beta(x_2)\pm \psi_\beta(x_1)\psi_\alpha(x_2)\right ]$

Symmetric function (+ sign): $\psi_{\alpha\beta}=\psi_{\beta\alpha}$

→ obey Bose-Einstein statistics → bosons

Anti-symmetric function (−sign): $\psi_{\alpha\beta}=-\psi_{\beta\alpha}$
→ obey Fermi-Dirac statistics → fermion

Electrons and positron (anti-elecron) → fermion

Protons and neutron → fermionic hadrons

composite particles (hadron)
composed of 3 fermionic quarks

Deuterons → bosons

nucleus of deuterium, an isotope of hydrogen
composed of a neutron (spin 1/2) and a proton (spin 1/2), total spin = 1

Answer: E

Quantum Mechanics - Spin

Let $|\alpha\rangle$ represent the state of an electron with spin up and $|\beta\rangle$ the state of an electron with spin down. Valid spin eigenfunctions for a triplet state (^3S)

of a two-electron atom include which of the following?

$\\$I. $ \hspace{5 mm} |\alpha\rangle_1|\alpha\rangle_2 \\\\$II. $ \hspace{4 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) \\\\$III. $ \hspace{3 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right )$

A. I only
B. II only
C. III only
D. I and III
E. II and III

GR0177 #82)

Solution:

Singlet state is anti-symmetric: ψ(1,2) = −ψ(2,1)
Triplet state is symmetric: ψ(1,2) = ψ(2,1)

$$I. $ \hspace{5 mm} |\alpha\rangle_1|\alpha\rangle_2$ is symmetric, because

$\\ \psi(1,2)=|\alpha\rangle_1|\alpha\rangle_2 \\ \psi(2,1)=|\alpha\rangle_2|\alpha\rangle_1=|\alpha\rangle_1|\alpha\rangle_2=\psi(1,2)$

$$II. $ \hspace{4 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right )$ is antisymmetric, because
$\begin{aligned} \\ \psi(1,2)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) \\ \psi(2,1)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_2|\beta\rangle_1-|\alpha\rangle_1|\beta\rangle_2\right )\\&=-\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) =-\psi(1,2) \end{aligned}$

$$III. $ \hspace{3 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right )$ is symmetric, because
$\begin{aligned} \\ \psi(1,2)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right ) \\ \psi(2,1)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_2|\beta\rangle_1+|\alpha\rangle_1|\beta\rangle_2\right )\\&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right ) =\psi(1,2) \end{aligned}$

Thus, II is singlet state, I and III are triplet state.

Answer: D

Quantum Mechanics - Spin operator

The state of a spin ½ particle can be represented using the eigenstate $\mid\uparrow\rangle$ and $\mid\downarrow\rangle$ of the S_z

operator.

$\\S_z \mid\uparrow\rangle = \frac{1}{2}\hbar \mid\uparrow\rangle \\S_z \mid\downarrow\rangle = -\frac{1}{2}\hbar \mid\downarrow\rangle$

Given the Pauli matrix $\sigma_x=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ which of the following is an eigenstate of

with eigenvalue $-\frac{1}{2} \hbar$ ?

$\\$A.$ \hspace{5 mm} \mid\uparrow\rangle \\$B.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle+\mid\downarrow\rangle\right ) \\$C.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle-\mid\downarrow\rangle\right ) \\$D.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle+i\mid\downarrow\rangle\right ) \\$E.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle-i\mid\downarrow\rangle\right )$

(GR0177 #83)

Solution:

Spin Operator:    $S_x=\frac{\hbar}{2}\sigma_x=\frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Eigenvalue:    $\alpha=-\frac{\hbar}{2}$

Eigenfunction:    $S_x\chi =\alpha \chi$

$\begin{aligned} \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=\begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix} \\&=\begin{bmatrix} -\hbar/2 & 0 \\ 0 & -\hbar/2 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix} \end{aligned}$

$\begin{aligned} \left(\begin{bmatrix} 0 & \hbar/2 \\ \hbar/2 & 0 \end{bmatrix}-\begin{bmatrix} -\hbar/2 & 0 \\ 0 & -\hbar/2 \end{bmatrix} \right ) \begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=0 \\\frac{\hbar}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=0 \end{aligned}$

$\begin{aligned} \chi_1&=-\chi_2\\ \chi&=\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}= \chi_1\begin{bmatrix} 1 \\ -1 \end{bmatrix} \end{aligned}$

Normalized $\chi$ :
$\chi_1^2+\chi_1^2=1\rightarrow 2\chi_1^2=1\rightarrow\chi_1=\frac{1}{\sqrt{2}}$
$\chi= \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Answer: C

Nuclear & Particle Physics - Selection Rules

An energy-level diagram of the n = 1 and n = 2 levels of atomic hydrogen (including the effect of spin-orbit coupling and relativity) is shown in the figure. Three transitions are labeled A, B, and C. Which of the transitions will be possible electric-dipole transition?

A. B only
B. C only
C. A and C only
D. B and C only
E. A, B, and C

(GR0177 #84)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

Transition A:
l = 0 to l = 0 → ∆l = 0 → NOT ALLOWED

Transition B:
l = 1 to l = 0 → ∆l = −1 → ALLOWED
j = 3/2 to j = 1/2 → ∆j = −1 → ALLOWED

Transition C:
j = 1/2 to j = 1/2 → ∆j = 0 → ALLOWED

Answer: D

Quantum Mechanics – Selection Rules

Which of the following is NOT compatible with the selection rule that controls electric dipole emission of photons by excited states of atoms?

A. ∆n may have any negative integral value
B. ∆l = ±1
C. ∆m_l = 0, ±1
D. ∆s = ±1
E. ∆j = ±1

(GR8677 #92)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

Answer: D

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