Angular Momentum and Spin Questions and Answers

Q: If a particle has the wave function \( \psi = e^{ikz} \), the z-component of its angular momentum is :

The \( z \)-component of angular momentum is zero.In cylindrical coordinates \( L_{z} = -i\hbar \frac{\partial}{\partial \phi} \). The function \( e^{ikz} \) depends only on \( z \), not on the azimuthal angle \( \phi \), so\[ \begin{aligned} L_{z}\, e^{ikz} &= -i\hbar \frac{\partial}{\partial \phi} e^{ikz} \\ &= 0 \end{aligned} \]The state is an eigenstate of \( L_{z} \) with eigenvalue 0. \( \hbar k \) is the trap: it is the eigenvalue of the linear momentum \( p_{z} \), not of the angular momentum. The imaginary values cannot be eigenvalues of a Hermitian operator.

Q: The orbital angular momentum of a single 2s electron is (h is the Planck's constant) :

The orbital angular momentum of a 2s electron is zero.The magnitude of orbital angular momentum is \( \sqrt{l(l+1)}\,\frac{h}{2\pi} \), and an s electron has \( l = 0 \) whatever its principal quantum number, so the result is 0. The 2 in 2s is n, which does not enter.\( h/2\pi \) is the Bohr-model value \( n\hbar \) for \( n = 1 \), which the quantum theory replaces. \( h\sqrt{2}/2\pi \) is the value for a p electron (\( l = 1 \)), and \( h/4\pi \) is the size of the spin component, not of orbital motion.

Q: The following Pauli spin matrices\[ \sigma_{x} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \]and\[ \sigma_{y} = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \]

The Pauli matrices anti-commute.Multiplying them both ways:\[ \sigma_{x}\sigma_{y} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = i\sigma_{z} \]\[ \sigma_{y}\sigma_{x} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} = -i\sigma_{z} \]So \(\sigma_{x}\sigma_{y} + \sigma_{y}\sigma_{x} = 0\): they anti-commute. They do not commute (their commutator is \(2i\sigma_{z}\), not zero), and they are not inverses of each other, since their product is \(i\sigma_{z}\), not the identity; each is its own inverse.

Q: An electron is in a state \(\chi = \begin{pmatrix} \frac{1+i}{\sqrt{6}} \\ \sqrt{2}/\sqrt{3} \end{pmatrix}\) where the basis functions are \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). The respective probabilities that a measurement of \(S_{z}\) yields values \(+\hbar/2\) and \(-\hbar/2\) are :

The probabilities are 1/3 and 2/3.The probability of each outcome is the squared modulus of the corresponding component:\[ \begin{aligned} P(+\hbar/2) &= \frac{|1+i|^{2}}{6} = \frac{2}{6} = \frac{1}{3} \\ P(-\hbar/2) &= \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^{2} = \frac{2}{3} \end{aligned} \]They add to 1, so the state is normalised. The trap is \(|1+i|^{2}\): it is 2, not 1 (which would give 1/6), and not \((1+i)^{2} = 2i\). Swapping the two values gives 2/3 and 1/3.

Q: Given the operators \(\hat{A} = i(xp_{y} - yp_{x})\) and \(\hat{B} = (yp_{z} + zp_{y})\) the commutator \([\hat{A}, \hat{B}]\) is :

The commutator is \(\hbar(xp_{z} + p_{x}z)\).Write \(\hat{A} = iL_{z}\) with \(L_{z} = xp_{y} - yp_{x}\), and use \([L_{z}, y] = -i\hbar x\), \([L_{z}, p_{y}] = -i\hbar p_{x}\), while \(L_{z}\) commutes with z and \(p_{z}\):\[ \begin{aligned} [L_{z}, yp_{z}] &= [L_{z}, y]p_{z} = -i\hbar xp_{z} \\ [L_{z}, zp_{y}] &= z[L_{z}, p_{y}] = -i\hbar zp_{x} \\ [\hat{A}, \hat{B}] &= i(-i\hbar)(xp_{z} + zp_{x}) \\ &= \hbar(xp_{z} + p_{x}z) \end{aligned} \]Since z and \(p_{x}\) commute, \(zp_{x} = p_{x}z\). The options with a minus sign in front come from dropping the factor i in \(\hat{A}\), and those with \(xp_{z} - p_{x}z\) from a sign slip in one of the two commutators.

Q: Two Stern-Gerlac apparatus \(S_{1}\) and \(S_{2}\) are kept along the z-axis with the directions of their magnetic fields along the positive x and y directions respectively. Each apparatus allows only those particles to pass which have their spins aligned in the direction of its magnetic field. If an unpolarised beam of spin ½ particles is made incident on \(S_{1}\), the ratio of the intensities \(I_{i} : I_{f}\) of the initial and final beams is :

The ratio \(I_{i} : I_{f}\) is 4 : 1.An unpolarised beam of spin-1/2 particles splits equally along any axis, so \(S_{1}\), passing only spin +x, transmits half the beam. Those particles are in the state \(|+x\rangle\). In the y basis\[ |\langle +y|+x\rangle|^{2} = \frac{1}{2} \]so \(S_{2}\) passes half of what reaches it. The final intensity is \(\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\) of the initial, a ratio of 4 : 1.2 : 1 counts only the first apparatus. 1 : 0 would follow if x and y spin states were orthogonal, but only opposite directions, such as +y and -y, are orthogonal for spin 1/2. 8 : 1 would need a third filter.

Q: The commutator of the orbital angular momentum operators [L+, L-], where L± = Lx ± iLy is :

\([L_{+}, L_{-}] = 2\hbar L_{z}\).Expand using \([L_{x}, L_{y}] = i\hbar L_{z}\):\[ \begin{aligned} [L_{x} + iL_{y}, L_{x} - iL_{y}] &= -i[L_{x}, L_{y}] + i[L_{y}, L_{x}] \\ &= -2i[L_{x}, L_{y}] \\ &= -2i(i\hbar L_{z}) = 2\hbar L_{z} \end{aligned} \]The commutator is not zero because \(L_{x}\) and \(L_{y}\) do not commute, and the cyclic algebra produces \(L_{z}\), not \(L_{x}\) or \(L_{y}\).

Q: Spin-orbit coupling leads to :

Spin-orbit coupling leads to the Anomalous Zeeman effect.Spin-orbit coupling combines L and S into J, splitting levels into fine-structure components with different Landé g factors. In a weak magnetic field each J level splits into \(2J + 1\) sublevels spaced by \(g_{J}\mu_{B}B\), and because \(g_{J}\) differs between levels, the lines split into more than three components: the anomalous Zeeman effect.The normal Zeeman effect (three lines) is seen when spin plays no part (singlet states). The Stark effect is due to an electric field, and hyperfine splitting comes from coupling with the nuclear spin.

Q: Consider a spin-1 particle in the state \(|S_{z}, +1\rangle\). Let \(P_{1}^{x}, P_{0}^{x}\) and \(P_{-1}^{x}\) denote the probabilities that the measurement of \(S_{x}\) gives the values \(\hbar\), 0 and \(-\hbar\), respectively. Then :

The probabilities are \(P_{1}^{x} = 1/4, P_{0}^{x} = 1/2, P_{-1}^{x} = 1/4\).For spin 1, the eigenvectors of \(S_{x}\) written in the \(S_{z}\) basis \((|{+1}\rangle, |0\rangle, |{-1}\rangle)\) are\[ \begin{aligned} |{+1}\rangle_{x} &= \tfrac{1}{2}(1, \sqrt{2}, 1) \\ |0\rangle_{x} &= \tfrac{1}{\sqrt{2}}(1, 0, -1) \\ |{-1}\rangle_{x} &= \tfrac{1}{2}(1, -\sqrt{2}, 1) \end{aligned} \]The probability of each outcome is the square of the first component (the overlap with \(|S_{z}, +1\rangle\)): \((1/2)^{2} = 1/4\), \((1/\sqrt{2})^{2} = 1/2\) and \(1/4\). They add to 1, and the symmetry between \(+\hbar\) and \(-\hbar\) is expected because \(|S_{z}, +1\rangle\) has \(\langle S_{x} \rangle = 0\).The equal split 1/3 each and the asymmetric splits fail that symmetry check.

11.

If a particle has the wave function \( \psi = e^{ikz} \), the z-component of its angular momentum is :

A.\( \hbar k \)
B.Zero
C.\( i\hbar k \)
D.\( -i\hbar k \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The \( z \)-component of angular momentum is zero.

In cylindrical coordinates \( L_{z} = -i\hbar \frac{\partial}{\partial \phi} \). The function \( e^{ikz} \) depends only on \( z \), not on the azimuthal angle \( \phi \), so

\[ \begin{aligned} L_{z}\, e^{ikz} &= -i\hbar \frac{\partial}{\partial \phi} e^{ikz} \\ &= 0 \end{aligned} \]

The state is an eigenstate of \( L_{z} \) with eigenvalue 0. \( \hbar k \) is the trap: it is the eigenvalue of the linear momentum \( p_{z} \), not of the angular momentum. The imaginary values cannot be eigenvalues of a Hermitian operator.

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12.

The orbital angular momentum of a single 2s electron is (h is the Planck's constant) :

A.\( h/2\pi \)
B.\( h/4\pi \)
C.\( h\sqrt{2}/2\pi \)
D.Zero

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The orbital angular momentum of a 2s electron is zero.

The magnitude of orbital angular momentum is \( \sqrt{l(l+1)}\,\frac{h}{2\pi} \), and an s electron has \( l = 0 \) whatever its principal quantum number, so the result is 0. The 2 in 2s is n, which does not enter.

\( h/2\pi \) is the Bohr-model value \( n\hbar \) for \( n = 1 \), which the quantum theory replaces. \( h\sqrt{2}/2\pi \) is the value for a p electron (\( l = 1 \)), and \( h/4\pi \) is the size of the spin component, not of orbital motion.

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13.

The following Pauli spin matrices

\[ \sigma_{x} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \]

and

\[ \sigma_{y} = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \]

A.Commute
B.Anti-commute
C.Do not possess commutation relation
D.Are inverses of each other.

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The Pauli matrices anti-commute.

Multiplying them both ways:

\[ \sigma_{x}\sigma_{y} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = i\sigma_{z} \]

\[ \sigma_{y}\sigma_{x} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} = -i\sigma_{z} \]

So \(\sigma_{x}\sigma_{y} + \sigma_{y}\sigma_{x} = 0\): they anti-commute. They do not commute (their commutator is \(2i\sigma_{z}\), not zero), and they are not inverses of each other, since their product is \(i\sigma_{z}\), not the identity; each is its own inverse.

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14.

An electron is in a state \(\chi = \begin{pmatrix} \frac{1+i}{\sqrt{6}} \\ \sqrt{2}/\sqrt{3} \end{pmatrix}\) where the basis functions are \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). The respective probabilities that a measurement of \(S_{z}\) yields values \(+\hbar/2\) and \(-\hbar/2\) are :

A.\(\frac{1}{2}\) and \(\frac{1}{2}\)
B.\(\frac{2}{3}\) and \(\frac{1}{3}\)
C.\(\frac{1}{4}\) and \(\frac{3}{4}\)
D.\(\frac{1}{3}\) and \(\frac{2}{3}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The probabilities are 1/3 and 2/3.

The probability of each outcome is the squared modulus of the corresponding component:

\[ \begin{aligned} P(+\hbar/2) &= \frac{|1+i|^{2}}{6} = \frac{2}{6} = \frac{1}{3} \\ P(-\hbar/2) &= \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^{2} = \frac{2}{3} \end{aligned} \]

They add to 1, so the state is normalised. The trap is \(|1+i|^{2}\): it is 2, not 1 (which would give 1/6), and not \((1+i)^{2} = 2i\). Swapping the two values gives 2/3 and 1/3.

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15.

Given the operators \(\hat{A} = i(xp_{y} - yp_{x})\) and \(\hat{B} = (yp_{z} + zp_{y})\) the commutator \([\hat{A}, \hat{B}]\) is :

A.\(-\hbar(xp_{z} - p_{x}z)\)
B.\(-\hbar(xp_{z} + p_{x}z)\)
C.\(\hbar(xp_{z} + p_{x}z)\)
D.\(\hbar(xp_{z} - p_{x}z)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The commutator is \(\hbar(xp_{z} + p_{x}z)\).

Write \(\hat{A} = iL_{z}\) with \(L_{z} = xp_{y} - yp_{x}\), and use \([L_{z}, y] = -i\hbar x\), \([L_{z}, p_{y}] = -i\hbar p_{x}\), while \(L_{z}\) commutes with z and \(p_{z}\):

\[ \begin{aligned} [L_{z}, yp_{z}] &= [L_{z}, y]p_{z} = -i\hbar xp_{z} \\ [L_{z}, zp_{y}] &= z[L_{z}, p_{y}] = -i\hbar zp_{x} \\ [\hat{A}, \hat{B}] &= i(-i\hbar)(xp_{z} + zp_{x}) \\ &= \hbar(xp_{z} + p_{x}z) \end{aligned} \]

Since z and \(p_{x}\) commute, \(zp_{x} = p_{x}z\). The options with a minus sign in front come from dropping the factor i in \(\hat{A}\), and those with \(xp_{z} - p_{x}z\) from a sign slip in one of the two commutators.

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16.

Two Stern-Gerlac apparatus \(S_{1}\) and \(S_{2}\) are kept along the z-axis with the directions of their magnetic fields along the positive x and y directions respectively. Each apparatus allows only those particles to pass which have their spins aligned in the direction of its magnetic field. If an unpolarised beam of spin ½ particles is made incident on \(S_{1}\), the ratio of the intensities \(I_{i} : I_{f}\) of the initial and final beams is :

A.8 : 1
B.4 : 1
C.1 : 0
D.2 : 1

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The ratio \(I_{i} : I_{f}\) is 4 : 1.

An unpolarised beam of spin-1/2 particles splits equally along any axis, so \(S_{1}\), passing only spin +x, transmits half the beam. Those particles are in the state \(|+x\rangle\). In the y basis

\[ |\langle +y|+x\rangle|^{2} = \frac{1}{2} \]

so \(S_{2}\) passes half of what reaches it. The final intensity is \(\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\) of the initial, a ratio of 4 : 1.

2 : 1 counts only the first apparatus. 1 : 0 would follow if x and y spin states were orthogonal, but only opposite directions, such as +y and -y, are orthogonal for spin 1/2. 8 : 1 would need a third filter.

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17.

The commutator of the orbital angular momentum operators [L₊, L_-], where L± = L_x ± iL_y is :

A.0
B.2 \(\hbar L_{x}\)
C.2 \(\hbar L_{y}\)
D.2 \(\hbar L_{z}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

\([L_{+}, L_{-}] = 2\hbar L_{z}\).

Expand using \([L_{x}, L_{y}] = i\hbar L_{z}\):

\[ \begin{aligned} [L_{x} + iL_{y}, L_{x} - iL_{y}] &= -i[L_{x}, L_{y}] + i[L_{y}, L_{x}] \\ &= -2i[L_{x}, L_{y}] \\ &= -2i(i\hbar L_{z}) = 2\hbar L_{z} \end{aligned} \]

The commutator is not zero because \(L_{x}\) and \(L_{y}\) do not commute, and the cyclic algebra produces \(L_{z}\), not \(L_{x}\) or \(L_{y}\).

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18.

Spin-orbit coupling leads to :

A.Normal Zeeman effect
B.Anomalous Zeeman effect
C.Stark effect
D.Hyperfine splitting

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Spin-orbit coupling leads to the Anomalous Zeeman effect.

Spin-orbit coupling combines L and S into J, splitting levels into fine-structure components with different Landé g factors. In a weak magnetic field each J level splits into \(2J + 1\) sublevels spaced by \(g_{J}\mu_{B}B\), and because \(g_{J}\) differs between levels, the lines split into more than three components: the anomalous Zeeman effect.

The normal Zeeman effect (three lines) is seen when spin plays no part (singlet states). The Stark effect is due to an electric field, and hyperfine splitting comes from coupling with the nuclear spin.

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19.

Which of the following statements is not correct ?

A.In the presence of spin-orbit coupling, the Hamiltonian does not commute with \(\bar{L}\) and \(\bar{S}\) operators
B.Even in presence of spin-orbit coupling, the Hamiltonian commutes with \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) operators
C.The spin-orbit interaction for an electron is described by \(H_{so} = (e^{2}/8\pi\epsilon_{0}))\,1/(m^{2}c^{2}r^{3})\,\bar{S}.\bar{L}\).
D.In the presence of spin-orbit interaction, the physical quantities measured by \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) are not conserved.

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The statement that is not correct is that, with spin-orbit interaction, \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) are not conserved.

Spin-orbit coupling is \(\xi(r)\bar{L}\cdot\bar{S} = \frac{\xi}{2}(J^{2} - L^{2} - S^{2})\). This commutes with \(L^{2}\), \(S^{2}\), \(J^{2}\) and the components of \(\bar{J}\), so those quantities remain conserved, while \(\bar{L}\) and \(\bar{S}\) separately are not, since the coupling transfers angular momentum between them.

So statements A and B are correct, and C is the standard form of the spin-orbit Hamiltonian (including the Thomas factor). Statement D claims the opposite of what the commutators show.

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20.

Consider a spin-1 particle in the state \(|S_{z}, +1\rangle\). Let \(P_{1}^{x}, P_{0}^{x}\) and \(P_{-1}^{x}\) denote the probabilities that the measurement of \(S_{x}\) gives the values \(\hbar\), 0 and \(-\hbar\), respectively. Then :

A.\(P_{1}^{x} = 1/2, P_{0}^{x} = 1/4, P_{-1}^{x} = 1/4\)
B.\(P_{1}^{x} = 1/3, P_{0}^{x} = 1/3, P_{-1}^{x} = 1/3\)
C.\(P_{1}^{x} = 1/4, P_{0}^{x} = 1/2, P_{-1}^{x} = 1/4\)
D.\(P_{1}^{x} = 1/4, P_{0}^{x} = 1/4, P_{-1}^{x} = 1/2\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The probabilities are \(P_{1}^{x} = 1/4, P_{0}^{x} = 1/2, P_{-1}^{x} = 1/4\).

For spin 1, the eigenvectors of \(S_{x}\) written in the \(S_{z}\) basis \((|{+1}\rangle, |0\rangle, |{-1}\rangle)\) are

\[ \begin{aligned} |{+1}\rangle_{x} &= \tfrac{1}{2}(1, \sqrt{2}, 1) \\ |0\rangle_{x} &= \tfrac{1}{\sqrt{2}}(1, 0, -1) \\ |{-1}\rangle_{x} &= \tfrac{1}{2}(1, -\sqrt{2}, 1) \end{aligned} \]

The probability of each outcome is the square of the first component (the overlap with \(|S_{z}, +1\rangle\)): \((1/2)^{2} = 1/4\), \((1/\sqrt{2})^{2} = 1/2\) and \(1/4\). They add to 1, and the symmetry between \(+\hbar\) and \(-\hbar\) is expected because \(|S_{z}, +1\rangle\) has \(\langle S_{x} \rangle = 0\).

The equal split 1/3 each and the asymmetric splits fail that symmetry check.

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