Identical Particles Questions and Answers

Q: Two projectile protons \(P_{1}\) and \(P_{2}\) both with spin up (along the \(+z\) direction) are scattered from another fixed target proton \(T\) with spin up at rest in the \(xy\) plane, as shown in the figure. They scatter one at a time. The nuclear interaction potential between both the projectiles and the target proton is \(\lambda\vec{L} \cdot \vec{S}\), where \(\vec{L}\) is the orbital angular momentum of the system with respect to the target, \(\vec{S}\) is the spin angular momentum of the system and \(\lambda\) is a negative constant in appropriate units. Which one of the following is correct?

Both protons are scattered upward, in the \(+y\) direction.Both spins point along \(+z\), so \(\vec{S}\) is along \(+z\). The orbital angular momentum about the target is \(\vec{L} = \vec{r} \times \vec{p}\) with \(\vec{p}\) along \(+x\): for \(P_{1}\), passing above (\(y \gt 0\)), \(L_{z} = -yp \lt 0\); for \(P_{2}\), passing below, \(L_{z} \gt 0\).With \(\lambda \lt 0\), the potential \(\lambda\vec{L}\cdot\vec{S}\) is positive (repulsive) for \(P_{1}\), which is pushed away from the target, that is upward. For \(P_{2}\) it is negative (attractive), so \(P_{2}\) is pulled towards the target, which is also upward.This left-right asymmetry is how spin-orbit coupling polarises a scattered beam. The options that send the two protons in opposite directions, or both downward, get the sign of \(\lambda\vec{L}\cdot\vec{S}\) wrong for at least one of them.

Q: Minimum energy of a collection of six non-interacting electrons of spin \(\frac{1}{2}\) and mass m, placed in a one-dimensional infinite potential well of width L is :

The minimum energy is \(\frac{14\pi^{2}\hbar^{2}}{mL^{2}}\).The levels of the well are \(E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\). By the Pauli principle each level holds two electrons of opposite spin, so in the ground state the six electrons fill n = 1, 2 and 3:\[ \begin{aligned} E &= 2(1^{2} + 2^{2} + 3^{2})\frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &= 28\frac{\pi^{2}\hbar^{2}}{2mL^{2}} = \frac{14\pi^{2}\hbar^{2}}{mL^{2}} \end{aligned} \]Putting all six in the lowest level, as for bosons, gives \(3\pi^{2}\hbar^{2}/mL^{2}\). Placing one electron in each of the six lowest levels, ignoring spin, gives a sum of squares of 91. Seven is half the correct coefficient.

Q: \(\psi(x, y) = \exp[|\sin(x - y)|] + (x - y)^{2}/2 + 3\) is :

The function is symmetric under interchange of x and y.Swapping \(x\) and \(y\) turns \(x - y\) into \(-(x - y)\). Each term depends on \(x - y\) only through an even function:\(|\sin(-(x - y))| = |\sin(x - y)|\), because the modulus removes the sign;\((-(x - y))^{2} = (x - y)^{2}\);the constant 3 is unchanged.So \(\psi(y, x) = \psi(x, y)\). It is not antisymmetric, since that would need \(\psi(y, x) = -\psi(x, y)\), impossible for a function that is always positive. A function cannot be both unless it is zero.

Two projectile protons \(P_{1}\) and \(P_{2}\) both with spin up (along the \(+z\) direction) are scattered from another fixed target proton \(T\) with spin up at rest in the \(xy\) plane, as shown in the figure. They scatter one at a time. The nuclear interaction potential between both the projectiles and the target proton is \(\lambda\vec{L} \cdot \vec{S}\), where \(\vec{L}\) is the orbital angular momentum of the system with respect to the target, \(\vec{S}\) is the spin angular momentum of the system and \(\lambda\) is a negative constant in appropriate units. Which one of the following is correct?

Target T on the positive x axis; projectile P1 above the x axis and P2 below it, both moving in the plus x direction

A.\(P_{1}\) will be scattered in the \(+y\) direction (upward) and \(P_{2}\) will be scattered in the \(-y\) direction (downward)
B.\(P_{1}\) will be scattered in the \(+y\) direction (upward) and \(P_{2}\) will be scattered in the \(+y\) direction (upward)
C.\(P_{1}\) will be scattered in the \(-y\) direction (downward) and \(P_{2}\) will be scattered in the \(+y\) direction (upward)
D.\(P_{1}\) will be scattered in the \(-y\) direction (downward) and \(P_{2}\) will be scattered in the \(-y\) direction (downward)

Exam Year(s)

GATE Physics - 2025,

Answer: B

Both protons are scattered upward, in the \(+y\) direction.

Both spins point along \(+z\), so \(\vec{S}\) is along \(+z\). The orbital angular momentum about the target is \(\vec{L} = \vec{r} \times \vec{p}\) with \(\vec{p}\) along \(+x\): for \(P_{1}\), passing above (\(y \gt 0\)), \(L_{z} = -yp \lt 0\); for \(P_{2}\), passing below, \(L_{z} \gt 0\).

With \(\lambda \lt 0\), the potential \(\lambda\vec{L}\cdot\vec{S}\) is positive (repulsive) for \(P_{1}\), which is pushed away from the target, that is upward. For \(P_{2}\) it is negative (attractive), so \(P_{2}\) is pulled towards the target, which is also upward.

This left-right asymmetry is how spin-orbit coupling polarises a scattered beam. The options that send the two protons in opposite directions, or both downward, get the sign of \(\lambda\vec{L}\cdot\vec{S}\) wrong for at least one of them.

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Minimum energy of a collection of six non-interacting electrons of spin \(\frac{1}{2}\) and mass m, placed in a one-dimensional infinite potential well of width L is :

A.\(\frac{3\pi^{2}\hbar^{2}}{mL^{2}}\)
B.\(\frac{14\pi^{2}\hbar^{2}}{mL^{2}}\)
C.\(\frac{7\pi^{2}\hbar^{2}}{mL^{2}}\)
D.\(\frac{91\pi^{2}\hbar^{2}}{mL^{2}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The minimum energy is \(\frac{14\pi^{2}\hbar^{2}}{mL^{2}}\).

The levels of the well are \(E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\). By the Pauli principle each level holds two electrons of opposite spin, so in the ground state the six electrons fill n = 1, 2 and 3:

\[ \begin{aligned} E &= 2(1^{2} + 2^{2} + 3^{2})\frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &= 28\frac{\pi^{2}\hbar^{2}}{2mL^{2}} = \frac{14\pi^{2}\hbar^{2}}{mL^{2}} \end{aligned} \]

Putting all six in the lowest level, as for bosons, gives \(3\pi^{2}\hbar^{2}/mL^{2}\). Placing one electron in each of the six lowest levels, ignoring spin, gives a sum of squares of 91. Seven is half the correct coefficient.

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Sixteen non-interacting electrons are confined in an infinite box of side a. The energies of the least energetic and the most energetic electrons in the ground state of the system are :

A.\(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}, \frac{32\hbar^{2}\pi^{2}}{ma^{2}}\)
B.\(0, \frac{32\hbar^{2}\pi^{2}}{ma^{2}}\)
C.\(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}, \frac{\hbar^{2}\pi^{2}}{2ma^{2}}\)
D.\(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}, \frac{128\hbar^{2}\pi^{2}}{ma^{2}}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The energies are \(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}\) and \(\frac{32\hbar^{2}\pi^{2}}{ma^{2}}\).

For a box of side a treated in one dimension, the levels are \(E_{n} = n^{2}\pi^{2}\hbar^{2}/2ma^{2}\). By the Pauli principle each level holds two electrons (spin up and down), so 16 electrons fill n = 1 to 8 in the ground state.

\[ \begin{aligned} E_{1} &= \frac{\pi^{2}\hbar^{2}}{2ma^{2}} \\ E_{8} &= \frac{64\pi^{2}\hbar^{2}}{2ma^{2}} = \frac{32\pi^{2}\hbar^{2}}{ma^{2}} \end{aligned} \]

Zero is not an allowed energy in an infinite well, equal energies would require all 16 electrons in the lowest level, which Pauli forbids, and \(128\hbar^{2}\pi^{2}/ma^{2}\) is the n = 16 level, reached only if each level held one electron.

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\(\psi(x, y) = \exp[|\sin(x - y)|] + (x - y)^{2}/2 + 3\) is :

A.symmetric under interchanged of x and y.
B.antisymmetric under interchanged x and y.
C.neither symmetric nor antisymmetric under the interchange of x and y
D.both symmetric and antisymmetric under the interchange of x and y respectively

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The function is symmetric under interchange of x and y.

Swapping \(x\) and \(y\) turns \(x - y\) into \(-(x - y)\). Each term depends on \(x - y\) only through an even function:

\(|\sin(-(x - y))| = |\sin(x - y)|\), because the modulus removes the sign;
\((-(x - y))^{2} = (x - y)^{2}\);
the constant 3 is unchanged.

So \(\psi(y, x) = \psi(x, y)\). It is not antisymmetric, since that would need \(\psi(y, x) = -\psi(x, y)\), impossible for a function that is always positive. A function cannot be both unless it is zero.

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