Identical Particles Questions and Answers

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.

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.

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.

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.

Answer: C

The ratio is 2.

The levels of a one-dimensional box are \(E_{n} = n^{2}E_{1}\). In the ground state, bosons can all occupy the lowest level, while spin-1/2 fermions obey the Pauli principle: at most two (spin up and spin down) per level.

\[ \begin{aligned} E_{\text{bosons}} &= 3E_{1} \\ E_{\text{fermions}} &= 2E_{1} + 4E_{1} = 6E_{1} \\ \text{ratio} &= 6/3 = 2 \end{aligned} \]

A ratio of 1 ignores the Pauli principle. 14/3 is what one gets by putting the three fermions in three different levels, \(1 + 4 + 9 = 14\), which forgets the spin degeneracy. 1/3 inverts the comparison in a different way and has no basis.

Answer: C

The ground state energy is \(\frac{\hbar^{2}\pi^{2}}{mL^{2}}\) with \(s = 0\).

Both fermions can sit in the lowest level \(n = 1\) only if their total wavefunction is antisymmetric. The spatial part \(u_{1}(x_{1})u_{1}(x_{2})\) is symmetric, so the spin part must be the antisymmetric singlet, \(s = 0\). The energy is

\[ \begin{aligned} E &= 2E_{1} = 2\cdot\frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &= \frac{\pi^{2}\hbar^{2}}{mL^{2}} \end{aligned} \]

A triplet (\(s = 1\)) needs an antisymmetric spatial state, so the lowest triplet puts one particle in \(n = 1\) and one in \(n = 2\), with energy \(5\pi^{2}\hbar^{2}/2mL^{2}\). That is higher, and it belongs to \(s = 1\), not \(s = 0\) as the last option states. The option \(\pi^{2}\hbar^{2}/mL^{2}\) with \(s = 1\) breaks the Pauli principle, and \(5\pi^{2}\hbar^{2}/4mL^{2}\) has half the correct triplet energy.