Question:
Two identical particles with a fixed total energy \(E = 2\hbar\omega\) are in thermal equilibrium in a one-dimensional harmonic oscillator potential \(\frac{1}{2}m\omega^2x^2\). Let the entropy of the particles be denoted by \(S_F\) if they are fermions with spin \(\frac{1}{2}\) (\(S_z = \pm\frac{\hbar}{2}\)) and by \(S_B\) if they are bosons with spin 0. Then, which of the following options is correct? (\(k_B\) is the Boltzmann constant)
Two identical particles with a fixed total energy \(E = 2\hbar\omega\) are in thermal equilibrium in a one-dimensional harmonic oscillator potential \(\frac{1}{2}m\omega^2x^2\). Let the entropy of the particles be denoted by \(S_F\) if they are fermions with spin \(\frac{1}{2}\) (\(S_z = \pm\frac{\hbar}{2}\)) and by \(S_B\) if they are bosons with spin 0. Then, which of the following options is correct? (\(k_B\) is the Boltzmann constant)
Show Hint
Total energy fixes \(n_1+n_2=1\), so the two particles must sit in levels 0 and 1. Count the allowed spin/label combinations for fermions and for bosons separately, then use \(S=k_B\ln\Omega\).
Updated On: Jul 28, 2026
- \(S_F = k_B\ln 2,\ S_B = k_B\ln 2\)
- \(S_F = 2k_B\ln 2,\ S_B = 0\)
- \(S_F = 4k_B\ln 2,\ S_B = 0\)
- \(S_F = 2k_B\ln 2,\ S_B = k_B\ln 2\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A single particle in this 1D oscillator can sit in any level \(n=0,1,2,\dots\) with energy \(\varepsilon_n = (n+\tfrac12)\hbar\omega\). With two particles the total energy is \(\varepsilon_{n_1}+\varepsilon_{n_2} = (n_1+n_2+1)\hbar\omega\).
Step 2: Key Formula or Approach:
Setting this equal to the given \(E=2\hbar\omega\) gives \(n_1+n_2 = 1\). Since \(n_1,n_2\) are non-negative integers, the only way to split 1 quantum between two levels is \(n_1=0, n_2=1\) (in either order); there is no solution with \(n_1=n_2\). So one particle always occupies \(n=0\) and the other occupies \(n=1\); the number of accessible microstates is found from the entropy formula \(S = k_B\ln\Omega\), where \(\Omega\) counts the distinct many-body quantum states with this energy.
Step 3: Detailed Explanation:
For fermions with spin \(\tfrac12\), each spatial level \(n=0\) and \(n=1\) can be paired with spin up or spin down. Because the two particles sit in different spatial levels (\(n=0\) and \(n=1\)), the Pauli principle places no restriction on their spins at all, so a valid antisymmetric (Slater determinant) state exists for every combination of the two spins.
Number of combinations: 2 spin choices for the particle in \(n=0\) times 2 spin choices for the particle in \(n=1\) gives \(\Omega_F = 2\times2 = 4\).
\[ S_F = k_B\ln\Omega_F = k_B\ln 4 = 2k_B\ln 2 \]
For bosons with spin 0, there is no spin label to combine with. The two particles still must occupy the different spatial levels \(n=0\) and \(n=1\) (there is no other way to make \(n_1+n_2=1\)), and for identical bosons there is exactly one symmetric combination of "one particle in 0, one particle in 1", not two. So \(\Omega_B = 1\).
\[ S_B = k_B\ln \Omega_B = k_B\ln 1 = 0 \]
Step 4: Why the other options are wrong.
Option (A) gives the bosons a nonzero entropy, but a unique, non-degenerate boson microstate has \(S_B=0\), not \(k_B\ln2\).
Option (C) overcounts the fermion microstates as if there were 16 of them (\(4k_B\ln2 = k_B\ln16\)); only 4 spin combinations are actually allowed here.
Option (D) again wrongly gives the bosons a nonzero entropy.
Final Answer:
The spin combinatorics of two spin-\(\tfrac12\) fermions in different orbitals gives \(S_F = 2k_B\ln2\), while the non-degenerate boson state gives \(S_B=0\). \[ \boxed{S_F = 2k_B\ln 2,\ S_B = 0} \]
A single particle in this 1D oscillator can sit in any level \(n=0,1,2,\dots\) with energy \(\varepsilon_n = (n+\tfrac12)\hbar\omega\). With two particles the total energy is \(\varepsilon_{n_1}+\varepsilon_{n_2} = (n_1+n_2+1)\hbar\omega\).
Step 2: Key Formula or Approach:
Setting this equal to the given \(E=2\hbar\omega\) gives \(n_1+n_2 = 1\). Since \(n_1,n_2\) are non-negative integers, the only way to split 1 quantum between two levels is \(n_1=0, n_2=1\) (in either order); there is no solution with \(n_1=n_2\). So one particle always occupies \(n=0\) and the other occupies \(n=1\); the number of accessible microstates is found from the entropy formula \(S = k_B\ln\Omega\), where \(\Omega\) counts the distinct many-body quantum states with this energy.
Step 3: Detailed Explanation:
For fermions with spin \(\tfrac12\), each spatial level \(n=0\) and \(n=1\) can be paired with spin up or spin down. Because the two particles sit in different spatial levels (\(n=0\) and \(n=1\)), the Pauli principle places no restriction on their spins at all, so a valid antisymmetric (Slater determinant) state exists for every combination of the two spins.
Number of combinations: 2 spin choices for the particle in \(n=0\) times 2 spin choices for the particle in \(n=1\) gives \(\Omega_F = 2\times2 = 4\).
\[ S_F = k_B\ln\Omega_F = k_B\ln 4 = 2k_B\ln 2 \]
For bosons with spin 0, there is no spin label to combine with. The two particles still must occupy the different spatial levels \(n=0\) and \(n=1\) (there is no other way to make \(n_1+n_2=1\)), and for identical bosons there is exactly one symmetric combination of "one particle in 0, one particle in 1", not two. So \(\Omega_B = 1\).
\[ S_B = k_B\ln \Omega_B = k_B\ln 1 = 0 \]
Step 4: Why the other options are wrong.
Option (A) gives the bosons a nonzero entropy, but a unique, non-degenerate boson microstate has \(S_B=0\), not \(k_B\ln2\).
Option (C) overcounts the fermion microstates as if there were 16 of them (\(4k_B\ln2 = k_B\ln16\)); only 4 spin combinations are actually allowed here.
Option (D) again wrongly gives the bosons a nonzero entropy.
Final Answer:
The spin combinatorics of two spin-\(\tfrac12\) fermions in different orbitals gives \(S_F = 2k_B\ln2\), while the non-degenerate boson state gives \(S_B=0\). \[ \boxed{S_F = 2k_B\ln 2,\ S_B = 0} \]
Was this answer helpful?
0
0
Top GATE PH Physics Questions
- An infinite one-dimensional lattice extends along the x-axis. At each lattice site, there exists an ion with spin 1/2. The spin can point either in the +z or -z direction only. Let \( S_P \), \( S_F \), and \( S_A \) denote the entropies of paramagnetic, ferromagnetic, and antiferromagnetic configurations, respectively. Which of the following relations is/are true?
- Two point charges of charge \( +q \) each are placed a distance \( 2d \) apart. A grounded solid conducting sphere of radius \( a \) is placed midway between them. Assume \( a^2 \ll d^2 \). Which of the following statements is/are true?
- A particle of mass $m$ is moving in the potential
\[ V(x) = \begin{cases} V_0 + \frac{1}{2} m \omega_{0P}^2 x^2 & \quad \text{if } x > 0 \\ \infty & \quad \text{if } x \leq 0 \end{cases} \]
Figures P, Q, R, and S show different combinations of the values of $ \omega_0 $ and $ V_0 $. - Decays of mesons and baryons can be categorized as weak, strong and electromagnetic decays depending upon the interactions involved in the processes. Which of the following option is/are true?
- The temperature \( T \) dependence of magnetic susceptibility \( \chi \) (Column I) of certain magnetic materials (Column II) are given below. Which of the following options is/are correct?
Column I Column II (1) 
(P) Diamagnetic (2) 
(Q) Paramagnetic (3) 
(R) Ferromagnetic (4) 
(S) Antiferromagnetic
View More Questions
Top GATE PH Thermodynamic and Statistical Physics Questions
- For a finite system of Fermions where the density of states increases with energy, the chemical potential
- Consider a system of three distinguishable particles, each having spin \( S = \frac{1}{2} \) such that \( S_z = \pm \frac{1}{2} \) with corresponding magnetic moments \( \mu_z = \pm \mu \). When the system is placed in an external magnetic field \( H \) pointing along the z-axis, the total energy of the system is \( \mu H \). Let \( x \) be the state where the first spin has \( S_z = \frac{1}{2} \). The probability of having the state \( x \) and the mean magnetic moment (in the +z direction) of the system in state \( x \) are
Consider a single one-dimensional harmonic oscillator of angular frequency \( \omega \), in equilibrium at temperature \( T = \left( k_B \beta \right)^{-1 }\). The states of the harmonic oscillator are all non-degenerate having energy \( E_n = \left( n + \frac{1}{2} \right) \hbar \omega \) with equal probability, where \( n \) is the quantum number. The Helmholtz free energy of the oscillator is
A system of two atoms can be in three quantum states having energies 0, $\epsilon$ and $2\epsilon$. The system is in equilibrium at temperature \( T = (k_B\beta)^{-1} \). Match the following Statistics with the Partition function.
- Consider the Otto cycle for an ideal gas engine consisting of two quasistatic adiabatic and two quasistatic isochoric processes. The correct temperature-entropy (T-S) phase diagram for the cycle is:
View More Questions
Top GATE PH Questions
- If \( F_1(Q, q) = Qq \) is the generating function of a canonical transformation from \((p, q)\) to \((P, Q)\), then which one of the following relations is correct?
- When an unpolarized plane electromagnetic wave in a dielectric medium with refractive index \( n_1 \) is incident on a plane interface separating it from another dielectric medium with refractive index \( n_2 \) (where \( n_2 > n_1 \)), and the angle of incidence is \(\tan^{-1} \left( \frac{n_2}{n_1} \right) \), the following statement is true:
- The wavefunction of a particle in an infinite one-dimensional potential well at time \( t \) is
\[ \Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1 t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6} e^{-iE_2 t/\hbar} \psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4} e^{-iE_3 t/\hbar} \psi_3(x) \]where \(\psi_1\), \(\psi_2\), and \(\psi_3\) are the normalized ground state, the normalized first excited state, and the normalized second excited state, respectively. \(E_1\), \(E_2\), and \(E_3\) are the eigen-energies corresponding to \(\psi_1\), \(\psi_2\), and \(\psi_3\), respectively. The expectation value of energy of the particle in state \(\Psi(x,t)\) is - If a thermodynamical system is adiabatically isolated and experiences a change in volume under an externally applied constant pressure, then the thermodynamical potential minimized at equilibrium is the
- The mean distance between the two atoms of an HD molecule is \( r \), where H and D denote hydrogen and deuterium, respectively. The mass of the hydrogen atom is \( m_H \). The energy difference between the two lowest lying rotational states of HD in multiples of \(\frac{h^2}{m_H r^2}\) is:
View More Questions