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Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

by Pravegaa Education
Chapter 10

Identical Particles: Quantum Statistics

Theory, worked examples and 22 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards. Click Show solution under any question.

Theory

10.1 Why identity matters

In classical mechanics, every particle follows its own trajectory, so particles can always be told apart, even if they are physically identical. In quantum mechanics, wavefunctions of identical particles overlap, and no experiment can label them. The many-particle wavefunction must then be symmetric under exchange of any two particles, for bosons (integer spin: photons, phonons, ⁴He, deuteron), or antisymmetric, for fermions (half-integer spin: electrons, protons, neutrons, ³He).

Antisymmetry forbids two fermions from being in the same single-particle state (the Pauli principle). Symmetry places no limit on bosons, and in fact enhances multiple occupation.

When does it matter? Quantum effects become important when the thermal de Broglie wavelength \(\lambda=h/\sqrt{2\pi mk_BT}\) is comparable to the interparticle spacing \(n^{-1/3}\):

\[n\lambda^3\gtrsim1\quad\Longrightarrow\quad\text{quantum (degenerate) regime};\qquad n\lambda^3\ll1\quad\Longrightarrow\quad\text{classical regime}.\]

Electrons in a metal have \(n\lambda^3\sim10^4\) and are therefore strongly degenerate even at room temperature, whereas air at STP has \(n\lambda^3\sim10^{-6}\) and is completely classical.

Which composite particles are bosons? Count the total number of fermion constituents (protons + neutrons + electrons). An even total gives a boson; an odd total gives a fermion.

Particle Constituents Total Type
¹H atom \(1p+1e\) 2 boson
⁴He atom \(2p+2n+2e\) 6 boson
³He atom \(2p+1n+2e\) 5 fermion
deuteron (nucleus) \(1p+1n\) 2 boson
α particle \(2p+2n\) 4 boson
⁹Be nucleus \(4p+5n\) 9 fermion
²³Na atom \(11p+12n+11e\) 34 boson
³⁰K atom (hypothetical) \(19p+11n+19e\) 49 fermion

10.2 Counting microstates

Group the single-particle states into levels: level \(i\) has energy \(E_i\) and contains \(g_i\) states (its degeneracy). A macrostate is specified by the occupation numbers \(\{n_i\}\), with

\[\sum_in_i=N,\qquad \sum_in_iE_i=U .\]

(a) Maxwell–Boltzmann (distinguishable particles)

We first choose which \(n_1\) particles go into level 1, which \(n_2\) into level 2, and so on, which can be done in \(\dfrac{N!}{n_1!\,n_2!\cdots}\) ways. Inside level \(i\), each of the \(n_i\) particles can then be in any of the \(g_i\) states, giving a further \(g_i^{n_i}\) ways. Hence

\[W_{\rm MB}=N!\prod_i\frac{g_i^{n_i}}{n_i!}.\]

(The “corrected Boltzmann” count divides by \(N!\) for identical particles in the classical limit.)

(b) Bose–Einstein (indistinguishable, any occupancy)

Put \(n_i\) identical balls into \(g_i\) boxes. Represent the arrangement as a row of \(n_i\) balls and \(g_i-1\) partitions. Each distinct ordering of this row is a distinct arrangement:

\[W_{\rm BE}=\prod_i\frac{(n_i+g_i-1)!}{n_i!\,(g_i-1)!}.\]

(c) Fermi–Dirac (indistinguishable, at most one per state)

Choose which \(n_i\) of the \(g_i\) states are occupied:

\[W_{\rm FD}=\prod_i\frac{g_i!}{n_i!\,(g_i-n_i)!},\qquad n_i\le g_i .\]

(d) Small-number counts

For two particles in \(g\) states:

Statistics Formula \(g=2\) \(g=3\) \(g=6\)
MB \(g^2\) 4 9 36
BE \(\binom{g+1}{2}=\dfrac{g(g+1)}{2}\) 3 6 21
FD \(\binom{g}{2}=\dfrac{g(g-1)}{2}\) 1 3 15

Several other counts are standard. \(N\) bosons in 2 states can be arranged in \(N+1\) ways, since the number in state 1 can be \(0,1,\dots,N\). More generally, \(N\) bosons in \(g\) states can be arranged in \(\binom{N+g-1}{g-1}\) ways; for 11 bosons in 3 levels this gives \(\binom{13}{2}=78\). Finally, three spin-½ fermions in two non-degenerate orbital levels have 4 spin-orbitals available, so there are \(\binom43=4\) ways.

10.3 Most probable distributions

Method. Maximise \(\ln W\) subject to fixed \(N\) and \(U\), using Lagrange multipliers \(\alpha\) and \(\beta\):

\[\delta\Big[\ln W-\alpha\sum_in_i-\beta\sum_in_iE_i\Big]=0 .\]

FD. With Stirling, \(\ln W_{\rm FD}=\sum_i\big[g_i\ln g_i-n_i\ln n_i-(g_i-n_i)\ln(g_i-n_i)\big]\). Then

\[\frac{\partial\ln W}{\partial n_i}=\ln\frac{g_i-n_i}{n_i}=\alpha+\beta E_i\quad\Rightarrow\quad n_i=\frac{g_i}{e^{\alpha+\beta E_i}+1}.\]

BE. Similarly, \(\partial\ln W_{\rm BE}/\partial n_i=\ln\dfrac{n_i+g_i}{n_i}\) (for \(g_i\gg1\)), giving \(n_i=\dfrac{g_i}{e^{\alpha+\beta E_i}-1}\).

MB. \(\partial\ln(W_{\rm MB}/N!)/\partial n_i=\ln(g_i/n_i)\), giving \(n_i=g_ie^{-\alpha-\beta E_i}\).

Identifying the multipliers. Comparison with thermodynamics gives \(\beta=1/k_BT\) and \(\alpha=-\mu/k_BT\). Writing \(A=e^\alpha\) and the fugacity \(z=1/A=e^{\beta\mu}\):

\[\boxed{\bar n_i=\frac{g_i}{e^{\beta(E_i-\mu)}+a}},\qquad a=+1\ (\text{FD}),\quad 0\ (\text{MB}),\quad -1\ (\text{BE}).\]

The same result follows exactly, without Stirling, from the grand canonical ensemble (Ch. 9).

Classical limit. All three distributions agree when \(e^{\beta(E-\mu)}\gg1\) for all \(E\), i.e. when \((E-\mu)\gg k_BT\). Then \(\bar n\ll1\): states are sparsely occupied, so the question of double occupancy never arises. This is equivalent to \(z=n\lambda^3/g\ll1\).

Restrictions on \(\mu\). For bosons, the requirement \(\bar n\ge0\) demands \(\mu<E_{\min}\) (or \(\mu\le0\) if \(E_{\min}=0\)). For fermions, \(\mu\) can have any sign; at \(T=0\), \(\mu=E_F>0\). For photons and phonons the number is not conserved, so there is no constraint on \(N\) and no multiplier \(\alpha\); hence \(\mu=0\).

Occupation of a single state (\(g=1\)), with \(q=e^{-\beta(\epsilon-\mu)}\):

Quantity FD BE
Allowed \(n\) 0, 1 0, 1, 2, …
\(P(n)\) \(P(1)=\bar n\), \(P(0)=1-\bar n\) \(P(n)=(1-q)q^n\) (geometric)
Mean \(\bar n=\dfrac{1}{q^{-1}+1}\) \(\bar n=\dfrac{q}{1-q}\)
Variance \(\bar n(1-\bar n)\) \(\bar n(1+\bar n)\)

For bosons, \(P(0)=1-q\) directly gives \(\bar n=\dfrac{1-P(0)}{P(0)}\).

10.4 Density of states

Counting in \(k\)-space. In a box of side \(L\) with periodic boundary conditions, \(\vec k=\dfrac{2\pi}{L}(n_x,n_y,n_z)\). Each state occupies a volume \((2\pi/L)^d\) in \(k\)-space. So the number of states with wavevector less than \(k\) in \(d\) dimensions is

\[\Sigma(k)=g\,\frac{V_d\,k^d}{(2\pi/L)^d},\]

where \(V_d k^d\) is the volume of a \(d\)-sphere (\(\frac43\pi k^3\), \(\pi k^2\), \(2k\) for \(d=3,2,1\)) and \(g\) is the internal (spin or polarisation) degeneracy.

Non-relativistic particles (\(\epsilon=\hbar^2k^2/2m\)). Then \(g(\epsilon)=d\Sigma/d\epsilon\):

Dimension \(g(E)\,dE\) Energy dependence
3D \(2\pi gV\Big(\dfrac{2m}{h^2}\Big)^{3/2}E^{1/2}dE\) \(\propto E^{1/2}\)
2D \(\dfrac{2\pi gmA}{h^2}dE\) constant
1D \(gL\Big(\dfrac{2m}{h^2}\Big)^{1/2}E^{-1/2}dE\) \(\propto E^{-1/2}\)

General dispersion \(\epsilon=Ak^s\) in \(d\) dimensions. \(\Sigma\propto k^d\propto\epsilon^{d/s}\), so

\[g(\epsilon)\propto\epsilon^{d/s-1}.\]

Pressure–energy relation. For an ideal gas with \(\epsilon\propto k^s\propto V^{-s/d}\) (since \(k\propto L^{-1}\)):

\[P=-\Big(\frac{\partial U}{\partial V}\Big)_S=-\sum_i\bar n_i\frac{d\epsilon_i}{dV}=\frac{s}{d}\frac{U}{V}\quad\Longrightarrow\quad PV=\frac sd\,U .\]

This holds for any statistics and any temperature. For a non-relativistic gas in 3D (\(s=2\)) it gives \(PV=\frac23U\); for a photon gas, or any ultra-relativistic gas (\(s=1\)), \(PV=\frac13U\); and for \(\epsilon\propto k^3\) in 3D, \(P=U/V\).

Photons. \(\epsilon=\hbar ck\), with 2 polarisations:

\[g(\nu)\,d\nu=\frac{8\pi V\nu^2}{c^3}d\nu,\qquad g(\omega)\,d\omega=\frac{V\omega^2}{\pi^2c^3}d\omega .\]

10.5 Maxwell–Boltzmann gas

Normalisation. \(N=\displaystyle\int\frac{g(E)\,dE}{A\,e^{\beta E}}\). In 3D with \(g=1\):

\[N=\frac{2\pi V}{A}\Big(\frac{2m}{h^2}\Big)^{3/2}\int_0^\infty E^{1/2}e^{-\beta E}dE=\frac{V}{A\lambda^3}\quad\Rightarrow\quad A=e^\alpha=\frac{V}{N\lambda^3}.\]

Energy distribution.

\[n(E)\,dE=\frac{2\pi N}{(\pi k_BT)^{3/2}}E^{1/2}e^{-E/k_BT}dE .\]

The mean energy is \(\langle E\rangle=\frac32k_BT\) in 3D. In 2D, \(g\) is constant, so \(n(E)\propto e^{-E/k_BT}\) and \(\langle E\rangle=k_BT\). The peak of \(n(E)\) follows from \(\dfrac{d}{dE}\big(E^{1/2}e^{-E/k_BT}\big)=0\), which gives \(E=\frac12k_BT\). Note that this peak of the energy distribution, \(\frac12k_BT\), is not \(\frac12mv_p^2=k_BT\), because the Jacobian \(dE=mv\,dv\) changes the shape.

Speed distribution (Ch. 1). \(f(v)=4\pi v^2\Big(\dfrac{m}{2\pi k_BT}\Big)^{3/2}e^{-mv^2/2k_BT}\). The characteristic speeds are \(v_p=\sqrt{2k_BT/m}\), \(\bar v=\sqrt{8k_BT/\pi m}\) and \(v_{\rm rms}=\sqrt{3k_BT/m}\), in the ratios \(v_p:\bar v:v_{\rm rms}=\sqrt2:\sqrt{8/\pi}:\sqrt3=1:1.128:1.225\).

10.6 Fermi–Dirac gas

(a) The Fermi energy at \(T=0\)

At \(T=0\), \(f(E)=1\) for \(E<E_F\) and \(0\) for \(E>E_F\). All states up to \(E_F\) are filled. In 3D:

\[N=\int_0^{E_F}2\pi gV\Big(\frac{2m}{h^2}\Big)^{3/2}E^{1/2}dE=\frac{4\pi gV}{3}\Big(\frac{2mE_F}{h^2}\Big)^{3/2}.\]

Solving,

\[\boxed{E_F=\frac{h^2}{2m}\Big(\frac{3N}{4\pi gV}\Big)^{2/3}=\frac{\hbar^2}{2m}\Big(\frac{6\pi^2n}{g}\Big)^{2/3}},\qquad p_F=\hbar\Big(\frac{6\pi^2n}{g}\Big)^{1/3},\qquad T_F=\frac{E_F}{k_B}.\]

For electrons, \(g=2\) and \(k_F=(3\pi^2n)^{1/3}\).

Ground-state energy.

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