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Identical particles

Symmetric and antisymmetric states, bosons and fermions, the Pauli exclusion principle, the exchange effect, and the Fermi energy of metals.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to build symmetric and antisymmetric states, fill levels with fermions, compute exchange effects, and estimate the Fermi energy of a metal.

2. What you already have

You know single-particle states in wells and atoms, spin-½, and from the lesson on adding angular momentum that two spins form a symmetric triplet and an antisymmetric singlet. So far every problem has had one particle. This lesson adds a second identical one and finds that its mere presence changes the allowed states, even without any force between them.

3. Words for this lesson

4. Swapping identical particles

For two distinguishable particles in states $\psi_a$ and $\psi_b$, the combined state is $\psi_a(x_1)\psi_b(x_2)$: particle 1 in $a$, particle 2 in $b$. For identical particles that statement is meaningless — nothing can tell which is which. Every probability must be unchanged when the labels are swapped, so $|\psi(x_2, x_1)|^2 = |\psi(x_1, x_2)|^2$. Nature uses only the two simplest solutions:

$$\psi_\pm(x_1, x_2) = \frac{1}{\sqrt{2}}\left[\psi_a(x_1)\psi_b(x_2) \pm \psi_b(x_1)\psi_a(x_2)\right].$$

Bosons, with integer spin, take the symmetric $+$ combination; fermions, with half-integer spin, take the antisymmetric $-$ one. The spin–statistics theorem of relativistic quantum field theory proves that the link between spin and symmetry must hold.

Setting $a = b$ in the fermion combination gives zero: two identical fermions cannot occupy the same state, the Pauli exclusion principle. With spin included, the whole state — space times spin — must be antisymmetric. Two electrons in a spin singlet (antisymmetric) have a symmetric spatial state and can share an orbital; in a triplet (symmetric) they must have an antisymmetric spatial state and different orbitals. Antisymmetric spatial states keep the particles apart, and symmetric ones draw them together: the exchange effect, which looks like a force but is not one.

5. The method, step by step, and how to check it

1. Identify the particles. Electrons, protons, neutrons, helium-3: fermions. Photons, helium-4, deuterons, pions: bosons. A composite particle is a boson if it contains an even number of fermions.
2. Fill levels. For noninteracting fermions, place them in single-particle levels from the bottom up, at most one per state; with spin-½, two per spatial orbital. Bosons all go into the lowest level.
3. Symmetrize. Write $\psi_\pm$ for two particles in different states. For many fermions, a Slater determinant does the antisymmetrizing automatically.
4. Compute averages. Cross terms involving $\langle x\rangle_{ab}$ appear only for identical particles and only when the two states overlap.

Checks. Exchanging labels must change the sign of a fermion state; setting two fermion states equal must give zero. The total energy of noninteracting fermions must exceed that of the same number of bosons. And exchange effects vanish when the particles' wave functions do not overlap: two electrons in different atoms far apart behave as distinguishable.

6. Why identical particles bunch or avoid each other

For distinguishable particles in states $a$ and $b$, $\langle(x_1 - x_2)^2\rangle_d = \langle x^2\rangle_a + \langle x^2\rangle_b - 2\langle x\rangle_a\langle x\rangle_b$. For the symmetrized states an extra term appears:

$$\langle(x_1 - x_2)^2\rangle_\pm = \langle(x_1 - x_2)^2\rangle_d \mp 2|\langle x\rangle_{ab}|^2,$$

where $\langle x\rangle_{ab} = \int x\,\psi_a^*\psi_b\,dx$ measures the overlap. Symmetric states bring the particles closer; antisymmetric states push them apart. For the two lowest states of a box, the rms separation is $0.32a$ for distinguishable particles, $0.20a$ in the symmetric state, and $0.41a$ in the antisymmetric state.

No force appears in the Hamiltonian. The effect comes from interference between the two ways of assigning particles to states, and it is often called the exchange force for convenience. Its consequences are real: in the hydrogen molecule the symmetric spatial state piles electron density between the nuclei and binds them; in iron, the antisymmetric state keeps parallel-spin electrons apart, lowering their repulsion and favoring aligned spins — ferromagnetism.

7. Atoms and the periodic table

In an atom, each orbital $(n, l, m_l)$ holds at most two electrons, one of each spin. Filling orbitals from the lowest energy upward builds the periodic table: two electrons fill $1s$ (helium), eight more fill $2s$ and $2p$ (neon), and so on. Without the exclusion principle, every electron would sit in $1s$, all atoms would be small and chemically similar, and chemistry as we know it would not exist.

Helium shows the spin–space link directly. In its excited states, one electron in $1s$ and one in $2s$, the spins can form a singlet (parahelium) or triplet (orthohelium). The triplet's antisymmetric spatial state keeps the electrons apart, lowering their repulsion, so orthohelium lies about $0.8$ eV below parahelium for the $1s2s$ configuration — the origin of Hund's rule that electrons in partly filled shells prefer parallel spins.

8. Degeneracy pressure

Fermions filling a box cannot all sit at the bottom, so even at absolute zero they have a large kinetic energy, and they push outward. Compressing the box raises every level as $1/L^2$, so the total energy rises and resists compression. This degeneracy pressure has nothing to do with temperature or with forces between the particles; it comes purely from antisymmetry.

Degeneracy pressure is why solids resist compression, since squeezing atoms pushes their electrons into higher states. It holds up white dwarf stars, where electrons packed a million times more densely than in ordinary matter support a star's mass against gravity. Above about $1.4$ solar masses, the Chandrasekhar limit, electrons become relativistic and can no longer resist, and the star collapses to a neutron star, now held up by neutron degeneracy pressure.

9. Bosons that crowd together

Bosons do the opposite. Their symmetric states enhance the probability of sharing: a boson is more likely to enter a state that already contains others, with a factor $N + 1$ for a state holding $N$. This is the root of stimulated emission — a photon entering a mode already filled with photons is emitted preferentially into it — and so of the laser.

At low temperature, bosonic atoms pile into the lowest single-particle state in large numbers, forming a Bose–Einstein condensate: millions of atoms described by one wave function. Helium-4 becomes superfluid below $2.17$ K, flowing without friction, while helium-3, a fermion, does so only below a few millikelvin, where its atoms pair up into bosons. The difference of one neutron changes the statistics and the physics completely.

10. Slater determinants

For $N$ fermions in single-particle states $\psi_1, \ldots, \psi_N$, the antisymmetric state is the determinant of the $N \times N$ matrix whose entry in row $i$, column $j$ is $\psi_j(x_i)$, divided by $\sqrt{N!}$. Swapping two particles swaps two rows and flips the sign; putting two particles in the same state makes two columns equal and the determinant zero. The exclusion principle is built in.

Slater determinants are the starting point of nearly all many-electron calculations in chemistry and solid-state physics, including the Hartree–Fock method mentioned in the variational lesson. They capture exchange exactly but miss the correlations produced by the electrons' Coulomb repulsion, which more advanced methods add as combinations of many determinants.

11. In the world: why metals conduct and resist squeezing

A copper wire contains about $10^{29}$ conduction electrons per cubic meter. If they were bosons they would all sit in the lowest state at low temperature. Being fermions, they fill states up to the Fermi energy of $7.0$ eV, and the highest of them move at $1.6 \times 10^6$ m/s even at absolute zero. Only electrons within about $k_BT$ of the Fermi energy can change state, which explains why the electrons contribute so little to a metal's heat capacity — a puzzle for classical physics that Arnold Sommerfeld solved in 1928 using Fermi–Dirac statistics.

The same filled sea sets metals' stiffness. Much of the bulk modulus of simple metals such as sodium comes from the degeneracy pressure of their conduction electrons, and the free-electron estimate $\tfrac{2}{3}nE_F$ comes within a factor of two of measured values. Engineers designing high-pressure equipment and geophysicists modeling Earth's iron core both rely on this quantum resistance to compression.

12. In the world: white dwarfs and neutron stars

When a star like the Sun exhausts its fuel, gravity crushes its core to about the size of Earth, a white dwarf, with densities near $10^9$ kg/m³. What stops further collapse is the degeneracy pressure of electrons forced into high-momentum states by the exclusion principle. Subrahmanyan Chandrasekhar showed in 1930 that above about $1.4$ solar masses the electrons become relativistic and the pressure can no longer win; he shared the 1983 Nobel Prize for it.

Heavier cores collapse further, electrons combining with protons into neutrons, until neutron degeneracy pressure halts the collapse at a radius of about $12$ km: a neutron star. NASA's NICER instrument on the International Space Station has measured the radii of several neutron stars by timing X-rays from hot spots on their surfaces, constraining how neutron matter resists compression. Both kinds of star are giant demonstrations of antisymmetric wave functions.

13. The exclusion principle is not a force

Because identical fermions avoid each other and resist compression, it is tempting to picture a repulsive force between them. There is none in the Hamiltonian. The effects come from the requirement that the wave function change sign under exchange, which removes states from the list of possibilities. Degeneracy pressure is kinetic, not potential: compressing fermions forces them into states of higher momentum.

A second misconception is that identical particles can be tracked by following their paths. In quantum mechanics particles have no definite paths, so when two identical particles' wave functions overlap, there is no fact about which one went where. This is not a limit of our instruments; experiments such as the Hong–Ou–Mandel effect, in which two identical photons meeting at a beam splitter always leave together, show that the two possibilities genuinely interfere.

14. Two particles in a box

1. Write the single-particle energies.

   $E_n = n^2E_1$

   Each particle alone would have these.

2. Find the ground state for distinguishable particles.

   $\psi_1(x_1)\psi_1(x_2), \qquad E = 2E_1$

   Both in the lowest level.

3. Find the ground state for identical bosons.

   $\psi_1(x_1)\psi_1(x_2) \text{ is already symmetric}, \qquad E = 2E_1$

   Bosons may share.

4. Find the ground state for identical fermions with parallel spins.

   $\tfrac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) - \psi_2(x_1)\psi_1(x_2)], \qquad E = 5E_1$

   They must use two different levels.

5. Find the first excited state for bosons.

   $\tfrac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2)], \qquad E = 5E_1$

   The symmetric partner of the fermion ground state.

15. The exchange effect in numbers

1. Take states $n = 1$ and $n = 2$ in a box of width $a$. Find $\langle x^2\rangle$ for each.

   $\langle x^2\rangle_1 = a^2\left(\tfrac{1}{3} - \tfrac{1}{2\pi^2}\right) = 0.2827a^2, \quad \langle x^2\rangle_2 = 0.3207a^2$

   From $\langle x^2\rangle_n = a^2(\tfrac{1}{3} - \tfrac{1}{2n^2\pi^2})$.

2. Find the distinguishable mean squared separation.

   $\langle(x_1 - x_2)^2\rangle_d = 0.2827a^2 + 0.3207a^2 - 2\left(\tfrac{a}{2}\right)^2 = 0.1034a^2$

   Both states have $\langle x\rangle = a/2$.

3. Evaluate the overlap integral.

   $\langle x\rangle_{12} = \dfrac{2}{a}\displaystyle\int_0^a x\sin\dfrac{\pi x}{a}\sin\dfrac{2\pi x}{a}\,dx = -\dfrac{16a}{9\pi^2} = -0.1801a$

   Nonzero because the states overlap.

4. Find the symmetric result.

   $\langle(x_1 - x_2)^2\rangle_+ = 0.1034a^2 - 2(0.1801a)^2 = 0.0385a^2$

   Bosons bunch.

5. Find the antisymmetric result.

   $\langle(x_1 - x_2)^2\rangle_- = 0.1034a^2 + 2(0.1801a)^2 = 0.1683a^2$

   Fermions with parallel spins avoid each other.

6. Take square roots.

   $d_+ = 0.196a, \qquad d_d = 0.322a, \qquad d_- = 0.410a$

   The same states, three different separations.

16. Electrons in a metal

1. Model copper's conduction electrons as a free gas in a cube of side $L$. Write the allowed wave vectors.

   $\vec{k} = \dfrac{2\pi}{L}(n_x, n_y, n_z)$

   Periodic boundary conditions.

2. Find the volume per state in $k$-space.

   $\left(\dfrac{2\pi}{L}\right)^3 = \dfrac{(2\pi)^3}{V}$

   One allowed $\vec{k}$ per cube of that size.

3. Fill a sphere of radius $k_F$ with two electrons per state.

   $N = 2 \cdot \dfrac{\tfrac{4}{3}\pi k_F^3}{(2\pi)^3/V} = \dfrac{Vk_F^3}{3\pi^2}$

   The exclusion principle forces the sphere to grow.

4. Solve for the Fermi wave number.

   $k_F = (3\pi^2n)^{1/3}$

   With $n = N/V$.

5. Evaluate for copper.

   $n = 8.47 \times 10^{28}\ \text{m}^{-3} = 84.7\ \text{nm}^{-3}, \quad k_F = (3\pi^2 \times 84.7)^{1/3} = 13.6\ \text{nm}^{-1}$

   One conduction electron per atom.

6. Find the Fermi energy.

   $E_F = 0.0381 \times 13.6^2 = 7.0\ \text{eV}$

   With $\hbar^2/2m = 0.0381$ eV nm².

7. Find the Fermi velocity.

   $v_F = \dfrac{\hbar k_F}{m} = 1.158 \times 10^{-4} \times 1.36 \times 10^{10} = 1.57 \times 10^{6}\ \text{m/s}$

   Half a percent of light speed, even at absolute zero.

8. Compare with thermal energy.

   $\dfrac{E_F}{k_BT} = \dfrac{7.0}{0.0259} \approx 270$

   Room temperature barely disturbs the filled sea.

17. Your turn: six noninteracting electrons fill a one-dimensional well with $E_1 = 10$ meV. What is the total energy?

1. Fill the levels two at a time.

   $n = 1, 2, 3$

   Two electrons per level.

2. Add the level energies twice.

   $E = 2 \times 10 \times (1 + 4 + 9)$

   $2n^2E_1$ per level.

3. Your turn: work this step out. Its working is at the end of the packet.

   Evaluate the total.

18. Guided practice

Two electrons, both spin up, are in an infinite well whose ground level is $E_1 = 14$ meV. Ignoring their repulsion, what is the lowest total energy?

• $70$ meV
• $28$ meV
• $56$ meV
• $112$ meV

19. Guided practice

Complete the worked solution: four noninteracting electrons fill an infinite well with $E_1 = 21$ meV. Find the energy of the lowest pair, of the next pair, and the total, in meV.

1. Place two electrons in the ground level.

   $2 \times E_1 =$ _a

   Opposite spins share $n = 1$.

2. Place the other two in the next level.

   $2 \times 4E_1 =$ _b

   Level $n = 2$ has energy $4E_1$.

3. Add the two pairs together.

   $E =$ _c

   Four bosons would total only $4E_1$.

20. Guided practice

Match each statement about identical particles to its content.

21. Practice

Two noninteracting particles are in an infinite well with $E_1 = 21$ meV. Fill in the lowest total energy, in meV, for each case.

22. Practice

$12$ noninteracting electrons are placed in a one-dimensional infinite well with $E_1 = 13$ meV. What is the lowest total energy, in meV?

Answer: meV

23. Practice

Two electrons with parallel spins are in an infinite well of width $a = 5$ nm, one in $n = 1$ and one in $n = 2$, so their spatial state is antisymmetric. Find their rms separation $\sqrt{\langle(x_1 - x_2)^2\rangle}$, in nm.

Answer: nm

24. Somewhere new

The conduction electrons in copper behave like a free gas with density $n = 8.47 \times 10^{28}$ m⁻³. Because of the exclusion principle they fill states up to the Fermi energy. What is it, in eV?

Answer: eV

25. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

26. Test question

Two noninteracting particles are in an infinite well with $E_1 = 27$ meV. Fill in the lowest total energy, in meV, for each case.

27. What you can do now

You can handle identical particles. Explain to someone why two electrons with the same spin cannot share the lowest level of a box.

Working for the steps left to you

17. Your turn: six noninteracting electrons fill a one-dimensional well with $E_1 = 10$ meV. What is the total energy?, step 3