core32 min studyLesson 30 in Physics

Identical particles and Fock space

Replace fictitious particle labels with mode occupations and distinguish bosonic enhancement, fermionic exclusion, and field-theory interpretation.

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What you’ll understand

  • Construct symmetric and antisymmetric two-particle states from distinct orbitals.
  • Count allowed occupations for bosonic and fermionic modes.
  • Explain how Fock space extends fixed-number quantum mechanics without itself specifying particle-creation dynamics.

When particles are identical, attaching permanent names to them introduces distinctions the experiment cannot observe. Quantum states must encode the appropriate exchange symmetry. Occupation numbers provide an efficient alternative: specify how many quanta occupy each single-particle mode. This bridges several-particle wavefunctions and quantum-field notation without yet specifying a field theory.

Exchange changes the description, not the physical labeling

For two identical particles in ordinary three-dimensional space, bosonic states are symmetric and fermionic states antisymmetric under exchange of the complete single-particle labels. If ∣a⟩|a\rangle and ∣b⟩|b\rangle are distinct orthonormal orbitals, the normalized combinations are (∣a⟩∣b⟩±∣b⟩∣a⟩)/2(|a\rangle|b\rangle\pm|b\rangle|a\rangle)/\sqrt2. The signs refer to exchange, not to particles carrying ordinary positive or negative probabilities. Spin and spatial labels must both be included in the complete state.

Antisymmetrizing two identical copies of the same complete single-particle state gives zero. This is Pauli exclusion. It does not forbid two electrons from occupying the same spatial orbital with opposite spin because those are different complete states. In nonrelativistic many-particle mechanics we impose the appropriate statistics for the species. The connection between integer spin and bosonic statistics, and half-integer spin and fermionic statistics, needs assumptions of relativistic quantum field theory; it is not derived merely by permuting a two-particle wavefunction.

∣a,b⟩±=∣a⟩⊗∣b⟩±∣b⟩⊗∣a⟩2,⟨a∣b⟩=0,P12∣a,b⟩±=±∣a,b⟩±|a,b\rangle_{\pm}=\frac{|a\rangle\otimes|b\rangle\pm|b\rangle\otimes|a\rangle}{\sqrt2},\qquad \langle a|b\rangle=0,\qquad P_{12}|a,b\rangle_{\pm}=\pm|a,b\rangle_{\pm}
This normalization assumes distinct orthonormal one-particle states. For two bosons in the same state, the normalized product is simply |a⟩⊗|a⟩; the displayed distinct-state formula must not be reused unchanged.↔ Scroll wide equations horizontally.

Count occupations instead of fictitious names

Choose a one-particle mode basis and write a many-particle state using occupations ∣n1,n2,…⟩|n_1,n_2,\ldots\rangle. Bosonic mode occupations can be any nonnegative integer; each fermionic mode has occupation zero or one. For noninteracting particles with one-particle energies ϵj\epsilon_j, the fixed-number energy is the sum ∑jϵjnj\sum_j\epsilon_jn_j, with any separately specified vacuum convention. Interactions generally make occupation-basis states mix and require additional Hamiltonian terms.

Bosonic creation and annihilation operators change a mode's occupation with square-root factors. Fermionic operators obey anticommutation relations instead, giving exclusion and ordering-dependent signs across different modes. The sign convention depends on a chosen order for the modes, while measurable predictions must be consistent with that convention. “Identical” therefore does not mean all particles have the same state; it means the species provides no physical particle labels that distinguish otherwise exchanged descriptions.

aj†∣…,nj,…⟩=nj+1∣…,nj+1,…⟩,[ai,aj†]=δij,{ci,cj†}=δij,(cj†)2=0a_j^\dagger|\ldots,n_j,\ldots\rangle=\sqrt{n_j+1}|\ldots,n_j+1,\ldots\rangle,\quad [a_i,a_j^\dagger]=\delta_{ij},\quad \{c_i,c_j^\dagger\}=\delta_{ij},\quad (c_j^\dagger)^2=0
The a operators are bosonic and c operators fermionic. The braces denote an anticommutator AB+BA. Fermionic cross-mode signs require a consistent mode-order convention.↔ Scroll wide equations horizontally.

What Fock space adds—and what it does not

Fock space is the direct sum of the zero-particle, one-particle, two-particle, and higher appropriately symmetrized or antisymmetrized sectors. A fixed-number problem lives in one sector. Including all sectors makes a variable particle number representable, but does not force it to change: a number-conserving Hamiltonian keeps an initial state inside its fixed-number sector. The vacuum means zero occupation in the chosen mode description, not necessarily zero energy or the absence of every kind of physical field.

For one mechanical oscillator, the number ladder originally counted excitations of one coordinate. In a quantized free field, independent modes behave like oscillators and their excitation numbers acquire a particle interpretation appropriate to the setup. This needs a field Hamiltonian, commutation rules, and a choice of modes; the oscillator analogy alone does not establish relativistic locality or an interacting theory.

Two particles and two modes

There are two orthonormal complete single-particle modes, a and b, with energies ϵa=1\epsilon_a=1 and ϵb=2\epsilon_b=2 in chosen units. List the occupation states for two noninteracting bosons, then for two identical fermions without any extra spin label.

  1. Fix the total occupation na+nb=2n_a+n_b=2. Bosons allow all three nonnegative solutions: (2,0)(2,0), (1,1)(1,1), and (0,2)(0,2). They are already exchange-adapted occupation states.

  2. Use the additive Hamiltonian: their energies are 2ϵa=22\epsilon_a=2, ϵa+ϵb=3\epsilon_a+\epsilon_b=3, and 2ϵb=42\epsilon_b=4. No extra factor of two appears for swapping fictitious particle labels.

  3. For fermions each complete mode can hold at most one, leaving only (1,1)(1,1) with energy 3. Adding independent spin labels would create more single-particle modes and change the counting; that is a different specified model.

The result

The bosonic two-particle sector has three states in this two-mode model; the fermionic sector has one. Exchange symmetry changes the allowed state space before any dynamics is solved.

Explore the model

Interactive lab

Interactive lab · spectral

Mode basis versus mode occupation

Change the number and coefficient decay of a finite classical sine expansion. This illustrates selecting modes for a basis only. Its continuous coefficients are not occupation numbers, and it contains no bosonic enhancement, fermionic exclusion, particle creation, or quantum exchange simulation.

Mode basis versus mode occupationA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFirst choose what the modes are; quantum occupation rules are an additional structure that a classical mode picture does not supply.

Check your understanding

Which statement correctly applies Pauli exclusion to two electrons?

Try it yourself

Start with one bosonic mode in normalized ∣2⟩|2\rangle. Apply a†a^\dagger twice without intermediate normalization. Find the resulting coefficient and norm squared. Does this alone give a transition probability?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply the factors 3\sqrt3 and then 4\sqrt4. An operator action need not be a normalized time-evolution or measurement map.

Compare with the solution

The result is 12∣4⟩\sqrt{12}|4\rangle with norm squared 12. That number is not a probability: creation operators are not unitary and this calculation supplies neither an interaction strength nor an evolution time or measurement normalization. A physical transition probability requires a fully specified process.

Take it with you

  • Identical-particle states obey exchange symmetry using complete labels, including spin.
  • Mode occupations avoid overcounting fictitious particle names and expose bosonic versus fermionic rules.
  • Fock space permits variable particle number; the Hamiltonian determines whether and how it changes.

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