Identical particles and Fock space
Replace fictitious particle labels with mode occupations and distinguish bosonic enhancement, fermionic exclusion, and field-theory interpretation.
On this page
The question for today
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 and are distinct orthonormal orbitals, the normalized combinations are . 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.
Count occupations instead of fictitious names
Choose a one-particle mode basis and write a many-particle state using occupations . Bosonic mode occupations can be any nonnegative integer; each fermionic mode has occupation zero or one. For noninteracting particles with one-particle energies , the fixed-number energy is the sum , 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.
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.
Work it through
Two particles and two modes
There are two orthonormal complete single-particle modes, a and b, with energies and in chosen units. List the occupation states for two noninteracting bosons, then for two identical fermions without any extra spin label.
Fix the total occupation . Bosons allow all three nonnegative solutions: , , and . They are already exchange-adapted occupation states.
Use the additive Hamiltonian: their energies are , , and . No extra factor of two appears for swapping fictitious particle labels.
For fermions each complete mode can hold at most one, leaving only with energy 3. Adding independent spin labels would create more single-particle modes and change the counting; that is a different specified model.
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.
The living notebook
Explore the model
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.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
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.
Pause & reason
Check your understanding
Your turn
Try it yourself
Start with one bosonic mode in normalized . Apply 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 and then . An operator action need not be a normalized time-evolution or measurement map.
Compare with the solution
The result is 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.
Keep the idea
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.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.