Physical problem
Exchange symmetry becomes an algebraic sign rule
Identical fermions occupy antisymmetric multiparticle states, so writing a state in an occupation basis requires a precise ordering convention. A creation operator at a given orbital must be moved past earlier occupied orbitals, generating a sign equal to the parity of those crossings. Here you can inspect that sign without hiding the intermediate state behind symbolic shorthand.
Step-by-step operator evaluation
Signs, blocking and anticommutators
Input
Occupied orbitals before the operator product
Each orbital has occupation zero or one (Pauli exclusion). The basis order is essential for computing parity.
Output
The occupied orbitals after AB
If either step violates Pauli exclusion or tries to remove an absent particle, the entire ordered product produces the zero vector.
Algebra
Why reversing the operator order can reverse the sign
AB applies B then A; BA applies A then B. For canonical fermionic operators, their sum is the anticommutator, with δjk when one is creation and the other annihilation. When both are the same type, the sum always vanishes.
Fock-space derivation
How many particles must a fermionic operator cross?
Choose a fixed ordered set of single-particle orbitals, labeled 0,1,2,3. Define the many-body basis |n₀n₁n₂n₃〉=(a†₀)n₀ (a†₁)n₁(a†₂)n₂(a†₃)n₃|vac〉. Every nⱼ is either zero or one. Fermi anticommutation gives a minus sign when an operator passes an already occupied lower-index orbital.
The step-by-step readout evaluates the earlier-occupation parity separately at both operator applications, since the first action can change the sign encountered by the second. The square of a creation operator on the same orbital is zero by antisymmetry.
Demonstrate opposite creation order on vacuum
Apply a†₁a†₀|vac〉 from right to left. First create orbital 0 with sign +1, giving |1000〉. Next create orbital 1: one earlier orbital is occupied, so the factor is −1 and the result is −|1100〉. Reversing the product to a†₀a†₁|vac〉 gives +|1100〉. Their sum is zero, demonstrating {a†₁,a†₀}=0.
Check the mixed anticommutator
For the same orbital j, aⱼa†ⱼ|n〉=(1−nⱼ)|n〉, while a†ⱼaⱼ|n〉=nⱼ|n〉. Adding both gives |n〉 for every occupation pattern. For distinct orbitals, changing the application order produces amplitudes of opposite parity and the same final occupation state, so they cancel.
Relation to bosonic creation operators
For canonical bosonic occupations nⱼ∈{0,1,2,…}, b†ⱼ|n〉=√(nⱼ+1)|n+eⱼ〉 and bⱼ|n〉=√nⱼ|n−eⱼ〉. They commute between distinct orbitals; their ladder factors replace the fermionic parity factor. A separate bosonic lab explores these number-dependent amplitudes.
Controlled experiments
Explore how the algebra follows from ordering
- Creation order. On |0000〉, select a†₁a†₀. Compare AB and BA. What is the relative sign? How many occupied orbitals does each second operation cross?
- Pauli blocking. Select |1000〉 and a†₀. Verify that attempting a second creation in orbital 0 produces the zero vector, irrespective of any signs.
- Canonical relation. Select A=a₀ and B=a†₀; compare AB and BA on an empty and an occupied orbital. In both cases, verify that their sum is the original state.
Self-check: a†₁a†₀|vac〉 equals which state in the ordered occupation basis?
Exact scope and connections
The model uses a four-orbital fermionic Fock space with occupation-basis amplitudes ±1 or zero. Spin labels are absorbed into orbital indices, so the same algebra applies to spin-orbitals of a many-electron system. Many-body Hamiltonians, hopping operators and interaction matrix elements can be assembled from these creation and annihilation rules; this first laboratory isolates the sign convention used to construct such matrix elements correctly.
Read the current Chapter 6 fermion operators, together with permutation symmetry.