Second Quantization: Fock States, Ladder Operators, (Anti)commutators

What you are seeing: a single-mode Fock space, a state written as amplitudes over the number basis |0>, |1>, |2>, and so on. The creation operator raises a state (a-dagger|n> = sqrt(n+1)|n+1>) and the annihilation operator lowers it (a|n> = sqrt(n)|n-1>). Switch between bosons, which climb the ladder without limit and obey [a, a-dagger] = 1, and fermions, which saturate at |1> by the Pauli principle and obey {a, a-dagger} = 1. One panel shows the ladder rungs and current amplitudes as you pump the state, a second verifies the (anti)commutator identity by overlaying (a a-dagger -/+ a-dagger a) acting on the state, and a third shows the occupation distribution, which is Poissonian for the bosonic coherent state.

Figure 1. The occupation-number (Fock) representation: ladder operators raise and lower number states by sqrt(n+1) and sqrt(n), bosons obey [a, a-dagger] = 1 and fermions {a, a-dagger} = 1 (Pauli, so a-dagger|1> = 0), and the bosonic coherent state is a Poissonian eigenstate of a. Method: exact Fock-space linear algebra; deterministic.

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coherent alpha2.40

WHAT TO TRY

Apply the creation operator repeatedly: a boson climbs the number ladder without limit (a-dagger|n> = sqrt(n+1)|n+1>), while a fermion saturates at |1> by the Pauli principle. The statistics toggle shows both.
Switch to the bosonic coherent state and tune alpha: the amplitudes settle into a Poisson distribution over number states, the closest a quantum field comes to a classical wave.
Read the commutation versus anticommutation relation: [a, a-dagger] = 1 for bosons, {a, a-dagger} = 1 for fermions. That one sign is the entire difference between the two kinds of matter.