|q⟩ Bad Qubits

intermediate · Physics · The Harmonic Oscillator: Ladder Operators

Raising and Lowering States

We know aa and a†a^\dagger shift the number eigenvalue by ∓1\mp 1, but by how much — what is the exact scalar coefficient? Fixing the normalization is what turns the qualitative "ladder" picture into a calculational tool.

Why the eigenvalue shifts

From the algebra N^a†=a†(N^+1)\hat{N}a^\dagger = a^\dagger(\hat{N} + 1) (a rearrangement of [N^,a†]=a†[\hat{N}, a^\dagger] = a^\dagger), apply both sides to ∣n⟩|n\rangle:

N^ (a†∣n⟩)=a†(N^+1)∣n⟩=(n+1) (a†∣n⟩).\hat{N}\,(a^\dagger|n\rangle) = a^\dagger(\hat{N} + 1)|n\rangle = (n+1)\,(a^\dagger|n\rangle).

So a†∣n⟩a^\dagger|n\rangle is an eigenstate of N^\hat{N} with eigenvalue n+1n+1; it must be proportional to ∣n+1⟩|n+1\rangle. Likewise a∣n⟩∝∣n−1⟩a|n\rangle \propto |n-1\rangle. It remains to find the proportionality constants.

Fixing the coefficients by normalization

Write a†∣n⟩=cn∣n+1⟩a^\dagger|n\rangle = c_n |n+1\rangle for some constant cnc_n. Take the squared norm of both sides. The right side gives ∣cn∣2|c_n|^2 (since ∣n+1⟩|n+1\rangle is normalized). The left side is

⟨n∣a a†∣n⟩=⟨n∣(a†a+1)∣n⟩=⟨n∣(N^+1)∣n⟩=n+1,\langle n|a\,a^\dagger|n\rangle = \langle n|(a^\dagger a + 1)|n\rangle = \langle n|(\hat{N}+1)|n\rangle = n + 1,

using the commutator aa†=a†a+1aa^\dagger = a^\dagger a + 1. Therefore ∣cn∣2=n+1|c_n|^2 = n+1, and choosing the phase so the coefficient is real and positive,

 a†∣n⟩=n+1 ∣n+1⟩ .\boxed{\ a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle\ }.

The same argument for the lowering operator uses ⟨n∣a†a∣n⟩=n\langle n|a^\dagger a|n\rangle = n, giving

 a∣n⟩=n ∣n−1⟩ .\boxed{\ a|n\rangle = \sqrt{n}\,|n-1\rangle\ }.

Notice that a∣0⟩=0 ∣−1⟩=0a|0\rangle = \sqrt{0}\,|{-1}\rangle = 0 automatically — the ladder simply stops at the bottom, exactly as the spectrum argument required.

Building states from the ground state

Iterating the raising relation generates every number state from ∣0⟩|0\rangle:

∣n⟩=(a†)nn! ∣0⟩.|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}}\,|0\rangle.

For example, a†∣0⟩=1 ∣1⟩a^\dagger|0\rangle = \sqrt{1}\,|1\rangle, then a†∣1⟩=2 ∣2⟩a^\dagger|1\rangle = \sqrt{2}\,|2\rangle, so (a†)2∣0⟩=2 ∣2⟩=2! ∣2⟩(a^\dagger)^2|0\rangle = \sqrt{2}\,|2\rangle = \sqrt{2!}\,|2\rangle, consistent with the 1/n!1/\sqrt{n!} normalization. The factorial appears because each step contributes its own n+1\sqrt{n+1}.

Try it

This is a numerical exercise — your code should return a number. Apply a†a^\dagger three times to the ground state ∣0⟩|0\rangle to reach ∣3⟩|3\rangle, and return the cumulative coefficient 1⋅2⋅3=6\sqrt{1}\cdot\sqrt{2}\cdot\sqrt{3} = \sqrt{6}.

Run your code to see the quantum state.

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