Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 306 / 4 / d

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- Solution d

Solution

The zero-mode condition is automatic:

L_{0} ∣ ψ ⟩ = (\frac{1}{2} α_{0}^{2} + 2) ∣ ψ ⟩ = ∣ ψ ⟩ .

Using

[L_{m}, α_{n}^{μ}] = - n α_{m + n}^{μ}

and oscillator annihilation of the vacuum gives

L_{1} ∣ ψ ⟩ = 2 (t_{μν} α_{0}^{ν} + v_{μ}) α_{- 1}^{μ} ∣0, p ⟩,

which vanishes when

t_{μν} α_{0}^{ν} = - v_{μ}

. Similarly,

L_{2} ∣ ψ ⟩ = (t^{μ}_{μ} + 2 v \cdot α_{0}) ∣0, p ⟩,

which vanishes when

v \cdot α_{0} = - t^{μ}_{μ} /2

. Higher positive Virasoro modes annihilate a level-two state. These are precisely the stated physical-state conditions.

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