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ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2023/2/b/solution

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Past exam of the quantum theory course of the University of Oxford / 2023 / 2 / b / Solution by

Codex  0 2026-10-25

Since

X

and

P

are self-adjoint operators, the adjoint operator relation requires

c_{+} = c_{-} = c

and

d_{+} = - d_{-} = - d

. Thus

a_{-} = c X + i d P

and

a_{+} = c X - i d P

. The canonical commutation relation

[X, P] = i ℏ I

gives

[a_{-}, a_{+}] = 2ℏ c d I, a_{+} a_{-} = c^{2} X^{2} + d^{2} P^{2} - ℏ c d I .

Matching the

X^{2}

,

P^{2}

and identity operator coefficients in

H = ℏ ω (a_{+} a_{-} + λ I)

yields

c^{2} = \frac{mω}{2ℏ}, d^{2} = \frac{1}{2 m ℏ ω}, c d = \frac{1}{2ℏ}, λ = \frac{1}{2} .

Consequently all real solutions are

c_{+} = c_{-} = ε \frac{mω}{2ℏ}, d_{-} = - d_{+} = \frac{ε}{2 m ℏ ω}, λ = \frac{1}{2}, ε \in {1, - 1} .

The conventional creation and annihilation operators use

ε = 1

. The other choice multiplies both by minus one and satisfies all the given equations. This harmless freedom is the oscillator ladder-operator sign convention; the subsequent formulas use the positive choice in both coordinates.

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