 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2015/2/c/ii/solution

Twice applying the creation operator or twice applying the annihilation operator changes the energy eigenvalue by

2ℏ ω

, unless they give zero. The same orthogonality argument as in part (i) therefore gives

E_{ψ} (a_{+}^{2}) = 0 = E_{ψ} (a_{-}^{2}) .

Also the factorization in part (a) gives

E_{ψ} (a_{+} a_{-}) = 2 m (E - ℏ ω /2), E_{ψ} (a_{-} a_{+}) = 2 m (E + ℏ ω /2) .

Now

P = (a_{+} + a_{-}) /2

and

X = (a_{+} - a_{-}) / (2 imω)

. Squaring in the correct operator order yields

E_{ψ} (P^{2}) E_{ψ} (X^{2}) = \frac{1}{4} E_{ψ} (a_{+}^{2} + a_{+} a_{-} + a_{-} a_{+} + a_{-}^{2}) = m E, = - \frac{1}{4 m ^{2} ω ^{2}} E_{ψ} (a_{+}^{2} - a_{+} a_{-} - a_{-} a_{+} + a_{-}^{2}) = \frac{E}{m ω ^{2}} .

By part (i), the means are zero, so these are the variances. Taking positive square roots gives the requested standard deviations:

Δ_{ψ} (X) = \frac{E}{m ω ^{2}}, Δ_{ψ} (P) = m E .

E /2

. Their uncertainty product is

E / ω = (n + \frac{1}{2}) ℏ

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