Discussion
The operator \(\left(\frac{d}{dx} - x\right)\left(\frac{d}{dx} + x\right)\) is equivalent to :
Answer: B
The operator is \(\frac{d^{2}}{dx^{2}} - x^{2} + 1\).
Act on a test function \(f\), remembering that \(d/dx\) also differentiates the \(x\) in \(xf\):
\[ \begin{aligned} \left(\frac{d}{dx} - x\right)(f' + xf) &= f'' + (xf)' - xf' - x^{2}f \\ &= f'' + f + xf' - xf' - x^{2}f \\ &= f'' - x^{2}f + f \end{aligned} \]
The extra 1 comes from the commutator \([d/dx, x] = 1\). This is the same algebra that gives the \(\pm\frac{1}{2}\hbar\omega\) in the harmonic-oscillator ladder operators.
\(d^{2}/dx^{2} - x^{2}\) treats the two factors as commuting numbers and drops the commutator. The options containing a first-derivative term keep the \(xf'\) pieces, which actually cancel.
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