Discussion
The potential of a quantum harmonic oscillator is modified from \(\frac{1}{2}kx^{2}\) to \(\frac{1}{2}kx^{2} + 3ax\), where \(k\) and \(a\) are constants and x is the position variable. When, the values of \(a = 2\) and \(k = 1\), which of the following options gives the change in the ground state energy?
(\(a\) and \(k\) are in appropriate units)
Answer: D
The ground state energy changes by \(-18\).
Completing the square shows the new potential is the same oscillator, shifted sideways and downwards:
\[ \begin{aligned} \tfrac{1}{2}kx^{2} + 3ax &= \tfrac{1}{2}k\left(x + \tfrac{3a}{k}\right)^{2} \\ &\quad - \tfrac{9a^{2}}{2k} \end{aligned} \]
The shift in \(x\) does not change the spectrum, since \(k\) and so \(\omega\) are unchanged; every level, including the ground state \(\frac{1}{2}\hbar\omega\), drops by the constant \(\frac{9a^2}{2k}\).
\[ \begin{aligned} \Delta E &= -\frac{9a^{2}}{2k} \\ &= -\frac{9 \times 4}{2 \times 1} = -18 \end{aligned} \]
The shift grows with the square of the linear coefficient \(3a\), so the smaller options each leave out part of \(\frac{9a^2}{2k}\). The result is exact, not a first-order estimate, because a linear term in a harmonic potential can be removed completely.
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