Discussion
Consider the differential operator
\[ \begin{aligned} L = a\frac{d^{3}}{dx^{3}} &+ b\frac{d^{2}}{dx^{2}} \\ &+ c\frac{d}{dx} \end{aligned} \]
acting on the space of functions \(\{u_{i}(x) \,|\, 1 \le x \le 2\}\) satisfying the boundary conditions \(u_{i}(1) = 0 = u_{i}(2)\). If \(L = L^{\dagger}\), then :
Answer: B
The operator is self-adjoint for \(a = i, b = 1, c = i\).
Integrating by parts with the boundary terms dropped, each derivative changes sign when moved onto the other function, and the inner product conjugates the coefficient:
\[ \begin{aligned} \left(a\frac{d^{3}}{dx^{3}}\right)^{\dagger} &= -a^{*}\frac{d^{3}}{dx^{3}} \\ \left(b\frac{d^{2}}{dx^{2}}\right)^{\dagger} &= b^{*}\frac{d^{2}}{dx^{2}} \\ \left(c\frac{d}{dx}\right)^{\dagger} &= -c^{*}\frac{d}{dx} \end{aligned} \]
\(L = L^{\dagger}\) therefore needs \(a = -a^{*}\) and \(c = -c^{*}\) (purely imaginary) and \(b = b^{*}\) (real). This is the familiar fact that \(-i\,d/dx\) is Hermitian while \(d/dx\) is not.
Only \(a = i, b = 1, c = i\) meets all three conditions. Real values \(a = \pm 1\) fail the first condition, and \(c = 1\) fails the third.
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