A second-order differential operator, denoted as $ L = P_2(x)\frac{d^2}{dx^2} + P_1(x)\frac{d}{dx} + P_0(x) $, is formally self-adjoint if its coefficients satisfy the relationship $ P_1(x) = \frac{dP_2}{dx}(x) $. This condition ensures the operator equals its formal adjoint.
We will examine each operator provided to determine if it meets this criterion.
| Operator ID | Operator Expression | Coefficient $P_2(x)$ | Coefficient $P_1(x)$ | Derivative $P_2'(x)$ | Self-Adjoint Condition Met? |
| 1 | $x^2 \frac{d^2}{dx^2} + 3x \frac{d}{dx} + x^2$ |
$x^2$ | $3x$ | $2x$ | No (since $3x \neq 2x$) |
| 2 (B) | $(1 - x^2)\frac{d^2}{dx^2} - 2x\frac{d}{dx} + 3x$ |
$1 - x^2$ | $-2x$ | $-2x$ | Yes (since $-2x = -2x$) |
| 3 (C) | $(3x - 4x^3)\frac{d^2}{dx^2} + (3 - 12x^2)\frac{d}{dx} + 12$ |
$3x - 4x^3$ | $3 - 12x^2$ | $3 - 12x^2$ | Yes (since $3 - 12x^2 = 3 - 12x^2$) |
| 4 | $x\frac{d^2}{dx^2} + x^2\frac{d}{dx} + \frac{5x}{3}$ |
$x$ | $x^2$ | $1$ | No (since $x^2 \neq 1$) |
The analysis shows that operators 2 (B) and 3 (C) satisfy the self-adjoint condition $ P_1(x) = P_2'(x) $. Therefore, these are the self-adjoint operators among the choices.
The wavefunction of a particle in one dimension is given by
$\psi(x) = \begin{cases} M, & -a < x < a \\ 0, & \text{otherwise.} \end{cases}$
Here $M$ and $a$ are positive constants. If $\phi(p)$ is the corresponding momentum space wavefunction, which one of the following plots best represents $|\phi(p)|^2$ ?