Question

For the wavefunction \(\Phi = x(a-x)\), where \(a\) is a constant, the correct statement(s) is(are)

• A. It represents a stationary state
• B. It is an odd function
• C. It represents the wavefunction of a particle moving on a circular ring of radius \(a\)
• D. It is an eigen function of the momentum operator

Hint:

Check each claim directly: test \(\Phi(-x)\) against \(-\Phi(x)\) for oddness, apply \(\hat{p}=-i\hbar\,d/dx\) for the momentum test, and recall that ring wavefunctions must be periodic in angle; a purely time-independent \(\Phi(x)\) is the spatial part of a stationary state.

Answer 1

The Correct Option is A

This question checks four different properties of a given spatial function: whether it is a stationary quantum state, whether it has odd symmetry, whether it belongs to a particle on a ring, and whether it is a momentum eigenfunction.

1. Stationary state (A): $\Phi = x(a-x)$ is a function of $x$ only, it carries no time label $t$. Time-independent spatial wavefunctions like this are exactly what go into the stationary-state ansatz $\Psi(x,t)=\Phi(x)e^{-iEt/\hbar}$, whose probability density $|\Psi|^2=\Phi(x)^2$ stays fixed as $t$ runs. Since $\Phi$ here has no time dependence at all, the state it describes is stationary. True.
2. Odd function (B): Expand $\Phi = ax - x^2$. Odd functions have only odd powers of $x$ and satisfy $\Phi(-x)=-\Phi(x)$. Here $\Phi(-x) = -ax - x^2$, but $-\Phi(x) = -ax + x^2$; the $x^2$ terms carry opposite signs, so the two are not equal anywhere except $x=0$. Shifting to the natural center $x=a/2$ also turns $\Phi$ into $a^2/4-\xi^2$, an even function of $\xi$. So $\Phi$ is not odd. False.
3. Particle on a ring (C): Wavefunctions on a ring must repeat every $2\pi$ in the angle $\phi$, so they take the exponential form $e^{im\phi}$. $\Phi=x(a-x)$ is a plain polynomial in a straight-line coordinate, it has no periodicity and is not even written in an angular variable. So it cannot be a ring wavefunction. False.
4. Momentum eigenfunction (D): Apply $\hat{p}=-i\hbar\,d/dx$: $$\hat{p}\Phi = -i\hbar(a-2x)$$ For $\Phi$ to be a momentum eigenfunction, this result would need to equal a constant times $\Phi$, but $(a-2x)$ is not a constant multiple of $x(a-x)$. Only plane waves $e^{ikx}$ are momentum eigenfunctions. False.

Working through the algebra on all four statements leaves only the stationary-state claim standing.

Let's summarize:

• A wavefunction written with no time variable, like $\Phi=x(a-x)$, is the time-independent part of a stationary state.
• $\Phi$ is neither odd, nor periodic like a ring wavefunction, nor an eigenfunction of momentum, since applying $\hat p$ to it does not return a multiple of itself.

The correct statement is A.