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

Step 1: Recall what a stationary state means.
A stationary state is a quantum state whose probability density \(|\Psi|^2\) does not change with time. Such states are written as \(\Psi(x,t) = \Phi(x)e^{-iEt/\hbar}\), where \(\Phi(x)\) is the time-independent spatial wavefunction. Since \(\Phi = x(a-x)\) is given purely as a function of position, with no time variable in it at all, it is exactly this kind of time-independent spatial part, so it represents a stationary state. Statement (A) is correct.

Step 2: Test whether it is an odd function.
An odd function satisfies \(f(-x) = -f(x)\). Here \(\Phi(-x) = -x(a+x) = -ax - x^2\), while \(-\Phi(x) = -ax + x^2\). These are not equal, so \(\Phi\) is not odd about \(x=0\). It is also not odd about the natural midpoint \(x=a/2\): writing \(\xi = x - a/2\) gives \(\Phi = \tfrac{a^2}{4} - \xi^2\), which is even in \(\xi\), not odd. Statement (B) is wrong.

Step 3: Check the circular ring claim.
A particle moving freely on a ring has wavefunctions of the form \(e^{im\phi}\), periodic in the angular coordinate \(\phi\), where \(m\) is the quantum number. \(\Phi = x(a-x)\) is a plain quadratic in a linear coordinate \(x\), with no angular periodicity at all, so it cannot describe a particle on a ring. Statement (C) is wrong.

Step 4: Test whether \(\Phi\) is a momentum eigenfunction.
The momentum operator is \(\hat{p} = -i\hbar \dfrac{d}{dx}\). Acting on \(\Phi\),
\[ \hat{p}\Phi = -i\hbar\frac{d}{dx}\big[x(a-x)\big] = -i\hbar(a-2x) \]
This result is proportional to \((a-2x)\), not to \(\Phi = x(a-x)\) itself, so \(\Phi\) is not an eigenfunction of \(\hat{p}\) (true momentum eigenfunctions are the complex exponentials \(e^{ikx}\)). Statement (D) is wrong.

Final Answer:
Only statement (A) is correct. \[ \boxed{\text{A}} \]