The given wavefunction is $\Phi(x) = x(a - x)$, which can be expanded as $\Phi(x) = ax - x^2$. We need to evaluate the truthfulness of the provided statements about this function.
Wavefunction Properties Analysis
Stationary State Possibility
A stationary state in quantum mechanics corresponds to a wavefunction that is an eigenfunction of the Hamiltonian operator ($H\Phi = E\Phi$). Such states have probability densities ($|\Psi|^2$) that are independent of time. The form $\Phi(x) = x(a-x)$ resembles the ground state wavefunction of a particle confined to an infinite potential well between $x=0$ and $x=a$. The ground state of this system is indeed a stationary state. Thus, the statement that it represents a stationary state is plausible in typical quantum mechanics contexts.
Parity Assessment
To determine if $\Phi(x)$ is an odd function, we check the condition $\Phi(-x) = -\Phi(x)$.
- Substitute $-x$ into the wavefunction: $\Phi(-x) = (-x)(a - (-x)) = (-x)(a + x) = -ax - x^2$.
- Calculate the negative of the original wavefunction: $-\Phi(x) = -(ax - x^2) = -ax + x^2$.
Comparing $\Phi(-x)$ and $-\Phi(x)$, we see that $-ax - x^2 \neq -ax + x^2$ (unless $x=0$). Therefore, $\Phi(x)$ is not an odd function.
Circular Ring Wavefunction Relevance
Wavefunctions for a particle on a circular ring typically depend on the angular coordinate $\phi$, often in the form of $e^{im\phi}$. The given function $\Phi(x) = x(a - x)$ is dependent solely on the Cartesian coordinate $x$. It does not possess the characteristics required for describing a particle moving on a circular path, nor does the parameter $a$ represent the radius in this context.
Momentum Operator Eigenfunction Test
The one-dimensional momentum operator is given by $\hat{p} = -i\hbar \frac{d}{dx}$. For $\Phi(x)$ to be an eigenfunction of $\hat{p}$, the action of $\hat{p}$ on $\Phi(x)$ must yield a constant multiple (the eigenvalue, $p$) of $\Phi(x)$ itself, i.e., $\hat{p}\Phi(x) = p\Phi(x)$.
Applying the momentum operator:
$ \hat{p}\Phi(x) = -i\hbar \frac{d}{dx} [x(a - x)] $ $ \hat{p}\Phi(x) = -i\hbar \frac{d}{dx} [ax - x^2] $Performing the differentiation:
$ \hat{p}\Phi(x) = -i\hbar (a - 2x) $The result, $-i\hbar (a - 2x)$, is not directly proportional to the original wavefunction $\Phi(x) = ax - x^2$. Therefore, $\Phi(x)$ is not an eigenfunction of the momentum operator.
Summary of Findings
Based on the analysis:
- The wavefunction $\Phi(x) = x(a - x)$ is likely a stationary state.
- It is not an odd function.
- It does not represent a particle on a circular ring.
- It is not an eigenfunction of the momentum operator.
Thus, the statement "It represents a stationary state" is the correct characterization.
Important Questions from Particle in a Box
Consider two non-interacting particles confined to a one-dimensional box with infinite potential barriers. Their wavefunctions are $\psi_1$ and $\psi_2$ and energies are $E_1$ and $E_2$, respectively. The INCORRECT statement(s) about this system is/are
- Wavefunctions and energies for a particle confined in a cubic box are $\psi_{n_x,n_y,n_z}$ and $E_{n_x,n_y,n_z}$, respectively. The functions $\Phi_1$, $\Phi_2$, $\Phi_3$, and $\Phi_4$ are written as linear combinations of $\psi_{n_x,n_y,n_z}$. Among these functions, the eigenfunction(s) of the Hamiltonian operator for this particle is/are
$\Phi_1 = \frac{1}{\sqrt{2}}\psi_{1,4,1} - \frac{1}{\sqrt{2}}\psi_{2,2,3}$
$\Phi_2 = \frac{1}{\sqrt{2}}\psi_{1,5,1} + \frac{1}{\sqrt{2}}\psi_{3,3,3}$
$\Phi_3 = \frac{1}{\sqrt{2}}\psi_{1,3,8} + \frac{1}{\sqrt{2}}\psi_{3,8,1}$
$\Phi_4 = \frac{1}{2}\psi_{3,3,1} + \frac{\sqrt{3}}{2}\psi_{2,4,1}$ The wave function of a particle in a cubic box (of side L) is given by
$\psi(x, y, z) = \sqrt{32/L^3} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L}$.
The ratio of the energy of the state corresponding to the above wave function to the ground state energy is ________.
(rounded off to the nearest integer)- The wavelength associated with a particle in one-dimensional box of length $L$ is ($n$ refers to the quantum number)
- Assume 1,3,5-hexatriene to be a linear molecule and model the $\pi$ electrons as particles in a one-dimensional box of length 0.70 nm. The wavelength (in nm) corresponding to the transition from the ground-state to the first excited-state is ________
