Question

A particle of mass m is subjected to a potential V(x). If its wavefunction is given by
\(\psi(x,t) = \alpha x^2 e^{-\beta x} e^{i\gamma t/\hbar}, x>0\)
\(\psi(x,t) = 0, x \le 0\),
then V(x) is (\(\alpha\), \(\beta\) and \(\gamma\) are constants of appropriate dimensions)

• A. \(-\gamma + \frac{\hbar^2}{2m} \left( \frac{2}{x^2} - \frac{4\beta}{x} + \beta^2 \right)\)
• B. \(-\gamma + \frac{\hbar^2}{2m} \left( \frac{2}{x^2} + \frac{4\beta}{x} + \beta^2 \right)\)
• C. \(-\gamma + \frac{\hbar^2}{2m} \left( \frac{2}{x^2} - \frac{4\beta}{x} - \beta^2 \right)\)
• D. \(-\gamma + \frac{\hbar^2}{2m} \left( -\frac{2}{x^2} - \frac{4\beta}{x} + \beta^2 \right)\)

Hint:

When asked to find the potential V(x) given a stationary state wavefunction \(\psi(x,t) = \phi(x)T(t)\), the first step is always to identify the energy E from the time-dependent part \(T(t)\). Then, plug \(\phi(x)\) and E into the time-independent Schrödinger equation and solve for V(x). Be careful with the signs and differentiation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The potential \(V(x)\) experienced by a particle can be determined from its wavefunction \(\psi(x,t)\) using the time-dependent Schrödinger equation (TDSE). Since the given wavefunction is a product of a spatial part and a time-dependent part, it represents a stationary state. This allows us to use the time-independent Schrödinger equation (TISE) to find \(V(x)\).
Step 2: Key Formula or Approach:
The time-dependent Schrödinger equation is:
\[ i\hbar \frac{\partial \psi(x,t)}{\partial t} = \left[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right] \psi(x,t) \] For a stationary state, \(\psi(x,t) = \phi(x) e^{-iEt/\hbar}\). The equation simplifies to the TISE:
\[ \left[ -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right] \phi(x) = E \phi(x) \] We can rearrange this to solve for \(V(x)\):
\[ V(x) = E + \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2\phi(x)}{dx^2} \] Step 3: Detailed Explanation:
The given wavefunction is \(\psi(x,t) = \alpha x^2 e^{-\beta x} e^{i\gamma t/\hbar}\).
Let's compare the time-dependent part \(e^{i\gamma t/\hbar}\) with the standard form \(e^{-iEt/\hbar}\).
We have \(-iEt/\hbar = i\gamma t/\hbar\), which implies \(E = -\gamma\).
The spatial part of the wavefunction is \(\phi(x) = \alpha x^2 e^{-\beta x}\).
Now we need to find the second derivative of \(\phi(x)\) with respect to x.
First derivative:
\[ \frac{d\phi}{dx} = \frac{d}{dx} (\alpha x^2 e^{-\beta x}) = \alpha \left[ (2x)e^{-\beta x} + x^2(-\beta e^{-\beta x}) \right] = \alpha e^{-\beta x} (2x - \beta x^2) \] Second derivative:
\[ \frac{d^2\phi}{dx^2} = \frac{d}{dx} \left[ \alpha e^{-\beta x} (2x - \beta x^2) \right] \] \[ = \alpha \left[ (-\beta e^{-\beta x})(2x - \beta x^2) + e^{-\beta x}(2 - 2\beta x) \right] \] \[ = \alpha e^{-\beta x} \left[ -2\beta x + \beta^2 x^2 + 2 - 2\beta x \right] \] \[ = \alpha e^{-\beta x} (2 - 4\beta x + \beta^2 x^2) \] Now, substitute this into the rearranged TISE to find \(V(x)\):
\[ V(x) = E + \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2\phi(x)}{dx^2} \] \[ V(x) = -\gamma + \frac{\hbar^2}{2m} \frac{1}{\alpha x^2 e^{-\beta x}} \left[ \alpha e^{-\beta x} (2 - 4\beta x + \beta^2 x^2) \right] \] Cancel the common terms \(\alpha e^{-\beta x}\):
\[ V(x) = -\gamma + \frac{\hbar^2}{2m} \frac{2 - 4\beta x + \beta^2 x^2}{x^2} \] Finally, split the fraction:
\[ V(x) = -\gamma + \frac{\hbar^2}{2m} \left( \frac{2}{x^2} - \frac{4\beta}{x} + \beta^2 \right) \] Step 4: Final Answer:
The potential \(V(x)\) is \(-\gamma + \frac{\hbar^2}{2m} \left( \frac{2}{x^2} - \frac{4\beta}{x} + \beta^2 \right)\). This matches option (A).