Consider a particle in a one-dimensional infinite potential well with its walls at \( x = 0 \) and \( x = L \). The system is perturbed as shown in the figure. The first order correction to the energy eigenvalue is
Consider a particle in a one-dimensional infinite potential well with its walls at \( x = 0 \) and \( x = L \). The system is perturbed as shown in the figure. The first order correction to the energy eigenvalue is
Show Hint
- \( \dfrac{V_0}{4} \)
- \( \dfrac{V_0}{3} \)
- \( \dfrac{V_0}{2} \)
- \( \dfrac{V_0}{5} \)
The Correct Option is C
Solution and Explanation
Step 1: The particle is inside an infinite potential well, with the potential \( V(x) \) changing at the boundaries, from 0 at \( x = 0 \) to \( V_0 \) at \( x = L \).
Step 2: The first-order energy correction due to the perturbation is given by the expectation value of the potential in the unperturbed state. The perturbation is a linear increase in potential from \( V(0) = 0 \) to \( V(L) = V_0 \), so the energy correction is proportional to \( V_0 \).
Step 3: The calculation leads to a correction of \( \frac{V_0}{2} \).
Thus, the correct answer is (C).
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