Discussion
A particle in a one-dimensional infinite potential well of width L is acted on by a perturbation
H′ = b(L - x) \(0 \le x \le L\)
= 0 otherwise
First order correction to the first excited state is :
Answer: A
The first-order correction is \(\frac{bL}{2}\).
The first excited state of the well is \(\psi_{2} = \sqrt{\frac{2}{L}}\sin\frac{2\pi x}{L}\). The first-order correction is the expectation value of the perturbation:
\[ \begin{aligned} E^{(1)} &= \langle \psi_{2}|b(L - x)|\psi_{2} \rangle \\ &= bL - b\langle x \rangle \\ &= bL - b\frac{L}{2} = \frac{bL}{2} \end{aligned} \]
Every stationary state of the infinite well has \(|\psi|^{2}\) symmetric about the centre, so \(\langle x \rangle = L/2\) and the answer is the same for any level. Options bL and 2bL ignore the averaging of x, and \(bL/\sqrt{2}\) would come from mixing up the normalisation constant.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development