Discussion
Using the triangular trial wave function :
\[ \psi(x) = \begin{cases} Ax & \text{for } 0 \le x \le a/2 \\ A(a - x) & \text{for } a/2 \le x \le a \\ 0 & \text{otherwise} \end{cases} \]
an upper bound on the ground state energy of the one-dimensional infinite square well is :
Answer: C
The upper bound is \(\frac{6\hbar^{2}}{ma^{2}}\).
Normalisation: \(2A^{2}\int_{0}^{a/2}x^{2}dx = A^{2}a^{3}/12 = 1\), so \(A^{2} = 12/a^{3}\). The slope of \(\psi\) is +A on the left half and -A on the right, so \(\psi''\) is a delta function of strength -2A at the peak x = a/2:
\[ \begin{aligned} \langle H\rangle &= -\frac{\hbar^{2}}{2m}\int\psi\psi''\,dx \\ &= -\frac{\hbar^{2}}{2m}\psi\!\left(\tfrac{a}{2}\right)(-2A) \\ &= \frac{\hbar^{2}}{2m}\cdot\frac{Aa}{2}\cdot 2A = \frac{\hbar^{2}A^{2}a}{2m} \end{aligned} \]
\[ \begin{aligned} \langle H\rangle &= \frac{\hbar^{2}}{2m}\cdot\frac{12}{a^{2}} \\ &= \frac{6\hbar^{2}}{ma^{2}} \end{aligned} \]
This is above the exact ground state \(\pi^{2}\hbar^{2}/2ma^{2} \approx 4.93\hbar^{2}/ma^{2}\), as the variational principle requires. \(5\hbar^{2}/ma^{2}\) is close to the exact value but is not what this trial function gives, and 7 and 8 come from arithmetic slips in the normalisation.
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