Question

The possible value(s) of energy ($E$) obtained from setting the following determinant equal to zero is/are

• A. $\alpha + \sqrt{2}\,\beta$
• B. $\alpha + \beta$
• C. $\alpha - \sqrt{2}\,\beta$
• D. $\alpha$

Hint:

Tridiagonal symmetric matrices often give eigenvalues of the form $\alpha \pm \sqrt{n}\beta$—look for patterns instead of expanding fully in exams

Answer 1

The Correct Option is A, C, D

Step 1: Expand the secular determinant.
Working out the three by three determinant gives
\[ (\alpha - E)\left[(\alpha - E)^2 - 2\beta^2\right] = 0 \]

Step 2: Solve the two factors.
\[ E = \alpha \quad \text{and} \quad E = \alpha \pm \sqrt{2}\,\beta \]

Step 3: Match the options.
These are $\alpha + \sqrt{2}\beta$, $\alpha$ and $\alpha - \sqrt{2}\beta$, which are choices A, C and D.

Step 4: Answer.
\[ \boxed{\text{A, C and D}} \]