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: Write the determinant.
\[ \left| \begin{matrix} \alpha - E & \beta & 0 \\ \beta & \alpha - E & \beta \\ 0 & \beta & \alpha - E \end{matrix} \right| \]

Step 2: Expand the determinant.
Expanding along the first row:
\[ (\alpha - E) \left| \begin{matrix} \alpha - E & \beta \\ \beta & \alpha - E \end{matrix} \right| - \beta \left| \begin{matrix} \beta & \beta \\ 0 & \alpha - E \end{matrix} \right| \]

Step 3: Compute sub-determinants.
\[ (\alpha - E)\left[(\alpha - E)^2 - \beta^2\right] - \beta \left[\beta(\alpha - E)\right] \]
\[ = (\alpha - E)\left[(\alpha - E)^2 - \beta^2\right] - \beta^2(\alpha - E) \]

Step 4: Simplify expression.
\[ = (\alpha - E)\left[(\alpha - E)^2 - 2\beta^2\right] \]

Step 5: Set equal to zero.
\[ (\alpha - E)\left[(\alpha - E)^2 - 2\beta^2\right] = 0 \]

Step 6: Solve for $E$.
\[ \alpha - E = 0 \Rightarrow E = \alpha \]
\[ (\alpha - E)^2 = 2\beta^2 \Rightarrow \alpha - E = \pm \sqrt{2}\beta \]
\[ E = \alpha \pm \sqrt{2}\beta \]

Step 7: Conclusion.
\[ \boxed{E = \alpha,\; \alpha + \sqrt{2}\beta,\; \alpha - \sqrt{2}\beta} \]