Question

The eigen value for the wave function \( \psi = A e^{ikx} + B e^{-ikx} \) of the operator \( \frac{d^2}{dx^2} \) is:

• A. -k
• B. -k^2
• C. k^2
• D. k

Hint:

The second derivative of an exponential wave function gives an eigenvalue that corresponds to the square of the wave number multiplied by -1.

Answer 1

The Correct Option is B

To find the eigenvalue of the operator \( \frac{d^2}{dx^2} \), we apply it to the wave function \( \psi = A e^{ikx} + B e^{-ikx} \).

Step 1: Apply the operator \( \frac{d^2}{dx^2} \) to the wave function. Compute the first derivative of \( \psi \): \[ \frac{d}{dx} (A e^{ikx} + B e^{-ikx}) = A ik e^{ikx} - B ik e^{-ikx} \] Compute the second derivative: \[ \frac{d^2}{dx^2} (A e^{ikx} + B e^{-ikx}) = - A k^2 e^{ikx} - B k^2 e^{-ikx} \] \[ = - k^2 (A e^{ikx} + B e^{-ikx}) \]

Step 2: Interpret the result. The result is \( \frac{d^2}{dx^2} \psi = - k^2 \psi \). This indicates that \( \psi \) is an eigenfunction with eigenvalue \( -k^2 \). Therefore, the eigenvalue is \( -k^2 \).

Final Answer: \[ \boxed{\text{(2) } -k^2} \]