Question

A particle of mass \(m\) is in an infinite square potential of length \(L\). The wave function is superimposed state of the first two energy eigenstates, given by:

\[ \Psi(x) = \sqrt{\frac{1}{3}} \Psi_{n=1}(x) + \sqrt{\frac{2}{3}} \Psi_{n=2}(x) \]
Identify the correct statements: 
A. \( \langle p \rangle = 0 \) 
B. \( \Delta p = \frac{\sqrt{3}h}{2L} \) 
C. \( \langle E \rangle = \frac{3h^2}{8mL^2} \) 
D. \( \Delta x = 0 \) 
Choose the correct answer from the options given below:

• A. A, B and D only
• B. A, B and C only
• C. A, B, C and D
• D. B, C and D only

Answer 1

The Correct Option is B

Step 1: Analyze the given wave function.

The state is \( \Psi = c_1 \Psi_1 + c_2 \Psi_2 \) with \( c_1 = \sqrt{1/3} \) and \( c_2 = \sqrt{2/3} \). The state is normalized because \( |c_1|^2 + |c_2|^2 = \frac{1}{3} + \frac{2}{3} = 1 \). The energy eigenvalues are \( E_n = \frac{n^2 h^2}{8m L^2} \).

Step 2: Evaluate each statement.

A. Expectation value of momentum \( \langle p \rangle \): For any energy eigenstate in an infinite square well, \( \langle p \rangle = 0 \). For a superposition of such states, the expectation value of momentum is also 0. So, statement A is correct.

C. Expectation value of energy \( \langle E \rangle \): \[ \langle E \rangle = |c_1|^2 E_1 + |c_2|^2 E_2 = \left(\frac{1}{3}\right)\left(\frac{1^2 h^2}{8m L^2}\right) + \left(\frac{2}{3}\right)\left(\frac{2^2 h^2}{8m L^2}\right) \] \[ \langle E \rangle = \frac{h^2}{8m L^2} \left( \frac{1}{3} + \frac{2 \cdot 4}{3} \right) = \frac{h^2}{8m L^2} \left( \frac{1 + 8}{3} \right) = \frac{3 h^2}{8m L^2} \] Statement C is correct.

B. Uncertainty in momentum \( \Delta p \): \[ (\Delta p)^2 = \langle p^2 \rangle - \langle p \rangle^2 \] Since \( \langle p \rangle = 0 \), \[ (\Delta p)^2 = \langle p^2 \rangle \] We use the relation \( \hat{H} = \frac{\hat{p}^2}{2m} \), so \[ \langle p^2 \rangle = 2m \langle E \rangle \] \[ \langle p^2 \rangle = 2m \left( \frac{3 h^2}{8m L^2} \right) = \frac{3 h^2}{4 L^2} \] \[ \Delta p = \sqrt{\langle p^2 \rangle} = \sqrt{\frac{3 h^2}{4 L^2}} = \frac{\sqrt{3} h}{2 L} \] Statement B is correct.

D. Uncertainty in position \( \Delta x \): \( \Delta x \) is the standard deviation in position. It cannot be zero for a particle in a box, as this would violate the Heisenberg Uncertainty Principle. Statement D is incorrect.

Step 3: Conclude the correct set of statements.

Statements A, B, and C are correct. This corresponds to option (2).