Physics Problems & Solutions: Quantum Mechanics

The energy eigenstates for a particle of mass m in a box of length L have wave functions

$\phi_n (x)= \sqrt{\frac{2}{L}} \sin \left(\frac{n\pi x}{L} \right )$

and energies

$E_n=\frac{n^2 \pi^2 x^2}{2mL^2}$

where =1,2,3,... . At time t = 0, the particle is in a state described as follows.

$\psi(t=0)=\frac{1}{\sqrt{14}} (\phi_1+2\phi_2+3\phi_3 )$

Which of the following is a possible result of a measurement of energy for the state

A. 2E₁
B. 5E₁
C. 7E₁
D. 9E₁
E. 14E₁

(GR0177 #44)

Solution:

The energy:

$E_n=\frac{n^2 \pi^2 x^2}{2mL^2} \rightarrow E_1=\frac{\pi^2 x^2}{2mL^2} \rightarrow E_n=n^2 E_1$

So, the possible result of a measurement of energy is a squared quantity:

$\\E_2=2^2 E_1=4E_1 \\E_3=3^2 E_1=9E_1$
... and so on.

The only choice is (D)

Answer: D

Physics Problems & Solutions

Quantum Mechanics - Particle in a Box

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