Answer:
The probability is [tex]\dfrac{1}{2}[/tex]
Explanation:
Given that,
The energy at time t will yield the result [tex]h^2pi^2/2ml^2[/tex]
Suppose, At time t = 0 the normalized wave function for a particle of mass m in the one-dimensional infinite well is given by,
[tex]V(x)={0 0<x<L,\ \infty\ elsewhere[/tex]
The normalized wave function is,
[tex]\psi(x)=\dfrac{1+i}{2}\sqrt{\dfrac{2}{L}}\sin\dfrac{\pi x}{L}+\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{2}{L}}\sin\dfrac{2\pi x}{L}\ 0<x<L, 0\ elsewhere[/tex]
What is the probability that a measurement of the energy at time t will yield the result h^2 pi^2/2mL^2?
We need to find the probability
Using given data
[tex]probability=(\dfrac{1+i}{2})(\dfrac{1-i}{2})[/tex]
[tex]probability=\dfrac{1-i^2}{4}[/tex]
[tex]probability=\dfrac{1+1}{4}[/tex]
[tex]probability=\dfrac{1}{2}[/tex]
Hence, The probability is [tex]\dfrac{1}{2}[/tex]
