Question

Please answer the question in full and show all work.

We have seen that the absolute square of the wave function VI,t) can be interpreted as the probability density for the location of the particle at time t. We have also seen that a particle's quantum state can be represented as a linear combination of eigenstates of a physical observable Q: V) SIT) where Q n ) = qn|n) and represents the probability to find the particle in the eigenstate (1) (and thus to measure the value 4n for the observable Q. 1. Calculate the expectation value of Q for the state ) given above. " ). Suppose 2. Consider a particle in the infinite square well system (with energy eigenvalues E, the state ) can be written as |) cm) where n) is an energy eigenstate with eigenvalue E, and C = A/n?. Find the appropriate value for A to normalize this state. (Ilint: 0) 3. Now determine the expectation value of energy for the state described in the previous problem. (Ilint: m .)

Answer 1

1) Expectation value of Q is

<Q>=<\/Q/v>= 26cm<n]@]m>= ccmm < nm > n.m n.mn

since n and m are orthonormal states

«Q>-Σε συνθεκ. < nn >-Σεσουβωδειοι - Σοσιθι - Σι Ρο. η. η, η n.m

2) The normalisation condition is

ΣIcal? = 1

or

Α? Σ1/n' = 1

or

4² x an

or

V90 A =

Therefore

1 = ج = لجاج F 06

or

90 = 744

c) The expectation value of Energy can be written using the result of part a

«E»-ΣΙ- Ε. - Σ30 Ε- Σ20 και 22 η

or

$<E>=\frac{45\hbar^2}{\pi^2 m a^2}\sum_n \frac{1}{ n^2}=\frac{45\hbar^2}{\pi^2 m a^2}\times \frac{\pi^2}{6}=\frac{45\hbar^2}{6 m a^2}$