Irishdoug
Homework Statement:
I'm self studying so I just want to ensure my answers are correct so I know I truly understand the material as it's easy to trick yourself in thinking you do!

A particle of mass m is in a 1-D infinite potential well of width a given by the potential:

V= ##\sqrt{4]/{a}## for 0 ##\leq## x ##\leq## a/4
= ##\infty## elsewhere
Relevant Equations:
See below.
I'm self studying so I just want to ensure my answers are correct so I know I truly understand the material as it's easy to trick yourself in thinking you do!

A particle of mass m is in a 1-D infinite potential well of width a given by the potential:

V= 0 for 0##\leq## x ##\leq## a
= ##\infty## for x<a ; x>a

The particle is located in the leftmost quarter of the well at time t = 0 and is described by the wave-function:

##\Psi (x, t=0)## = ##\sqrt{\frac{4}{a}}## for 0 ##\leq## x ##\leq## a/4
= 0 elsewhere

The eigen-functions and eigen-energies are given by ##\phi##(x) = ##\sqrt{\frac{2}{a}}## sin(##\frac{n\pi x}{a}##) and ##E_{n}## = ##{{\hbar}^2 {\pi}^2 {n}^2}/{2m {a}^2}##

Q.1 Write the expansion of the wavefunction ##\Psi ##(x,0) in terms of the eigenfunctions and explicitly compute the expansion co-efficients ##c_{n}##

##\Psi##(x,0) = ##\sum_{n=1}^{\infty} c_{n} \phi_{n}## = ##c_{1} \sqrt{\frac{2}{a}}## sin(##\frac{\pi x}{a}##) + ##c_{2} \sqrt{\frac{2}{a}}## sin(##\frac{2 \pi x}{a}##)... etc.

##c_{n}## = ##\int_{0}^{a/4} \sqrt{\frac{2}{a}} sin( \frac{n\pi x}{a}) \Psi (x,0)dx ##

##\rightarrow## ##\sqrt{\frac{2}{a}} \sqrt{{4}/{a}}## ##\int_{0}^{a/4} sin( \frac{n\pi x}{a}) dx ##

##\rightarrow## -##\frac{8}{a} ( \frac{a}{n \pi} cos (\frac{n\pi x}{a}##))

##\rightarrow## carry out substitution for x

##\rightarrow## -##\frac{8}{n\pi}( cos (\frac{n\pi}{4})) - 1##

so; ##c_{n}## = ##\frac{8-4\sqrt{2}}{n\pi }## for n = odd ; ##\frac{8}{n\pi }## for n = even

Q.2 Give an expression for ##\Psi##(x,t) for some later arbitrary time t

##\Psi##(x,t) = ##c_{n}## ##\phi_{n}(x)## ##exp(\frac{-i Ent}{\hbar})## = ##c_{n}## ## \sqrt{\frac{2}{a}}## ## sin( \frac{n\pi x}{a}) ## ##exp(\frac{-i Ent}{\hbar}##)

Q.3 If an energy measurement is made, what values will be observed and with what probabilities?

I said multiples of ##E_{0}## i.e ##n^{2}## ##(\frac{{\hbar}^2 {\pi}^2}{2m {a}^2}##) where n= 1, 2, 3 etc.

I'm not sure if this is correct. The probabilities are then given by ##|c_{n}|^{2}##

Q.4. What is the probability of measuring ##E_{2}##?

##{|c_{2}|^2}## = ##{\frac{-8}{2\pi}}^2## ##({cos\frac{2\pi}{4} -1})^2##

##\rightarrow## (##{\frac{-8}{2\pi}})^2##(##{-1}^2##) = ##\frac{16}{\pi \pi}##

Q.5 Are there energy egenvalues that will never be observed?

In theory, all eigenvalues can be observed, however in reality some will not as there is not enough time to make measurements to observe all eigen-values.

**I presume this is incorrect but I am unsure how to answer it**

Q.6 Give an expression for the expectation value of the inverse energy ##<\frac{1}{E}>## as a function of time.

##\int_{0}^{a/4} \phi^{*} E \phi dx##

##\frac{\hbar^{2}}{2m}## ##\frac{\partial^2 sin(\frac{n \pi x}{a} )exp(\frac{iEt}{\hbar})}{\partial x^2}##

= ##\frac{\hbar^{2}}{2m}## (-##\frac{a}{n \pi}^{2} ##) ##sin(\frac{n \pi x}{a} )exp(\frac{-iEt}{\hbar})##

so, -##\frac{4 \hbar^{2}}{2m n^{2} \pi^{2}}## ##\int_{0}^{a/4}## ## sin(\frac{n \pi x}{a} )exp(\frac{iEt}{\hbar}) ## ##sin(\frac{n \pi x}{a} )exp(\frac{-iEt}{\hbar})## dx

The exponentials cancel are left with:

-##\frac{4 \hbar^{2}}{2m n^{2} \pi^{2}}## ##\int_{0}^{a/4}## ## sin^{2}(\frac{n \pi x}{a} )##

= -##\frac{4 \hbar^{2}}{2m n^{2} \pi^{2}}## (##\frac{ -2a sin(\frac{n\pi}{2}) - \pi an}{8\pi n}##)

= +##\frac{4 \hbar^{2}}{2m n^{2} \pi^{2}}## (##\frac{ 2a sin(\frac{n\pi}{2}) - \pi an}{8\pi n}##)

You then put this over 1 to get the answer.

I believe this is incorrect however as I have no time dependency! Can someone point out where I have gone wrong. Thankyou.

Abhishek11235
I think you need to find ##<1/E>## whereas you have found ##<E>##. You can use following:
$$<E^{-1}>= \sum_n|c_n ^2| E_n ^{-1}$$

Irishdoug
What I did was find <E>, and then I put this over 1. My answer should read:

##\frac{2mn^{2}\pi^{2}}{4\hbar^{2}} \frac{8 \pi n}{2a sin(\frac{n\pi}{2}) - \pi n}##

Is this method incorrect?