- #1

Irishdoug

- 102

- 16

- 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}##

##\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

##\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}##)

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}##

##{|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}##

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**

##\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})## dxThe 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.

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})## dxThe 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.