Solving the Schrödinger Equation for Particle in a Box

[mrguru34]

I have a problem:
Let V0(x) denote the potential corresponding to the innite
square well (`box') extending from x = 0 to x = a . Let
us replace it by the potential
V (x) = 0 ; |x| \leq a/2
\infty; |x| \geq a/2
i want to find the solutions of the Schrodinger equation associated
with V using the knowledge of the system described by V0.

(i)What is the mathematical relationship between V0(x) and V (x)?
Sketch both to illustrate this relationship.

(ii) Write down the time-independent Schrodinger equation for a particle
of mass m moving in the potential V (x) described by wave
functions \Phi(x).

(iii) Let \Psi(x) denote the solutions of the time-independent Schrodinger
equation with potential V0(x) derived in class. Using (i) express
the `new' wave functions \Phi(x) in terms of the `old' ones, \Psi(x).

(iv) Write down explicit expressions for the wave functions \Psi(x).
[Hint: distinguish between n even and n odd.]

(v) Show that the wave functions \Phi obey the appropriate boundary
conditions for motion in the potential V (x).

(vi) Write down the energy eigenvalues En associated with the eigenfunctions
\Phi (no calculation required).

Thanks for your help.
It would be much appreciated if someone could actually explain very simply what's going on here aswell
Thanks again

 

[The_Duck]

It would be much appreciated if someone could actually explain very simply what's going on here aswell

Presumably you have treated the infinite square well in class where the well extends from x=0 to x=a. This problem is asking to you adapt your solution to that to a situation where everything is the same except the well has been shifted by a/2 to the left, so that the well extends from x=-a/2 to x = +a/2.

This is perhaps not as complicated as it looks. An analagous question from classical mechanics would be, "Suppose we know that a ball fired from the origin has some given trajectory. What does the trajectory look like if we instead fire the ball from ten feet to the right of the origin?"

 

[mrguru34]

I see ok!
So basically the graph of the box has move a/2 to the left

 

[SaveFerris]

Hello! I have this exact same question and was wondering if you'd gotten anywhere with it since posting? I think I've done the first and second part but am pretty stuck on the rest.
Thanks!

 

[mrguru34]

Hiya mate!
Na I've got literally no where
WHat bout you ?
Howd you manage to do the 1st and 2nd bits?

 

[vela]

Remember that your choice of coordinate system can't affect the physics. You could set the origin x=0 at the left end of the potential, the right end, in the middle, or anywhere, and it won't change how the solution fits into the potential. The ground state will always be zero at the endpoints of the well and will have the same shape.

The actual function used to describe a state does depend on your choice of coordinate system, and the problem is asking you to find the new functions from the old when the potential shifts by a/2 to the left. If the potential shifts to the left by a/2, the solutions must also shift to the left by a/2.

 

[mrguru34]

Vela!

That really helps! :)
Thanks! :D

The only problem i have now is i don't understand part iv)
Could you help with that?

 

[vela]

What do you have for the original solutions, and what do you get when you shift them to the left?

 

[mrguru34]

Ok so for basically part iii) i got
\Psi(x) = \sqrt{\frac{2}{a}} sin \frac{n*Pi*x}{a}
\Phi(x) = \sqrt{\frac{2}{a}} sin \frac{n*Pi*(x+\frac{a}{2}}){a}

 

[mrguru34]

that a on the end of the bottom equation is suposed to be on the bottom where the bracket is

 

[vela]

Good! Now if you multiply out the inside, you get

\Phi(x) = \sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}+\frac{n\pi}{2}\right)

You want to simplify that. You'll get different results depending on whether n is even or odd.

 

[mrguru34]

So basically simplifying it you get
\sqrt{\frac{2}{a}}* sin * (\frac{n*Pi*(2x+a)}{2a}
So then how do you do incorporate the odd an even?

 

[vela]

No, you want to use a trig identity to get rid of the sum in the argument of sine. Try using the angle addition formula for sine.

 

[mrguru34]

so using the sine addition formula you get
sin(\frac{n*Pi*x}{a})*cos(\frac{n*pi}{2})+sin(\frac{n*pi}{2})*cos(\frac{n*Pi*x}{a})