Finite/Infinite Potential Well Problems

  • #1
Hi, (first ever post)

I have the following three problems

1) A particle is in the ground state of an infinte well from 0<x<a. Find the probability that it is in the middle half of the well.

I think I have managed this, its just the integral of the wave function squared then I put in the values 3a/4 and a/4 to get the probability. It all cancels out nicely to get 0.82 which seems about right

psi = (2/a)^0.5 * sin(pi*x/a)
psi^2 = (2/a) * sin^2(pi*x/a)
Integral = ((2/a)((x/2)-((a/4pi)sin(2pix/a)
then put in the limits 3a/4 and a/4 and get 0.82
Is this the right method

2) A particle is in second state n=3 of an infinite well from -a/2<x<a/2. Find the position of the two nodes.

psi = (2/a)^0.5 * sin(pi*x*3/a)
probability density = psi^2 = (2/a) * sin^2(pi*x*3/a)
At the nodes i thought this must be equal to zero
so (2/a) * sin^2(pi*x*3/a)=0
therefore the nodes must be at x=a/3 and x=-a/3

3) This is what I am stuck on. Consider a well of finite depth that is just deep enough to contain three states. Sketch n=3 wave function and give the two values of the two nodes in this case.

I have sketched it from a textbook but I don't know how to find the nodes because I am confused about what the wave function is. Would it just be Asin(kx)+Bcos(kx). Any thoughts appreciated.

Thanks
 

Answers and Replies

  • #2
286
0
for question no. (2):

the wavefunction for the boundaries from x= -a/2 to x= a/2 (with n = 3) is sqrt(2/a) cos(npix/a) ..

how will that change your answer?
 
  • #3
Really? I cant see why, looking at this diagram surely it describes a sine function at n=3?

http://www.vectorsite.net/tpqm_02_05.png

They all appear to be sine functions, because at the walls there are nodes. Maybe I am looking at it wrong actually, I can see how the symmetry of the n=3 graph would suggest it is a cos function.
 
  • #4
sqrt(2/a) cos(npix/a) if it is this then

psi^2= (2/a)*cos^2 (3pix/a)
for this to equal zero x=a/6 or -a/6

so is that the correct answer for part 2?

Does anyone have any bright ideas for part 3?

thanks for the help by the way
 
  • #5
286
0
sqrt(2/a) cos(npix/a) if it is this then

psi^2= (2/a)*cos^2 (3pix/a)
for this to equal zero x=a/6 or -a/6

so is that the correct answer for part 2?

Does anyone have any bright ideas for part 3?

thanks for the help by the way

that is correct .. sorry for the late reply ..

look .. when the boundaries run from -a/2 to a/2 .. then the function will alternate between being symmetric and antisymmetric ( cosine and sine ), then n=1 is cosine function, n=2 is a sine, n=3 is a cosine .. etc

I can see that you are wondering why it is so ( I used to have the same question in my mind when I first studied it too ) ..

you know that the function for the boundaries x = 0 to x= a is sqrt(2/a) sin(npix/a) quick comparison between it and the one between x= -a/2 to x= a/2 you would see that they have the same length (which is a) and the only difference is that the other one is shifted a/2to the left .. so change x to x-a/2 then the sine in the wavefunction would change from
sin(npix/a) to sin(npix/a - npi/2) = sin(npix/a)cos(npi/2) - cos(npix/a)sin(npi/2)
from there you can conclude that for n=1,3,5.. the solution would just be sqrt(2/a)cos(npix/a)..


hopefully this is more clear now :) ..
 

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