Infinite square well centered at the origin

  • Thread starter Kaguro
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  • #1
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Moved from a technical forum
The problem is:
Solve the time independent Schrodinger Equation for infinite square well centered at origin. Show that the energy is same as in the original case(well between x=0 and x=L). Also show that the solution to the this case can be obtained by setting x to x-L/2 in ##\psi## in the original one.

Now, in the original problem, E = ## \frac{n^2\pi^2ħ^2}{2mL^2}##.

Here, I got my answer as 4 times that value:

From the Schrodinger Equation, put V=0 in (-L/2,L/2):
##\frac{d^2\psi}{dx^2}## + ##\frac{2mE}{ħ^2}\psi## = 0

So,
##\psi## = Acos(kx) + Bsin(kx) where ##k^2 = \frac{2mE}{ħ^2}##

Now the boundary conditions require ##\psi## be 0 at x=-L/2 and x=+L/2
Putting these and solving them I get:
0 = Acos(kL/2) - Bsin(kL/2) and 0 = Acos(kL/2) + Bsin(kL/2)

From them:
tan(kL/2) = A/B and also A/B = -tan(kL/2)

which means: 2tan(kL/2) = 0
and so kL/2 = n##\pi##

From the definition of k, I have:
E = ##\frac{2n^2\pi^2ħ^2}{mL^2}##

What mistake did I make?
 

Answers and Replies

  • #2
PeterDonis
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From them:
tan(kL/2) = A/B and also A/B = -tan(kL/2)
This assumes that B is nonzero. What if B is zero?
 
  • #3
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This assumes that B is nonzero. What if B is zero?
Actually I am having trouble finding the constants A and B. I tried various ways and got 4 possibilities:
A=0,B=0(not together), cos(kL/2) = 0, or sin(kL/2)=0.

Except the condition that A and B together must not be zero, and the condition that kL/2 can't have two different values, I don't see any condition which will tell me which constant to set to 0.
 
  • #4
PeroK
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Actually I am having trouble finding the constants A and B. I tried various ways and got 4 possibilities:
A=0,B=0(not together), cos(kL/2) = 0, or sin(kL/2)=0.

Except the condition that A and B together must not be zero, and the condition that kL/2 can't have two different values, I don't see any condition which will tell me which constant to set to 0.
Either ##A = 0## or ##B = 0##. You need to solve both cases to get the full list.

You solved for ##A = 0## only, hence got every other solution, hence your ##n## was out of step with the full set of energy states.
 
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  • #5
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Either ##A = 0## or ##B = 0##. You need to solve both cases to get the full list.

You solved for ##A = 0## only, hence got every other solution, hence your ##n## was out of step with the full set of energy states.
If I do that, I get two different values of k:
Set A=0:
k = 2n##\pi##/L

Set B=0:
k = (2n-1)##\pi##/L
 
  • #6
PeroK
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If I do that, I get two different values of k:
Set A=0:
k = 2n##\pi##/L

Set B=0:
k = (2n-1)##\pi##/L
There's nothing wrong with that. That's an either/or. There would be nothing wrong with finding that, say: ##k =0## or ##k =1##.
 
  • #7
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There's nothing wrong with that. That's an either/or. There would be nothing wrong with finding that, say: ##k =0## or ##k =1##.
Well... then can you tell me how to proceed from here? I have 2 possible values of k and so using the definition of k in terms of E I can get two possible values of E:

E = ##\frac{2n^2\pi^2ħ^2}{L^2}##
Or
E = ##\frac{(2n-1)^2\pi^2ħ^2}{2mL^2}##

Which one to take? And why?
 
  • #8
PeterDonis
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I have 2 possible values of k
Both of which are actually special cases of one general formula for ##k##. Can you see what it is?

(Hint: ##2n## is even and ##2n - 1## is odd. When you combine the set of even numbers with the set of odd numbers, what do you get?)
 
  • #9
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Both of which are actually special cases of one general formula for ##k##. Can you see what it is?

(Hint: ##2n## is even and ##2n - 1## is odd. When you combine the set of even numbers with the set of odd numbers, what do you get?)
Okay okay... I can take just n. So this works up pretty well and I get the correct value for energy. But I still don't know ##\psi##.
 
  • #10
Vanadium 50
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But I still don't know ψ\psi.
Can you write down an expression for ψ?
 
  • #11
PeroK
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Well... then can you tell me how to proceed from here? I have 2 possible values of k and so using the definition of k in terms of E I can get two possible values of E:

E = ##\frac{2n^2\pi^2ħ^2}{L^2}##
Or
E = ##\frac{(2n-1)^2\pi^2ħ^2}{2mL^2}##

Which one to take? And why?
I think you lost sight of what you were trying to do. You were looking for ALL solutions to the equation:

##\psi'' + k^2 \psi =0##

With the boundary conditions that ##\psi(L/2) = \psi(-L/2) = 0##

First, you found a sequence of solutions for odd positive integers ##n##:

##k = n \pi/L##

and

##\psi(x) = A \cos(\frac{n\pi x}{L})##

And, you found a sequence of solutions for even positive integers ##n##:

##k = n \pi/L##

and

##\psi(x) = B \sin(\frac{n\pi x}{L})##

It's not clear why you think that you can only have solutions for odd ##n## or even ##n## but not both. If you try these functions in the origin equation, you'll see that both sets of functions are indeed solutions.

In any case, you clearly have a solution for every positive integer.
 
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  • #12
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Can you write down an expression for ψ?
##\psi##= Acos (kx) + Bsin (kx).
I now know k but not A or B.
 
  • #13
PeroK
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##\psi##= Acos (kx) + Bsin (kx).
I now know k but not A or B.
In general, ##A## and ##B## can be any complex numbers. But, wavefunctions must be normalised, so you need to calculate the magnitude of ##A## and ##B## to get a well-defined eigenfunction in each case.
 
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  • #14
PeterDonis
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I now know k but not A or B.
For any particular value of ##k## (i.e., any particular ##n##), won't one of ##A## or ##B## be zero?
 
  • #15
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In general, ##A## and ##B## can be any complex numbers. But, wavefunctions must be normalised, so you need to calculate the magnitude of ##A## and ##B## to get a well-defined eigenfunction in each case.
If I try to normalize then I get this:
##A^2+B^2=\frac {2}{L}##
 
  • #16
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For any particular value of ##k## (i.e., any particular ##n##), won't one of ##A## or ##B## be zero?
If you mean at the boundaries then I would say it depends on whether n is odd or even.
 
  • #17
PeterDonis
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If you mean at the boundaries then I would say it depends on whether n is odd or even.
The values of ##A## and ##B## for a particular value of ##n## can't be different at the boundaries than elsewhere; the solution for a particular value of ##n## has to have the same ##A## and ##B## everywhere within the well.
 
  • #18
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The values of ##A## and ##B## for a particular value of ##n## can't be different at the boundaries than elsewhere; the solution for a particular value of ##n## has to have the same ##A## and ##B## everywhere within the well.
Yes but I don't have any information within the well. Only that V=0...
 
  • #19
PeterDonis
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I don't have any information within the well.
So what? Did you read what I said? Did you understand that, since the boundary conditions are sufficient to find ##A## and ##B## at the boundaries for each ##n## (one of the two is zero for every ##n##, so the other can be found easily by normalization), that gives you the solution for each ##n## within the entire well?
 
  • #20
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The boundary conditions are sufficient to find ##A## and ##B## at the boundaries for each ##n## (one of the two is zero for every ##n##
Evidently I don't see how...

I think this is about time I give up and look up the solution from somewhere. Thanks for help.
 
  • #21
PeterDonis
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Evidently I don't see how...
You already knew that either ##A = 0## or ##B = 0## for every ##n## when you made your post #5. So for every ##n##, you only have one nonzero term, and you can find its coefficient (##A## or ##B##, whichever is nonzero for that value of ##n##) by normalization (all you have to do is take the formula you gave in post #15 and remove whichever of ##A## or ##B## is zero for that ##n##). So for every ##n## you know the exact solution.

I think this is about time I give up
You should not give up now, when you are almost there (you've already done all the pieces, as above, you just need to put them together).
 
  • #22
PeroK
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If I try to normalize then I get this:
##A^2+B^2=\frac {2}{L}##
That looks right. But, remember that for every solution either ##A = 0## or ##B = 0##.
 
  • #23
hutchphd
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That looks right. But, remember that for every solution either ##A = 0## or ##B = 0##.
Does the OP understand that all of the A and B should really carry a subscript n (or k) and that specifying a particular solution for a given problem requires more information (perhaps the wavefunction everywhere at t=0)?
 
  • #24
PeterDonis
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specifying a particular solution for a given problem requires more information (perhaps the wavefunction everywhere at t=0)?
The problem asks for solutions to the time-independent Schrodinger Equation, so it's asking for energy eigenstates, not a general time-dependent state (which would be an arbitrary linear combination of energy eigenstates). Enough information is given to find the energy eigenstates.
 

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