Infinite square well solution - periodic boundary conditions

  • #1

Main Question or Discussion Point

If we have an infinite square well, I can follow the usual solution in Griffiths but I now want to impose periodic boundary conditions. I have

[tex]\psi(x) = A\sin(kx) + B\cos(kx)[/tex]

with boundary conditions [tex]\psi(x) = \psi(x+L)[/tex]

In the fixed boundary case, we had [tex]\psi(0) = 0[/tex] which meant [tex]B=0[/tex] and [tex]\psi(L)=0[/tex] which allows discrete values of k. I'm a little stuck with how to proceed with this in the periodic boundary condition case.

I think that [tex]k = n\pi/L[/tex] must still be true to satisfy the boundary condition (though I'm unable prove it). But now, I think that negative n also matter and they're different to the positive n case.

What is the general wavefunction solution in this case?
 

Answers and Replies

  • #2
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Your starting point should be ## \psi(x+L)=\psi(x) ##. Writing it explicitly, you'll have:
## A\sin(kx+kL)+B\cos(kx+kL)=A\sin(kx)+B\cos(kx) ##.
Trigonometric identities should help you proceed further.
 
  • #3
This yields, substituting ##x=0##
##B = A\sin(kL) +B\cos(kL)##

The equivalence of the first derivative at ##0## and ##L## yields
##A = A\cos(kL) + B\sin(kL)##

Is this correct so far? I'm not really sure how to solve these two simultaneous equations for A and B. So far, I also have no restriction on ##k##
 
  • #4
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Its not correct. The condition should be correct for any x so you can't set x=0. As I said, look for the proper trigonometric identity!
 
  • #5
Sorry, I'm quite stuck and really don't know how to proceed, so could you elaborate please? Just using sin(A+B) = sin(A)cos(B) + cos(A)sin(B) and similar for cos(A+B) is not illuminating it for me (note that no assumption is made so far about values of k).
 
  • #6
vanhees71
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No go on, it's very illuminating! BTW you should post such homework(-like) questions to the homework forum.
 
  • #7
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Sorry, I'm quite stuck and really don't know how to proceed, so could you elaborate please? Just using sin(A+B) = sin(A)cos(B) + cos(A)sin(B) and similar for cos(A+B) is not illuminating it for me (note that no assumption is made so far about values of k).
Yeah, those are the identities you should use. So now you have two linear combinations of sines and cosines with unknown coefficients that are equal. What does the equality imply for the coefficients?
 
  • #8
Okay, let's see:

##A\sin(kx)\cos(kL) + A\cos(kx)\sin(kL) + B\cos(kx)\cos(kL) - B\sin(kx)\sin(kL) = A\sin(kx) + B\cos(kx)##

I can see that ##\cos(kL) = 1## does give a valid solution and this would quantize ##k## but I can't see why this is necessarily the only solution to this equation. And, also proceeding with this, it seems like there is no restriction on A and B apart from normalization i.e. ##A^2 +B^2 = 1##
 
  • #9
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Okay, let's see:

##A\sin(kx)\cos(kL) + A\cos(kx)\sin(kL) + B\cos(kx)\cos(kL) - B\sin(kx)\sin(kL) = A\sin(kx) + B\cos(kx)##

I can see that ##\cos(kL) = 1## does give a valid solution and this would quantize ##k## but I can't see why this is necessarily the only solution to this equation. And, also proceeding with this, it seems like there is no restriction on A and B apart from normalization i.e. ##A^2 +B^2 = 1##
## \cos(kL)=1 ## is not the most general thing you can conclude. Bring all the sines and cosines to the same side and collect the coefficients. How can this expression be zero?

And it is wrong to expect a restriction on A and B at this level. These constants are there so you can match your solution to different initial conditions but now you don't have any!
 
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  • #10
Okay, a little algebra later, we have

##\sin(kx)(A\cos(kL)-B\sin(kL) - A) +\cos(kx)(A\sin(kL) + B\cos(kL) -B) = 0## (Eq 1)

I assume that since this is valid for all x, each term in the brackets must be 0 independently. Thus, we have

##(1-\cos(kL)) = -\sin(kL).B/A## (Eq 2a)
##(1-\cos(kL)) = \sin(kL).A/B## (Eq 2b)

There are possibilities that A or B are zero in which case I get ##cos(kL)=1## from Eq. 1. Otherwise, I multiply the last two equations and get
##(1-\cos(kL))^2 = -\sin^2(kL)## which also gives ##\cos(kL) = 1##.

Er is this correct?
 
  • #11
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Yeah that's correct!
And...I don't know why I said its not the most general thing you can write! Sorry about that!
 
  • #12
Yeah that's correct!
And...I don't know why I said its not the most general thing you can write! Sorry about that!
Thank you. It's better to prove it so I'm glad I went through it anyway.

I still have some questions though. The general solution is now ##A\sin(kx) + B\cos(kx)## with ##k = n\pi/L## for integer ##n##. Compare this to the old solution for fixed boundaries where we have ##A\sin(kx)## with ##k = n\pi/L##. In the fixed boundary case, negative ##n## was simply absorbed into the normalization constant ##A## but now, I cannot do that anymore.

So a given energy eigenstate has forward and backward moving waves and is correctly described by any superposition of them? Can you comment on this because this is very different from the fixed boundary case? Thank you.
 
  • #13
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## \cos(kL)=1 ## implies ## k=\frac{2n\pi}{L} ##! Because ## \cos[(2n+1)\pi]=-1## which is not acceptable.

The normalization constant for the fixed boundary case is ## \sqrt{\frac 2 L} ## regardless of the value(and sign) of k. So its not correct that the sign of k can be absorbed in the normalization constant.
 
  • #14
## \cos(kL)=1 ## implies ## k=\frac{2n\pi}{L} ##! Because ## \cos[(2n+1)\pi]=-1## which is not acceptable.

The normalization constant for the fixed boundary case is ## \sqrt{\frac 2 L} ## regardless of the value(and sign) of k. So its not correct that the sign of k can be absorbed in the normalization constant.
Sorry, yes it's ##2n\pi/L##.

But my earlier point was that ##n## can be any integer but in the fixed boundary case where ##\psi(x) = A\sin(kx)##, we absorb all the negative signs using ##A##. This is also mentioned in Griffiths (just above equation 2.26, second edition). Now, that isn't true for periodic boundaries, correct? So for a given energy ##E##, isn't the fully general solution something like (I may be wrong here but just illustrating the difference due to our inability to absorb negative k values)

##A\sin(k_{E}x) + B\cos(k_{E}x) + A'\sin(-k_{E}x) + B'\cos(-k_{E}x)##, where ##k_E = \sqrt{2mE}/\hbar## and we can take any values for ##A##, ##B##, ##A'## and ##B'## upto an overall normalization?

EDIT: Okay, I see that's incorrect because ##A\sin(k_{E}x) + B\cos(k_{E}x) + A'\sin(-k_{E}x) + B'\cos(-k_{E}x)## simply becomes ##A''\sin(k_{E}x) + B''\cos(k_{E}x) ##. And now, only positive values of ##k_E## need to be considered, just like we did with fixed boundaries. Thanks for the help ShayanJ
 
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  • #15
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Actually the most general solution is ##\displaystyle \psi(x)=\sum_{n=-\infty}^\infty [A_n \sin(\frac{2n\pi}{L}x)+B_n \cos(\frac{2n\pi}{L}x)] ##.
 
  • #16
Actually the most general solution is ##\displaystyle \psi(x)=\sum_{n=-\infty}^\infty [A_n \sin(\frac{2n\pi}{L}x)+B_n \cos(\frac{2n\pi}{L}x)] ##.
I meant the stationary states so only energy is allowed
 

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