Infinite square well solution - periodic boundary conditions

In summary: In the fixed boundary case, the energy eigenstate is simply the sum of the energy eigenstates for the discrete values of k. In the periodic boundary case, the energy eigenstate is a sum of energy eigenstates for different values of k that are all repeated infinitely many times.
  • #1
McLaren Rulez
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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?
 
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  • #2
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.
 
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  • #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
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!
 
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  • #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
No go on, it's very illuminating! BTW you should post such homework(-like) questions to the homework forum.
 
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  • #7
McLaren Rulez said:
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?
 
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  • #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
McLaren Rulez said:
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
Yeah that's correct!
And...I don't know why I said its not the most general thing you can write! Sorry about that!
 
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  • #12
Shayan.J said:
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
## \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
Shayan.J said:
## \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
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
Shayan.J said:
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
 

FAQ: Infinite square well solution - periodic boundary conditions

What is the infinite square well solution?

The infinite square well solution is a mathematical model used in quantum mechanics to describe the behavior of a particle confined within a potential well with infinite potential walls on either side.

What are periodic boundary conditions in the context of the infinite square well solution?

Periodic boundary conditions refer to the requirement that the wave function of a particle in the infinite square well must be continuous and periodic at the boundaries of the well. This means that the wave function must have the same value at both ends of the well and must repeat itself after a certain distance.

How do periodic boundary conditions affect the energy levels in the infinite square well solution?

Periodic boundary conditions lead to quantization of energy levels in the infinite square well. This means that the energy of the particle can only take on certain discrete values, rather than being able to take on any value as in classical mechanics.

Can the infinite square well solution with periodic boundary conditions be applied to real-life systems?

While the infinite square well solution is a simplified model, it can be used to approximate the behavior of certain systems in real life, such as atoms or molecules confined within a potential well.

How does the width of the well affect the energy levels in the infinite square well solution with periodic boundary conditions?

The width of the well affects the energy levels in the infinite square well solution by changing the distance between the boundaries and thus altering the allowed wavelengths of the particle. A wider well will have higher energy levels and a smaller spacing between them, while a narrower well will have lower energy levels and a larger spacing between them.

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