Until now, solving the Schrodinger Equation for a particle in a box was relatively easy because the boundaries conditions imposed zero value on the wave function at the boundaries. But now I must find the normalized wave function of the same problem imposing just these periodic boundaries conditions:

$$Y(x,y,z)=Y(x+L,y,z),\\ Y(x,y,z)=Y(x,y+L,z),\\ Y(x,y,z)=Y(x,y,z+L).$$

I got stuck in the normalization process. Before, using the boundary condition (one dimension) $Y(0)=Y(L)=0$, I could get just one constant to solve for, in

$$Y(x)= A\sin{kx} + B\cos{kx}.$$

Applying the conditions above, I get

$$Y(x)= A\sin{kx}\quad \text{where}\quad k=\pi n/L$$

which is easy normalize. But now, with this periodic boundary condition, $Y(0)=Y(L)$. How could I find it?

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    $\begingroup$ Where exactly are you running into trouble? Right now you just have the definition of periodic boundary conditions in your question and we have to guess why you're having difficulties. Please show us the work you've done already and indicate where you got stuck. $\endgroup$ – Wouter May 18 '13 at 19:01
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    $\begingroup$ Hint (which might be obvious): Think of it as a particle on a "ring". That gives you a natural coordinate to talk about and things might be easier. $\endgroup$ – Siva May 18 '13 at 19:04
  • $\begingroup$ Both instances of "$Y(x) = Y(L)$" should be "$Y(0) = Y(L)$"? $\endgroup$ – user10851 May 23 '13 at 2:32
  • $\begingroup$ Does this answer your question? Energy dependence on boundary conditions for particle in a box $\endgroup$ – EverydayFoolish Oct 2 at 16:15

When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate $x$ must match that at coordinate $x+L$, so we have:


In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be helpful to remember that this can also be expressed as an exponential with the form:

$$\Psi(x)\propto e^{ikx}$$

Hopefully that should get you off the starting block.

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In general, what we use for periodic boundary conditions is defined by $Y(0)=Y(L)$ and $\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition $Y(0) = Y(L)$ I believe is not sufficient to impose conditions on $k$.

This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the schrodinger equation is of order 2 in derivative, there should be at least 2 boundary conditions.

By imposing the two stated periodc boundary conditions, you obtain the following system : \begin{align} A = A\cos(kL)+B\sin(kL) \\ B = -A\sin(kL)+B\cos(kL) \end{align} Rearraging \begin{align} A(1-\cos(kL))+B(-\sin(kL)) = 0\\ A(\sin(kL))+B(1-\cos(kL)) = 0 \end{align}

Now, the system yields non trivial solutions (trivial solution being A=B=0) only if it's determinant is 0. This gives the conditions $\cos(kL)=1$ which gives the desired $k=\frac{2n\pi}{L}$

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  • $\begingroup$ The system of equations has non-trivial solutions only if the determinant is NOT 0! So this derivation seems to have a flaw in there and is probably more complicated... $\endgroup$ – Guiste Apr 23 at 5:42
  • $\begingroup$ No, for a homogenous linear system of equations as we have in the problem, if determinant is NOT 0 we only have the trivial solution as stated. When the determinant is 0, we either have no solution, or we have an infinite amount of solutions, which is in this case the "non-trivial" solutions. $\endgroup$ – Frotaur Apr 24 at 16:01
  • $\begingroup$ My bet, did not consider it to be homogeneous. Then I have one more question. Since we have 2 linear independent equations, we should get information about two coefficients. One of them is the k, do we also learn something about the relation of A and B? $\endgroup$ – Guiste Apr 26 at 22:30
  • $\begingroup$ Yes you get two equations : the $det=0$ condition gives you the possible $k$ for a non-trivial solution. However, if you plug this $k$ in the boundary condition equations, you can see they are automatically satisfied. This means the parameters $A$ and $B$ are arbitrary, and to fix them you need more boundary conditions/informations about your problem. $\endgroup$ – Frotaur Apr 27 at 16:49

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