0
$\begingroup$

Consider 3d box of size $L$ with periodic boundary. Then the Schrodinger equation gives

\begin{align} \frac{d^2 \Psi}{d x_i^2} = -k_i^2 \Psi \end{align} thus we can set the solution in the following way.

Let consider $x$ axis only \begin{align} \Psi(x) = Ae^{ikx} + Be^{-ikx} \end{align} The periodic boundary condition gives $\Psi(0) = \Psi(L)$ which is lacks of solving schrodinger equation $i.e$ \begin{align} A+B = Ae^{ikL} + Be^{-ikL} \end{align} there are two variables and boundary condition only gives one constraints so we can not determine $A$, $B$.

What i know, for free particle, the boundary conditions $\Psi(0)= \Psi(L)=0$ gives two coefficients $A$, $B$ independently, and find quantization as \begin{align} kL = \pi m \quad \rightarrow \quad k = \frac{\pi m}{L} \end{align} for integer $m$.


But prof. says we can set plane waves as a solution \begin{align} \Psi(x) = A e^{ikx} \end{align} where $k$ runs negative to positive values. and imposing boundary condition it gives \begin{align} kL = 2\pi m \quad k = \frac{2\pi m}{L} \end{align} where $m$ is integer.

I don't understand why we only choose one particular basis.

Via web-searching i found similar content in see note section 8.1.d.

Can you explain why this works?

$\endgroup$
  • $\begingroup$ It is unclear what "which is lacks of solving schrodinger equation" is supposed to mean. $\endgroup$ – ACuriousMind Oct 31 '15 at 23:19
2
$\begingroup$

In the case of periodic boundary conditions it is necessary to ensure the continuity of the solution not only in value on the boundary, $\Psi(0) = \Psi(L)$, but also in derivative, $\partial_x \Psi(0) = \partial_x \Psi(L)$. This is simply to rule out kinks and ensure that the solution is translationally invariant.

The derivative condition as applied to $$ \Psi(x) = Ae^{ikx} + B e^{-ikx} $$ gives $$ A - B = Ae^{ikL} - B e^{-ikL} $$ and together with $$ A + B = Ae^{ikL} + B e^{-ikL} $$ implies $$ A = Ae^{ikL} \;\; \text{and} \;\; B = Be^{-ikL} $$ This is basically equivalent to considering the forward and backward traveling waves directly on grounds of translational symmetry and imposing only the continuity condition at the boundary, since the derivative condition is implicit in the assumption of a traveling wave.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.