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I do not understand this piece of my professor's lectures about the calculation of wavefunction of an electron inside a cubic semiconductor with side length L.

It expresses the solution of the Schrodinger equation in this way:

\begin{align} \Psi_k(r) & = A_k(r) \sin k_xx \sin k_y y \sin k_z z \\ \text{with} \qquad k_x L &= 2\pi n_x,\qquad k_y L = 2\pi n_y,\qquad k_z L = 2\pi n_z,\qquad \end{align} for $n_x,n_y,n_z$ integers.

For the boundary conditions: $k_x=\frac{2\pi n_x}{L} = \frac{p_x}{\hbar} = 2\pi \frac{p_x}{\hbar} \implies hn_x = p_x L$, and similarly $hn_y = p_y L$ and $hn_z = p_z L$.

I have these doubts:

  1. The Bloch theorem states that in a periodic structure (such as a crystal) the wavefunction is expressed as the product of a complex exponential term and a periodic term: $$ \psi(\mathbf r) = e^{i\mathbf k\cdot \mathbf r} u(\mathbf r) $$ Well, I do not see the link between this expression and the previous solution. Is Ak(r) the exponential function (and if yes, why not to express it directly like in the Bloch theorem)? Moreover, why are there three factorized periodic functions?

  2. The boundary conditions are that at the edges of the cube (x,y,z = 0,L) the wavefunction must be 0. Which is the physical reason of this constraint?

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    $\begingroup$ It seems that the correct solution is $\psi_k(x,y,z) = A_k ...$ and not $\psi_k(r) = A_k(r) ...$ $\endgroup$ Jun 23 '20 at 16:40
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax. You've been around long enough that these should not come as surprises to you; if you want others to put in the effort to answer your questions, you should put in a corresponding effort to typeset your question in a readable way. $\endgroup$ Jun 24 '20 at 14:52
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The TISE:

$$\Big[-\frac{\hbar}{2m}\nabla^2+U(x,y,z)\Big]\psi(x,y,z)=E\psi(x,y,z)$$

We assume a cuboid crystal with length $L$ and zero $U$ inside and $U=+\infty$ outside of it, so:

$$-\frac{\hbar}{2m}\nabla^2\psi(x,y,z)=E\psi(x,y,z)$$

Or:

$$\nabla^2\psi(x,y,z)=-k^2\psi(x,y,z)$$

where:

$$k=\sqrt{\frac{2Em}{\hbar}}$$

This is solved by means of the Ansatz:

$$\psi(x,y,z)=X(x)Y(y)Z(z)$$

and separation of variables.

Using the boundary condition where the sides of the cuboid are at $U=+\infty$, so the wavefunction $\psi$ must be $\text{zero}$ there, we get the solution you cited.

The boundary conditions are that at the edges of the cube $(x,y,z = > 0,L)$ the wavefunction must be 0. Which is the physical reason of this constraint?

As stated above: it's assumed the potential on the boundaries and beyond is $+\infty$, so that $\psi$ is $\text{zero}$ there.


But it's quite clear from [this document here][1] that $\psi(\mathbf{r})=e^{i\mathbf{k.r}}u(\mathbf{r})$ is the solution for a very different system than an 'empty' crystal. That solution applies for potentials with the shape:

Periodic potential

The wave functions satisfying the TISE in that case are called Bloch waves.

http://www.unesco.org/new/fileadmin/MULTIMEDIA/FIELD/Cairo/images/FNS-15n.pdf

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  • $\begingroup$ Thank you for your answer. Which is the reason that allows us to assume that U is 0 inside the cuboid and infinite outside? $\endgroup$
    – Kinka-Byo
    Jun 23 '20 at 17:48
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    $\begingroup$ Hi. Because those are the ONLY conditions that deliver the solution you cited at the start of your question. To be fair, it also works for $U=U_0$ (where $U_0$ is some finite value) inside the crystal but that's a trivial difference mathematically speaking. It's possible that the solution you presented is not for the same case where $\psi(r)=e^{ik.r}u(r)$. $\endgroup$
    – Gert
    Jun 23 '20 at 18:02
  • $\begingroup$ You mention a periodic structure but the simple case that leads to your cited solution (start of your question) is not for a periodic structure, even though the wave function is clearly periodic ($\sin$). $\endgroup$
    – Gert
    Jun 23 '20 at 18:05
  • $\begingroup$ (If $U(x,y,z)$ is a periodic function, then $\psi$ will reflect that of course) $\endgroup$
    – Gert
    Jun 23 '20 at 18:08
  • $\begingroup$ Example of a periodic potential: unesco.org/new/fileadmin/MULTIMEDIA/FIELD/Cairo/images/… $\endgroup$
    – Gert
    Jun 23 '20 at 18:19

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