I am interesting in solving the time-independent Schrodinger equation (TISE) for the scenario where we have an electron plane wave of fixed energy incident upon a potential that is infinite and periodic only the transverse plane (i.e. a very thin or monolayer crystal). This corresponds to the physical situation of transmission electron microscopy (TEM) with a thin specimen. My question relates to the assumptions we make about the form of the wave function when solving the TISE (see below).

As some background, the TISE can be written as

$$ \{\nabla^2 + 4\pi^2 [U(\textbf{r}) + K^2]\}\Psi(\textbf{r}) = 0$$

and I believe the boundary conditions in this scenario can be roughly specified by dividing the system into three regions in the $z$ direction:

  1. Above the specimen: contains the incident plane wave plus back-scattered plane waves
  2. Within the specimen: contains wave functions as dictated by the potential $U(\textbf{r})$
  3. Below the specimen: contains only transmitted plane waves

I assume that requiring continuity in $\Psi(\textbf{r})$ across the region boundaries would fully specify this problem (up to a phase factor), although I'd appreciate any clarity on that if I'm mistaken there.

It is conventional in TEM theory that a Bloch wave ansatz is made for the functional form of the solution, for the same reason that Bloch waves are used in solid state physics, viz. the periodic nature of the crystal potential.We assume that the wave function $\Psi(\textbf{r})$ can be expanded in Bloch states $B_i(\textbf{r})$ as follows:

$$ \begin{align} \Psi(\textbf{r}) & = \sum_i \alpha_i B_i(\textbf{r}) \\ B_i(\textbf{r}) & = \sum_\textbf{g} C_\textbf{g} \exp[2\pi i (\textbf{k}_i + \textbf{g})\cdot \textbf{r}] \end{align} $$

If we substitute this into the TISE we end up with a linear algebra problem in the coefficients $C_\textrm{g}$ and $\alpha_i$, which is reasonably straightforward to solve using some approximations and matrix diagonalisation.

I am comfortable with the use of the Bloch wave solutions for a crystal which is infinite in all directions, where it can be proved that the Bloch waves form a complete basis set. In my scenario however, the cystal is finite in the $z$ dimension, and so the Bloch theorem won't strictly hold. There will probably be some surface states possible and the boundary conditions in the $z$ direction will affect the bulk properties, so on a physical basis we expect some departure from the ideal Bloch states.

For moderately thick specimens where there is some periodicity in the $z$ dimension, it seems at least plausible that the Bloch wave ansatz is not a bad starting point, and might let us solve the TISE to an acceptable approximation (particularly where the incident electrons are high-energy and perhaps aren't impacted by the fine details).

But for a very thin specimen, e.g. a single plane of atoms, it doesn't seem reasonable to assume the Bloch wave ansatz, because there is no periodicity in $z$ at all. However, I could imagine we might arrive at a similar ansatz through some hand wavy argument about the high energy of the incident electron letting us have a separable solution of this form:

$$\Psi_i(\textbf{r}) = \exp(2\pi i k_i z) \sum_{\textbf{g}_\perp} C_{\textbf{g}_\perp} \exp[2\pi i {\textbf{g}_\perp} \cdot \textbf{r}]$$

which reflects the periodicity in the transverse plane and assumes that the incident plane wave is effectively just "modulated" a bit by the specimen potential. I'm not entirely confident about this ansatz though for the single layer specimen.

My question

Is there some more general approach to solving the TISE in this scenario, which doesn't involve starting with the Bloch wave ansatz, but does make it clear how we might still arrive at that functional form to a decent approximation in the case of finite and very thin specimens? I'm expecting that at some step in the solution method, the high energy of the incident electron can be explicitly introduced to simplify the equations and recover something like the Bloch wave functional form.

  • $\begingroup$ "I assume that requiring continuity in $\Psi({\bf r})$ across the region boundaries would fully specify this problem (up to a phase factor)...." For a second-order equation like the Schrödinger, you actually need to have continuity of both the wave function and its derivatives." $\endgroup$
    – Buzz
    Jul 20, 2022 at 2:06
  • $\begingroup$ Thanks @Buzz, that's good to know, was wondering if that would be the case. When backscattering is neglected the PDE is reduced to first order in $z$ but I am indeed interested in the general solutions of the second order equation. $\endgroup$
    – bdforbes
    Jul 20, 2022 at 2:48

1 Answer 1


In my scenario however, the cystal is finite in the z dimension, and so the Bloch theorem won't strictly hold.

This is wrong. Bloch theorem holds in any periodic medium, including finite ones. In particular, it's easy to prove for the case of periodic (Born—von Karman) boundary conditions: there's a well-defined translation operator that commutes with the Hamiltonian operator and thus must share eigenfunctions with it—and these are the Bloch wavefunctions. The spectrum is purely discrete in this case, though the eigenvalues do clump in sorts of bands, becoming denser as number of cells in the lattice increases.

Now, your setting isn't a periodic one. But it works the same way as the use of the free-space solutions when solving Schrödinger's equation for a finite rectangular quantum well or barrier: by matching the solutions in the piecewise intervals.

Why does it work?

Time-independent Schrödinger's equation is a linear homogeneous second-order ODE. By itself, it doesn't know anything about boundary conditions. They may not even be given. This equation has exactly two linearly independent solutions. Any other solution is a linear combination of them.

Now, suppose we have a pair of Bloch solutions (one $+k$, another $-k$) for an infinite crystal with potential $V(x)$ for $x\in\mathbb R.$ If, for $x\in(\infty,a]\cup[b,\infty),$ we replace $V(x)$ with some other potential $U(x)$, these solutions will still satisfy the Schrödinger's equation for $x\in(a,b)$. Moreover, whatever solution you find for the Schrödinger's equation with new potential, given the same energy $E,$ it'll still be (in $x\in(a,b)$) a linear combination of these two Bloch solutions. This is what lets us simply match the functions at the boundaries by providing the appropriate pairs of coefficients for the infinite-space solutions. The requirement of continuity of the wavefunction and its derivative makes sure that the equation is also satisfied at the interfaces $x=a$ and $x=b.$

  • $\begingroup$ This is interesting; I had considered the idea of repeating the finite crystal in the z direction with a very long repeat distance. But I was bothered by the fact that in my system the incident scattering electron should only see the one unique instance of the finite crystal, not recurring instances in the z direction. $\endgroup$
    – bdforbes
    Jul 24, 2022 at 0:35
  • $\begingroup$ Does your argument still apply when it is a monolayer with zero periodicity in z? Do we just imagine it's a system where the periodicity is very infrequent in the z direction? $\endgroup$
    – bdforbes
    Jul 24, 2022 at 0:36
  • $\begingroup$ And another question, does anything about your answer substantially change when we consider that there are the x and y dimensions as well? Because you talk about a linear homogeneous ODE , but I'm dealing with a PDE. $\endgroup$
    – bdforbes
    Jul 24, 2022 at 0:38
  • $\begingroup$ Sorry, one last question: do we not expect new, qualitatively different types of solutions for a finite crystal, such as surface states, that would not be possible to find using the Born-von Karman boundary conditions? I had come across this book: amazon.co.uk/Electronic-States-Crystals-Finite-Size/dp/… which goes to great lengths to treat finite crystals from scratch rather than just as infinite crystals with BvK b.c.s. Would we still find those solutions in your solution approach? $\endgroup$
    – bdforbes
    Jul 24, 2022 at 0:54
  • $\begingroup$ @bdforbes one monolayer still works, but it's basically nothing interesting, because quasi-wavenumber will be zero for all stationary states.As for PDE instead of ODE: if your PDE is separable (e.g. for a sequence of infinite planes of different materials, making up a 1D superlattice grown by epitaxy), then your longitudinal equation is 1D, and the transverse ones are almost empty-space equations. $\endgroup$
    – Ruslan
    Jul 24, 2022 at 9:09

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