Editing Particle in a one-dimensional lattice

{{Short description|Model in Quantum Physics}}
{{more citations needed|date=December 2013}}
In [[quantum mechanics]], the '''particle in a one-dimensional lattice''' is a problem that occurs in the model of a periodic [[Crystal structure|crystal lattice]]. The potential is caused by [[ion]]s in the periodic structure of the crystal creating an [[electromagnetic field]] so electrons are subject to a regular potential inside the lattice.  It is a generalization of the [[free electron model]], which assumes zero potential inside the lattice.

==Problem definition==
When talking about solid materials, the discussion is mainly around crystals – periodic lattices. Here we will discuss a 1D lattice of positive ions. Assuming the spacing between two ions is {{mvar|a}}, the potential in the lattice will look something like this:
[[File:Potential-actual.PNG|center]]
The mathematical representation of the potential is a periodic function with a period {{mvar|a}}. According to [[Bloch's theorem]],<ref name="Bloch 1929 pp. 555–600">{{cite journal | last=Bloch | first=Felix | title=Über die Quantenmechanik der Elektronen in Kristallgittern | journal=Zeitschrift für Physik | publisher=Springer Science and Business Media LLC | volume=52 | issue=7–8 | year=1929 | issn=1434-6001 | doi=10.1007/bf01339455 | pages=555–600 | bibcode=1929ZPhy...52..555B | s2cid=120668259 | language=de}}</ref> the wavefunction solution of the [[Schrödinger equation]] when the potential is periodic, can be written as:

<math display="block"> \psi (x) = e^{ikx} u(x),</math>

where {{math|''u''(''x'')}} is a [[periodic function]] which satisfies {{math|1=''u''(''x'' + ''a'') = ''u''(''x'')}}. It is the Bloch factor with Floquet exponent <math> k</math> which gives rise to the band structure of the energy spectrum of the Schrödinger equation with a periodic potential like the Kronig–Penney potential or a cosine function as it was shown in 1928 by  Strutt.<ref>{{Cite journal |last=Strutt |first=M.J.O. |date=1928 |title=Zur Wellenmechanik des Atomgitters |url=https://onlinelibrary.wiley.com/doi/10.1002/andp.19283911006 |journal= Annalen der Physik|volume=86 |issue=10 |pages=319–324 |doi=10.1002/andp.19283911006 |bibcode=1928AnP...391..319S |via=|url-access=subscription }}</ref> The solutions can be given with the help of the [[Mathieu function|Mathieu functions]].

When nearing the edges of the lattice, there are problems with the boundary condition. Therefore, we can represent the ion lattice as a ring following the [[Born–von Karman boundary condition|Born–von&nbsp;Karman boundary conditions]]. If {{mvar|L}} is the length of the lattice so that {{math|''L'' ≫ ''a''}}, then the number of ions in the lattice is so big, that when considering one ion, its surrounding is almost linear, and the wavefunction of the electron is unchanged. So now, instead of two boundary conditions we get one circular boundary condition:
<math display="block"> \psi (0)=\psi (L).</math>

If {{mvar|N}} is the number of ions in the lattice, then we have the relation: {{math|1=''aN'' = ''L''}}. Replacing in the boundary condition and applying Bloch's theorem will result in a quantization for {{mvar|k}}:
<math display="block"> \psi (0) = e^{ik \cdot 0} u(0) = e^{ikL} u(L) = \psi (L)</math>
<math display="block"> u(0) = e^{ikL} u(L)=e^{ikL} u(N a) \to e^{ikL} = 1</math>
<math display="block"> \Rightarrow kL = 2\pi n \to k = {2\pi \over L} n \qquad \left( n=0, \pm 1, \dots, \pm \frac{N}{2} \right).</math>

==Kronig–Penney model==
The Kronig–Penney model (named after [[Ralph Kronig]] and [[William Penney, Baron Penney|William Penney]]<ref name="de L. Kronig Penney pp. 499–513">{{cite journal | last1=de L. Kronig | first1=R. | last2=Penney | first2=W. G. | title=Quantum Mechanics of Electrons in Crystal Lattices | journal=Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences | publisher=The Royal Society | volume=130 | issue=814 | date=3 February 1931 | issn=1364-5021 | doi=10.1098/rspa.1931.0019 | pages=499–513| bibcode=1931RSPSA.130..499D | doi-access=free }}</ref>) is a simple, idealized quantum-mechanical system that consists of an infinite periodic array of [[rectangular potential barrier]]s.

The potential function is approximated by a rectangular potential:
[[File:Periodic square potential 130707.png|centre|600px|alt=Rectangular potential graph of ions equally spaced a units apart. Rectangular areas of height v0 are drawn directly underneath each ion, starting at the x-axis and going downwards.]]

Using [[Bloch state|Bloch's theorem]], we only need to find a solution for a single period, make sure it is continuous and smooth, and to make sure the function {{math|''u''(''x'')}} is also continuous and smooth.

Considering a single period of the potential:<br>
We have two regions here. We will solve for each independently:
Let ''E'' be an energy value above the well (E>0)
* For <math> 0 < x < (a-b)</math>: <math display="block">\begin{align}
\frac{-\hbar^2}{2m} \psi_{xx} &= E \psi \\
\Rightarrow \psi &= A e^{i \alpha x} + A' e^{-i \alpha x} & \left( \alpha^2 = {2mE \over \hbar^2} \right)
\end{align}</math>
*For <math> -b <x < 0 </math>: <math display="block">\begin{align}
\frac{-\hbar^2}{2m} \psi_{xx} &= (E+V_0)\psi \\
\Rightarrow \psi &= B e^{i \beta x} + B' e^{-i \beta x} & \left( \beta^2 = {2m(E+V_0) \over \hbar^2} \right).
\end{align}</math>

To find ''u''(''x'') in each region, we need to manipulate the electron's wavefunction:
<math display="block">\begin{align}
 \psi(0<x<a-b) &= A e^{i \alpha x} + A' e^{-i \alpha x} = e^{ikx} \left( A e^{i (\alpha-k) x} + A' e^{-i (\alpha+k) x} \right) \\
 \Rightarrow u(0<x<a-b) &= A e^{i (\alpha-k) x} + A' e^{-i (\alpha+k) x}.
\end{align}</math>

And in the same manner:
<math display="block"> u(-b<x<0)=B e^{i (\beta-k) x} + B' e^{-i (\beta+k) x}.</math>

To complete the solution we need to make sure the probability function is continuous and smooth, i.e.:
<math display="block"> \psi(0^{-})=\psi(0^{+}) \qquad \psi'(0^{-})=\psi'(0^{+}).</math>

And that {{math|''u''(''x'')}} and {{math|''u′''(''x'')}} are periodic:
<math display="block"> u(-b)=u(a-b) \qquad u'(-b)=u'(a-b).</math>

These conditions yield the following matrix:
<math display="block"> \begin{pmatrix}
1 & 1 & -1 & -1 \\
\alpha & -\alpha & -\beta & \beta \\
e^{i(\alpha-k)(a-b)} & e^{-i(\alpha+k)(a-b)} & -e^{-i(\beta-k)b} & -e^{i(\beta+k)b} \\
(\alpha-k)e^{i(\alpha-k)(a-b)} & -(\alpha+k)e^{-i(\alpha+k)(a-b)} & -(\beta-k)e^{-i(\beta-k)b} & (\beta+k)e^{i(\beta+k)b} \end{pmatrix}
\begin{pmatrix} A \\ A' \\ B \\ B' \end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.</math>

For us to have a non-trivial solution, the determinant of the matrix must be 0. This leads us to the following expression:
<math display="block"> \cos(k a) = \cos(\beta b) \cos[\alpha(a-b)]-{\alpha^2+\beta^2 \over 2\alpha \beta} \sin(\beta b) \sin[\alpha(a-b)].</math>

To further simplify the expression, we perform the following approximations:
<math display="block"> b \to 0; \quad V_0 \to \infty; \quad V_0 b = \mathrm{constant}</math>
<math display="block"> \Rightarrow \beta^2 b = \mathrm{constant}; \quad \alpha^2 b \to 0</math>
<math display="block"> \Rightarrow \beta b \to 0; \quad \sin(\beta b) \to \beta b; \quad \cos(\beta b) \to 1.</math>

The expression will now be:
<math display="block"> \cos(k a) = \cos(\alpha a)+P \frac{\sin(\alpha a)}{\alpha a}, \qquad P= \frac{m V_0 ba}{\hbar^2}.</math>

For energy values inside the well (''E''&nbsp;<&nbsp;0), we get:
<math display="block"> \cos(k a) = \cos(\beta b) \cosh[\alpha(a-b)]-{\beta^2-\alpha^2 \over 2\alpha \beta} \sin(\beta b) \sinh[\alpha(a-b)],</math>
with <math>\alpha^2 = {2 m |E| \over \hbar^2}</math> and <math>\beta^2 = \frac{2 m (V_0-|E|)}{\hbar^2}</math>.

Following the same approximations as above (<math> b \to 0; \, V_0 \to \infty; \, V_0 b = \mathrm{constant}</math>), we arrive at
<math display="block"> \cos(k a) = \cosh(\alpha a) + P \frac{\sinh(\alpha a)}{\alpha a}</math>
with the same formula for ''P'' as in the previous case <math>\left(P = \frac{m V_0 b a}{\hbar^2}\right)</math>.

==Band gaps in the Kronig–Penney model==

[[File:Penney-Kronig Allowed Energy.svg|thumb|The value of the expression to which cos(k a) is equated in the dispersion relation, with P = 1.5.  The black bars denote regions of <math> \alpha a</math> for which k can be calculated.]]

[[File:Penney-Kronig Dispersion.svg|thumb|The dispersion relation for the Kronig–Penney model, with P = 1.5.]]

In the previous paragraph, the only variables not determined by the parameters of the physical system are the energy ''E'' and the crystal momentum ''k''.  By picking a value for ''E'', one can compute the right hand side, and then compute ''k'' by taking the <math>\arccos</math> of both sides.  Thus, the expression gives rise to the [[dispersion relation]].

The right hand side of the last expression above can sometimes be greater than 1 or less than –1, in which case there is no value of ''k'' that can make the equation true.  Since <math>\alpha a \propto \sqrt{E}</math>, that means there are certain values of ''E'' for which there are no eigenfunctions of the Schrödinger equation.  These values constitute the [[band gap]].

Thus, the Kronig–Penney model is one of the simplest periodic potentials to exhibit a band gap.

==Kronig–Penney model: alternative solution==
An alternative treatment <ref>{{cite journal |author=Surjit Singh|title=Kronig–Penney model in reciprocal lattice space| journal=American Journal of Physics|volume=51 |pages=179|year=1983|issue=2 |doi=  10.1119/1.13321|bibcode=1983AmJPh..51..179S }}</ref> to a similar problem is given. Here we have a ''delta'' periodic potential:
<math display="block">V(x) = A\cdot\sum_{n=-\infty}^{\infty}\delta(x - n a).</math>

{{mvar|A}} is some constant, and {{mvar|a}} is the lattice constant (the spacing between each site). Since this potential is periodic, we could expand it as a Fourier series:
<math display="block">V(x) = \sum_K \tilde{V}(K)\cdot e^{i K x},</math>
where
<math display="block">\tilde{V}(K) = \frac{1}{a}\int_{-a/2}^{a/2}dx\,V(x)\,e^{-i K x} = \frac{1}{a}\int_{-a/2}^{a/2} dx \sum_{n=-\infty}^{\infty} A\cdot \delta(x-na)\,e^{-i K x} = \frac{A}{a}.</math>

The wave-function, using Bloch's theorem, is equal to <math>\psi_k(x) = e^{i k x} u_k(x)</math> where <math>u_k(x)</math> is a function that is periodic in the lattice, which means that we can expand it as a Fourier series as well:
<math display="block">u_k(x)=\sum_{K} \tilde{u}_k(K)e^{i K x}.</math>

Thus the wave function is:
<math display="block">\psi_k(x)=\sum_{K}\tilde{u}_k(K)\,e^{i(k+K)x}.</math>

Putting this into the Schrödinger equation, we get:
<math display="block">\left[\frac{\hbar^2(k+K)^2}{2m}-E_k\right] \tilde{u}_k(K)+\sum_{K'}\tilde{V}(K-K')\,\tilde{u}_k(K') = 0</math>
or rather:
<math display="block">\left[\frac{\hbar^2(k+K)^2}{2m}-E_k\right] \tilde{u}_k(K)+\frac{A}{a}\sum_{K'}\tilde{u}_k(K')=0</math>

Now we recognize that:
<math display="block">u_k(0)=\sum_{K'}\tilde{u}_k(K')</math>

Plug this into the Schrödinger equation:
<math display="block">\left[\frac{\hbar^2(k+K)^2}{2m}-E_k\right] \tilde{u}_k(K)+\frac{A}{a}u_k(0)=0</math>

Solving this for <math>\tilde{u}_k(K)</math> we get:
<math display="block">\tilde{u}_k(K)=\frac{\frac{2m}{\hbar^2}\frac{A}{a}f(k)}{\frac{2mE_k}{\hbar^2}-(k+K)^2}=\frac{\frac{2m}{\hbar^2}\frac{A}{a}}{\frac{2mE_k}{\hbar^2}-(k+K)^2}\,u_k(0)</math>

We sum this last equation over all values of {{mvar|K}} to arrive at:
<math display="block">\sum_{K}\tilde{u}_k(K)=\sum_{K}\frac{\frac{2m}{\hbar^2}\frac{A}{a}}{\frac{2mE_k}{\hbar^2}-(k+K)^2}\,u_k(0)</math>

Or:
<math display="block">u_k(0)=\sum_{K}\frac{\frac{2m}{\hbar^2}\frac{A}{a}}{\frac{2mE_k}{\hbar^2}-(k+K)^2}\,u_k(0)</math>

Conveniently, <math>u_k(0)</math> cancels out and we get:
<math display="block">1=\sum_{K}\frac{\frac{2m}{\hbar^2}\frac{A}{a}}{\frac{2mE_k}{\hbar^2}-(k+K)^2}</math>

Or:
<math display="block">\frac{\hbar^2}{2m}\frac{a}{A}=\sum_{K}\frac{1}{\frac{2mE_k}{\hbar^2}-(k+K)^2}</math>

To save ourselves some unnecessary notational effort we define a new variable:
<math display="block">\alpha^2 := \frac{2 m E_k}{\hbar^2}</math>
and finally our expression is:
<math display="block">\frac{\hbar^2}{2m}\frac{a}{A}=\sum_{K}\frac{1}{\alpha^2-(k+K)^2}</math>

Now, {{mvar|K}} is a reciprocal lattice vector, which means that a sum over {{mvar|K}} is actually a sum over integer multiples of <math>\frac{2\pi}{a}</math>:
<math display="block">\frac{\hbar^2}{2m}\frac{a}{A}=\sum_{n=-\infty}^{\infty}\frac{1}{\alpha^2-(k+\frac{2\pi n}{a})^2}</math>

We can juggle this expression a little bit to make it more suggestive (use [[partial fraction decomposition]]):
<math display="block">\begin{align}
\frac{\hbar^2}{2m}\frac{a}{A} &= \sum_{n=-\infty}^{\infty}\frac{1}{\alpha^2-(k+\frac{2\pi n}{a})^2} \\
&=-\frac{1}{2\alpha}\sum_{n=-\infty}^{\infty}\left[\frac{1}{(k+\frac{2\pi n}{a})-\alpha}-\frac{1}{(k+\frac{2\pi n}{a})+\alpha}\right] \\
&=-\frac{a}{4\alpha}\sum_{n=-\infty}^{\infty}\left[\frac{1}{\pi n + \frac{k a}{2}-\frac{\alpha a}{2}}-\frac{1}{\pi n +\frac{k a}{2}+\frac{\alpha a} {2}} \right] \\
&=-\frac{a}{4\alpha}\left[\sum_{n=-\infty}^{\infty}\frac{1}{\pi n + \frac{k a}{2}-\frac{\alpha a}{2}} - \sum_{n=-\infty}^{\infty}\frac{1}{\pi n +\frac{k a}{2}+\frac{\alpha a}{2}} \right]
\end{align}</math>

If we use a nice identity of a sum of the cotangent function ([http://mathworld.wolfram.com/Cotangent.html Equation 18]) which says:
<math display="block">\cot(x)=\sum_{n=-\infty}^{\infty}\frac{1}{2 \pi n + 2x}-\frac{1}{2 \pi n - 2x}</math>
and plug it into our expression we get to:
<math display="block">\frac{\hbar^2}{2m}\frac{a}{A} = -\frac{a}{4\alpha}\left[\cot\left(\tfrac{k a}{2}-\tfrac{\alpha a}{2}\right) - \cot\left(\tfrac{k a}{2}+\tfrac{\alpha a}{2}\right)\right]</math>

We use the sum of {{math|cot}} and then, the product of {{math|sin}} (which is part of the formula for the sum of {{math|cot}}) to arrive at:
<math display="block">\cos(k a)=\cos(\alpha a)+\frac{m A}{\hbar^2 \alpha}\sin(\alpha a)</math>

This equation shows the relation between the energy (through {{mvar|α}}) and the wave-vector, {{mvar|k}}, and as you can see, since the left hand side of the equation can only range from {{math|−1}} to {{math|1}} then there are some limits on the values that {{mvar|α}} (and thus, the energy) can take, that is, at some ranges of values of the energy, there is no solution according to these equation, and thus, the system will not have those energies: energy gaps. These are the so-called band-gaps, which can be shown to exist in ''any'' shape of periodic potential (not just delta or square barriers).

For a different and detailed calculation of the gap formula (i.e. for the gap between bands) and the level splitting of eigenvalues of the one-dimensional Schrödinger equation see Müller-Kirsten.<ref>Harald J. W. Müller-Kirsten, Introduction to Quantum Mechanics: Schrödinger Equation and Path Integral, 2nd ed., World Scientific (Singapore, 2012), 325–329, 458–477.</ref> Corresponding results for the cosine potential (Mathieu equation) are also given in detail in this reference.

==Finite lattice==

In some cases, the Schrödinger equation can be solved analytically on a one-dimensional lattice of finite length<ref name="Ren2002">
{{cite journal | last = Ren | first = Shang Yuan |title = Two Types of Electronic States in One-dimensional Crystals of Finite length | year = 2002 | journal = Annals of Physics |volume = 301 | issue = 1 |pages = 22–30 | doi = 10.1006/aphy.2002.6298 | arxiv = cond-mat/0204211 | bibcode = 2002AnPhy.301...22R | s2cid = 14490431 }}</ref><ref name="Ren2017">
{{cite book | last = Ren | first = Shang Yuan |title = Electronic States in Crystals of Finite Size: Quantum Confinement of Bloch Waves | edition = 2 | year = 2017 | publisher = Singapore, Springer }}</ref> using the theory of periodic differential equations.<ref name="eastham1973">{{cite book | last = Eastham | first = M.S.P. |title = The Spectral Theory of Periodic Differential Equations | year = 1973 | publisher = Edinburgh, Scottish Academic Press}}</ref> The length of the lattice is assumed to be <math>L = N a</math>, where <math>a</math> is the potential period and the number of periods <math>N</math> is a positive integer. The two ends of the lattice are at <math>\tau</math> and <math>L + \tau</math>, where <math>\tau</math> determines the point of termination. The wavefunction vanishes outside the interval <math>[\tau,L+\tau]</math>.

The eigenstates of the finite system can be found in terms of the Bloch states of an infinite system with the same periodic potential. If there is a band gap between two consecutive energy bands of the infinite system, there is a sharp distinction between two types of states in the finite lattice. For each energy band of the infinite system, there are <math> N - 1 </math> bulk states whose energies depend on the length <math>N</math> but not on the termination <math>\tau</math>. These states are [[Standing wave|standing waves]] constructed as a superposition of two Bloch states with momenta <math>k</math> and <math>-k</math>, where <math>k</math> is chosen so that the wavefunction vanishes at the boundaries. The energies of these states match the energy bands of the infinite system.<ref name="Ren2002" />

For each band gap, there is one additional state. The energies of these states depend on the point of termination <math> \tau </math> but not on the length <math>N</math>.<ref name="Ren2002" /> The energy of such a state can lie either at the band edge or within the band gap. If the energy is within the band gap, the state is a [[surface state]] localized at one end of the lattice, but if the energy is at the band edge, the state is delocalized across the lattice.

==See also==
* [[Empty lattice approximation]]
* [[Nearly free electron model]]
* [[Crystal structure]]

==References==
{{Reflist}}

==External links==
* "[http://demonstrations.wolfram.com/TheKronigPenneyModel/ The Kronig–Penney Model]" by Michael Croucher, an interactive calculation of 1d periodic potential band structure using [[Mathematica]], from  [[The Wolfram Demonstrations Project]].

{{DEFAULTSORT:Particle In A One-Dimensional Lattice}}
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