**Elektronning davriy potensial maydondagi harakati**

Elektronning davriy {\displaystyle U(x)} potensial maydondagi harakatini koʻrib chiqaylik. Ushbu holat uchun Schr{\displaystyle {\ddot {\text{o}}}}dinger tenglamasining yechimini atom toʻlqin funksiyalarining kuchli bogʻlanishi usulidan foydalanib aniqlaymiz:

{\displaystyle \psi \_{n}(x),\ \qquad \ (n=1,2,\ldots ,\ N)}
bu yerda {\displaystyle N} — atom yacheykasining oʻlchami $a$ ga teng boʻlgan, kattaligi $L=na$ ga teng boʻlgan sohadagi atomlar soni.

Har bir atom yacheykasidagi to'lqin funksiya quyidagi koʻrinishdagi Schr{\displaystyle {\ddot {\text{o}}}}dinger tenglamasini qanoatlantiradi:

{\displaystyle \left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right)\psi \_{n}''+U(x)\psi \_{n}(x)=\varepsilon \psi \_{n}(x),\ \ \qquad c\_{n}\leq x\leq c\_{n+1}\ \ \ \ \ \ (1)}
Shuningdek, ushbu funksiya qiymatlari atom yacheykasining chegaraviy qiymatlarida yetarlicha darajada kichik. Toʻlqin funksiyaning normallash shartidan foydalangan holatda, kvaziklassik yaqinlashish uchun quyidagini hosil qilamiz[1]:

{\displaystyle \psi \_{n}(x)={\begin{cases}0,&x<c\_{n}\\{\sqrt {\dfrac {\omega \_{n}}{2\pi |\nu \_{n}(x)|}}}\exp {-\int \limits \_{x}^{a\_{n}}|k(x){\text{d}}x},&c\_{n}\leq x<a\_{n}\\{\sqrt {\dfrac {2\omega \_{n}}{2\pi |\nu \_{n}(x)|}}}\cos \left[\int \limits \_{a\_{n}}^{x}|k(x){\text{d}}x-{\dfrac {\pi }{4}}\right],&a\_{n}<x<b\_{n}\\{\sqrt {\dfrac {\omega \_{n}}{2\pi |\nu \_{n}(x)|}}}\exp {-\int \limits \_{b\_{n}}^{x}|k(x){\text{d}}x},&b\_{n}\leq x<c\_{n+1}\\0,&x\geq c\_{n+1}\end{cases}}\ \ \ \ \ \ (2)}
bu yerda {\displaystyle \omega \_{n}} — davriy harakat chastotasi, uni quyidagicha aniqlanadi:

{\displaystyle \omega \_{n}={\dfrac {\pi }{\int \limits \_{a}^{b}{\dfrac {1}{\nu (x)}}{\text{d}}x}}}
Kristall panjaradagi toʻlqin funksiyalar {\displaystyle \psi \_{m}(x)} ni atom toʻlqin funksiyalarining yigʻindisi koʻrinishida aniqlash mumkin:

{\displaystyle \psi \_{m}(x)=N^{-1/2}\sum \limits \_{n=1}^{N}e^{ik\_{m}na}\psi \_{n}(x)\ \ \ \ \ \ (3)}

  

(3) formulada {\displaystyle k\_{m}={\dfrac {2\pi }{N\_{a}}}m={\dfrac {2\pi }{L}}m} boʻladi va ushbu (3) ifoda {\displaystyle c\_{0}\leq x\leq c\_{N}} sohada Schr{\displaystyle {\ddot {\text{o}}}}dinger tenglamasini qanoatlantiradi:

{\displaystyle -{\dfrac {\hbar ^{2}}{2m\_{0}}}\psi ''(x)+U(x)\psi \_{m}(x)=E\_{m}\psi \_{m}(x)}
Bu holda Born-Karman davriylik sharti bajariladi, deb hisoblash mumkin:

{\displaystyle \psi \_{m}(x+La)=\psi \_{m}(x)}
{\displaystyle E\_{m}} ning qiymatini aniqlash uchun (2) tenglamani {\displaystyle \psi \_{m}(x)} ga koʻpaytiramiz, (1) tenglamani esa {\displaystyle \psi \_{n}(x)} ga koʻpaytiramiz. Soʻng hosil boʻlgan tenglamalarni toʻla fazo boʻyicha integrallab, birinchisidan ikkinchisini ayiramiz.

Integral ostida (1) ifoda turganligi uchun, amalda integrallash chegarasi sifatida {\displaystyle c\_{n}\leq x\leq c\_{n+1}} olinadi[2]:

{\displaystyle \left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right)\int (\psi \_{m}''(x)\psi \_{n}(x)-\psi \_{n}''(x)\psi \_{m}(x)){\text{d}}x=(E\_{m}-\varepsilon )\int \psi \_{m}(x)\psi \_{n}(x){\text{d}}x}
{\displaystyle \psi \_{m}(x)} uchun (2) ifodadan foydalangan holda, quyidagini hosil qilamiz:

{\displaystyle \left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right)(\psi \_{m}'(x)\psi \_{n}(x)-\psi '\_{n}(x)\psi \_{m}(x)){\bigg |}\_{c\_{n}}^{c\_{n+1}}=(E\_{m}-\varepsilon )N^{-1/2}e^{ik\_{m}na}}
Endi esa hosil boʻlgan tenglikning chap tarafini batafsil koʻrib chiqamiz:

{\displaystyle \left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right)(\psi '\_{m}(c\_{n+1})\psi \_{n}(c\_{n+1})-\psi '\_{n}(c\_{n+1})\psi \_{m}(c\_{n+1})-\psi \_{m}'(c\_{n})\psi \_{n}(c\_{n})+\psi \_{n}'(c\_{n})\psi \_{m}(c\_{n}))=\left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right){\dfrac {1}{\sqrt {N}}}e^{ik\_{m}na}{\bigg \{}\psi '\_{n}(c\_{n+1})\psi \_{n}(c\_{n+1})+\psi '\_{n+1}(c\_{n+1})\psi \_{n}(c\_{n+1})e^{ik\_{m}a}-\psi '\_{n}(c\_{n+1})\psi \_{n}(c\_{n+1})-\psi '\_{n}(c\_{n+1})\psi \_{n+1}(c\_{n+1})e^{ik\_{m}a}-\psi '\_{n}(c\_{n})\psi \_{n}(c\_{n})-\psi '\_{n-1}(c\_{n})\psi \_{n}(c\_{n})e^{-ik\_{m}a}+\psi \_{n}'(c\_{n})\psi \_{n}(c\_{n})+\psi '(c\_{n})\psi \_{n-1}(c\_{n})e^{-ik\_{m}a}{\bigg \}}=\left(-{\dfrac {\hbar ^{2}}{2m\_{0}}}\right){\dfrac {1}{\sqrt {N}}}e^{ik\_{m}na}{\bigg \{}\psi '\_{n+1}(c\_{n+1})\psi \_{n}(c\_{n+1})+\psi '\_{n}(c\_{n+1})\psi \_{n+1}(c\_{n+1})e^{ik\_{m}a}\psi \_{n}'(c\_{n})\psi \_{n-1}(c\_{n})+\psi \_{n-1}'(c\_{n})\psi \_{n}(c\_{n})e^{-ik\_{m}a}{\bigg \}}.}
(2) ga binoan:

{\displaystyle {\begin{cases}\psi '\_{n+1}(c\_{n+1})=|k(c\_{n+1})|\psi \_{n+1}(c\_{n+1})\\\psi '\_{n}(c\_{n+1})=-|k(c\_{n+1})|\psi \_{n}(c\_{n+1})\\\psi '\_{n}(c\_{n})=|k(c\_{n})\psi \_{n}(c\_{n})\\\psi \_{n-1}'(c\_{n})=-|k(c\_{n})|\psi \_{n-1}(c\_{n})\end{cases}}}
Bundan kelib chiqadiki

{\displaystyle -{\dfrac {\hbar ^{2}}{2m}}{\bigg \{}2|k(c\_{n+1}|\psi \_{n+1}(c\_{n+1})\psi (c\_{n+1})e^{ik\_{m}a}+2|k(c\_{n})\psi \_{n}(c\_{n})|\psi \_{n}(c\_{n})\psi \_{n-1}(c\_{n})e^{-ik\_{m}a}{\bigg \}}=E\_{m}-\varepsilon }
Kristallning davriylik xossasidan foydalansak

{\displaystyle |k(c\_{n})=|k(c\_{n+1})|;\ \qquad \ \psi \_{n}(c\_{n})\psi \_{n-1}(c\_{n})=\psi \_{n+1}(c\_{n+1})\psi \_{n}(c\_{n+1})}
Natijada quyidagi ifoda hosil boʻladi:

{\displaystyle E\_{m}=\varepsilon -{\dfrac {\hbar ^{2}}{m\_{0}}}\cdot 2|k(c\_{n})\psi \_{n+1}(c\_{n+1})\psi \_{n}(c\_{n+1})\cos k\_{m}a}
Yoki toʻlqin funksiya uchun (2)tenglamaga qoʻysak,

{\displaystyle E\_{m}=\varepsilon -{\dfrac {\hbar \omega }{\pi }}\exp {-\int \limits \_{b\_{n}}^{a\_{n+1}}|k(x){\text{d}}x}\cos k\_{m}a,\ \quad \ m=1,2,3,\ldots ,N\ \ \ \ \ \ (4)}
Shunday qilib, (4) dan koʻrinib turibdiki, har bir {\displaystyle \varepsilon } atom sathi, {\displaystyle N} ta atomning chiziqli zanjirga birlashishi natijasida {\displaystyle N} ta sathdan iborat boʻlgan polosalarga ajraladi. Har bir polosa kengligi {\displaystyle \Delta E} quyidagiga teng boʻladi:

{\displaystyle \Delta E={\dfrac {2\hbar \omega \_{n}}{\pi }}\exp {-\int \limits \_{b\_{n}}^{a\_{n+1}}|k(x)|{\text{d}}x}}
Agar zanjirdagi har bir atom tarkibida ikkitadan elektron boʻlsa, Pauli prinsipiga binoan, {\displaystyle N} ta sathlarning hammasi elektronlar bilan toʻlgan boʻladi. Bunday chiziqli zanjir oʻzini xuddi izolyator kabi tutadi. Yaʼni elektr maydon qoʻyilganda ham unda tok hosil boʻlmaydi. Agar zanjirdagi har bir atom bittadan elektronga ega boʻlsa, sathlar teng yarmigacha toʻlgan boʻladi. Bunday chiziqli model o'tkazgichlarga mos keladi. Elektr maydon qoʻyilganda, elektronlarning energiyasi ortadi, yaʼni elektr toki oqadi.

Manbalar
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1. *Tipler, Paul & Ralph Llewellyn*, Modern Physics (4th ed.). New York: W. H. Freeman and Company. ISBN 0-7167-4345-0
2. *Gerald Teschl*, Mathematical Methods in Quantum Mechanics; With Applications to Schrödinger Operators. American Mathematical Society. ISBN 978-0-8218-4660-5.
3. *iswesser, William J.* , „The Periodic System and Atomic Structure I. An Elementary Physical Approach". Journal of Chemical Education. 22 (7): 314-322.
4. *Gallagher, Thomas F.*, Rydberg Atoms. Cambridge University Press. ISBN 978-0-521-02166-1.
5. *Matveev A. N.*, Quantum mechanics and structure of atoms, Moscow, High School Press. 1985, (translated to eng. 1990), pp 119-122
6. *Kachnelson, M.I, Polikarpov A. F.* Introduction to Atomic Physics, M. 2017, pp 119-125