**Toʻgʻri burchakli potensial oʻra** 

  

Aytaylik, elektron cheksiz chuqur, toʻgʻri burchakli bir oʻlchamli, kengligi {\displaystyle \ell } ga teng boʻlgan potensial oʻrada asosiy holatda turgan boʻlsin ({\displaystyle 0<x<\ell }). Zarraning {\displaystyle 0<x<\ell /3}; {\displaystyle \ell /3<x<2\ell /3}; {\displaystyle 2\ell /3<x<\ell } oraliqlarda topilish ehtimolliklarini hisoblaylik.

Schr{\displaystyle {\ddot {\text{o}}}}dingerning statsionar tenglamasini yozib olamiz:

{\displaystyle {\hat {H}}\psi =E\psi ,\ \ \ \ \ (1)}
Potensial oʻra ichida zarraning potensial energiyasi nolga teng {\displaystyle U(x)=0}. Shu sababli

{\displaystyle -{\dfrac {\hbar ^{2}}{2m}}\Delta \psi =E\psi \ \qquad \ {\dfrac {\hbar ^{2}}{2m}}\cdot {\dfrac {\partial ^{2}\psi }{\partial x^{2}}}+E\psi =0}
Yuqoridagi tenglamani kanonik koʻrinishga keltiramiz:

{\displaystyle {\dfrac {\partial ^{2}\psi }{\partial x^{2}}}+{\dfrac {2mE}{\hbar ^{2}}}\psi =0\ \ \ \ \ (2)}
Quyidagicha belgilash kiritamiz:

{\displaystyle k^{2}={\dfrac {2mE}{\hbar ^{2}}}}
U holda (2) tenglama quyidagi koʻrinishga keladi:

{\displaystyle {\dfrac {\partial ^{2}\psi }{\partial x^{2}}}+k^{2}\psi =0}
U holda bu differensial tenglama uchun xarakteristik tenglama quyidagi koʻrinishga keladi:

{\displaystyle \lambda ^{2}+k^{2}=0}
Uning ildizlari:

{\displaystyle \lambda ^{2}=-k^{2},\ {\text{u holda,}}\ \lambda \_{1}=ik,\ \ \lambda \_{2}=-ik}
Hosil boʻlgan toʻlqin funksiyani oshkormas koʻrinishda quyidagicha yozishimiz mumkin:

{\displaystyle \psi =Ae^{\lambda \_{1}x}+Be^{\lambda \_{2}x}=Ae^{ikx}+Be^{-ikx}\ \ \ \ \ (3)}
Eiler formulasi {\displaystyle e^{iz}=\cos z+i\sin z} va {\displaystyle e^{-iz}=\cos z-i\sin z} dan foydalangan holda (3) tenglamani quyidagicha yozamiz:

{\displaystyle \psi =A(\cos kx+i\sin kx)+B(\cos kx-i\sin kx)=(A+B)\cos kx+(A-B)i\sin kx}
Chegaraviy shartlardan foydalangan holda, {\displaystyle k} va {\displaystyle E} larning xususiy qiymatlarini aniqlaymiz.

To'lqin funksiyaning asosiy xossalaridan biri — bu uzluksizlik, shu sababli, {\displaystyle \psi (0)=\psi (\ell )=0}

{\displaystyle {\begin{cases}\psi (0)=(A+B)\cos(k\cdot 0)+(A-B)i\sin(k\cdot 0)=A+B=0,\\\psi (\ell )=(A+B)\cos(k\ell )+(A-B)i\sin(k\ell )=(A-B)i\sin(k\ell )=0\end{cases}}}

Sistemaning 1-tenglamasidan koʻrinib turibdiki, {\displaystyle A=-B}. Ikkinchi tenglamadan esa, {\displaystyle (A-B)i\neq 0} boʻlgani uchun {\displaystyle \sin(k\ell )=0} ekanligini hosil qilamiz.

U holda {\displaystyle k\ell =\pi n}, yoki {\displaystyle k={\dfrac {\pi n}{\ell }}}, bu yerda {\displaystyle n} — bosh kvant soni deyiladi.

Endi esa, {\displaystyle k^{2}={\dfrac {\pi ^{2}n^{2}}{\ell ^{2}}}={\dfrac {2mE}{\hbar ^{2}}}} ekanligini hisobga olgan holda {\displaystyle E={\dfrac {n^{2}\hbar ^{2}\pi ^{2}}{2m\ell ^{2}}}} ifodani hosil qilamiz. Bu ifoda zarra energiyasining xususiy qiymati deyiladi.

Natijada zarra toʻlqin funksiyasining {\displaystyle \psi =2iA\sin \left({\dfrac {\pi n}{\ell }}x\right)} koʻrinishida ekanligini aniqlaymiz.

Normallash sharti
-----------------

Bundan soʻng, toʻlqin funksiyaning normallash shartidan foydalanib, {\displaystyle A} ni aniqlaymiz. Maʼlumki, zarra toʻlqin funksiyasidan toʻla fazo boʻyicha olingan integral qiymati 1 ga teng boʻlishi kerak. Bizning misol uchun esa {\displaystyle 0<x<\ell } sohada zarra topilishi ehtimolligi 1 ga teng, chunki, uning oʻradan tashqarida toʻlqin funksiyasi nolga teng. Demak,

{\displaystyle 1=\int \limits \_{0}^{\ell }4i^{2}A^{2}\sin ^{2}\left({\dfrac {\pi n}{\ell }}\right){\text{d}}x=4i^{2}A^{2}\cdot {\dfrac {\ell }{\pi n}}\int \limits \_{0}^{\ell }\sin ^{2}\left({\dfrac {\pi n}{\ell }}\right){\text{d}}\left({\dfrac {\pi n}{\ell }}x\right)={\dfrac {4i^{2}A^{2}e}{\pi n}}\times \int \limits \_{0}^{\ell }{\dfrac {1-\cos \left({\dfrac {2\pi n}{\ell }}x\right)}{2\cdot 2}}{\text{d}}\left({\dfrac {2\pi n}{\ell }}x\right)={\dfrac {2i^{2}A^{2}\ell }{2\pi n}}\left({\dfrac {2\pi n}{\ell }}x-\sin \left({\dfrac {2\pi n}{\ell }}x\right)\right){\bigg |}\_{0}^{\ell }={\dfrac {i^{2}A^{2}\ell }{\pi n}}\cdot {\dfrac {2\pi n}{\ell }}\cdot \ell =2i^{2}A^{2}\ell }
Bundan kelib chiqadiki,

{\displaystyle A={\dfrac {1}{i{\sqrt {2\ell }}}}}
Yuqoridagi yechimlardan foydalanib, toʻlqin funksiya uchun eng oxirgi ifodani quyidagicha yozish mumkin:

{\displaystyle \psi (x)={\sqrt {\dfrac {2}{\ell }}}\left({\dfrac {\pi n}{\ell }}x\right),\ \ {\text{bu yerda}}\ n=1,2,3,\ldots \ \ \ \ \ \ \ (4)}
### Misol

Endi esa oʻraning {\displaystyle 0<x<\ell /3} sohasida zarra topilish ehtimolligini hisoblaymiz. Asosiy holatda {\displaystyle n=1} uchun hisoblaylik:

{\displaystyle P(0<x<\ell /3)=\int \limits \_{0}^{\ell /3}{\dfrac {2}{\ell }}\sin ^{2}\left({\dfrac {\pi n}{\ell }}x\right){\text{d}}x=\int \limits \_{0}^{\ell /3}{\dfrac {2}{\ell }}\cdot {\dfrac {1-\cos \left({\dfrac {2\pi n}{\ell }}x\right)}{2}}{\text{d}}\left({\dfrac {2\pi n}{\ell }}x\right){\dfrac {\ell }{2\pi n}}={\dfrac {1}{2\pi n}}\left({\dfrac {2\pi n}{\ell }}x-\sin \left({\dfrac {2\pi n}{\ell }}x\right)\right){\bigg |}\_{0}^{\ell /3}={\dfrac {1}{2\pi n}}\left({\dfrac {2\pi n}{3}}-\sin {\dfrac {2\pi n}{3}}\right)={\dfrac {1}{3}}-{\dfrac {\sqrt {3}}{2}}\cdot {\dfrac {1}{2\pi }}={\dfrac {1}{3}}-{\dfrac {\sqrt {3}}{4\pi }}=0.33-0.14\approx 0.195\ {\text{yoki}}\ P\approx 19.5\%}
Xuddi shu usulda, {\displaystyle \ell <x<2\ell /3} va {\displaystyle 2\ell /3<x<\ell } sohalar uchun ham integral chegaralarini oʻzgartirgan holda hisoblaymiz va mos ravishda 61% va 19.5% qiymatlarni aniqlaymiz.

Manbalar
--------

1. Griffiths, David J.*. Introduction to Quantum Mechanics (2nd ed.)*. Prentice Hall, 2004. ISBN 0-13-111892-7.
2. Cohen-Tannoudji, Claude; Diu, Bernard; Laloë, Franck*. Quantum mechanics.*, transl. from the French by Susan Reid Hemley, Wiley-Interscience: Wiley, 1996 — 231–233 bet. ISBN 978-0-471-56952-7.