Toʻgʻri burchakli potensial oʻra
ℓ
{\displaystyle \ell }
0
<
x
<
ℓ
{\displaystyle 0<x<\ell }
0
<
x
<
ℓ
/
3
{\displaystyle 0<x<\ell /3}
ℓ
/
3
<
x
<
2
ℓ
/
3
{\displaystyle \ell /3<x<2\ell /3}
2
ℓ
/
3
<
x
<
ℓ
{\displaystyle 2\ell /3<x<\ell }
Schr
o
¨
{\displaystyle {\ddot {\text{o}}}}
H
^
ψ
=
E
ψ
,
(
1
)
{\displaystyle {\hat {H}}\psi =E\psi ,\ \ \ \ \ (1)}
Potensial oʻra ichida zarraning potensial energiyasi nolga teng
U
(
x
)
=
0
{\displaystyle U(x)=0}
−
ℏ
2
2
m
Δ
ψ
=
E
ψ
ℏ
2
2
m
⋅
∂
2
ψ
∂
x
2
+
E
ψ
=
0
{\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:
∂
2
ψ
∂
x
2
+
2
m
E
ℏ
2
ψ
=
0
(
2
)
{\displaystyle {\dfrac {\partial ^{2}\psi }{\partial x^{2}}}+{\dfrac {2mE}{\hbar ^{2}}}\psi =0\ \ \ \ \ (2)}
Quyidagicha belgilash kiritamiz:
k
2
=
2
m
E
ℏ
2
{\displaystyle k^{2}={\dfrac {2mE}{\hbar ^{2}}}}
U holda (2) tenglama quyidagi koʻrinishga keladi:
∂
2
ψ
∂
x
2
+
k
2
ψ
=
0
{\displaystyle {\dfrac {\partial ^{2}\psi }{\partial x^{2}}}+k^{2}\psi =0}
U holda bu differensial tenglama uchun xarakteristik tenglama quyidagi koʻrinishga keladi:
λ
2
+
k
2
=
0
{\displaystyle \lambda ^{2}+k^{2}=0}
Uning ildizlari:
λ
2
=
−
k
2
,
u holda,
λ
1
=
i
k
,
λ
2
=
−
i
k
{\displaystyle \lambda ^{2}=-k^{2},\ {\text{u holda,}}\ \lambda _{1}=ik,\ \ \lambda _{2}=-ik}
Hosil boʻlgan toʻlqin funksiyani oshkormas koʻrinishda quyidagicha yozishimiz mumkin:
ψ
=
A
e
λ
1
x
+
B
e
λ
2
x
=
A
e
i
k
x
+
B
e
−
i
k
x
(
3
)
{\displaystyle \psi =Ae^{\lambda _{1}x}+Be^{\lambda _{2}x}=Ae^{ikx}+Be^{-ikx}\ \ \ \ \ (3)}
Eiler formulasi
e
i
z
=
cos
z
+
i
sin
z
{\displaystyle e^{iz}=\cos z+i\sin z}
e
−
i
z
=
cos
z
−
i
sin
z
{\displaystyle e^{-iz}=\cos z-i\sin z}
ψ
=
A
(
cos
k
x
+
i
sin
k
x
)
+
B
(
cos
k
x
−
i
sin
k
x
)
=
(
A
+
B
)
cos
k
x
+
(
A
−
B
)
i
sin
k
x
{\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,
k
{\displaystyle k}
E
{\displaystyle E}
To'lqin funksiyaning asosiy xossalaridan biri — bu uzluksizlik, shu sababli,
ψ
(
0
)
=
ψ
(
ℓ
)
=
0
{\displaystyle \psi (0)=\psi (\ell )=0}
{
ψ
(
0
)
=
(
A
+
B
)
cos
(
k
⋅
0
)
+
(
A
−
B
)
i
sin
(
k
⋅
0
)
=
A
+
B
=
0
,
ψ
(
ℓ
)
=
(
A
+
B
)
cos
(
k
ℓ
)
+
(
A
−
B
)
i
sin
(
k
ℓ
)
=
(
A
−
B
)
i
sin
(
k
ℓ
)
=
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,
A
=
−
B
{\displaystyle A=-B}
(
A
−
B
)
i
≠
0
{\displaystyle (A-B)i\neq 0}
sin
(
k
ℓ
)
=
0
{\displaystyle \sin(k\ell )=0}
U holda
k
ℓ
=
π
n
{\displaystyle k\ell =\pi n}
k
=
π
n
ℓ
{\displaystyle k={\dfrac {\pi n}{\ell }}}
n
{\displaystyle n}
Endi esa,
k
2
=
π
2
n
2
ℓ
2
=
2
m
E
ℏ
2
{\displaystyle k^{2}={\dfrac {\pi ^{2}n^{2}}{\ell ^{2}}}={\dfrac {2mE}{\hbar ^{2}}}}
E
=
n
2
ℏ
2
π
2
2
m
ℓ
2
{\displaystyle E={\dfrac {n^{2}\hbar ^{2}\pi ^{2}}{2m\ell ^{2}}}}
xususiy qiymati deyiladi.
Natijada zarra toʻlqin funksiyasining
ψ
=
2
i
A
sin
(
π
n
ℓ
x
)
{\displaystyle \psi =2iA\sin \left({\dfrac {\pi n}{\ell }}x\right)}
Bundan soʻng, toʻlqin funksiyaning normallash shartidan foydalanib,
A
{\displaystyle A}
0
<
x
<
ℓ
{\displaystyle 0<x<\ell }
1
=
∫
0
ℓ
4
i
2
A
2
sin
2
(
π
n
ℓ
)
d
x
=
4
i
2
A
2
⋅
ℓ
π
n
∫
0
ℓ
sin
2
(
π
n
ℓ
)
d
(
π
n
ℓ
x
)
=
4
i
2
A
2
e
π
n
×
∫
0
ℓ
1
−
cos
(
2
π
n
ℓ
x
)
2
⋅
2
d
(
2
π
n
ℓ
x
)
=
2
i
2
A
2
ℓ
2
π
n
(
2
π
n
ℓ
x
−
sin
(
2
π
n
ℓ
x
)
)
|
0
ℓ
=
i
2
A
2
ℓ
π
n
⋅
2
π
n
ℓ
⋅
ℓ
=
2
i
2
A
2
ℓ
{\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,
A
=
1
i
2
ℓ
{\displaystyle A={\dfrac {1}{i{\sqrt {2\ell }}}}}
Yuqoridagi yechimlardan foydalanib, toʻlqin funksiya uchun eng oxirgi ifodani quyidagicha yozish mumkin:
ψ
(
x
)
=
2
ℓ
(
π
n
ℓ
x
)
,
bu yerda
n
=
1
,
2
,
3
,
…
(
4
)
{\displaystyle \psi (x)={\sqrt {\dfrac {2}{\ell }}}\left({\dfrac {\pi n}{\ell }}x\right),\ \ {\text{bu yerda}}\ n=1,2,3,\ldots \ \ \ \ \ \ \ (4)}
Endi esa oʻraning
0
<
x
<
ℓ
/
3
{\displaystyle 0<x<\ell /3}
n
=
1
{\displaystyle n=1}
P
(
0
<
x
<
ℓ
/
3
)
=
∫
0
ℓ
/
3
2
ℓ
sin
2
(
π
n
ℓ
x
)
d
x
=
∫
0
ℓ
/
3
2
ℓ
⋅
1
−
cos
(
2
π
n
ℓ
x
)
2
d
(
2
π
n
ℓ
x
)
ℓ
2
π
n
=
1
2
π
n
(
2
π
n
ℓ
x
−
sin
(
2
π
n
ℓ
x
)
)
|
0
ℓ
/
3
=
1
2
π
n
(
2
π
n
3
−
sin
2
π
n
3
)
=
1
3
−
3
2
⋅
1
2
π
=
1
3
−
3
4
π
=
0.33
−
0.14
≈
0.195
yoki
P
≈
19.5
%
{\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,
ℓ
<
x
<
2
ℓ
/
3
{\displaystyle \ell <x<2\ell /3}
2
ℓ
/
3
<
x
<
ℓ
{\displaystyle 2\ell /3<x<\ell }
