Penyelesaian Potensial Gendenshtein IV dengan PD Hypergeometry

Oleh :

ISMAIL (S911408001)

PROGRAM STUDI ILMU FISIKA

PROGRAM PASCASARJANAUNIVERSITAS SEBELAS MARET

SURAKARTA2015

1

Penyelesaian Potensial Gendenshtein IV dengan PD Hypergeometry

Persamaan potensial Gendenshtein IV dinyatakan dengan :

V eff =ℏ2

2m {b2+a (a−1 )cos2 x

+2b(a−

12 )sin x

cos2 x}

(1)Persamaan Schrodinger untuk potensial Gendenshtein IV dapat dinyatakan sebagai :

−ℏ2

2md2 χ

dx 2+{ ℏ2

2mb2+a (a−1 )cos2 x

+ℏ2

2m

2b(a−12 )sin x

cos2 x} χ=Eχ

(2)

dimisalkan

sin x=1−2 zd sin x=d (1−2 z )cos xdx=−2 dzdzdx

=−cos x2

=−2√ z (1−z )2

=−√ z (1−z )

sin2 x+cos2 x=1cos2 x=1−sin2 xcos2 x=1−(1−2 z )2

cos2 x=1−(1−4 z+4 z2)cos2 x=4 z−4 z2

cos2 x=4 z (1−z )cos x=√4 z (1−z )=2√z (1−z )

2

d2

dx 2=dz

dxddz

dzdx

ddz

d2

dx 2=(−√z (1−z ) )d

dz (−√ z (1−z )ddz )

d2

dx 2=(−√z (1−z ) )d

dz(−(z (1−z )

12))d

dz

d2

dx 2=(−√z (1−z ) ){12 (−z (1−z ) )

−12 (1−2 z )d

dz+(−z (1−z ) )

12 d2

dz 2 }d2

dx 2=(−√z (1−z ) )1

2(1−2 z )

(−√z (1−z ) )ddz

+ ( z (1−z ) ) d2

dz 2

d2

dx 2=1

2(1−2 z )d

dz+ ( z (1−z ) ) d

2

dz 2

d2

dx 2=( z (1−z ) ) d

2

dz2+(12 −z )ddz

Dengan substitusi persamaan (3) dan (4) ke persamaan (2) diperoleh hasil :

−ℏ2

2m {( z (1−z ) ) d2 χ

dz2+(1

2−z ) dχ

dz }+{ ℏ2

2mb2+a (a−1 )cos2 x

+ℏ2

2m

2b(a−12 )sin x

cos2 x} χ=Eχ

dengan mengalikan semua suku dengan −2 m

ℏ2diperoleh hasil :

( z (1−z ) ) d2 χ

dz2+(1

2−z ) dχ

dz−{b2+a (a−1 )

cos2 x+

2b(a−12 )sin x

cos2 x}χ=−

2m

ℏ2Eχ

( z (1−z ) ) d2 χ

dz2+(1

2−z ) dχ

dz−{b2+a (a−1 )

4 z (1−z )+

2b(a−12 ) (1−2 z )

4 z (1−z ) }χ=−2m

ℏ2Eχ

Dengan memisalkan

2m

ℏ2E=k 2

diperoleh hasil :

( z (1−z ) ) d2 χ

dz2+(1

2−z ) dχ

dz−{b2+a (a−1 )

4 z (1−z )+

2 b(a−12 ) (1−2 z )

4 z (1−z )⏟( I )

} χ=−k2 χ

Persamaan (I) dapat diubah menjadi

3

b2+a (a−1 )4 z (1−z )

+2b(a−1

2 ) (1−2 z )

4 z (1−z )

=b2+a (a−1 )4 z (1−z )

+2b {(a−

12 )−2 z (a−

12 )}

4 z (1−z )

¿b2+a (a−1 )4 z (1−z )

+2b(a−1

2 )4 z (1−z )

−2b2 z (a−1

2 )4 z (1−z )

¿b2+a (a−1 )4 z (1−z )

+2b(a−

12 )

4 z (1−z )−

4 b(a−12 )

4 (1−z )

¿b2+a (a−1 )4 z

+b2+a ( a−1 )4 (1−z )

+2b (a−1

2 )4 z

+2b(a−1

2 )4 (1−z )

−4b (a−1

2 )4 (1−z )

=b2+a (a−1 )4 z

+2b (a−1

2 )4 z

+b2+a (a−1 )4 (1−z )

+2b(a−1

2 )4 (1−z )

−4 b(a−1

2 )4 (1−z )

¿b2+a (a−1 )+2b(a−1

2 )4 z

+b2+a (a−1 )−2 b(a−1

2 )4 (1−z )

¿b2+((a−1

2 )2

−14 )+2b (a−1

2 )4 z

+b2+((a−1

2 )2

−14 )−2b(a−1

2 )4 (1−z )

¿b2+(a−1

2 )2

+2b(a−12 )−1

44 z

+b2+(a−1

2 )2

−2b(a−12 )−1

4

4 (1−z )

4

Catatan Penting1

z (1−z )=1

z+ 1

1−z

Catatan Penting

b2+(a−12 )

2

+2 b(a−12 )=(b+(a−

12 ))

2

b2+(a−12 )

2

−2 b(a−12 )=(b−(a−1

2 ))2

=(b+(a−

12 ))

2

−14

4 z+(b−(a−

12 ))

2

−14

4 (1−z )

Sehingga persamaan (6) menjadi

( z (1−z ) ) d2 χ

dz2+(12 −z )dχ

dz−{(b+(a−1

2 ))2

−14

4 z+(b−(a−1

2 ))2

−14

4 (1−z ) } χ=−k 2 χ

( z (1−z ) ) d2 χ

dz2+(12 −z )dχ

dz−{(b+(a−

12 ))

2

−14

4 z+(b−(a−

12 ))

2

−14

4 (1−z )+k 2}χ=0

( z (1−z ) ) d2 χ

dz2+(12 −z )dχ

dz+{k2−

(b+(a−12 ))

2

−14

4 z−

(b−(a−12 ))

2

−14

4 (1−z ) } χ=0

Persamaan (7) merupakan persamaan differensial orde dua homogeny yang mempunyai titik regular singular di titik z = 0 atau z = 1

Untuk titik z = 0

Suku k2

dan

(b−(a−12 ))

2

−14

4 (1−z ) diabaikan terhadap suku

(b+(a−12 ))

2

−14

4 z , sehingga persamaan (7) menjadi :

( z (1−z ) ) d2 χdz2

+(12−z ) dχ

dz−

(b+(a−12 ))

2

−14

4 zχ=0

Sedangkan untuk z = 1

Suku k2

dan

(b+(a−12 ))

2

−14

4 z diabaikan terhadap suku

(b−(a−12 ))

2

−14

4 (1−z ) sehingga persamaan (7) menjadi :

( z (1−z ) ) d2 χdz2

+(12−z ) dχ

dz−

(b−(a−12 ))

2

−14

4 z (1−z )χ=0

5

Fungsi Baru

χ=zα (1−z )β f (z )dχdz

=αzα−1 (1−z )β f ( z )−zα β (1−z )β−1 f ( z )+zα (1−z )β f ' ( z )

d2 χdz2

=α (α−1 ) zα−2 (1−z )β f ( z )−αzα−1 β (1−z )β−1 f ( z )+αzα−1 (1−z )β f ' ( z )−αβ zα−1 (1−z )β−1 f ( z )

+β ( β−1 ) za (1−z )β −2 f (z )−βzα (1−z )β−1 f ' ( z )+αzα−1 (1−z )β f ' ( z )−zα β (1−z )β−1 f ' (z )+zα (1−z )β f } } left (z right ) {} # =α left (α - 1 right )z rSup { size 8{α - 2} } left (1 - z right ) rSup { size 8{β} } f left (z right ) - 2αz rSup { size 8{α - 1} } β left (1 - z right ) rSup { size 8{β - 1} } f left (z right )+2αz rSup { size 8{α - 1} } left (1 - z right ) rSup { size 8{β} } f rSup { size 8{'} } left (z right )+z rSup { size 8{a} } β left (β - 1 right ) left (1 - z right ) rSup { size 8{β - 2} } f left (z right ) {} # - 2z rSup { size 8{α} } β left (1 - z right ) rSup { size 8{β - 1} } f rSup { size 8{'} } left (z right )+z rSup { size 8{α} } left (1 - z right ) rSup { size 8{β} } f rSup { size 8{ ( z )¿ zα (1−z )β¿

{α (α−1 ) z−2 f ( z )−2 αz−1 β (1−z )−1 f (z )+2 αz−1 f ' ( z )+ β ( β−1 ) (1−z )−2 f ( z ) ¿} {−2 β (1−z )−1 f ' ( z )+ f } } left (z right ) {} # right rbra } } lbrace rbrace {} # =z rSup { size 8{α} } left (1 - z right ) rSup { size 8{β} } left lbrace { {α left (α - 1 right )} over {z rSup { size 8{2} } } } f left (z right ) - { {2 ita l αβ } over {z rSup { size 8{1} } left (1 - z right ) rSup { size 8{1} } } } f left (z right )+ { {2α} over {z rSup { size 8{1} } } } f rSup { size 8{'} } left (z right )+ { {β left (β - 1 right )} over { left (1 - z right ) rSup { size 8{2} } } } f left (z right ) - { {2β} over { left (1 - z right ) rSup { size 8{1} } } } f rSup { size 8{'} } left (z right )+f rSup { size 8 { ( z ) } ¿

Dari persamaan fungsi baru di atas, persamaan (7) menjadi :

6

z (1−z )[ zα (1−z )β {α (α−1 )z2

f ( z )−2 αβ

z1 (1−z )1f ( z )+2 α

z1f ' ( z )+β ( β−1 )

(1−z )2f ( z )−2 β

(1−z )1f ' ( z )+ f } } left (z right ) right rbrace right ]} {} # + left ( { {1} over {2} } - z right ) left [αz rSup { size 8{α - 1} } left (1 - z right ) rSup { size 8{β} } f left (z right ) - z rSup { size 8{α} } β left (1 - z right ) rSup { size 8{β - 1} } f left (z right )+z rSup { size 8{α} } left (1 - z right ) rSup { size 8{β} } f rSup { size 8{'} } left (z right ) right ] {} # + left lbrace - { { left (b+ left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4z} } - { { left (b - left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4 left (1 - z right )} } +k rSup { size 8{2} } right rbrace left (z rSup { size 8{α} } left (1 - z right ) rSup { size 8{β} } f left (z right ) right )=0 {} # =z left (1 - z right ) left [z rSup { size 8{α} } left (1 - z right ) rSup { size 8{β} } left lbrace { {α left ( α - 1 r ight ) } ove r {z r Sup { siz e 8{2} } } } f le f t ( z r ight ) - { {2 ita l αβ } over {z rSup { size 8{1} } left (1 - z right ) rSup { size 8{1} } } } f left (z right )+ { {2α} over {z rSup { size 8{1} } } } f rSup { size 8{'} } left (z right )+ { {β left (β - 1 right )} over { left (1 - z right ) rSup { size 8{2} } } } f left (z right ) - { {2β} over { left (1 - z right ) rSup { size 8{1} } } } f rSup { size 8{'} } le f t ( z r ight ) + f r Sup { siz e 8{ ( z )}] ¿+(12 −z) [ zα (1−z )β {αz−1 f ( z )−β (1−z )−1 f ( z )+ f ' ( z ) }]

+{−(b+(a−12 ))

2

−14

4 z−

(b−(a−12 ))

2

−14

4 (1−z )+k2}(zα (1−z )β f ( z) )=0

¿ z (1−z )¿¿

¿

7

=(1−z ) α (α−1 )z⏟

(a )

f ( z )−2 αβ f ( z )+(1−z ) 2αf ' ( z )+ zβ ( β−1 )(1−z )⏟

(b)

f ( z )−2β zf ' ( z )+ z (1−z ) f } } left (z right )} {} # + { { { left ( { {1} over {2} } - z right )α} over {z} } } underbrace { size 8{ \( c \) } } f left (z right ) - { { {β left ( { {1} over {2} } - z right )} over { left (1 - z right )} } } underbrace { size 8{ \( d \) } } f left (z right ) - left ( { {1} over {2} } - z right )f rSup { size 8{'} } left (z right ) {} # + left lbrace - { { left (b+ left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4z} } - { { left (b - left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4 left (1 - z right )} } +k rSup { size 8{2} } right rbrace f left (z right )=0 {} } } { ¿ ¿¿

¿

¿

Persamaan (a)

=(1−z )α (α−1 )z

=α (α−1 ) (1−z )z

=α ( (α−1 )−z (α−1 ) )z

=α (α−1 )−αz (α−1 )z

=α (α−1 )z

−αz (α−1 )z

=α (α−1 )z

−α (α−1 )

Persamaan (b)

zβ ( β−1 )(1−z )

=zβ ( β−1 )+β (β−1 )−β (β−1 )(1−z )

=zβ ( β−1 ) ( β−1 )(1−z )

+β ( β−1 )(1−z )

=β ( z−1 ) (β−1 )(1−z )

+β ( β−1 )(1−z )

=−β ( β−1 )+β ( β−1 )(1−z )

Persamaan (c)

( 12−z)α

z=α

z (12−z )= α

2 z−α

Persamaan (d)

β (12 −z)(1−z )

=β1−z (12 −z)=β

1−z (12 +12

−12

−z)=β1−z (1−1

2−z)=β

1−z ((1−z )−12 )

=β−β2 (1−z )

8

Sehingga persamaannya menjadi :

=(α (α−1 )z

−α (α−1 )) f (z )−2αβf ( z )+ (1−z )2 αf ' ( z )+(−β ( β−1 )+β ( β−1 )(1−z ) ) f ( z )

+(α2 z−α) f ( z )−(β−β

2 (1−z ) ) f ( z )−(12−z ) f ' (z )−2 β zf ' ( z )+z (1−z ) f } } left (z right ) {} # + left lbrace - { { left (b+ left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4z} } - { { left (b - left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4 left (1 - z right )} } +k rSup { size 8{2} } right rbrace f left (z right )=0 {} # =alignl { stack { left lbrace left ( { {α left (α - 1 right )} over {z} } - α left (α - 1 right ) right ) - 2 ital αβ left ( - β left (β - 1 right )+ { {β left (β - 1 right )} over { left (1 - z right )} } right )+ left ( { {α} over {2z} } - α right ) - β+ { {β} over {2 left (1 - z right )} } k rSup { size 8{2} } {} # right rbrace left lbrace - { { left (b+ left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4z} } - { { left (b - left (a - { {1} over {2} } right ) right ) rSup { size 8{2} } - { {1} over {4} } } over {4 left (1 - z right )} } {} # right rbra } } lbrace rbrace f left (z right ) {} # + left lbrace left (1 - z right )2α - 2zβ+ { {1} over {2} } - z right rbrace f rSup { size 8{'} } left (z right )+z left (1 - z right )f rSup { size 8{ ( z )=0

Pengubahan parameter

(b+(a−12 ))

2

−14=2 α (2 α−1 )

(b−(a−12 ))

2

−14=2 β (2 β−1 )

Untuk

{(α (α−1 )z

−α (α−1 ))−2αβ (−β ( β−1 )+β ( β−1 )(1−z ) )+(α2 z

−α )−β+β2 (1−z )

k2 ¿}¿{}f ( z )

¿ {2α (2 α−1 )4 z

+2 β (2 β−1 )4 (1−z )

−α2−β2−2 αβ+k2−2 α (2α−1 )4 z

−2 β (2 β−1 )4 (1−z ) }f ( z )

¿ {−α2−β2−2αβ+k2 } f ( z)={k 2−( α+β )2} f ( z )

9

Untuk

{(1−z ) 2α−2 zβ+12− z}f ' ( z )+z (1−z ) f } } left (z right )} {} # = left lbrace 2α - 2αz - 2βz+ { {1} over {2} } - z right rbrace f rSup { size 8{'} } left (z right ) {} # = left lbrace 2α+ { {1} over {2} } - z left (2α+2β+1 right ) right rbrace f rSup { size 8{'} } left (z right ) {} } } { ¿ ¿¿

¿¿

Disubtitusikan kembali ke persamaan diperoleh :

z (1−z ) f } } left (z right )+ left lbrace 2α+ { {1} over {2} } - z left (2α+2β+1 right ) right rbrace f rSup { size 8{'} } left (z right )+ left lbrace k rSup { size 8{2} } - left (α+β right ) rSup { size 8{2} } right rbrace f left (z right )=0} {} # =z left (1 - z right ) { {d rSup { size 8{2} } f} over { ital dz rSup { size 8{2} } } } + left lbrace 2α+ { {1} over {2} } - left (2α+2β+1 right )z right rbrace { { ital df } over { ital dz } } + left lbrace k rSup { size 8{2} } - left (α+β right ) rSup { size 8{2} } right rbrace f=0 {} } } {¿ ¿¿¿

¿

Persamaan di atas identik dengan PD orde dua Fungsi Hypergeometry

z (1−z ) d2φdz 2

+ {c−(a+b+1 ) z } dφdz

−ab φ=0

dimana

−ab={k2−(α+β )2 }ab={( α+β )2−k2 }= {α+β+k }⏟

a

{α+β−k }⏟b

a={α+ β+k }b={α+ β−k }

Penentuan nilai α dan β

(b+(a−12 ))

2

−14

=2 α (2 α−1 )

b2+2b (a−12 )+(a−

12 )

2

−14

=2α (2 α−1 )

b2+2ab−b+a2−a+14

−14

=2 α (2 α−1 )

(a+b )2−a−b=4 α 2−2α

4 α 2=( a+b )2

α 2=(a+b )2

4

α=a+b2

−2 α=−a−b

α=−a−b−2

α=a+b2

10

(b−(a−12 ))

2

−14

=2 β (2 β−1 )

b2−2 b(a−12 )+(a−

12 )

2

−14

=2 β (2 β−1 )

b2−2 ab+b+a2−a+14

−14

=2 β (2 β−1 )

(a−b )2−a+b=4 β2−2 β

4 β2=( a−b )2

β2=(a−b )2

4

β=a−b2

−2 β=−a+b

β=−a+b−2

β=a−b2

Menentukan spectrum energy dengan memilih a = - n

{α +β+k }=−na+b2

+a−b2

+k=−n

2 a2

+k=−n

a+k=−nk=−a−nk=−(a+n )

√2 mℏ2

En=− (a+n )

2 mℏ2

En=(a+n )2

2 mℏ2

En=(a+n )2

En=ℏ2

2 m(a+n )2

11

Menentukan fungsi gelombang tingkat dasar

χ=zα (1−z )β f ( z )

¿(1−sin x2 )

a+b2(1−1−sin x

2 )a−b

2=(1−sin x2 )

a+b2(22 −1−sin x

2 )a−b

2

¿(1−sin x2 )

a+b2(1+sin x

2 )a−b

2=(1−sin x2 )

a2(1−sin x

2 )b

2 (1+sin x2 )

a2 (1+sin x

2 )−b

2

¿(1−sin x2 )

a2(1+sin x

2 )a

2(1−sin x2 )

b2 (1+sin x

2 )−b

2=(1−sin2 x4 )

a2((

1−sin x2 )

b2

(1+sin x2 )

b2 )

¿(1−sin2 x4 )a

2((1−sin x2 )(21+sin x ))

b2=(1−sin2 x

4 )a2 (1−sin x

1+sin x )b

2

¿(1−sin2 x4 )a

2(1−sin x1+sin x

×1+sin x1+sin x )

b2=(cos2 x

4 )a2 ((cos2 x

(1+sin x )2 ))b

2

¿((cos x2 )

2)a

2 ((cos x(1+sin x ) )

2)b

2=(cos x2 )

a

(cos x(1+sin x ) )

b

¿(cos x2 )

a

(cos x(1+sin x )

×22 )

b

=(cos x2 )

a

(cos x2

×21+sin x )

b

¿(cos x2 )

a

(cos x2 )

b

(21+sin x )

b

=(cos x2 )

a+b

(1+sin x2 )

−b

12