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Quantum Mechanics M. Khari Secario Page 1 Potential Barier : From Schroedinger equation : βˆ’ β„Ž 2 4 2 2 2 2 + = 1. For x<-a βˆ’ β„Ž 2 4 2 2 2 1 2 = 1 β„Ž 2 4 2 2 2 1 2 + 1 =0 2 1 2 + 4 2 2 β„Ž 2 1 =0 2 1 2 + 1 2 1 =0 Solution : 1 ( )= 1 + βˆ’ 1 2. For –a<x<a βˆ’ β„Ž 2 4 2 2 2 2 2 + 0 2 = 2 β„Ž 2 4 2 2 2 2 2 + 2 βˆ’ 0 2 =0 2 2 2 + 4 2 2 β„Ž 2 (βˆ’ 0 ) 2 =0 2 2 2 + 2 2 =0 Solution : 2 ( )= + βˆ’ If we use : = βˆ’ Then : 2 ()= βˆ’ + Vo -a a

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Mekanika kuantum mengenai potensial barier

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Page 1: Potential Barier

Quantum Mechanics

M. Khari Secario Page 1

Potential Barier :

From Schroedinger equation :

βˆ’β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆπœ•πœ•π‘₯π‘₯2 + π‘‰π‘‰π‘ˆπ‘ˆ = πΈπΈπ‘ˆπ‘ˆ

1. For x<-a

βˆ’β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ1

πœ•πœ•π‘₯π‘₯2 = πΈπΈπ‘ˆπ‘ˆ1

β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ1

πœ•πœ•π‘₯π‘₯2 + πΈπΈπ‘ˆπ‘ˆ1 = 0

πœ•πœ•2π‘ˆπ‘ˆ1

πœ•πœ•π‘₯π‘₯2 +4πœ‹πœ‹22π‘šπ‘šπΈπΈ

β„Ž2 π‘ˆπ‘ˆ1 = 0

πœ•πœ•2π‘ˆπ‘ˆ1

πœ•πœ•π‘₯π‘₯2 + π‘˜π‘˜12π‘ˆπ‘ˆ1 = 0

Solution : π‘ˆπ‘ˆ1(π‘₯π‘₯) = π‘’π‘’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯ +π‘…π‘…π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯

2. For –a<x<a

βˆ’β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ2

πœ•πœ•π‘₯π‘₯2 + 𝑉𝑉0π‘ˆπ‘ˆ2 = πΈπΈπ‘ˆπ‘ˆ2

β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ2

πœ•πœ•π‘₯π‘₯2 + πΈπΈπ‘ˆπ‘ˆ2 βˆ’ 𝑉𝑉0π‘ˆπ‘ˆ2 = 0

πœ•πœ•2π‘ˆπ‘ˆ2

πœ•πœ•π‘₯π‘₯2 +4πœ‹πœ‹22π‘šπ‘šβ„Ž2 (𝐸𝐸 βˆ’π‘‰π‘‰0)π‘ˆπ‘ˆ2 = 0

πœ•πœ•2π‘ˆπ‘ˆ2

πœ•πœ•π‘₯π‘₯2 + π‘žπ‘ž2π‘ˆπ‘ˆ2 = 0

Solution : π‘ˆπ‘ˆ2(π‘₯π‘₯) = π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ +π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯

If we use : πœ…πœ… = βˆ’π‘–π‘–π‘žπ‘ž

Then : π‘ˆπ‘ˆ2(π‘₯π‘₯) = π΄π΄π‘’π‘’βˆ’πœ…πœ…π‘₯π‘₯ +π΅π΅π‘’π‘’πœ…πœ…π‘₯π‘₯

Vo

-a a

khari
Sticky Note
In this case, we assume that incident wave is come from left to right. Incident Energy (E) is lower than barier potential (Vo) At left of barrier is region 1 At barrier is region 2 At right of barrier is region 3
khari
Sticky Note
Actually, there is a constant at front of first order. For simplification, we use that constant is 1
Page 2: Potential Barier

Quantum Mechanics

M. Khari Secario Page 2

3. For x>a

βˆ’β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ3

πœ•πœ•π‘₯π‘₯2 = πΈπΈπ‘ˆπ‘ˆ3

β„Ž2

4πœ‹πœ‹22π‘šπ‘šπœ•πœ•2π‘ˆπ‘ˆ3

πœ•πœ•π‘₯π‘₯2 + πΈπΈπ‘ˆπ‘ˆ3 = 0

πœ•πœ•2π‘ˆπ‘ˆ3

πœ•πœ•π‘₯π‘₯2 +4πœ‹πœ‹22π‘šπ‘šπΈπΈ

β„Ž2 π‘ˆπ‘ˆ3 = 0

πœ•πœ•2π‘ˆπ‘ˆ3

πœ•πœ•π‘₯π‘₯2 + π‘˜π‘˜32π‘ˆπ‘ˆ3 = 0

Solution : π‘ˆπ‘ˆ3(π‘₯π‘₯) = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜3π‘₯π‘₯ +π‘Šπ‘Šπ‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜3π‘₯π‘₯

Remember, at Region 3, there is only transmitted wave (T)

Then : π‘ˆπ‘ˆ3(π‘₯π‘₯) = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜3π‘₯π‘₯

Now, we have 3 eq. :

β€’ π‘ˆπ‘ˆ1(π‘₯π‘₯) = π‘’π‘’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯ + π‘…π‘…π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯ β€’ π‘ˆπ‘ˆ2(π‘₯π‘₯) = π΄π΄π‘’π‘’βˆ’πœ…πœ…π‘₯π‘₯ + π΅π΅π‘’π‘’πœ…πœ…π‘₯π‘₯ = π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ + π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ β€’ π‘ˆπ‘ˆ3(π‘₯π‘₯) = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜3π‘₯π‘₯

With π‘˜π‘˜1 = π‘˜π‘˜3 = π‘˜π‘˜

Remember at x=-a and x=a :

β€’ At x=-a

π‘ˆπ‘ˆ1(βˆ’π‘Žπ‘Ž) = π‘ˆπ‘ˆ2(βˆ’π‘Žπ‘Ž) ...Β©

π‘‘π‘‘π‘ˆπ‘ˆ1𝑑𝑑π‘₯π‘₯

(βˆ’π‘Žπ‘Ž) = π‘‘π‘‘π‘ˆπ‘ˆ2𝑑𝑑π‘₯π‘₯

(βˆ’π‘Žπ‘Ž) ... ©©

β€’ At x=a

π‘ˆπ‘ˆ2(π‘Žπ‘Ž) = π‘ˆπ‘ˆ3(π‘Žπ‘Ž) ... ©©©

π‘‘π‘‘π‘ˆπ‘ˆ2𝑑𝑑π‘₯π‘₯

(π‘Žπ‘Ž) = π‘‘π‘‘π‘ˆπ‘ˆ3𝑑𝑑π‘₯π‘₯

(π‘Žπ‘Ž) ... ©©©©

khari
Sticky Note
At this point, the value of incident wave and transmitted wave must continue. Derivative is used so the wave function is smooth from region 1 to region 2 Same from region 2 to region 3
Page 3: Potential Barier

Quantum Mechanics

M. Khari Secario Page 3

Β©... π‘ˆπ‘ˆ1(βˆ’π‘Žπ‘Ž) = π‘ˆπ‘ˆ2(βˆ’π‘Žπ‘Ž)

[π‘’π‘’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯ + π‘…π‘…π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜1π‘₯π‘₯ ](βˆ’π‘Žπ‘Ž) = οΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ + π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ οΏ½(βˆ’π‘Žπ‘Ž)

π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž + π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž = π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž + π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž ...

©©... π‘‘π‘‘π‘ˆπ‘ˆ1𝑑𝑑π‘₯π‘₯

(βˆ’π‘Žπ‘Ž) = π‘‘π‘‘π‘ˆπ‘ˆ2𝑑𝑑π‘₯π‘₯

(βˆ’π‘Žπ‘Ž)

𝑑𝑑𝑑𝑑π‘₯π‘₯

οΏ½π‘’π‘’π‘–π‘–π‘˜π‘˜π‘₯π‘₯ + π‘…π‘…π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘₯π‘₯οΏ½ =𝑑𝑑𝑑𝑑π‘₯π‘₯

(π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ +π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ )

π‘–π‘–π‘˜π‘˜οΏ½π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž βˆ’ π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘ŽοΏ½ = π‘–π‘–π‘žπ‘ž(π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž) ...

©©©... π‘ˆπ‘ˆ2(π‘Žπ‘Ž) = π‘ˆπ‘ˆ3(π‘Žπ‘Ž)

οΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ +π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ οΏ½(π‘Žπ‘Ž) = οΏ½π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘₯π‘₯ οΏ½(π‘Žπ‘Ž)

π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž +π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž ...

©©©©... π‘‘π‘‘π‘ˆπ‘ˆ2𝑑𝑑π‘₯π‘₯

(π‘Žπ‘Ž) = π‘‘π‘‘π‘ˆπ‘ˆ3𝑑𝑑π‘₯π‘₯

(π‘Žπ‘Ž)

𝑑𝑑𝑑𝑑π‘₯π‘₯

οΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ + π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘₯π‘₯ οΏ½ =𝑑𝑑𝑑𝑑π‘₯π‘₯

(π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘₯π‘₯ )

π‘–π‘–π‘žπ‘žοΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘ŽοΏ½ = π‘–π‘–π‘˜π‘˜(π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž) ...

Thus we have 4 eq (look at smiley) :

β€’ π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž +π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž = π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž + π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž β€’ π‘–π‘–π‘˜π‘˜οΏ½π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž βˆ’ π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘ŽοΏ½ = π‘–π‘–π‘žπ‘ž(π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž) β€’ π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž + π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž β€’ π‘–π‘–π‘žπ‘žοΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘ŽοΏ½ = π‘–π‘–π‘˜π‘˜οΏ½π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘ŽοΏ½

With a β€œlittle” algebra, we’ll find out T. But we’ll using Gauss Elimination instead.

Page 4: Potential Barier

Quantum Mechanics

M. Khari Secario Page 4

Now, let see :

β€’ π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž +π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž = π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž + π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž β€’ π‘–π‘–π‘˜π‘˜οΏ½π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž βˆ’ π‘…π‘…π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘ŽοΏ½ = π‘–π‘–π‘žπ‘ž(π΄π΄π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž) β€’ π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž + π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž = π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž β€’ π‘–π‘–π‘žπ‘žοΏ½π΄π΄π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž βˆ’ π΅π΅π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘ŽοΏ½ = π‘–π‘–π‘˜π‘˜οΏ½π‘‡π‘‡π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘ŽοΏ½

For simplification :

β€’ π‘’π‘’βˆ’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž = 𝐿𝐿 β€’ π‘’π‘’π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž = 𝑀𝑀 β€’ π‘’π‘’βˆ’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž = 𝑁𝑁 β€’ π‘’π‘’π‘–π‘–π‘žπ‘žπ‘Žπ‘Ž = 𝑃𝑃

So we have :

β€’ 𝐿𝐿 +𝑅𝑅𝑀𝑀 = 𝐴𝐴𝑁𝑁 +𝐡𝐡𝑃𝑃 β€’ π‘–π‘–π‘˜π‘˜(𝐿𝐿 βˆ’π‘…π‘…π‘€π‘€) = π‘–π‘–π‘žπ‘ž(π΄π΄π‘π‘βˆ’ 𝐡𝐡𝑃𝑃) β€’ 𝐴𝐴𝑃𝑃 +𝐡𝐡𝑁𝑁 = 𝑇𝑇𝑀𝑀 β€’ π‘–π‘–π‘žπ‘ž(𝐴𝐴𝑃𝑃 βˆ’π΅π΅π‘π‘) = π‘–π‘–π‘˜π‘˜(𝑇𝑇𝑀𝑀)

Put in some order :

β€’ π‘…π‘…π‘€π‘€βˆ’ π΄π΄π‘π‘βˆ’ π΅π΅π‘ƒπ‘ƒβˆ’ 0 = βˆ’πΏπΏ β€’ βˆ’π‘–π‘–π‘˜π‘˜π‘…π‘…π‘€π‘€βˆ’ π‘–π‘–π‘žπ‘žπ΄π΄π‘π‘+ π‘–π‘–π‘žπ‘žπ΅π΅π‘ƒπ‘ƒ + 0 = βˆ’π‘–π‘–π‘˜π‘˜πΏπΏ β€’ 0 +𝐴𝐴𝑃𝑃 + 𝐡𝐡𝑁𝑁 βˆ’π‘‡π‘‡π‘€π‘€ = 0 β€’ 0 + π‘–π‘–π‘žπ‘žπ΄π΄π‘ƒπ‘ƒβˆ’ π‘–π‘–π‘žπ‘žπ΅π΅π‘π‘βˆ’ π‘–π‘–π‘˜π‘˜(𝑇𝑇𝑀𝑀) = 0

Make a Matrices based on coefficient :

00

0 00 0

00

0 00 0

M N P R LikM iqN iqP A ikL

P N M BiqP iqN ikM T

M N P LikM iqN iqP ikL

P N MiqP iqN ikM

βˆ’ βˆ’ βˆ’ βˆ’ = βˆ’ βˆ’ βˆ’

βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’

khari
Sticky Note
For simplification, I use 1 matices with disabled coefficient (R,A,B,T)
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Quantum Mechanics

M. Khari Secario Page 5

After a β€œfew” steps, we’ll get :

1 0

0 1 0

0 0 1

0 0 0

N P LM M M

M PN P N P

M PikM iqPN P N P

Ξ³ Ξ²Ξ²

Ξ³ Ξ³Ξ± Ξ²Ξ±Ξ²

Ξ³ Ξ³

βˆ’ βˆ’

βˆ’ βˆ’ βˆ’ βˆ’

βˆ’ + βˆ’ +

βˆ’ βˆ’

So,

οΏ½βˆ’π‘–π‘–π‘˜π‘˜π‘€π‘€ +𝛼𝛼𝑀𝑀

𝑁𝑁 βˆ’π‘ƒπ‘ƒπ›Ύπ›ΎοΏ½ 𝑇𝑇 = βˆ’π‘–π‘–π‘žπ‘žπ‘ƒπ‘ƒπ›½π›½+

𝑃𝑃𝛼𝛼𝛽𝛽𝑁𝑁 βˆ’π‘ƒπ‘ƒπ›Ύπ›Ύ

With :

β€’ 𝛾𝛾 = π‘–π‘–π‘žπ‘žπ‘ƒπ‘ƒβˆ’π‘–π‘–π‘˜π‘˜π‘ƒπ‘ƒβˆ’π‘–π‘–π‘žπ‘žπ‘π‘βˆ’π‘–π‘–π‘˜π‘˜π‘π‘

β€’ 𝛽𝛽 = 2π‘–π‘–π‘˜π‘˜πΏπΏβˆ’π‘–π‘–π‘žπ‘žπ‘π‘βˆ’π‘–π‘–π‘˜π‘˜π‘π‘

β€’ 𝛼𝛼 = βˆ’π‘–π‘–π‘žπ‘žπ‘π‘ βˆ’ π‘–π‘–π‘žπ‘žπ‘ƒπ‘ƒπ›Ύπ›Ύ

After a β€œlittle” solving :

𝑇𝑇 = π‘’π‘’βˆ’2π‘–π‘–π‘˜π‘˜π‘Žπ‘Ž 2π‘˜π‘˜πœ…πœ…2π‘˜π‘˜πœ…πœ… cosh 2πœ…πœ…π‘Žπ‘Ž + 𝑖𝑖(π‘˜π‘˜2 βˆ’ πœ…πœ…2) sinh 2πœ…πœ…π‘Žπ‘Ž

|𝑇𝑇|2 =(2π‘˜π‘˜πœ…πœ…)2

(π‘˜π‘˜2 βˆ’πœ…πœ…2)2 sinh2 2πœ…πœ…π‘Žπ‘Ž + (2π‘˜π‘˜πœ…πœ…)2

khari
Sticky Note
This result came from Gauss Elimination
khari
Sticky Note
Look at last line matrices. It's a value for coefficient T
khari
Sticky Note
It's a probability for transmitted wave can occur at region 3

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