"Step function potential scattering"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) (→메타데이터) |
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(사용자 3명의 중간 판 19개는 보이지 않습니다) | |||
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− | + | ==introduction== | |
− | * [[ | + | * [[Schrodinger equation]] |
+ | :<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math> | ||
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− | + | ||
− | + | ==step potential 1:== | |
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− | + | * From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p | |
− | + | * Let the potential is given by | |
+ | :<math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math> | ||
+ | * solution: | ||
+ | <math> | ||
+ | \begin{cases} | ||
+ | \psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\ | ||
+ | \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ | ||
+ | \psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a | ||
+ | \end{cases} | ||
+ | </math> where $k_0=\sqrt{2m E/\hbar^{2}}$ if $x<0$ or $x>a$ and $k_1=\sqrt{2m (E-V_0)/\hbar^{2}}$ if $0<x<a$ | ||
+ | * we impose two conditions on the wave function | ||
+ | ** the wave function be continuous in the origin | ||
+ | ** the first derivative of the wave function be continuous in the origin | ||
+ | * the coefficient must satisfy: | ||
+ | <math> | ||
+ | \begin{cases} | ||
+ | A_r + A_l - B_r - B_l = 0 \\ | ||
+ | ik_0(A_r - A_l) = ik_1(B_r - B_l) \\ | ||
+ | B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\ | ||
+ | ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0}) | ||
+ | \end{cases} | ||
+ | </math> | ||
− | + | ||
− | + | ===scattering=== | |
− | * special case of scattering problem <math>A_r=1, | + | * special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math> |
− | * wave function | + | * wave function |
+ | :<math> | ||
+ | \begin{cases} | ||
+ | \psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\ | ||
+ | \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ | ||
+ | \psi_R(x)= t e^{i k_0 x} \quad x>a | ||
+ | \end{cases} | ||
+ | </math> | ||
+ | * coefficient satisfy | ||
+ | :<math> | ||
+ | \begin{cases} | ||
+ | 1 + r - B_r - B_l = 0 \\ | ||
+ | ik_0 (1-r)= ik_1(B_r - B_l) \\ | ||
+ | B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\ | ||
+ | ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0}) | ||
+ | \end{cases} | ||
+ | </math> | ||
− | + | * <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math> | |
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− | + | ==step potential 2 : not corrected== | |
− | < | + | * Let the potential is given by:<math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math> |
− | * | + | * solution of the stationary S |
+ | :<math> | ||
+ | \psi(x) = | ||
+ | \begin{cases} | ||
+ | \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ | ||
+ | \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0, | ||
+ | \end{cases} | ||
+ | </math> | ||
+ | * we impose two conditions on the wave function | ||
+ | ** the wave function be continuous in the origin | ||
+ | ** integrate the Schrödinger equation around x = 0, over an interval [−ε, +ε] and In the limit as ε → 0, the right-hand side of this equation vanishes; the left-hand side becomes | ||
+ | * first condition:<math>\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l</math>:<math>A_r + A_l - B_r - B_l = 0</math> | ||
+ | * second condition:<math> -\frac{\hbar^2}{2 m} \int_{-\epsilon}^{+\epsilon} \psi''(x) \,dx + \int_{-\epsilon}^{+\epsilon} V(x)\psi(x) \,dx = E \int_{-\epsilon}^{+\epsilon} \psi(x) \,dx</math> LHS becomes <math>-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)</math> RHS becomes 0:<math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math> | ||
+ | * the coefficient must satisfy | ||
+ | :<math> | ||
+ | \begin{cases} | ||
+ | A_r + A_l - B_r - B_l = 0 \\ | ||
+ | -A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l) | ||
+ | \end{cases} | ||
+ | </math> | ||
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− | < | + | ===delta potential scattering=== |
+ | * special case of scattering problem <math>A_r=1, A_l=r, B_r=t , B_l = 0</math> | ||
+ | * wave function:<math>\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}</math> | ||
+ | * <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math> | ||
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− | + | ==history== | |
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− | + | * http://www.google.com/search?hl=en&tbs=tl:1&q= | |
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− | + | ==related items== | |
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− | + | ==encyclopedia== | |
− | + | * http://en.wikipedia.org/wiki/Rectangular_potential_barrier | |
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− | + | [[분류:개인노트]] | |
+ | [[분류:physics]] | ||
+ | [[분류:math and physics]] | ||
+ | [[분류:migrate]] | ||
− | * [ | + | ==메타데이터== |
− | * [ | + | ===위키데이터=== |
− | + | * ID : [https://www.wikidata.org/wiki/Q2279049 Q2279049] | |
− | + | ===Spacy 패턴 목록=== | |
+ | * [{'LOWER': 'solution'}, {'LOWER': 'of'}, {'LOWER': 'schrödinger'}, {'LOWER': 'equation'}, {'LOWER': 'for'}, {'LOWER': 'a'}, {'LOWER': 'step'}, {'LEMMA': 'potential'}] |
2021년 2월 17일 (수) 01:04 기준 최신판
introduction
\[E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi\]
step potential 1:
- From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
- Let the potential is given by
\[\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}\]
- solution\[ \begin{cases} \psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a \end{cases} \] where $k_0=\sqrt{2m E/\hbar^{2}}$ if $x<0$ or $x>a$ and $k_1=\sqrt{2m (E-V_0)/\hbar^{2}}$ if $0<x<a$
- we impose two conditions on the wave function
- the wave function be continuous in the origin
- the first derivative of the wave function be continuous in the origin
- the coefficient must satisfy\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ ik_0(A_r - A_l) = ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0}) \end{cases} \]
scattering
- special case of scattering problem \(A_r=1, A_l=r, C_r=t , C_l = 0\)
- wave function
\[ \begin{cases} \psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= t e^{i k_0 x} \quad x>a \end{cases} \]
- coefficient satisfy
\[ \begin{cases} 1 + r - B_r - B_l = 0 \\ ik_0 (1-r)= ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0}) \end{cases} \]
- \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]
step potential 2 : not corrected
- Let the potential is given by\[\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}\]
- solution of the stationary S
\[ \psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0, \end{cases} \]
- we impose two conditions on the wave function
- the wave function be continuous in the origin
- integrate the Schrödinger equation around x = 0, over an interval [−ε, +ε] and In the limit as ε → 0, the right-hand side of this equation vanishes; the left-hand side becomes
- first condition\[\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l\]\[A_r + A_l - B_r - B_l = 0\]
- second condition\[ -\frac{\hbar^2}{2 m} \int_{-\epsilon}^{+\epsilon} \psi''(x) \,dx + \int_{-\epsilon}^{+\epsilon} V(x)\psi(x) \,dx = E \int_{-\epsilon}^{+\epsilon} \psi(x) \,dx\] LHS becomes \(-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)\) RHS becomes 0\[-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)\]
- the coefficient must satisfy
\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ -A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l) \end{cases} \]
delta potential scattering
- special case of scattering problem \(A_r=1, A_l=r, B_r=t , B_l = 0\)
- wave function\[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\]
- \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]
history
encyclopedia
메타데이터
위키데이터
- ID : Q2279049
Spacy 패턴 목록
- [{'LOWER': 'solution'}, {'LOWER': 'of'}, {'LOWER': 'schrödinger'}, {'LOWER': 'equation'}, {'LOWER': 'for'}, {'LOWER': 'a'}, {'LOWER': 'step'}, {'LEMMA': 'potential'}]