Boundary conditions for two dimensional problems in Quantum mechanics

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SUMMARY

The discussion focuses on defining boundary conditions for two-dimensional problems in quantum mechanics (QM). It specifies the wavefunction within the domain as ψ(𝑟) = e^{i𝑘_i·𝑟} + r e^{-i𝑘_f·𝑟} and outside the domain as ψ(𝑟) = e^{i𝐾_s S - 𝐾_n N}. The boundary conditions require continuity of the wavefunction and its derivatives along both normal and tangential directions. Specifically, the conditions are expressed as ψ(𝑅) = e^{i𝐾_s S - 𝐾_n N} and involve derivatives with respect to normal and tangential coordinates.

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I am stuck at the problems of Boundary conditions for two dimensional problem in QM.
iIf we have a two-dimensional domain,
along the boundary, we can define two directions, one is tangential, the other is normal,
assuming that there is no current flowing in and out along the normal direction.
How can we define the boundary conditions?
To be specific, we have the following wavefunction in the domain
[tex]\psi({\vec r})=e^{i{\vec k}_i\cdot {\vec r}}+r e^{-i{\vec k}_f\cdot {\vec r}}[/tex]
while outside the domain, we have
[tex]\psi({\vec r})=e^{i{K}_s S-K_n N}[/tex]
[tex]K_s, K_n[/tex] is the tangential and normal components of the momentum [tex]{\vec K}[/tex] outside the domain.
S,N are the coordinates of the position vector [tex]{\vec R}[/tex]
 
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outside the domain.The boundary conditions for a two-dimensional problem in quantum mechanics are typically expressed as a combination of continuity of the wavefunction at the boundary and a condition on the derivatives of the wavefunction along the normal and tangential directions. For the example given, the boundary conditions can be written as:In the domain:\psi({\vec r})=e^{i{\vec k}_i\cdot {\vec r}}+r e^{-i{\vec k}_f\cdot {\vec r}}At the boundary:\psi({\vec R})=e^{i{K}_s S-K_n N}Continuity of the wavefunction:\psi(R)=e^{i{\vec k}_i\cdot {\vec R}}+r e^{-i{\vec k}_f\cdot {\vec R}} = e^{i{K}_s S-K_n N}Derivative along the normal direction:\frac{\partial}{\partial N}\psi(R) = -K_n e^{i{K}_s S-K_n N}Derivative along the tangential direction:\frac{\partial}{\partial S}\psi(R) = K_s e^{i{K}_s S-K_n N}
 

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