- #1
amjad-sh
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We know that the solution of the schroedinger equation for a free particle in position representation is a plane wave:
[itex]\psi(x)=Ae^{ikx}+Be^{-ikx}[/itex] which means that the particle has a probability to move either to the left or to the right.
"Now let's take the potential step problem.
V(x)=0 for x<0
V(x)=V for x>0
The dynamics of the particle is regulated by the schroedinger equation which is given in those two regions by:
[itex](\frac{d^{2}}{dx^{2}}+k_{1}^{2})\psi_{1}(x)=0[/itex] (x<0)
[itex](\frac{d^{2}}{dx^{2}}+k_{2}^{2})\psi_{2}(x)=0[/itex] (x>0)
where [itex]k_{1}^{2}=2mE/\hbar^{2}[/itex] and [itex]K_{2}^{2}=2m(E-V)/\hbar^{2}[/itex]
The most general solutions of these two equations are plane waves:
[itex]\psi_{1}(x)=Ae^{ik_{1}x}+Be^{-ik_{1}x}[/itex] (x<0)
[itex]\psi_{2}(x)=Ce^{ik_{2}x}+De^{-ik_{2}x}[/itex] (x>0)
Where [itex]Ae^{ik_{1}x}[/itex]and [itex]Ce^{ik_{2}x}[/itex]represents waves moving in the positive x direction, but [itex]Be^{-ik_{1}x}[/itex]and[itex]De^{-ik_{2}x}[/itex] corresponds to waves moving in the negative x direction.We are interested in the case where the particles are initially incident on the potential step from the left:they can be reflected and transmitted at x=0.Since no wave is reflected from the region x>0 to the left,the constant D must vanish"(zetteli text).
I have two questions here:
1)Why we are talking about reflection here? I mean even in case where there is no potential
like in the free particle case, its wave function is divided into two components one that moves to the right and the other moves to the left.
2)"constant D must vanish" why there is no reflected wave in the region x>0?
Thanks!
[itex]\psi(x)=Ae^{ikx}+Be^{-ikx}[/itex] which means that the particle has a probability to move either to the left or to the right.
"Now let's take the potential step problem.
V(x)=0 for x<0
V(x)=V for x>0
The dynamics of the particle is regulated by the schroedinger equation which is given in those two regions by:
[itex](\frac{d^{2}}{dx^{2}}+k_{1}^{2})\psi_{1}(x)=0[/itex] (x<0)
[itex](\frac{d^{2}}{dx^{2}}+k_{2}^{2})\psi_{2}(x)=0[/itex] (x>0)
where [itex]k_{1}^{2}=2mE/\hbar^{2}[/itex] and [itex]K_{2}^{2}=2m(E-V)/\hbar^{2}[/itex]
The most general solutions of these two equations are plane waves:
[itex]\psi_{1}(x)=Ae^{ik_{1}x}+Be^{-ik_{1}x}[/itex] (x<0)
[itex]\psi_{2}(x)=Ce^{ik_{2}x}+De^{-ik_{2}x}[/itex] (x>0)
Where [itex]Ae^{ik_{1}x}[/itex]and [itex]Ce^{ik_{2}x}[/itex]represents waves moving in the positive x direction, but [itex]Be^{-ik_{1}x}[/itex]and[itex]De^{-ik_{2}x}[/itex] corresponds to waves moving in the negative x direction.We are interested in the case where the particles are initially incident on the potential step from the left:they can be reflected and transmitted at x=0.Since no wave is reflected from the region x>0 to the left,the constant D must vanish"(zetteli text).
I have two questions here:
1)Why we are talking about reflection here? I mean even in case where there is no potential
like in the free particle case, its wave function is divided into two components one that moves to the right and the other moves to the left.
2)"constant D must vanish" why there is no reflected wave in the region x>0?
Thanks!