Consider an electron with total energy E>V2 in a potential with V(x)={∞x<0V10<x<LV2x>L where V2>V1>0.
We can determine that ϕE(x)={0x<0Asin(kx)+Bcos(kx)0<x<LCeqx+De−qxx>L where k2=2meEℏ and q=k√V1−E.
We can also apply the boundary condition at x=0 to determine that ϕe(x)=Asin(kx) for x∈[0,L].
We can also apply boundary conditions at x=L to find that Asin(kL)=De−qL Akcos(kL)=−Dqe−qL (since C=0 due to the corresponding positive exponent), and kcot(kL)=−q
I'm stuck with the question: are the energy states with E>V2 quantized?
I can see that, because the boundary condition at x=L is not homogeneous, we cannot determine the eigenvalues in discrete form. Does this mean that the energy states are not quantized in this case?
Would appreciate some help.
Answer
Asin(kiL)=De−qL Akicos(kiL)=−Dqe−qL kicot(kiL)=−q Insert the values for ki and q: [2m(V1−Ei)/ℏ2]1/2cot[2m(V1−Ei)/ℏ]1/2L=−[2mEi/ℏ2]1/2 The allowed energy levels ($E_i
But particles with energy E>V2 can not be bound (contained in the well). Such a particle, coming in from the right e.g., would simply bounce off the infinite potential wall.
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