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Degeneracy of energy eigenstates when $E > V$

Physics Asked by Jaime_mc2 on February 10, 2021

Reading my text book on quantum physics, I found the following statement:

Let’s suppose we have a short-scale force, so we have a potential energy such that
$$
V(x) = V_- quad text{if} quad x ll 0
V(x) = V_+ quadtext{if}quad x gg 0
$$

with $V_+ ge V_-$, the wavefunction on these regions is
$$
Phi_E = A e^{ik_-x} + B e^{-ik_-x} quadtext{if}quad x ll 0 qquad text{with}quad k_- = sqrt{2m(E-V_-)}/hbar
Phi_E = C e^{ik_+x} + D e^{-ik_+x} quadtext{if}quad x gg 0 qquad text{with}quad k_+ = sqrt{2m(E-V_+)}/hbar
$$

Since there exist a linear relationship between $Phi_E(xll 0)$ and $Phi_E(xgg 0)$, the coefficients also have a linear relationship
$$
left(begin{array}{c}ABend{array}right) = left(begin{array}{cc}t_{11}&t_{12}t_{21}&t_{22}end{array}right) left(begin{array}{c}CDend{array}right)
$$

and only two coefficients are indepent. Thus, all the energies $E>V_+$ have a multiplicity of two.

I understand that the degeneracy must be two because of the two independent coefficients, but I don’t see where the linear relationship between $Phi_E(xll 0)$ and $Phi_E(xgg 0)$ comes from to explain this result. Is there any easy proof of this?

One Answer

  1. The point is that the 1D TISE for fixed energy $E>V_{pm}$ is a 2nd-order linear homogeneous ODE. We assume that the solutions are globally well-defined on the whole real line $mathbb{R}$. So the general solution $psi$ is a complex linear combination of 2 linearly independent solutions $psi_{1/2}$.

  2. We next choose $psi_{1/2}$ such that they satisfy the boundary condition $$ lim_{xto infty}e^{(-1)^a ik_+ x}psi_a(x)~=~1, qquad a~in~{1,2}.$$ Note that $psi_1$ corresponds to $(C,D)=(1,0)$ while $psi_2$ corresponds to $(C,D)=(0,1)$.

  3. We finally define the transfer matrix elements as $$psi_a(x)~sim~t_{1a}e^{ ik_- x}+t_{2a}e^{ -ik_- x} quad{rm for}quad xto -infty.$$ Note that $psi_1$ corresponds to $(A,B)=(t_{11},t_{21})$ while $psi_2$ corresponds to $(A,B)=(t_{12},t_{22})$.

  4. This explains the linear relationship between $(C,D)$ and $(A,B)$. See also e.g. my related Phys.SE answer here.

Answered by Qmechanic on February 10, 2021

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