Potential step at a Barrier in Quantum mechanics

  • #1
LagrangeEuler
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In quantum mechanics in books authors discuss only cases ##E<V_0## and ##E>V_0##, where ##E## is energy of the particle and ##V_0## is height of the barrier. Why not ##E=V_0##?

In that case for ##x<0##
[tex]\psi_1(x)=Ae^{ikx}+Be^{-ikx}[/tex]
and for ##x\geq 0##
[tex]\psi_2(x)=Cx+D [/tex]
and then from ##\psi_1(0)=\psi_2(0)## and ##\psi_1'(0)=\psi_2'(0)## I got a system
[tex]1+\frac{B}{A}=\frac{D}{A}[/tex]
[tex]ik-ik\frac{B}{A}=\frac{C}{A}[/tex]
and I can not solve this. Maybe is necessary to take ##C=0##? But why?
 

Answers and Replies

  • #2
jtbell
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Maybe is necessary to take ##C=0##? But why?
If ##C \ne 0##, what happens to ##\psi_2(x)## as ##x \rightarrow +\infty##?
 
  • #3
Dr_Nate
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I like to suggest people try virtual simulations in addition to going through the math. When you try the simulation at that link make sure to check the boxes for 'show energy levels' and 'show transmission and reflection probabilities'.
 
  • #4
LagrangeEuler
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Well in that case I will not have bound state. So wave function will go to either to ##+\infty## or ##-\infty##. But why I should have bound state?
 
  • #5
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in that case I will not have bound state. So wave function will go to either to ##- \infty## or ##\infty##.

If a wave function does not go to zero at ##- \infty## or ##\infty##, is it valid?
 
  • #6
jtbell
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But why I should have bound state?
We're not dealing with bound states here.

Did you study the derivation for ##E < V_0##? In that case, for ##x > 0##, the general solution is something like $$\psi_2(x) = Ce^{\kappa x} + De^{-\kappa x}$$ Here we must set ##C=0## so that ##\psi_2(x)## doesn't "blow up" as ##x \rightarrow +\infty##.
 
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  • #7
vanhees71
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If a wave function does not go to zero at ##- \infty## or ##\infty##, is it valid?
For a scattering state it may be valid with some qualifications. The most simple case is the free particle, i.e., ##\hat{H}=\hat{p^2}/(2m)##. The energy-eigensolutions are the plane waves ##u_p(x)=N_p \exp(\mathrm{i} p x)## with eigenvalues ##E(p)=p^2/(2m)## (using natural units with ##\hbar=1##).

These are not Hilbert-space vectors though since ##|u_p|^2## is not integrable over ##x \in \mathbb{R}##, but they are generalized functions (distributions) living the in the dual of the domain of the position and momentum operators. They are "normalizable to a ##\delta## distribution", i.e.,
$$\int_{\mathbb{R}} \mathrm{d} x u_{p'}^*(x) u_{p}(x)=2 \pi |N_p|^2 \delta(p-p').$$
The usual choice for the normalization constants thus is ##N_p=1/\sqrt{2 \pi}##.

This should hold true for all scattering states in problems with a potential too. Now think again about what this implies for your constant ##C##!

For a thorough discussion of how to treat the normalization of scattering states (particularly also in this one-dimensional case) look in the textbook by Messiah, who does this very carefully.
 
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