Why must A2 be zero in Schaum's Quantum Mechanics?

[atomqwerty]

Hello,
In http://img96.imageshack.us/img96/2825/sinttulobdd.png" from Schaum's Quantum Mechanics, I don't get why they say that A₂ (with ' ) must be zero; I think in that case $\Phi_{II}(x)$ will diverge if x=+infinite, and we don't want that function to diverge, do we?

Thanks!

 

[vanhees71]

With this step you simply fulfill the boundary condition that there should be incoming particles only from the left and not from the right.

 

[atomqwerty]

With this step you simply fulfill the boundary condition that there should be incoming particles only from the left and not from the right.

Yes, but it you put the condition that when x->infinite, the wave function must be zero, the it doesn't match!
Maybe this condition that I say is just when E<V...

 

[vanhees71]

If you are in the continuous part of the spectrum your "eigenstates" are not Hilbert-space vectors but generalized eigenstates. They belong to the dual of the nuclear space in the sense of the G'elfand triple (="rigged Hilbert space").

The most simple example are the "eigenstates" of the momentum operator. These are (in position representation) just the plane waves (here for simplicity written for the one-dimensional motion of a particle)

$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p).$$

This is not a square-integrable function. It can not be normalized to 1 in the usual sense but only to a $\delta$ distribution, according to

$$\langle p_1|p_2 \rangle = \int_{-\infty}^{\infty} \mathrm{d} x \; u_{p_1}^*(x) u_{p_2}(x)=\delta(p_1-p_2).$$