Delta fuction potential general solution

[Danny Boy]

Hi, in the book 'Introduction to Quantum Mechanics' by Griffiths, on page 71 in the section 'The Delta-Function Potential' he states that the general solution to time independent Schrödinger Equation is $$\psi(x) = Ae^{-\kappa x} + B e^{\kappa x}$$

he then notes that the first term blows up as $$x \to -\infty,$$ so we must choose $$A=0.$$ Why is it that we are rejecting a wave function that goes to infinity? Is it simply because we are looking for normalizable solutions?

Thanks.

 

[vanhees71]

For $x<0$ you must choose $A=0$ and for $x>0$ you must have $B=0$, if $\kappa>0$. That's indeed, because you want to have bound states in this case, leading to normalizable wave functions. Note that there may be solutions with $\kappa \in \mathrm{i} \mathbb{R}$, leading to scattering states with energy eigenvalues in the continuum, and these wave functions are generalized eigenfunctions that are "normalizable to a $\delta$ distribution" only.

 

[Danny Boy]

For $x<0$ you must choose $A=0$ and for $x>0$ you must have $B=0$, if $\kappa>0$. That's indeed, because you want to have bound states in this case, leading to normalizable wave functions. Note that there may be solutions with $\kappa \in \mathrm{i} \mathbb{R}$, leading to scattering states with energy eigenvalues in the continuum, and these wave functions are generalized eigenfunctions that are "normalizable to a $\delta$ distribution" only.

So my reasoning is correct then that we want normalizable solutions hence the requirement that the wave function is bounded?

 

[vanhees71]

If you restrict yourself to the bound-state solutions, it's correct.

 

[Danny Boy]

If you restrict yourself to the bound-state solutions, it's correct.

Okay thanks. One quick question, do you maybe know why the delta function is said to have units 1/length?

 

[Avalanche_]

Simple explanation: in electrodynamics when you want to write (for example) charge density function for point charge in the origin. You would write that as ρ $= q \delta (x) \delta (y) \delta (z)$ . You will write it in that way so that ∫ $\rho dV$ over the whole space gives you the result $q$. But ρ must have units coulomb per volume, but in your formula you have coulombs in $q$, so delta functions have units 1/length