Bound state of Hydrogen atom at large $r$

Question

When the radial equation of SE is solved for Hydrogen atom, to see the asymptotic behavior, we assume $r$ tends to infinity. The differential equation we are left with is:

$$ d^2U/dr^2 = -\frac{2mE}{\hbar^2} $$ where $U = R(r)/r$. If we assume $E$ to be positive, we get an oscillating term which is reasonable. But if we assume $E$ to be negative, we get an exponentially decaying term which I don't get. Here we've ignored the potential so what is $E$ negative relative to? How could the wave function possibly decay radially, if we are considering a region where potential is negligible and essentially free?

Answer 1

If there is no potential then there can be no bound state and the energy is positive. Only for a bound state the Schrödinger energy is negative. Therefore your choice of negative energy is inconsistent with your assumption of no potential.

Answer 2

The properties of a physical system depends on the boundary condition as well as the governing equations. All eigen vlaues are results from a giving boundary condition.

Therefore, it is to be kept in mind that a bound state is sujected to the boundary condition: $$ \lim_{r\to\infty} \Psi(\vec{r}) = 0. $$

Without this condition, the wavefuncion would not be able to be normalized. An oscilatory tail (without decaying) is certainly not allowed for bound state.

Answer 3

I'm not sure where your DE comes from. The full, radial part of the hydrogenic TISE is:

$$\frac{\rm d}{{\rm d}r}\left(r^2\frac{{\rm d}R}{{\rm d}r}\right)+\frac{2\mu r^2}{\hbar^2}\left(E+\frac{Ze^2}{4\pi\epsilon_0r}\right)R-l(l+1)R=0$$ With $l=0,1,2,...$

Developed: $$r^2\frac{{\rm d}^2R}{{\rm d}r^2}+2r\frac{{\rm d}R}{{\rm d}r}+\frac{2\mu r^2}{\hbar^2}\left(E+\frac{Ze^2}{4\pi\epsilon_0r}\right)R-l(l+1)R=0$$ and divided by $r^2$: $$\frac{{\rm d}^2R}{{\rm d}r^2}+\bbox[lightblue]{\frac{2}{r}\frac{{\rm d}R}{{\rm d}r}}+\left(\frac{2\mu}{\hbar^2}\left(E+\bbox[lightblue]{\frac{Ze^2}{4\pi\epsilon_0r}}\right)\bbox[lightblue]{-\frac{l\left(l+1\right)}{r^2}}\right)R=0$$ We can't solve this straight away, but for very large $r$, the highlighted terms are forced to zero because they go reciprocal with $r$.

That leaves us with an asymptotic equation: $$\frac{{\rm d}^2R_{\infty}}{{\rm d}r^2}+\frac{2\mu E}{\hbar^2}R_{\infty}=0$$

$$R_{\infty}=c_3\exp{-\left(\sqrt{-\frac{2\mu E}{\hbar^2}}r\right)}$$

Here we've taken $E\to 0$ for $r \to +\infty$ and $E<0$ (so there is no imaginary part in the exponent).

This results in the radial solution: $$R_{n,l}(r)=R_{\infty}(r)b_0\exp-{\left(\frac{\mu Ze^2r}{2\pi\epsilon_0\hbar^2n}\right)}$$

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Re. your DE:

$$d^2U/dr^2 = -\frac{2mE}{\hbar^2}$$

with $U(r)=\frac{R(r)}{r}$

solves to:

$$U(r)=-\frac{2mE}{\hbar^2}r^2+c_1 r+c_2$$ So that

$$R(r)=-\frac{2mE}{\hbar^2}r^3+c_1 r^2+c_2 r$$

which does not go asymptotically to $0$ for $r \to +\infty$.