Number of electron of a Hydrogen atom (molecule)

[torehan]

Hi,
If the normalized 1D wave-function of hydrogen atom for n=1, l=0, m_l=0;

$$\psi_{1s}(x)=\frac{1}{\sqrt{\pi} a_{0}^{3/2}}e^{-x/a_{0}}$$

and probability distribution of wave-function,

$$\mid\psi_{1s}(x)^2\mid$$

so integration of rho over all x should give the number of electrons which is equal to 1

$$\int{\mid\psi_{1s}(x)^2\mid}dx=1$$

theory is simple and understandable. But when I apply a numerical procedure to this mathematical aspect I couldn't get the right result.
With a very simple computer program,hoping to find number of electrons, I use the easiest integration technique as converting integral to a discrete sum of each value of the charge density function multiplied by dx which is increment of variable x,

a0=0.53;
x=-10*a0:0.001:10*a0; 
R=-0.36;
NORM=1/sqrt(%pi*a0^3);
dx=20*a0/size(x,2);
nelect=0;

psi1s=NORM*exp(-abs((x-R))/a0);
rho1s=psi1s^2;
for i=1:size(x,2)
  nelect=nelect+rho(i)*dx;
end
return nelect;

> nelect= 1.1331234

But this doesn't return the correct value of nelect.

Am I making a fundamental mistake which I couldn't see right now?

Thanks

 

[alxm]

If the normalized 1D wave-function of hydrogen atom for n=1, l=0, m_l=0;

Well, there's your problem: There's no 1-dimensional wave function here, it's a 3-dimensional function in spherical coordinates, where your 'x' is the radial distance (usually denoted r). It's just that you don't have an angle-dependency for s-type functions..

So the normalization is:
$$\int |\psi(\vec{x})|^2 dV = \int_0^{2\pi}\int_{-\pi}^{\pi}\int_0^\infty |\psi(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 1$$

If you don't have an angle-dependency then you just integrate over the surface of a sphere, so:
$$\int_0^{2\pi}\int_{-\pi}^{\pi}\int_0^\infty |\psi(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 2\pi \int_0^\infty r^2 |\psi(r)|^2 dr = 1$$

 

[phyzguy]

That last line should read:

$$\int_0^{2\pi}\int_{-\pi}^{\pi}\int_0^\infty |\psi(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 4\pi \int_0^\infty r^2 |\psi(r)|^2 dr = 1$$

 

[alxm]

(1-(-1))*2*pi = 4*pi.. you're right! :)

 

[torehan]

Thanks for the replies.

But I'm not sure about the boundaries of the integration with respect to $$r$$ and $$\varphi$$

I think the integration over all space must be like,

$$\int_0^{2\pi}\int_{0}^{\pi}\int_{-\infty}^\infty |\psi(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 2\pi \int_{-\infty}^\infty r^2 |\psi_{1s}|^2 dr = 1$$

Please correct me if I'm wrong.

 

[torehan]

Assuming that

$$\int_0^{2\pi}\int_{0}^{\pi}\int_{-\infty}^\infty |\psi(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 2\pi \int_{-\infty}^\infty r^2 |\psi_{1s}|^2 dr = 1$$

is correct,
If I use a Gaussian as a hydrogen 1s wave-function what would be the normalization constant?

(And here is the question one. Which one is the correct form for a Gaussian wave-function? 2nd. one?)
1.
$$\psi_{1s}(r)=Ne^{-r/a_{0}}$$
2.
$$\psi_{1s}(r)=Ne^{-r^2/a_{0}}$$
$$\int\mid\psi_{1s}(r)\mid^2dV=N^2\int_{-\infty}^{\infty}e^{r^2/a_{0}^2}dV=1$$
$$\int_0^{2\pi}\int_{0}^{\pi}\int_{-\infty}^\infty |\psi_{1s}(r,\theta,\varphi)|^2 r^2 sin(\theta) dr d\theta d\varphi = 2\pi \int_{-\infty}^\infty r^2 |\psi_{1s}(r)|^2 dr = 1$$

$$2\pi N^2\int_{-\infty}^{\infty}r^2dre^{r^2/a0^2}=1$$

where $$a=1/a_0^2$$
$$\int_{-\infty}^{\infty}r^2dre^{-ar^2}= \frac{1}{2}\sqrt{\frac{\pi}{a^3}}$$$$\frac{1}{2}\sqrt{\frac{\pi}{a^3}}=\frac{1}{2}\sqrt{\pi}a_0^3$$

$$N^2=\frac{1}{a_0^3 \pi^{3/2}}$$

I think derivation of normalization constant like that. Can you see anything nonsense?Even I integrate normalized wave-function I couldn't get the result 1. ouf!

 

[phyzguy]

You're right that theta goes from 0 to pi. I missed that before. But r goes from 0 to infinity. A radius by definition must be greater than zero. So the angular integrals give:
$$\int_0^{2\pi}\int_0^\pi sin(\theta)d\theta d\phi = 2\pi (-cos(\pi)+cos(0)) = 4\pi$$
Of course we know that the surface area of a sphere is 4 pi r^2, so the angular integral mustgive 4 pi. As far as the radial integral, the radial wavefunction goes like exp(-r/a0), not exp(-r^2/a0). Try this reference:

http://en.wikipedia.org/wiki/Hydrogen_atom

 

[phyzguy]

Trying again to get the LateX right. Note that:$$(e^{-\frac{r}{a_0}})^2 = e^{-\frac{2 r}{a_0}}, NOT e^{-\frac{r^2}{a_0^2}}$$

 

[torehan]

I have problems on getting used to latex.

Thanks