# Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?

Apologies if this is stating the obvious, but I'm a non-physicist trying to understand Griffiths' discussion of the hydrogen atom in chapter 4 of Introduction to Quantum Mechanics. The wave equation for the ground state (I believe) is:$$\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$$

where $a$ is the Bohr radius $0.529\times10^{-10}$m. If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius $x$? I've done this for $x=1\textrm{ m}$ and $x=\infty$ and got the answer $1$ (I guess a two metre diameter sphere is pretty big compared to a hydrogen atom). For $x=9\times10^{-11}\textrm{ m}$, the answer is $0.66077$. Is this interpretation correct?

-
–  Kitchi Jan 21 '13 at 19:41
You're interpretation is exactly right. The integral of $|\psi (x)|^2$ from $0$ to $x$ will give you the probability that the particle is in that region.The reason you get a probability of $1$ is because $a$ is really really small, so the exponential falls of quite quickly. If you calculate with enough precision, you'll find the answer to be just slightly less that one. –  Kitchi Jan 21 '13 at 19:44

No, you are wrong. Particular for the following statement:

If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius x?

The probability density at any points is given by $|\psi(r,\theta,\phi)|^2$. Certainly, the probability is for any region $V$ is given by $$P=\int_V|\psi|^2 dxdydz$$ However, if you are considering a sphere of radius $R$, you should change the integral into the spherical coordinate using the relation $dxdydz=r^2 \sin(\theta)drd\theta d\phi$ as follow $$P(R)=\int_V|\psi|^2 dxdydz=\int_0^R \int_0^\pi \int_0^{2\pi} |\psi|^2 r^2 \sin(\theta) dr d\theta d\phi = 4\pi \int_0^R r^2|\psi|^2 dr$$ Note that there is an extra $r^2$ in the equation and an extra constant $4 \pi$. Simply integrate the $|\psi|^2$ from $0$ to $R$ is wrong. It is very easy to check that this integral lead to the normalization condition of the wavefunction, i.e. $P(R\to\infty)=1$.

-
From his question it is not clear whether he integrates over the volume $dV$ or over the radius $dr$. We suppose he does it correctly, following the textbook. –  Vladimir Kalitvianski Jan 22 '13 at 10:02
I cut a few corners in my question and have now been found out. My original calculation of the probability $P$ of finding the electron in a sphere radius $x$ was along the lines of $$P=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{x}\left|\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}\right|^{2}\left(r^{2}\sin\theta drd\theta d\phi\right)=\frac{4\pi}{\pi a^{3}}\int_{0}^{x}r^{2}e^{-2r/a}dr.$$ continued ... –  Peter4075 Jan 22 '13 at 10:36
The $\pi$ 's cancelled, so I didn't mention them. Nor did I mention that I meant the complex square of the wave function. If $x=\infty$ then $P=1$, and if $x=0.00000000009$ then $P=0.66077$. I now know my interpretation was right, but my maths was a bit ragged. Thanks. –  Peter4075 Jan 22 '13 at 10:38
@Peter4075 Remember to mention the unit you use, it would be make things clear :) –  hwlau Jan 22 '13 at 16:36
Detail is important, that's what science is about. The need for angular integration is shown in the normalisation constant of the wavefunction. This is why I gave you +1. –  JKL Feb 3 '13 at 0:51

Yes, you are practically right. The only subtlety is that your wave function argument $\mathbf{r}$ is a relative distance between the nucleus and the electron. Nucleus itself "turns around" the atomic center of mass, but with a shorter radius $(m_e/M_N)\cdot a$.

-
Well actually you should integrate $|\psi|^2$, not $\psi$. –  Bzazz Jan 21 '13 at 22:42
@Bzazz: In the original post he integrates a square because $\psi$ is real function in his case. –  Vladimir Kalitvianski Jan 22 '13 at 8:15

In my opinion, if you want to know the probability of a particle being in a sphere, there should be an integration over a sphere, therefore I would add an additional factor of $4\pi$.

The thing I don't understand is the normalization of the wave function, it doesn't yield 1 if I integrate it over whole space.

My last issue is just a small one, I wouldn't call this a wave equation, it is already a wave function, the corresponding wave equation is the Schrödinger equation.

-