Probability to find electron in ground state

[skrat]

Homework Statement

An electron in hydrogen atom can be described with normalized wavefunction $\psi (r,\vartheta ,\varphi )=(\frac{1}{64\pi r_B^3})^{\frac{1}{2}}exp(-\frac{r}{4r_B})$.

Calculate the probability that the electron is in ground state of hydrogen atom. How much is the energy of electron in state $\psi$?
Ground state is $\psi (r,\vartheta ,\varphi )=\frac{1}{\sqrt{4\pi }}\frac{2}{r_B^{3/2}}exp(-\frac{r}{r_B})$

Homework Equations

The Attempt at a Solution

I have no idea on how to even start with the fist part - calculating the probability...

But the second part should be relatively easy to do: $\left \langle \psi\left | \hat{H} \right | \psi \right \rangle=\left \langle \psi\left | \frac{\hbar^2}{2m}\hat{p}^2+\frac{\hat{l}^2}{2m\left \langle r \right \rangle^2}-\frac{e^2}{4\pi \varepsilon _0}\frac{1}{\hat{r}} \right | \psi \right \rangle$

But still... what is the idea behind the first question?

 

[TSny]

Let $\psi$ be your given normalized wavefunction and let $\phi_1, \phi_2, \phi_3,...$ be the normalized wavefunctions for the energy eigenstates of the atom. So, $\phi_1$ is the ground state, $\phi_2$ is the first excited state, etc.

The set of eigenfunctions $\{\phi_1, \phi_2, \phi_3,...\}$ form a complete, orthonormal set of functions. So, you could imagine expanding $\psi$ in terms of the eigenfunctions:

$\psi = a_1\phi_1+a_2\phi_2+a_3\phi_3...$

How is the coefficient $a_1$ related to the probability that you are asked to find, and how can you calculate the value of $a_1$?

 

[skrat]

$\left | a_1 \right |^2$ is by definition the probability that the electron will be in state $\phi _1$.

The sum of those coefficients is therefore $\sum_{i=1}^{n}\left | a_i \right |=1$.

Now how to calculate them... hmm

If I am not wrong $a_n=\left \langle \psi _n \right | \psi \rangle$ or is it not?

 

[skrat]

$a_1=\left \langle \psi _1 \right | \psi \rangle=\int_{0}^{2\pi }d\varphi \int_{0}^{\pi }d\vartheta sin\vartheta \int_{0}^{\infty}(\frac{1}{64\pi r_B^3})^{\frac{1}{2}}exp(-\frac{r}{4r_B})\frac{1}{\sqrt{4\pi }}\frac{2}{r_B^{3/2}}exp(-\frac{r}{r_B})r^2dr$

$a_1=\frac{1}{2r_B^3}\int_{0}^{\infty}r^2exp(-\frac{5r}{4r_B})dr$

which gives me $a_1=\frac{16}{25}$

Therefore the probability should be 0.4096.

I hope.

 

[TSny]

$a_1=\left \langle \psi _1 \right | \psi \rangle=\int_{0}^{2\pi }d\varphi \int_{0}^{\pi }d\vartheta sin\vartheta \int_{0}^{\infty}(\frac{1}{64\pi r_B^3})^{\frac{1}{2}}exp(-\frac{r}{4r_B})\frac{1}{\sqrt{4\pi }}\frac{2}{r_B^{3/2}}exp(-\frac{r}{r_B})r^2dr$

$a_1=\frac{1}{2r_B^3}\int_{0}^{\infty}r^2exp(-\frac{5r}{4r_B})dr$

Looks good.

which gives me $a_1=\frac{16}{25}$

I get a different value.

 

[skrat]

Of course you do, $5^2$ was never $20$.

$a_1=64/125$.

Thank you TSny!

 

[TSny]

Of course you do, $5^2$ was never $20$.