# Hydrogen Ground State

Consider the ground state of Hydrogen.Treat the nucleus as infinitely heavy. Show that the fine structure constant $\alpha =e^2/(4\pi\epsilon_0\hbar c)$ determines the typical speed of the electron (relative to $c$) $<v^2>^{1/2}\sim \alpha c$. To do so:

a) Explain why the mean momentum and position of the electron vanish $<\overrightarrow{p}>=<\overrightarrow{r}>=0$, implying that the mean square momentum $<\overrightarrow{p}^2>$ is directly determined by the uncertainty in momentum $|\Delta p|$, and similarly the mean inverse radius $<|r|^{-1}|$ will be comparable to the inverse of the position uncertainty $|\Delta r|$ up to some factor of order unity.

b) Using these estimates and the uncertainty principle, find the value of $|\Delta p|$ (or $|\Delta r|$ which minimizes the expectation energy, $<H>$.

c) Show that your results imply that the typical electron speed, relative to $c$, is given by the fine structure constant (up to some factor of order unity).

d) How does the binding energy of hydrogen compare to the rest energy of the electron?

e) the above analysis uses a non-relativistic treatment of the dynamics of the electron. How large are the relativistic effects? In other words, how should the size of neglected relativistic corrections compare to the binding energy, or to the electron rest energy?

f) How do the above results change if the proton is replaced by a nucleus with charge $|Z|e$ (but there is still just a single electron)?

a)

The physical explanation of why $<r ⃗>$ vanishes is that the electrons is moving in a circle (in the ground state) so that the average of position vector on a complete circle is zero.$<r ⃗>=0$. For the average momentum from the its definition $<p ⃗> =<m*(d r ⃗)/dt>=m*(d<r ⃗>)/d t=0$.

Observation: in the ground state the average value of $<r>$ is nonzero and is equal with

$<r>=3/2 a$ where $a=0.53*10^{-10} m$ (first Bohr radius)

This is computed by taking the wave function of the ground state

$ψ(r)=A*exp(-r/a)$ and the finding $<r>$.

Just the average of the position vector $<r ⃗>=0$ because

$r ⃗=x*i ⃗+y*j ⃗+z*k ⃗$ and $x=r*\sin θ*\cos φ,y=r*\sin θ*\sin φ$ and $z=r*\cos θ$

$<x>=∫_0^∞ r*exp(-2r/a)*(r^2 d r)∫_0^π \sin θ(\sin θ dθ)∫_0^2π \cos φ dφ =0$

since $∫_0^2π \cos φ dφ=0$

$<y>=⋯=0$ and $<z>=⋯=0$

Since

$|Δp ⃗ |=\sqrt{((p ⃗-<p ⃗>)^2 )}=\sqrt{p ⃗^2}$ or alternately $σ_p=\sqrt{(<p^2>-<p>^2)}=\sqrt{<p^2>}$

And

$|Δr ⃗ |=..=\sqrt{r ⃗^2}=|r|$ or $1/|r ⃗ | ≅1/|Δr ⃗ |$

b) Uncertainty principle is

$|Δp|*|Δr|≅ℏ/2$

From

$<H>=(<p^2>)/2m-(k e^2)/(|r ⃗|)(=T+U)$

$<H>=|Δp ⃗ |^2/2m-(k e^2)/|Δr ⃗ | =ℏ^2/(4|Δr|^2 )*1/2m-(k e^2)/|Δr ⃗ |$

$f(x)=ℏ^2/(8mx^2 )-(k e^2)/x$ with $df/dx=-ℏ^2/(4mx^3 )+(k e^2)/x^2 =0$ or $ℏ^2=4mke^2 x$ or $x=|Δr ⃗|=ℏ^2/(4mke^2)$

c)

$|Δp ⃗ |≅|p ⃗ |=ℏ/2|Δr ⃗ | =(2mke^2)/ℏ$

$m|v ⃗ |=(2mke^2)/ℏ$ or $|v ⃗ |/c=(2ke^2)/ℏc=2e^2/(4πϵ_0*ℏc)=2α$

with $|v|=√(<v^2>)$

d)

Binding energy for electron on the first level is

$E_1=-13.6 eV$ known!

Rest energy of electron is

$E_0=m_0 c^2=(9.1*10^{-31}*(3*10^8 )^2)/(1.6*10^{-19} )=511.87 keV=0.511 MeV$

$E_0/E_1 ≈40*10^3$ (!)

e)

relativistic energy is

$E^2=p^2 c^2+m_0^2 c^4=p^2 c^2+E_0^2≈E_k^2+E_0^2$

And since

$E_1=2E_k$ the total energy electron in atom is twice its kinetic energy

One has

$E^2=4 E_1^2+E_0^2$

So that the correction in relativistic energy is

$ΔE=√(ΔE^2 )=\sqrt{(E^2-E_0^2)}=2E_1≈26 eV$ very small‼

e)

If nucleus charge $e→Ze$ then $E_1→Z^2*E_1$ (energy of first level in atom) and $r_1→r_1/Z$

Since $E_1=m|v|^2$ then $|v|→Z|v|$ and $|v|/c=2Zα$