# Hydrogen first energy (Quantum Physics)

The average energy of the hydrogen atom in arbitrary bound state f(r) is given by the integral

$<E> = <f(r)|H|f(r)>$

Show that

$<E> = <100|H|100> = E1$

$ψ_{100} (r,θ,φ)=R_{10} (r)*Y_0^0 (θ,φ)=1/\sqrt{π}*1/a^{3/2} *exp⁡(-r/a)=A*exp⁡(-r/a)$  where $A=1/√π*1/a^{3/2}$

Schroedinger equation is

$H |ψ>=E|ψ>$

$H=-ℏ^2/2m*∇^2+ke^2/r^2$

$∇^2 ψ=1/r^2 *d/dr*(r^2*dψ/dr)=1/r^2 *d/dr*[-r^2*A/a*exp⁡(-r/a) ]=$

$=-2/r*A/a*exp⁡(-r/a)+A/a^2 *exp⁡(-r/a)=A/a*exp⁡(-r/a)*(1/a-2/r)$

Having this computed one has

$H |ψ>=-ℏ^2/2m*(1/a-2/r)*A/a*exp⁡(-r/a)+(ke^2)/r*A*exp⁡(-r/a)=$

$=-A*exp⁡(-r/a)*[ℏ^2/(2ma^2 )-ℏ^2/(amr)+(ke^2)/r]$

$a=ℏ^2/(kme^2)$    or equivalent $ℏ^2/(amr)=(ke^2)/r$

We have
$H|ψ>=-A*exp⁡(-r/a)*[ℏ^2/2m*(k^2 m^2 e^4)/ℏ^4 ]=$

$=(-(k^2 me^4)/(2ℏ^2 ))*A*exp⁡(-r/a)$

Back into the original equation by completing the equation with the vector $<ψ|$ we have

$<ψ|H ̂ |ψ>=<ψ|E|ψ>=<E>$

$<100|H|100>=<E>$

The element of volume is spherical coordinates is

$dV=r^2 dr*dΩ=r^2 dr* sin⁡ θdθ*dφ$

$∫_0^∞ A*exp⁡(-r/a)*(-(k^2 me^4)/(2ℏ^2 ))*A*exp⁡(-r/a)*r^2 dr∯dΩ=<E>$

$A^2*(-(k^2 me^4)/(2ℏ^2 ))*∫_0^∞ exp⁡(-2r/a)*r^2 dr∯dΩ=<E>$

$1/(πa^3 )*(-(k^2 me^4)/(2ℏ^2 ))*a^3/4*4π=<E>$

$E_1=-(k^2 me^4)/(2ℏ^2 )=<E>$