# Hydrogen |nlm> Eigenstates

A hydrogen atom is in a state described by the wave function $\Psi_{nlm}$ with $n=4$, $l=3$ and $m=3$.

a) What is the magnitude of the orbital momentum of the electron in this state?

b) What is the angle between the orbital moment and the $z$ axis? can this angle be reduced by changing $n$ or $m$ while keeping $l$ constant?

$Ψ_nlm (r ⃗)$ with $n=4$,$l=3$ and $m=3$

a,b)

Orbital momentum is

$|L|^2=l(l+1) ℏ^2$ so that $L=\sqrt{l(l+1)}*ℏ=\sqrt{12}*ℏ$

Projection of orbital momentum on z axis is

$L_z=mℏ$ so that $L_z=3ℏ$

Angle between $L_z$ and $L$ is

$\cos(θ)=L_z/L=3/\sqrt{12}=3/(2\sqrt{3}=\sqrt{3}/2$ so that $θ=π/6$

Since $L_z$ depends on $m$ and $L$ depends on $l$, than this angle can be modified by changing $m$ while keeping $l$ constant.

c)

Radial wave function is

$R_{43}=(r/a)^3*\exp(-r/4a)*H_4 (r/a)=[16(r/a)^4-48(r/a)^2+12]*(r/a)^3*\exp(-r/4a)$

The probability of finding the electron at a given distance $r$ is space $dr$ is

$P_{43}=r^2 R_{43}^2$

An electron in a Hydrogen atom is in the energy eigenstates $\psi_2,1,-1(r,\theta,\phi)=Ne^{-r/2a}Y_{1,-1}(\theta,\phi)$

a) Find the normalization constant

b) What is the probability density of finding the particle within intervals $d r$, $dθ$, $dφ$ at $r=a$, $θ=45°$, $φ=60°$?

c) What is the probability of finding the particle within interval $d r$ at $r=a$?

d) For the measurements of $L^2$ and $L_z$ what will be the result?

a)

Normalization constant

$1=N^2 ∫_0^∞ \exp(-r/a)*∬ |Y_{1,-1} |^2*dΩ=N^2*1*1=N^2$ so that $N=1$

b)

For $r=a$,$θ=45°$,$φ=60°$ the probability to find the particle in $d r*dθ*dφ$ is

$d V=r^2 d r*(sinθ*dθ*dφ)$

$P=r^2 \exp(-r/a)*\sinθ*|Y_{(1,-1)} |^2$ with $Y_{(1,-1)}=1/2*\sqrt{(3/2π)}*\sinθ*e^{-iϕ}$

$P=a^2*1/e*1/4*3/2π*\sin^345=0.0155a^2$ where $a=0.53*10^{-10} m$

c)

For $r=2a$ the probability to find the particle in $d r$ is

$P=r^2*exp(-r/a)=4a^2*1/e^2 =0.541a^2$ where $a=0.53*10^{-10} m$

d)

The quantum numbers are

$n=2$, $l=1$, $m=-1$

So that

$L^2=l(l+1) ℏ^2=2ℏ^2$ and $L_z=mℏ=-ℏ$