# Hydrogen Mix of States (6-325)

Consider a Hydrogen atom in a state described by the following wave function

$\psi(r)=\frac{A}{\sqrt{\pi}}(1/a)^{3/2}e^{-r/a} +\frac{1}{2\pi}\frac{z-\sqrt{2}x}{r}*R_{21}(r)$

where $A$ is a real constant, $a$ is the Bohr radius, and R_{21}(r) is the radial wave function:

$R_{21}(r)=\frac{1}{\sqrt{6}}(1/a)^{3/2}(r/2a)e^{-r/2a}$

a) Rewrite this function as a linear combination of normalized Hydrogen wave functions.

b) Find value of $A$ such that the function is normalized.

c) If you measure $L^2$ and $L_z$ what results can you get and with what probabilities?

d) If you measure energy find the possible outcomes and their probabilities.

e) Find the expectation value of the radial coordinate in this state.

$ψ(r ⃗ )=(A/√π)*(1/a)^{3/2}*e^{-r/a}+(1/√2π) [(z-√2 x)/r] R_{21} (r)$

We have

$z=r*\cos⁡ θ$  and $x=r*\sin θ*\cos⁡ φ$   with $\cos φ=(e^{iφ}+e^{-iφ})/2$

$ψ(r)=\frac{A}{√π}*(\frac{1}{a})^{3/2}*e^{-r/a}+(\frac{1}{√2π}) R_{21} (r)*\cos⁡(θ)-\frac{1}{2√π} R_{21} \sin⁡θ e^{iφ}-\frac{1}{(2√π)} R_{21} \sin θ*e^{-iφ}$

From tables we have:

$R_{10}=2(1/a)^{3/2} e^{-r/a}$    ;
$Y_{1,1}=-(1/2) \sqrt{(3/2π)}*\sin θ*e^{iφ}$  ;

$Y_{1,-1}=(1/2) \sqrt{(3/2π)}*\sin θ*e^{-iφ}$  ;$Y_{1,0}=(1/2) \sqrt{(3/π)} \cos ⁡θ$

$Y_{0,0}=(1/2) \sqrt{(1/π)}$  ;

$Y_{1,0}=(1/2) \sqrt{(3/π)} \cos ⁡θ$;

$Y_{1,1}=-(1/2) \sqrt{(3/2π)}*\sin θ*e^{iφ}$;

$Y_{(1,-1)}=(1/2) \sqrt{(3/2π)}*\sin θ*e^{-iφ}$;

So that

$ψ(r ⃗ )=\frac{A}{√π} \frac{R_{10}}{2}*(2√π) Y_{0,0}+\frac{1}{√2π}*2\sqrt{\frac{π}{3}} R_{21} Y_{1,0}-\frac{1}{2√π}-2*\sqrt{\frac{2π}{3}}R_{21} Y_{1,1}-$

$-\frac{1}{(2√π)} (2\sqrt{\frac{2π}{3}})R_{21} Y_{(1,-1)}$

$ψ(r ⃗ )=A*R_{10} Y_{0,0}+√(2/3) R_{21} Y_{1,0}+√(2/3) R_{21} Y_{1,1}-√(2/3) R_{21} Y_{(1,-1)}$

$ψ(r ⃗ )=Aψ_{100}+√(2/3) ψ_{210}+√(2/3) ψ_{211}-√(2/3) ψ_{(21-1)}$

b)

All $ψ_{nlm}$  are orthonormal so that

$1=∫ |ψ(r) |^2 dr=A^2+2/3+2/3+2/3 =A^2 +2$  so that $A^2=-1$ and $A=i$

c)

If you measure $L^2$ you get

$L^2=l(l+1) ℏ^2$   with $l=0$ or $1$ (present in combination of $ψ_{nlm}$  above)

$L^2=0$  or $L^2=2ℏ^2$

Probabilities are the (sum of) square of their coefficients in the combination above.

$P(0)=A^2=-1$  and $P(2ℏ^2 )=2/3+2/3+2/3=2$

If you measure $L_z$ you get

$L_z=mℏ$ with $m=0,1,-1$ (present in combination $ψ_{nlm}$  above)

$L_z=0$,$L_z=ℏ$  and $L_z=-ℏ$

Probabilities are

$P(0)=A^2+2/3=-1+2/3=-1/3$   ;$P(ℏ)=2/3$   and $P(-ℏ)=2/3$

d)

For the possible energies measurement values we have:

$E=E_1/n^2$    with $n=1$ or $n=2$  and $E_1=-13.6 eV$  so that $E=E_1$  or $E=E_1/4$
The probabilities are

$P(E_1 )=A^2=-1$  and $P(E_2 )=2/3+2/3+2/3=2$

e)

The expectation value for r is

$<r>=<ψ|r|ψ>=(-1)<r_1>+(2/3+2/3+2/3) <r_2>=-1*a+2*(2^2 a)=7a$