Can the Hydrogen Wavefunction Be Normalized?

[natugnaro]

Homework Statement

At time t=0 hydrogen atom is in state

\psi(r,0)=\frac{4}{(2a)^{3/2}}[e^{-r/a}+iA\frac{r}{a}e^{-r/2a}(-iY^{1}_{1}+Y^{-1}_{1}+\sqrt{7}Y^{0}_{1})]

a) Is it possible to normalize wave function ?
b) Find \psi(r,t) if at time t=0 measuring L_{z} we find \hbar

Homework Equations

The Attempt at a Solution

a)
Using eigenstates of hydrogen I can write\psi(r,0) as
\psi(r,0) = \frac{4\sqrt{\pi}}{2^{^3/2}}\varphi_{100} + 4\sqrt{3}A\varphi_{211}-4\sqrt{21}Ai\varphi_{210}-4\sqrt{3}Ai\varphi_{21-1}

Normalization condition \sum|c_{n}|=1 gives me A^{2}=\frac{1-2\pi}{432} or A=\sqrt{\frac{2\pi-1}{432}}i , but this does not satisfy normalization condition since I assumed that A is real.
I could assume that A is complex, but then I would get two unknowns (A=x+iy).
So I would say that it is not possible to normalize wave function.
If my answer is correct can someone explain this to me on practical example, do I need more information for normalizing ? where do I get it ? by measurment ?

 

[pam]

I didn't check your numbers, but you must have made a mistake somewhere.
Your coefficient for \phi_110 has to be less than 1.

 

[natugnaro]

The problem statement is an exact copy. Multiplying eigenstates by my coefficients I can get starting state.
But if the first term in parenthesis is \frac{e^{-r/a}}{\sqrt{4\pi}} instead of e^{-r/a} wave function can be normalized, A=\frac{1}{\sqrt{864}}
Maybe it is just a typo error.

 

[natugnaro]

Hi liorda , here is my solution:
Correcting some +/- signs and assuming that the first term in parenthesis is as I stated.
First multiply eigenstates of hydrogen by this constants:
\varphi_{100}*(\frac{1}{\sqrt{2}})
\varphi_{211}*(4\sqrt{3}A)
\varphi_{21-1}*(4\sqrt{3}Ai)
\varphi_{210}*(4\sqrt{21}Ai)

now we can write starting wave function as linear combination of hydrogen eigenstates:

\psi(r,0) = \frac{1}{\sqrt{2}}\varphi_{100} + 4\sqrt{3}A\varphi_{211}+4\sqrt{21}Ai\varphi_{210}+4\sqrt{3}Ai\varphi_{21-1}

Normalization:
\left|\frac{1}{\sqrt{2}}\right|^{2}+\left|4\sqrt{3}A\right|^{2}+\left|4\sqrt{21}Ai\right|^{2}+\left|4\sqrt{3}Ai\right|^{2}=1
gives A=1/sqrt(864)

b)
You can write hydrogen eigenstates as radial part multiplied by spherical harmonics:
\varphi_{nlm}=R_{nl}(r)*Y^{m}_{l}(\theta,\phi)

so using this we can write Psi(r,0) as:

\psi(r,0) = \frac{1}{\sqrt{2}}R_{10}*Y^{0}_{0} + 4\sqrt{3}AR_{21}*Y^{1}_{1}+4\sqrt{21}AiR_{21}*Y^{0}_{1}+4\sqrt{3}AiR_{21}*Y^{-1}_{1}

In spherical harmonics Y^{m}_{l}(\theta,\phi) m is integer value of \hbar (m=-1*\hbar, 0*\hbar, 1*\hbar) you can obitain by measurment of Lz.
By measuring Lz system is left in eigenstate of operator Lz.
Since measurment gives L_{z}=\hbar , system is left in state:
\psi(r,0) = 4\sqrt{3}AR_{21}*Y^{1}_{1}
Now i'ts easy to find Psi(r,t) , just multiply Psi(r,0) by Exp(-i*E2*t/(hbar)).
E2 is because energy is determined by principal quantum number , which is 2 in this case.
(in Dirac notation state is written as |nlm> , so our state is |211>) .
Again this is my own solution, I don't have the "official" solution.

 

[natugnaro]

In spherical harmonics Y^{m}_{l}(\theta,\phi) m is integer value of \hbar (m=-1*\hbar, 0*\hbar, 1*\hbar) you can obitain by measurment of Lz.

correction:
In spherical harmonics Y^{m}_{l}(\theta,\phi) m=-l,...,0,...,+l.
Since l=1 => m=-1,0,1 , possible values for measurment of Lz are Lz=-hbar,0,+hbar