1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normalizing Hydrogen wavefunction

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data
    At time t=0 hydrogen atom is in state


    a) Is it possible to normalize wave function ?
    b) Find [tex]\psi(r,t)[/tex] if at time t=0 measuring [tex]L_{z}[/tex] we find [tex]\hbar[/tex]
    2. Relevant equations

    3. The attempt at a solution

    Using eigenstates of hydrogen I can write[tex]\psi(r,0)[/tex] as
    [tex] \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}[/tex]

    Normalization condition [tex]\sum|c_{n}|=1[/tex] gives me [tex]A^{2}=\frac{1-2\pi}{432}[/tex] or [tex]A=\sqrt{\frac{2\pi-1}{432}}i[/tex] , 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 ?
  2. jcsd
  3. Feb 15, 2008 #2


    User Avatar

    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.
  4. Feb 17, 2008 #3
    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 [tex]\frac{e^{-r/a}}{\sqrt{4\pi}}[/tex] instead of [tex]e^{-r/a}[/tex] wave function can be normalized, [tex]A=\frac{1}{\sqrt{864}}[/tex]
    Maybe it is just a typo error.
  5. Aug 5, 2008 #4
    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:

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

    [tex] \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}[/tex]

    gives A=1/sqrt(864)

    You can write hydrogen eigenstates as radial part multiplied by spherical harmonics:

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

    [tex] \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}[/tex]

    In spherical harmonics [tex]Y^{m}_{l}(\theta,\phi)[/tex] m is integer value of [tex]\hbar[/tex] ([tex]m=-1*\hbar, 0*\hbar, 1*\hbar[/tex]) you can obitain by measurment of Lz.
    By measuring Lz system is left in eigenstate of operator Lz.
    Since measurment gives [tex]L_{z}=\hbar[/tex] , system is left in state:
    [tex] \psi(r,0) = 4\sqrt{3}AR_{21}*Y^{1}_{1}[/tex]
    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.
  6. Aug 5, 2008 #5
    In spherical harmonics [tex]Y^{m}_{l}(\theta,\phi)[/tex] m=-l,...,0,...,+l.
    Since l=1 => m=-1,0,1 , possible values for measurment of Lz are Lz=-hbar,0,+hbar
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Normalizing Hydrogen wavefunction