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Homework Help: 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


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    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
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