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Hydrogen - adding a proton

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    2s9rng5.png

    Does adding a proton change the atom's orbital angular momentum? Find the probability that the energy is unchanged.

    2. Relevant equations



    3. The attempt at a solution

    The orbital angular momentum is unchanged, since ##<L^2> = l(l+1)\hbar^2## is independent of Z.

    To find the probability that the energy is unchanged, we overlap the old state (Z=1) with the new state (Z=2):

    [tex]|\langle\psi^{Z=1}|\psi^{Z=2}\rangle[/tex]

    Not sure how to find this value, without using brute force integration.
     
  2. jcsd
  3. May 7, 2014 #2

    Simon Bridge

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    I want to start by checking your understanding:
    The electron starts in an energy eigenstate of the old system - how would you describe it's state in the new system?

    Can you express the question in terms of performing a measurement?
     
  4. May 7, 2014 #3
    The Z=2 state will be a linear combination of Z=1 states?

    The probability of energy being unchanged = probability of measuring old energy
     
  5. May 7, 2014 #4

    Simon Bridge

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    Other way around.
    You can expand the Z=2 eigenstates that way but it does not help you.

    The old system no longer exists - all subsequent measurements will be done in the new system.

    Helps to talk about this if we define a notation... how about:
    If the old state is ##\psi_{o}## corresponding to energy ##E_{o}:\text H_o\psi_o=E_o\psi_o## of the Z=1 system; and you have a complete set of orthonormal eigenvectors of the Z=2 system ##\{\psi_n\}## then you know to write ##\psi_{old}=\sum_n c_n\psi_n## where ##\text H\psi_n = E_n\psi_n##

    This means that the electron is no longer in an energy eigenstate of the system - well it might be(!) - thus it's energy is uncertain until measured. Once measured, however, it will be found in an eigenstate of the new system with a specific value for energy and subsequent measurements will find the same thing - this should sound familiar to you.

    Just checking:
    In terms of the above expansion, what is the probability that the electron will be found with particular energy ##E_n##?
     
    Last edited: May 7, 2014
  6. May 8, 2014 #5
    Yes, it is due to the phenomenon of the collapse of the wavefunction.

    Probability to find electron with energy ##E_n## is ##|c_n|^2##. But none of this ##E_n## is equal to the old energy!
     
  7. May 9, 2014 #6

    Simon Bridge

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    So what is the probability that a subsequent measurement of energy will find the electron in a state ##\psi_n: E_n = E_{o}##?
    Therefore, what is the probability that the electron keeps the same energy?

    However - you will need to prove the statement "But none of this ##E_n## is equal to the old energy!".
     
    Last edited: May 9, 2014
  8. May 9, 2014 #7
    The probability is 0, since now all the energy eigenkets are different from the old enegrgy eigenket.

    Thus, it is impossible to find the atom in the old energy state.
     
  9. May 9, 2014 #8

    Simon Bridge

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    The probability reasoning here is good, but the rest is not: two different eigenkets can have the same eigenvalue - especially if they come from different Hamiltonians.

    You still need to prove that no energy eigenvalue of the new system has the same value as the old energy level.
    To do that you need to express the new energy levels in terms of the old one, and solve for n: En=Eo.
     
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