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Stark effect, mixing state of 2s and 2p

  1. Nov 17, 2011 #1
    hi all, can anyone help me understanding what is the meaning of the mixing state of 2p and 2s? thanks in advance.

    when an electron in n=2 sate in a hydrogen atom, experience a external em field, then the degenerate states split in to 3 classes, which are:

    1/sqrt2( 210 + 200 ) with higher energy
    2 1 +1 and 2 1 -1 with the same energy as before
    1/sqrt( 210 - 200) with lower energy

    my question is how to interpretate the upper and lower energy states?
    for upper level, i think it as there is a half probability to find the electron in 210 state and half in 200 state. but if i think in terms of this way, then the lower level becomes half probability in 210 state(which contridicts with it can also in uper level) and -1/2 probability in 200 state( wich is obviously wrong).

    thanks for any helps.
     
  2. jcsd
  3. Nov 21, 2011 #2

    Cthugha

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    Science Advisor

    No, it does not work that way. Note that you are adding the wavefunctions and then square to get the probability density. Adding the already squared wavefunctions would have the meaning you are thinking about. But that will not give the eigenstates of the problem.

    Now for visualization: Your Hamiltonian couples two degenerate levels. As a rough analogy think of a common spring pendulum in classical mechanics. If you have two identical ones of those and couple them with another spring, one of the standard problems in classical mechanics is to solve the coupled problem via some matrix eigenvalue equation and to show that it is a bad idea to express the solution in terms of the eigenstates of the bare single pendulum. Instead these two can exchange energy and the commonly chosen basis for describing all kinds of motion of the coupled pendulum are the states where the two masses oscillate in phase or exactly out of phase corresponding to the two masses showing symmetric or antisymmetric motion.

    So in the case of the linear Stark effect, the eigenstates are also the symmetric and antisymmetric combinations of the initial states. If you sketch them, you will find that they cannot be eigenstates of the parity operator and have a non-vanishing dipole moment. The latter also is the reason for the energy splitting.
     
  4. Nov 22, 2011 #3
    Thank you very much indeed, now I am happy now. Great answer.
     
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