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The Normal Zeeman Effect and Hydrogen States

  1. May 30, 2013 #1
    I'm studying the hydrogen atom and have this question. Apparently it can be solved without perturbation theory, however I'm having trouble justifying it.

    1. The problem statement, all variables and given/known data

    153wwgi.png

    2. The attempt at a solution

    Avoiding perturbation theory I simply get:

    E = E(n) - constant*(mh) where m is the angular momentum number not mass

    Which seems a little too easy.

    For the second part I can see the initial t=0 wavefunction is normalised no problem, but when I apply the time evolution operator e^(iEt)/h things get messy.

    Am I right in thinking the probability for the third state | 2 1 0 > is just zero for all time?
     
  2. jcsd
  3. May 30, 2013 #2

    TSny

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    Hello mrjohns. Welcome to PF!

    That looks good.

    Can you show a little more detail. It shouldn't be messy.

    Yes.
     
  4. May 31, 2013 #3
    Thanks for the reply.

    If I say the energies are E=E(n)-E(b)*mh I get by time evolution:

    |psi(t)> = 1/sqr(2) [ e^-i(E(n)+E(b))t | 2 1 -1 > - e^-i(E(n)-E(b))t | 2 1 1 > ]

    Which by Eulers formula:

    =1/sqrt(2) { [ cos[(-E(n)-E(b))t] + isin[(-E(n)-E(b))t] ] | 2 1 -1 > - [ cos[(-E(n)+E(b))t] + isin[(-E(n)+E(b))t] ] | 2 1 1 > }

    Which is normalised for all time, since (cos(x)+isin(x))(cos(x)-isin(x))=1

    Now obviously it can never get to the |psi(3)>=|210> state of |nlm> since this evolution doesn't change the value of m.

    But I can't see how it can evolve to the |psi(2)> state. Should I have left in m, rather than substituted 1 and -1 in for it?
     
  5. May 31, 2013 #4

    TSny

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    You have |ψ(t)> = 1/√2 [e-iω1t | 2 1 -1 > - e-iω2t | 2 1 1 > ] where ω1 = E(n)+E(b) and ω2 = E(n)-E(b)

    Note that the phase factors in front of | 2 1 -1 > and | 2 1 1 > oscillate with different frequencies. So, there will be instants of time when the phase factors are equal (which gives |ψ1>) and instants of time when the phase factors are exactly out of phase with each other (which gives |ψ2>).

    Now that you have an expression for |ψ(t)>, how would you get the probability amplitude for this state to be measured as |ψ1>?
     
  6. Jun 1, 2013 #5
    Using matrix multiplication and putting bras to kets i get:

    Prob amp for |ψ1> = 1/2 [e^-iω1t + e^-iω2t] => Probability is one when e^-iω1t = e^-iω2t = 1

    Prob amp for |ψ2> = 1/2 [e^-iω1t - e^-iω2t] => Probability is one when e^-iω1t = -e^-iω2t = 1

    Prob amp for |ψ3> = 0 for all time

    Now these probability amplitudes for |ψ1>,|ψ2>,|ψ3> sum to e^-iω1t which does not equal one. Is this correct?

    Or am I reading the question wrong, and is it simply asking whether <ψ(t)|ψ(t)> = 1 for all time? (Which it does.)

    [Edit-And I've also got the added condition for the phases e^-iω1t = e^-iω2t = 1 rather than when they're just equal, and e^-iω1t = -e^-iω2t = 1 rather than when they're just out of phase.

    I know I want to do <ψ1|ψ(t)> to find the probability amplitude for ψ1

    Do I want <ψ1| = 1/√2 [< 2 1 -1 | - < 2 1 1 |] or am I missing something?]
     
    Last edited: Jun 1, 2013
  7. Jun 1, 2013 #6

    TSny

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    A probability amplitude is a product of a bra with a ket. So, I'm not following your statements above. But you are correct in what you state below:

    Yes this is all you need to do, find <ψ1|ψ(t)>

    where <ψ1| = 1/√2 [< 2 1 -1 | - < 2 1 1 |] and |ψ(t)> = 1/√2 [e-1t | 2 1 -1 > - e-iω2t | 2 1 1 > ]
     
  8. Jun 1, 2013 #7
    So does the methodology here look right?

    <ψ1|ψ(t)> = {1/√2 [< 2 1 -1 | - < 2 1 1 |]}{1/√2 [e-iω1t | 2 1 -1 > - e-iω2t | 2 1 1 > ] }
    <ψ1|ψ(t)> = 1/2 [< 2 1 -1 | e-iω1t | 2 1 -1 > + < 2 1 1 | e-iω2t | 2 1 1 >]
    <ψ1|ψ(t)> = 1/2 [e^-iω1t + e^-iω2t] by orthonormality

    And similarly:

    <ψ2|ψ(t)> = 1/2 [e^-iω1t - e^-iω2t] by orthonormality

    <ψ3|ψ(t)> = 0

    To get the probability density I need to square by complex conjugate, so get:

    Prob:(ψ1) = 1/4 [e^-iω1t + e^-iω2t][e^+iω1t + e^+iω2t]
    Prob:(ψ1) = 1/4 [2+2e^(-i^2)ω1ω2(t^2)]
    Prob:(ψ1) = 1/2 [1+e^ω1ω2(t^2)]

    And similarly:

    Prob:(ψ2) = 1/2 [1-e^ω1ω2(t^2)]

    Thus Prob:(ψ1) + Prob:(ψ2) + Prob:(ψ3) = 1/2 [1+e^ω1ω2(t^2)] + 1/2 [1-e^ω1ω2(t^2)] = 1 for all time.

    And we get:

    Prob:(ψ1)=1 if e^ω1ω2(t^2)=1
    Prob:(ψ2)=1 if e^ω1ω2(t^2)=-1

    Does this seem ok? One part that's bugging me is e^ω1ω2(t^2) - how do I know it is bound between -1 and 1 in order for the probability density equations to make sense?
     
    Last edited: Jun 1, 2013
  9. Jun 1, 2013 #8

    TSny

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    Good up to the blue line. Rethink the product e-iω1te+iω2t
     
  10. Jun 1, 2013 #9
    Whoops, so I should have typed:

    Prob:(ψ1) = 1/4 [e^-iω1t + e^-iω2t][e^+iω1t + e^+iω2t]
    Prob:(ψ1) = 1/4 [1+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t+1]
    Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t]

    And similarly:

    Prob:(ψ2) = 1/4 [2-e^-i(ω1-ω2)t-e^-i(ω2-ω1)t]

    Thus Prob:(ψ1) + Prob:(ψ2) + Prob:(ψ3) = 1/4 [2+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t] + 1/4 [2-e^-i(ω1-ω2)t-e^-i(ω2-ω1)t] = 1 for all time.

    And we get:

    Prob:(ψ1)=1 if e^-i(ω1-ω2)t+e^-i(ω2-ω1)t=2
    Prob:(ψ2)=1 if e^-i(ω1-ω2)t+e^-i(ω2-ω1)t=-2

    So now I need | e^-i(ω1-ω2)t+e^-i(ω2-ω1)t | <= 2 - how do I know it is bound in this way? And isn't the complex part an issue?
     
    Last edited: Jun 1, 2013
  11. Jun 1, 2013 #10

    TSny

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    Use the identity eix + e-ix = 2cos(x)
     
  12. Jun 2, 2013 #11
    Ah, so:

    Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^+i(ω1-ω2)t]
    Prob:(ψ1) = 1/2 {1+cos[(ω1-ω2)t]}

    Prob:(ψ2) = 1/4 [2-(e^-i(ω1-ω2)t+e^+i(ω1-ω2)t)]
    Prob:(ψ2) = 1/2 {1-cos[(ω1-ω2)t]}

    Now I can see that looks a lot better - the probabilities for each are bound between 0 and 1, and Prob:(ψ1)=1 at t=0.

    When cos[(ω1-ω2)t=1, Prob:(ψ1)=1, and when cos[(ω1-ω2)t]=-1, Prob:(ψ2)=1.
     
  13. Jun 2, 2013 #12

    TSny

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    Yes. Good.
     
  14. Jun 2, 2013 #13
    Cheers, thanks so much for your help.
     
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