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Physical meaning of zero eigenvalue

  1. Dec 29, 2011 #1
    Hello,
    Given the hamiltonian :

    [itex] H = -( aS_z^2 + b(S_+^2 +S_-^2) ) [/itex]

    with S=1 and a,b>0 are constants.

    working with the base: { |m=1> , |m=-1> , |m=0> }
    The matrix form of H is:

    [itex]

    H = \left( \begin{array}{ccc}
    -ah^2 & -bh^2 & 0 \\
    -bh^2 & -ah^2 & 0 \\
    0 & 0 & 0 \end{array} \right)

    [/itex]

    or, in the base: { (1/sqrt(2))( |1> + |-1> ) , (1/sqrt(2))( |1> - |-1> ) , |0> }

    [itex]

    H = \left( \begin{array}{ccc}
    -(a+b)h^2 & 0 & 0 \\
    0 & -(a-b)h^2 & 0 \\
    0 & 0 & 0 \end{array} \right)

    [/itex]

    So, my question is- what is the physical meaning of '0' as an eigenvalue of a state (of m=0 in this case)? Does it mean this value will not be observed ?
    If thats the case- I am a bit confused, because the time-evolution of a state under this hamiltonian's dynamics, goes something like:
    cos(aht)|1> + isin(aht)|-1>
    meaning the "m" value pariodically shifts between +-1.
    Now, given that m=0 is basically an allowed state of the system, how can it shift from an m=+1 orientation to m=-1, without going thru m=0 ?
     
  2. jcsd
  3. Dec 30, 2011 #2

    strangerep

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    Science Advisor

    A state which has 0 eigenvalue of the Hamiltonian typically means the ground state. It's not clear from the info you provided whether the other two eigenvalues are +ve or -ve. If -ve, I guess a bound state is possible for the system -- but this doesn't matter very much for the answer...

    The time evolution of one of your non-ground states (+1, or -1) just shows that they evolve into linear combinations (superpositions) of the two. They do not need to "go through" the 0 state to do this. It just means that at one instant of time, the probability of finding the system in the +1 state is 1 (certain), whereas later there's a nonzero probability of finding the system either in state +1 or state -1. You'll never find the system in the ground state unless your initial state is a superposition of the ground state and one of the other states.

    HTH.
     
  4. Dec 30, 2011 #3
    Thank you very much for the reply!
    I still don't understand a couple of things..

    a. How could m=0 be the ground state, when its energy is 0, and the energy of the other two hamiltonian eigenstates are negative. shouldn't the ground state have the lowest energy?

    b. what physical mechanism can make the spin flip between m= +-1 orientations, without having any probability of being at m=0, although m=0 is basically a "legal" orientation?
    I'm okay with, say, an electron "jumping" from one orbital from another without even being in between orbitals, but this is because being "in between" is not an allowed state. But in our case, its like the "in between" zone is allowed, but the electron still ignores it.
     
  5. Dec 30, 2011 #4

    strangerep

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    Science Advisor

    I overlooked the part of your first post where you said a,b > 0. Yes, the ground state is the one with lowest energy.

    Let's go back to something you said earlier:

    It's a bit misleading to think that the "m" value periodically shifts deterministically. Rather, the expectation of measuring a particular value of m (over a large ensemble of identically-prepared copies of the system) changes. For any single measurement, you only get +1 or -1. But the relative probability of these two results changes over time.
     
  6. Dec 31, 2011 #5

    tom.stoer

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    Science Advisor

    b/c in QM you can add any constant E° to H, an eigenvalue E=0 need not mean anything special.

    In the hydrogen atom case E=0 is the boundary between discrete (bound) and continuous (unbound) eigenvalues (solutions).
    In the harmonic oscillator case E=0 is not a solution unless you shift H by -1/2.
     
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