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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 ?

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# Physical meaning of zero eigenvalue

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