- #1
dudy
- 18
- 0
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 that's 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 ?
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 that's 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 ?