Joschua_S
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Hi
Let [itex]D[/itex] be an anisotropic tensor. This means especially, that [itex]D[/itex] is traceless. [itex]\mathrm{tr}(D) = 0[/itex]
Apply the representating matrix of [itex]D[/itex] to a basis vector [itex]S[/itex], get a new vector and multiply this by dot product to your basis vector. Than you got a scalar function.
Now integrate this function over a symmetric region, for example the n-dimensional unit-sphere or a n-dimensional Cube or something other symmetric.
[itex]\int SDS ~ \mathrm{d} \Omega[/itex]
This integral vanish!
[itex]\int SDS ~ \mathrm{d} \Omega = 0[/itex]
My question:
Trace is for me like an average of something. The symmetric integration vanish also, like the trace.
Is there a link between trace zero and the vanishing integral? What is the math behind this.
If you wish one example. A part of a dipole-dipole coupling Hamiltonian in spectroscopy is given by [itex]H_{DD} = SDS[/itex], with the anisotropic zero field splitting tensor [itex]D[/itex] and spin [itex]S[/itex].
Greetings
Let [itex]D[/itex] be an anisotropic tensor. This means especially, that [itex]D[/itex] is traceless. [itex]\mathrm{tr}(D) = 0[/itex]
Apply the representating matrix of [itex]D[/itex] to a basis vector [itex]S[/itex], get a new vector and multiply this by dot product to your basis vector. Than you got a scalar function.
Now integrate this function over a symmetric region, for example the n-dimensional unit-sphere or a n-dimensional Cube or something other symmetric.
[itex]\int SDS ~ \mathrm{d} \Omega[/itex]
This integral vanish!
[itex]\int SDS ~ \mathrm{d} \Omega = 0[/itex]
My question:
Trace is for me like an average of something. The symmetric integration vanish also, like the trace.
Is there a link between trace zero and the vanishing integral? What is the math behind this.
If you wish one example. A part of a dipole-dipole coupling Hamiltonian in spectroscopy is given by [itex]H_{DD} = SDS[/itex], with the anisotropic zero field splitting tensor [itex]D[/itex] and spin [itex]S[/itex].
Greetings
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