Joschua_S
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Hi
Let D be an anisotropic tensor. This means especially, that D is traceless. \mathrm{tr}(D) = 0
Apply the representating matrix of D to a basis vector S, 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.
\int SDS ~ \mathrm{d} \Omega
This integral vanish!
\int SDS ~ \mathrm{d} \Omega = 0
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 H_{DD} = SDS, with the anisotropic zero field splitting tensor D and spin S.
Greetings
Let D be an anisotropic tensor. This means especially, that D is traceless. \mathrm{tr}(D) = 0
Apply the representating matrix of D to a basis vector S, 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.
\int SDS ~ \mathrm{d} \Omega
This integral vanish!
\int SDS ~ \mathrm{d} \Omega = 0
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 H_{DD} = SDS, with the anisotropic zero field splitting tensor D and spin S.
Greetings
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