Timeline for Defining the derivative of a Hermitian inner product symbolically
Current License: CC BY-SA 3.0
5 events
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Jan 12, 2017 at 1:53 | comment | added | Jack Moon |
Whilst X = HEvaluate@D[H[F, F], z] /. {Conjugate'[z] -> 0} HEvaluate@D[X, z] /. {Conjugate'[z] -> 0} should have a second order derivative term of the form $ f \cdot \overline{\frac{\partial^2 f}{\partial z^2}}$, instead it has the term f[z,Conjugate[z]].((Conjugate')[(f^(1,0))[z,Conjugate[z]]] (f^(2,0))[z,Conjugate[z]]) It's like this (Conjugate')[(f^(1,0))[z,Conjugate[z]]] term should be the conjugate operator on the following term.
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Jan 12, 2017 at 1:47 | comment | added | Jack Moon |
The problem is these Conjugate' terms in the previous expression. So the output for F = f[z, Conjugate[z]]; D[H[F, F], z] ; HEvaluate@D[H[F, F], z] /. {Conjugate'[z] -> 0} is as expected f[z,Conjugate[z]].Conjugate[(f^(1,0))[z,Conjugate[z]]]+(f^(1,0))[z,Conjugate[z]].Conjugate[f[z,Conjugate[z]]] , however preforming the operation twice, say for example; D[D[H[F, F], z] , z]; HEvaluate@D[D[H[F, F], z] , z] /. {Conjugate'[z] -> 0} results in terms such as (H^(0,1))[(f^(1,0))[z,Conjugate[z]],f[z,Conjugate[z]]]
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Jan 11, 2017 at 20:32 | comment | added | b3m2a1 |
Sorry I'm not exactly sure what you are getting at here. Is your problem that taking the derivative after applying HEvaluate gives a wonky result? Because that's not too surprising. Try applying HEvaluate only at the very end. That is, do all your Hermitian stuff symbolically, passing the operator itself around, then later apply HEvaluate .
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Jan 11, 2017 at 4:35 | comment | added | Jack Moon |
This works when you differentiate the dot product once, however if I feed the result back into the evaluation wrapper D[H[F, F], z] X = HEvaluate@D[H[F, F], z]; D[X, z] HEvaluate@D[X, z] gives $f\left(z,z^*\right).\left(f^{(2,0)}\left(z,z^*\right) \text{Conjugate}'\left(f^{(1,0)}\left(z,z^*\right)\right)\right)+f^{(1,0)}\left(z,z^*\right).\left(f^{(1,0)}\left(z,z^*\right) \text{Conjugate}'\left(f\left(z,z^*\right)\right)\right)+f^{(1,0)}\left(z,z^*\right).f^{(1,0)}\left(z,z^*\right)^*+f^{(2,0)}\left(z,z^*\right).f\left(z,z^*\right)^*$ (I don't know how to directly copy the output)
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Jan 11, 2017 at 4:20 | history | answered | b3m2a1 | CC BY-SA 3.0 |