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Jan
31
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
I added another paragraph on accidental degeneracies.
Jan
31
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Added more explanation
Jan
31
comment Incorrect Left and Right Eigenvectors in Mathematica
Possibly related: Orthonormalization of non-hermitian matrix eigenvectors
Jan
31
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Accidental degenracies
Jan
31
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Accidental degenracies
Jan
31
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Completed the formal projector treatment.
Jan
31
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Introduction added
Jan
30
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
The projector in the case funs[[5]] returns numerically zero, indicating that these states don't belong to the 2D irreducible representation onto which I project. But there are other (1D) representations, each with their own projection operators. You would have to construct them using the rules of group theory and apply them to each eigenstate to isolate the symmetrized components. You'll need a character table for the group and repeat what I did above - it can be automated, but I'll have to leave that to you or someone else for now.
Jan
30
revised Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
not on display of letters
Jan
30
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
You can do a lot by applying group theory if there are symmetries. I give an example for the square in my answer.
Jan
30
answered Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
Jan
30
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
The L shape is very different from the square because the latter is invariant under the symmetry group $C_{4v}$ which causes the degeneracies. There is no such symmetry in the L. So the goal of your question isn't clear to me. You're asking about a highly symmetric example, where the degeneracies can be removed by reducing the domain. But that's not a Mathematica problem, and it's also unrelated to what you'd see in the L shape. All I can say then is: degeneracies are something you see in the spectrum and not in the wave functions.
Jan
30
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
In more general domains, you won't even see this problem because degeneracies will be very rare. So maybe we aren't really talking about the square here... what shapes are you really interested in?
Jan
30
comment Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
The reason is the same as what I said in my comment to your earlier Q: degeneracies can lead to eigenvectors that are arbitrary orthogonal superpositions of the symmetrized states you expected. If your new Q is about achieving this symmetry given the solutions you plotted, then there are several ways to answer it. Is that what you're asking?
Jan
30
comment How can I solve a 3D heat transfer partial differential equation?
What exactly is the issue you're asking about? What have you tried? Without evidence of some effort, the question is likely to be closed.
Jan
30
awarded  Nice Answer
Jan
30
comment Angle between two vectors in spherical coordinates
Also see this Q&A for a pathological case where VectorAngle becomes inaccurate. The three answers there provide different workarounds.
Jan
30
comment How to deal with complicated Gaussian integrals in Mathematica?
I think it's worth mentioning that the slowness of Moment that would potentially affect this answer appears to have been fixed in Mathematica version 10.3. See also How to efficiently find moments of a multinormal distribution
Jan
30
comment How to efficiently find moments of a multinormal distribution?
In Mathematica version 10.3 on Mac OS X, I get a much better timing for the moments. Your example yields AbsoluteTiming of 0.004 here. It looks like some serious improvements have happened in the new version. This apparently makes my answer to How to deal with complicated gaussian integrals in Mathematica roughly as fast as yours now...
Jan
30
revised How to deal with complicated Gaussian integrals in Mathematica?
Combined my original piecemeal hints into a final function