The discussion centers on the linearity and Hermiticity of operators defined as T = iħ (d/dx) + γ and T' = -iħ (d/dx) + γ, where γ is a constant. It is established that the commutation of these operators does not imply linearity; rather, the presence of γ introduces non-linearity unless γ = 0. The operators are classified as affine-linear due to their behavior when applied to functions, specifically when γ is treated as a constant translation. The relationship between operator symmetry and linearity is also clarified, indicating that symmetry does not affect the linearity of operators.
Quantum physicists, mathematicians specializing in operator theory, and students studying linear algebra in the context of quantum mechanics will benefit from this discussion.
Your post doesn't make any sense. I am banning your from this thread.chiro said:The point is that it is invertible - which is what I said a long time ago.
Do you agree or not?
If there is no nullity in the matrix and it is square it must be invertible.
Do you agree or not?
This is not the right place to discuss this. The question about invertibility had already been off-topic, since it has nothing to do with linearity, which is the subject of this thread.mathwonk said:I agree it makes life harder but I suggest we not block people just becaue they don't understand whatn they are saying, even if they don't realize it. At some point they may get the message. Just a suggestion, of course I am suggesting work for other people! feel free to ignore.