Does this look right to any of you?
How does one differentiate with respect to an operator?
>>7588366
>Notation. Same as when you differentiate with respect to a vector.
ahh okay, just looked it up
but then either both vecors should be transposed, or none, since the tensor product of two vectors is defined as
(x\otimes y)(z) := \left \langle y,z \right \rangle x
It basically depends, if the matrix derivatives operates on the vector space or it's dual
>>7588330
Dirac notation would have made this obvious.
Not saying it's better - just putting it out there.
>>7588441
care to explain?
is it just physicists handwaving a la
\frac{\partial }{\partial B} \left \langle x |B| x \right \rangle = \langle x | \langle x |
where suddenly the B vanishes and a ket turns into a bra, or is there something more rigorous to it?
>>7588451
The ket wouldn't turn into a bra. Like people have been saying, only one should be transposed.
I wouldn't call it handwaving - it's more that you prove how these things behave, and then the notation makes things simpler. Or doesn't.
>>7588456
>I wouldn't call it handwaving - it's more that you prove how these things behave, and then the notation makes things simpler. Or doesn't.
This explains you physicists' mindset so well. What normal human beings see as retarded, senseless handwaving, you people see as helpful memorisation tools. It's why everyone hates your stupid fucking field.
Fascinating.
>>7588456
only one should be transposed if it would be
x^Tx^T and not x^T \otimes x^T
Just what I've got from the wikipedia article on matrix calculus:
https://en.wikipedia.org/wiki/Matrix_calculus#Scalar-by-matrix
>Notice that the indexing of the gradient with respect to X is transposed as compared with the indexing of X
so it seems like the matrix derivative is an operator, that acts on the dual space (or in the finite dimensional case on transposed vectors)
I guess I made a mistake - it should be
|x \rangle \langle x |
>>7588451
You can use the multiplication rule and since the bra isn't an explicit function of b, only the ket and bra term remains
>>7588482
by that reasoning I would be left with
<x|x> which is not an operator, but a scalar.
Also A|x> is a nonscalar function, so how is matrix differentiation defined on that?
>>7588330
I'd write the right hand side as x\otimes x = xx^T. Then it looks right to me.
>>7588511
in my opinion its perfectly fine.
I guess it gets more clear if you look at
\frac{\partial }{\partial B}x^TBy = (xy^T)^T = (y^T)^T(x^T) = y^T \otimes x^T
