Yeah, i know the solutions , thanks... Well i end up with (E - lamda - a)(E - lamda) + a(E - lamda - a) - a = 0 as the equation of the determinant. I am a bit rusty on my high school algebra, is there a way to factorise that? (You could multiply it all out and spend a lifetime dividing it by a...
Hope someone can help me out here.. I've found the eigenvalues (lamda) of this matrix, but through a very very long way, does anyone know of a quicker way (there must be a quicker way). The matrix is 3x3:
H = ( E, a, a; a, E, a; a, a, E ).
I can reduce the determinant to the following, but...
I hope someone can help me out here,
I am confused with a line of text I read - it is an example of a 2D Hilbert space with orthonormal basis e1, e2. The Hamiltonian of the system is the Pauli matrix in the y-direction. Given by the matrix:
\sigma_{y} = (\frac{0, -i}{i, 0})
The eigenvectors...
Yeah, the property is overlooked on many sites I have been on. However, I looked in Dirac's Principles of QM and he explains it there, which is fantastic. But it's a quite basic property of linear operators you can infer from simple derivatives.
Oh sorry, I understand what i did wrong.
Is it right to say: (A(hat) + B(hat)).|psi> = A(hat)|psi> + B(hat)|psi> ?
If so, then I believe that is the condition I was looking for... Well.. it has to be correct as it works for replacing the operator with real numbers.
Thanks for all of your posts. I have, since, extensively been reading about Dirac notation, dual space, projectors, etc.. but I still have one query regarding an operation with commutators.
How do you go from: <x'|x(hat)A(hat) - A(hat)x(hat)|x> to x'<x'|A(hat)|x> - <x'|A(hat)|x>x ?
(You may...
Okay, I didn't know you can multiply it from both sides. I know that A is a matrix, but the bra's and ket's are just vectors.. and i know that the matrix A can be found using the orthonormal basis, but i can't seem to make sense of it.
I have a basic question that I have overlooked in the past, given that you have
<psi2|A = lamda2<psi2|, where <| is a bra and lamda2 is the eigenvalue. If you were to multiply the equation by |psi1>, why do you get <psi2|A|psi1> = lamda2<psi2|psi1> and not |psi1><psi2|A = lamda2|psi1><psi2| ...
I was just looking at an expression (a dispersion relation, omega^2 = ...) similar to that of warm electron's in a plasma http://en.wikipedia.org/wiki/Plasma_oscillation expect with an extra imaginary term, which I think comes out from the full derivation of the dispersion relation for warm...
I just read something about the creation of deuteron in the first step of the pp cycle. Given that you have the reaction: p + p -> d + e^+ + v_e, where e^+ is a positron and v_e is an electron neutrino. Since there is a neutrino present, it is a weak interaction. In addition, as the interaction...