Recent content by ian2012

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    Determinant of this special matrix

    Okay, i may have canceled abit too early, the factorization I am looking for is of: (E - lamda - a)^2(E- lamda + a) = a(E - lamda - a)
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    Determinant of this special matrix

    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...
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    Determinant of this special matrix

    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...
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    Using the time evolution operator

    Oh right, of course, so it let's you simplify the expression.
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    Using the time evolution operator

    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...
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    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.
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    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.
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    What linearity condition do you use ?
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    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...
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    Do you know of a website or (intuitive) method i can use to learn the properties of bra-ket algebra? How did you learn it, for instance?
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    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.
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    Bra-Ket Operation: Multiplying Equation by |psi1>

    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| ...
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    Langmuir waves dispersion relation

    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...
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    The weak force and effective coupling

    Okay then, so weak interactions don't really proceed at low energies? (i am not too sure what the ideal energies are for strong interactions).
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    The weak force and effective coupling

    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...
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