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# Quantum Mechanics.
Do the eigen vectors of every hermitian operator form a complete orthonormal set of vectors? Like eigen vectors of Potential energy, Kinetic energy, orbital angular momentum, spin angular momentum etc. When we can say for sure that yes they do form a basis in Hilbert space any state vector can be expanded in any of these basis? How to check it?
I am so much confused about it. Please help.........
 9 months ago
 9 months ago
# Quantum Mechanics. Do the eigen vectors of every hermitian operator form a complete orthonormal set of vectors? Like eigen vectors of Potential energy, Kinetic energy, orbital angular momentum, spin angular momentum etc. When we can say for sure that yes they do form a basis in Hilbert space any state vector can be expanded in any of these basis? How to check it? I am so much confused about it. Please help.........
 9 months ago
 9 months ago

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wini_bosonBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr ,@Ghazi
 8 months ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
No idea, that was too long ago!
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.1
The answer to the first part is, yes. However I'm afraid that's all I can help you with on this question
 8 months ago

wini_bosonBest ResponseYou've already chosen the best response.0
thanks @ sarahusher
 8 months ago

ghaziBest ResponseYou've already chosen the best response.3
@wini_boson , sorry to interrupt you but how come you mentioned me ..lol its been 4 months since i left the OS...but as far as your question is concerned there are three basic properties of hermitian operators and that will answer your first question > Hermitian operators can be flipped over to the other side in inner products >Hermitian operators have only real eigenvalues > Hermitian operators have a complete set of orthonormal eigenfunctions (or eigenvectors) now second part of your question is quite lengthy to solve here but still i hope it helps you out, most of it is maths which you can ask @sarahusher , if you have the confusion and i hope you can interpret this maths in terms of physics :D Maths> to form a basis of a vector space you need to define a subset of the elements of vectors that are linearly independent and vector space span V... \[V= V _{1} i+V _{2} j+V _{3} k.....\] V1, V2 , V3 are the elements and therefore, your vector forms the basis iff V can be expressed as \[V=a _{1}V _{1}+a _{2}V _{2}+.....\], where a1, a2... are elements of base field A vector space V will have many different bases, but there are always the same number of basis vectors in each of them. The number of basis vectors in V is called the dimension of V. Every spanning list in a vector space can be reduced to a basis of the vector space. When a vector space is infinite dimensional, then a basis exists, as long as one assumes the axiom of choice. A subset of the basis which is linearly independent and whose span is dense is called a complete set, and is similar to a basis. When V is a Hilbert space, a complete set is called a Hilbert basis. *footnote> sorry couldnt think of an example at 5.20 in the morning :P ...hope that helps you out :D i shall try to get an example for you :D
 8 months ago

wini_bosonBest ResponseYou've already chosen the best response.0
@ghazi thanks for coming here. and please don't leave OS.
 8 months ago

wini_bosonBest ResponseYou've already chosen the best response.0
is it possible to represent same state vector in continuous basis and discrete basis? (ofcourse in infinite dimensional space.) And also please clear my confusion... since spin operator has only two eigen state "up" and "down" means only two basis. So how its going to represent a state which was initially represented in position or momentum or energy basis of infinite dimension.
 8 months ago
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