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Operator in nonorthogonal basis 
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#1
Oct407, 05:47 PM

P: 6

Hi, is possible make up a operator in a nonorthogonal basis, if is possible how I can contruct the operator.
thanks 


#2
Oct407, 08:26 PM

P: 701

why not form your operators as b><a



#3
Oct507, 10:11 AM

P: 6

which are the consequence of choice a basis nonorthogonal?



#4
Oct507, 11:25 AM

P: 46

Operator in nonorthogonal basis



#5
Oct507, 11:44 AM

Sci Advisor
HW Helper
P: 2,948

All you need to know is the effect of the operator on all the basis states. So if you know all the values of [itex] <a_iAa_j>[/itex] then you know everything about the operator. Alternatively, as quetzalcoatl9 pointed out, an arbitrary operators can be written as [itex] A = \sum c_{ij} a_i><a_j [/itex] One consequence of having a non orthonogonal basis is that you can't read off directly from the above expression what is the effect of applying the operator to a basis state gives. If the basis is orthogonal, then applying A to, say, [itex] a_3> [/itex] will simply give [itex] c_{13} a_1> + c_{23} a_2> + \ldots [/itex] (I am assuming that the labels of the states are discrete and start at 1). If the basis is not orthogonal, the expression is of course more complicated. 


#6
Oct507, 01:01 PM

P: 6

I can construct a basis depent of basis nonorthogonal, how might make up? and what happen with the eigenvalues and elements of the operator.
someone know if the situation present in some quantum system. 


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