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Matrix element (raising and lowering operators) 
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#1
May1105, 01:03 AM

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How to determine the matrix representation of position & momentum operator using the energy eigenstates as a basis



#2
May1105, 01:51 AM

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PF Gold
P: 6,236

In any case, you end up by finding an equivalence: f_E (x) <> E> ; so you can consider that f_E(x) = <xE> This means that f_E(x), now seen as F(E,x) is the matrix element of the basis transformation that maps the basis {x>} into the basis {E>}. That's sufficient to transform your representation of the position operator X (which is simply x delta(xx0) for <xXx0>) into the E> basis <E XE> ; and in the same way to transform the representation of P in the x> basis into a representation in the E> basis. What I've outlined above is the pedestrian method. Sometimes more elegant algebraic methods exist: for instance in the case of the harmonic oscillator, with the creation and annihilation operators. I don't know how general that approach is. cheers, Patrick. 


#3
May1105, 05:43 PM

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Funny,the title mentioned raising & lowering ladder operators.Could he possibly mean the SHO?
Daniel. 


#4
May1105, 06:37 PM

P: 13

Matrix element (raising and lowering operators)
yes i did mean sho.



#5
May1205, 05:09 AM

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Voilą.So you're interested in making a unitary transformation from the matrix
[tex] \langle i\hat{H}j\rangle [/tex] (i,j can run from 0 to +infinity) to [tex] \langle i\hat{x}j\rangle [/itex] and [tex] \langle i\hat{p}j\rangle [/itex] ,where,now [tex] i\rangle \longrightarrow \langle xi\rangle [/itex] ,i.e.u'll be needing the SHOs wavefunctions.U can use only coordinate ones,just as long as u express the momentum operator in the same basis [itex] \{\langle x \} [/itex]. Daniel. 


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