# Matrix element (raising and lowering operators)

1. May 11, 2005

### ayalam

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

2. May 11, 2005

### vanesch

Staff Emeritus
You need to know your hamiltonian as a function of the position and momentum operators. Once you know that, you can solve for the eigenstates of the hamiltonian in the position or the momentum representation (by solving the associated Schroedinger equation). Then you need to find a link between your specific given basis of energy eigenstates you mentionned, and the solutions of the Schroedinger equation you found in the (say) position representation. This mapping can be trivial (in the non-degenerate case) because for same eigenvalues, you have a clear identification between your solution of the Schroedinger equation with that eigenvalue and the given energy eigenstate (well, up to a phase factor...). It can also be more complicated in the degenerate case. Then you'll need to look at how exactly the energy eigenstates of your given basis are defined.

In any case, you end up by finding an equivalence:

f_E (x) <--> |E> ; so you can consider that f_E(x) = <x|E>

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(x-x0) for <x|X|x0>) into the |E> basis <E |X|E> ; 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. May 11, 2005

### dextercioby

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

Daniel.

4. May 11, 2005

### ayalam

yes i did mean sho.

5. May 12, 2005

### dextercioby

Voilà.So you're interested in making a unitary transformation from the matrix

$$\langle i|\hat{H}|j\rangle$$ (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 x|i\rangle [/itex]

,i.e.u'll be needing the SHO-s wavefunctions.U can use only coordinate ones,just as long as u express the momentum operator in the same basis $\{\langle x| \}$.

Daniel.

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