Expectation values for expanded wave functions

[betba]

So I'm a little confused on the notation when working with wave functions constructed as a linear combination of an orthornormal basis set. Like on the form:

$\Phi$=Ʃ_n c_nψ_n

If I want to find the expectation value represented by the operator O for the state described by $\Phi$, I would calculate the inner product between $\Phi$ and O$\Phi$, like:

<$\Phi$|O|$\Phi$> = ∫dq $\Phi$*(q)O$\Phi$(q) (assuming $\Phi$ is normalized so <$\Phi$|$\Phi$> = 1)

And now comes the question: When I insert the expanded wave function, why is 2 different indices used for the summations/basis functions:

<$\Phi$|O|$\Phi$> = ∫dq(Ʃ_n c_n*ψ_n*)O(Ʃ_m c_mψ_m)

This is how the derivations look like in most textbooks, and I don't understand the difference between n and m. I would think the indices should be the same, as it is the same wave function, $\Phi$.

Thanks in advance!

 

[cgk]

That is not an issue of wave functions or quantum mechanics. That is simply algebra: how you expand the product of a sum. Observe:
$$(a+b+c)*(a+b+c) = a*(a+b+c) + b*(a+b+c) + c*(a+b+c) = a*a + a*b + a*c + b*a + b*b + ...$$
That is exactly what you are doing in this case. Note that the result is not simply
$$a*b+b*b+c*c.$$

 

[betba]

But why use 2 different sums for the same wave function? $\Phi$ is equal to Ʃ_nc_nψ_n, not Ʃ_mc_mψ_m. Or cleary it is, but then n = m?? Then why not just only use n in the first place?

 

[cgk]

But why use 2 different sums for the same wave function? $\Phi$ is equal to Ʃ_nc_nψ_n, not Ʃ_mc_mψ_m. Or cleary it is, but then n = m?? Then why not just only use n in the first place?

$\Phi=\sum c_n \Psi_n$ is the short form for $\Phi=\Psi_1 + \Psi_2 + \Psi_3 + \ldots + \Psi_N$. Please think for a second about what happens when you insert this expression in two places into an expression like $\Phi\cdot\Phi$. Play it through with an example with two terms if you must. Again, this has nothing at all to do with quantum mechanics or wave functions: This is elementary algebra, expanding the product of two sums!

 

[betba]

Im not sure what your point is (maybe I am just very bone-headed). I know how to expand the product of 2 sums. Thats not the problem. The problem is inserting 2 different expressions for the same wave function. Like putting it your way:

$\Phi$ = $\psi$₁ + $\psi$₂ + $\psi$₃...$\psi$_n

When calculating the product $\Phi$*$\cdot$$\Phi$ I insert the expression for $\Phi$:

($\psi$₁ + $\psi$₂ + $\psi$₃...$\psi$_n)*$\cdot$($\psi$₁ + $\psi$₂ + $\psi$₃...$\psi$_n) and get n² terms. Here I clearly inserted the same expression for $\Phi$.
In all textbooks I've read so far, what is actually inserted is:

$\Phi$*$\cdot$$\Phi$ = ($\psi$₁ + $\psi$₂ + $\psi$₃...$\psi$_n)*$\cdot$($\psi$₁ + $\psi$₂ + $\psi$₃...$\psi$_m)

What does this m mean? Why don't they use n for both sums. Its the same wave function?

A link to a derivation of the variational method/principle http://vergil.chemistry.gatech.edu/notes/quantrev/node28.html

Here the indices i and j are used interchangeably, and I don't understand why.

 

[ZealScience]

I think that n and m are index of summation. If you have learned mathematics about Σ summation you should have known that these numbers mean sum over these numbers of terms.

In quantum mechanics, states are usually represented by these wave functions and bras and kets. Then these numbers are basically number of eigenkets or eigenfunctions that can form a complete set which is a basis. Basis are just like x y z in describing position. This is my personal view.

 

[ZealScience]

Basis in quantum mechanics are like tossing a quantum coin with eigenkets |head> and |tail>, which are sufficient to describe the whole system.