# Simple question about matrix mechanics

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1. Nov 7, 2014

### Jdraper

HI, I've been running through my lectures notes and have stumbled upon something i can't quite figure out.

I am given

Ψ(x)=∑a_iΨ_i(x)

Then

OΨ(x)=∑ a_i O Ψ_i(x) , where O is an operator acting upon Ψ

Then i am given something which i don't quite understand,

OΨ_i(x) = ∑ O_ji Ψ_j(x) , Where O_ji (i assume) is now a matrix

I understand why the a_i terms disappear in this second equation but i'm unsure why the operator turns into a matrix and why the sum is now over all j's rather than i's

2. Nov 7, 2014

### ShayanJ

Well, in QM, we say that for a complete set of eigenstates we have $I=\sum_j \psi_j^\dagger \psi_j$(where I is the identity matrix and $\psi^\dagger$ means complex conjugating the elements of the column matrix and also transposing it). So we can write:
$O \psi_i=O I \psi_i=O(\sum_j \psi_j^\dagger \psi_j) \psi_i=\sum_j \psi_j^\dagger O \psi_i \psi_j$
Now if we set $O_{ji}=\psi_j^\dagger O \psi_i$, we'll have the desired result.
I should add the explanation that the $\psi$s are column matrices and O was a matrix all along the way. The only difference is, when we write O without subscripts, it means we don't know(or don't write) O's elements and only know what O does to each column matrix. But after a certain point, we find out(or decide to write) O's elements.

Last edited: Nov 7, 2014
3. Nov 7, 2014

### Jdraper

Ok I think i understand, I am assuming it is a property of the identity matrix that you can input it between O and Ψ and it will still be equal to OΨ. If that is true then i understand everything, think i may brush up on my knowledge of matricies before progressing any further.

Thanks Shyan.

4. Nov 7, 2014

### ShayanJ

Oh god...sorry man. I was wrong!
At first, $I=\sum_j \psi_j \psi_j^\dagger$. The reversed product gives a number and can't be equal to an operator!
Second, I couldn't just send O through the column matrix!
So I should've written:
$O \psi_i=IO\psi_i=(\sum_j \psi_j \psi_j^\dagger) O \psi_i=\sum_j \psi_j \psi_j^\dagger O \psi_i=\sum_j O_{ji}\psi_j$