Why Does an Operator in Quantum Mechanics Transform into a Matrix?

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Discussion Overview

The discussion revolves around the transformation of operators in quantum mechanics into matrix representations, particularly in the context of applying operators to wave functions expressed as linear combinations of eigenstates. Participants explore the mathematical relationships and properties involved in this transformation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an equation involving an operator acting on a wave function expressed as a sum of eigenstates and questions why the operator appears as a matrix in a subsequent equation.
  • Another participant explains the role of the identity matrix in quantum mechanics and how it can be inserted into equations without changing their equality, suggesting this is key to understanding the transformation.
  • A later reply acknowledges a misunderstanding regarding the identity matrix and the manipulation of operators, leading to a corrected formulation of the operator's action on the wave function.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are corrections and differing interpretations of the mathematical properties involved in the transformation of operators into matrices.

Contextual Notes

Limitations include potential misunderstandings of matrix properties and operator manipulation, as well as the dependence on specific definitions of the identity matrix and operator forms.

Jdraper
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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

Thanks in advance for your help, John.
 
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Well, in QM, we say that for a complete set of eigenstates we have [itex]I=\sum_j \psi_j^\dagger \psi_j[/itex](where I is the identity matrix and [itex]\psi^\dagger[/itex] means complex conjugating the elements of the column matrix and also transposing it). So we can write:
[itex] 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[/itex]
Now if we set [itex]O_{ji}=\psi_j^\dagger O \psi_i[/itex], we'll have the desired result.
I should add the explanation that the [itex]\psi[/itex]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.
 
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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.
 
Oh god...sorry man. I was wrong!
At first, [itex]I=\sum_j \psi_j \psi_j^\dagger[/itex]. 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:
[itex] 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[/itex]
 
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