# Simultaneous Eigenkets?

## Homework Statement

If A and B are observables, suppose that the simultaneous eigenkets of A and B, {|a',b'>} form a complete orthonormal set of base kets.
Can we always conclude that [A,B]=0

A|a'> = a' |a'>
B|b'> = b' |b'>

## The Attempt at a Solution

I Honestly don't know where to start.
What does it mean that the are "Simultaneous Eigenkets"?

I do know that it implies that you can take a measurement of both without having to destroy the previous measurement. Everywhere i look seems to start at the opposite end assuming that they commute.
So if someone can explain what "Simultaneous eigenkets" means, I can probably get a bunch further..
I want to figure this out but i can't seem to really even get started.

## Answers and Replies

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ShayanJ
Gold Member
It means that you can write ## A|a',b'\rangle=a' |a',b'\rangle ## and ## B|a',b'\rangle=b'|a',b'\rangle ##, which means ##|a',b'\rangle## is the eigenket of both A and B, so their "simultaneous eigenket"!

Dick
Homework Helper

## Homework Statement

If A and B are observables, suppose that the simultaneous eigenkets of A and B, {|a',b'>} form a complete orthonormal set of base kets.
Can we always conclude that [A,B]=0

A|a'> = a' |a'>
B|b'> = b' |b'>

## The Attempt at a Solution

I Honestly don't know where to start.
What does it mean that the are "Simultaneous Eigenkets"?

I do know that it implies that you can take a measurement of both without having to destroy the previous measurement. Everywhere i look seems to start at the opposite end assuming that they commute.
So if someone can explain what "Simultaneous eigenkets" means, I can probably get a bunch further..
I want to figure this out but i can't seem to really even get started.
|a',b'> is a simultaneous eigenket of both A and B if it's an eigenket of BOTH the operators A and B. I.e. A|a',b'>=a'|a',b'> and B|a',b'>=b'|a',b'>. Think about what the matrix and A and B look like in that basis.