Question

Operators question

Answer 1

If I know that two operators commute (if necessary we can assume they are Hermitian):

\[[A,B]=0\]

Is this enough to imply that:
\[[\sqrt{A},B] = 0\]

Answer 2

They don't commute, that's why they're called commutators which is basically how badly they don't commute.

Answer 3

Nerd

Answer 4

!! The commutator tells you how much they commute, if their commutator is 0 they 100% commute, just look:

\[[A,B]=0\]
\[AB-BA=0\]
\[AB=BA\]
\[\uparrow\]commutes!

Answer 5

You can't do that

Answer 6

boom just did

Answer 7

fite me

Answer 8

\[AB \neq BA\]

Answer 9

Omg

Answer 10

Please don't say it's just a prank bro next

Answer 11

Wait "bra" get it

Answer 12

it's real life, son

Answer 13

\[[x,p]=0\]
\[xp-px=0\]
\[xp \neq px\]

Answer 14

put a function there and check

Answer 15

Hahaha stop trolling me

Answer 16

I'm serious lol

Answer 17

They don't commute :S

Answer 18

\[[x,p]=0\]
\[xp-px=0\]
add \(px\) to both sides
\[xp-px+px=px\]
use parenthesis to see the obvious:
\[xp+(-px+px)=px\]
if you agree that
\[-px+px= 0\]
then you disagree with:
\[xp \neq px\]

Answer 19

YOU CAN'T JUST DO THAT

Answer 20

You need to be careful with operators

Answer 21

Hahaha, no you are misguided somehow, probably been playing with x and p operators too much.

Answer 22

I think you're messing with me

Answer 23

Haha, I wish I was now, unfortunately you're wrong :P

Answer 24

if A and B commute, their commutator is 0, it's exactly why it's defined this way, give me an example of two operators that commute which don't have commutator 0.