Operators A & B Commute: Explain Why or Why Not?

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Homework Help Overview

The discussion revolves around the commutation of two mathematical operators, specifically the square function (SQR) and the square root function (SQRT). The original poster is trying to understand why these operators do not commute, as indicated by their textbook.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the definition of commutation using the equation ABf(x) - BAf(x) = 0. They express confusion over their calculations suggesting that the operators commute, despite the textbook stating otherwise. Other participants question the assumptions about the square root operation, particularly regarding its applicability to negative values.

Discussion Status

Participants are exploring the nuances of the square and square root operations, particularly in relation to their definitions and the conditions under which they can be applied. Some guidance has been offered regarding the implications of negative inputs for the square root function, contributing to the understanding of why the operators do not commute.

Contextual Notes

There is an underlying assumption that the functions involved are real-valued, which affects the discussion about the square root operation. The original poster references a textbook, indicating that there may be specific constraints or definitions provided therein that are influencing the discussion.

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Homework Statement


Determine whether or not the following pairs of operators commute...and there was one I could not solve...according to the back of the textbook, I do understand 14.c does NOT commute, but I don't understand...

(14)c.
A = SQR
B = SQRT

Homework Equations


ABf(x) - BAf(x) = 0

The Attempt at a Solution


ABf(x) = A[f(x)]1/2 = f(x)
BAf(x) = Bf2(x) = f(x)...so I thought they DO commute, but the textbook says NO! Could someone explain? Thanks!
 
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terp.asessed said:

Homework Statement


Determine whether or not the following pairs of operators commute...and there was one I could not solve...according to the back of the textbook, I do understand 14.c does NOT commute, but I don't understand...

(14)c.
A = SQR
B = SQRT

Homework Equations


ABf(x) - BAf(x) = 0

The Attempt at a Solution


ABf(x) = A[f(x)]1/2 = f(x)
BAf(x) = Bf2(x) = f(x)...so I thought they DO commute, but the textbook says NO! Could someone explain? Thanks!
I think you're getting lost in the symbolism. The question is asking whether
##(\sqrt{f(x)})^2 = \sqrt{(f(x))^2}##
Can you always take the square root of something?
 
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Wait, you mean,
Mark44 said:
##(\sqrt{f(x)})^2 = \sqrt{(f(x))^2}##
Can you always take the square root of something?

are different? I'm sorry, but I'm confused...
 
Well, if they're different, then the two operations aren't commutative. If you believe they are different, why are they different?

I'll ask again, can you always take the square root of something?
 
Come to think of it, one can't square root of the function that is negative...so NOT always...
 
terp.asessed said:
Come to think of it, one can't square root of the function that is negative...so NOT always...
Right. That's why the two operations don't commute. Good!
 
GOTCHA----thank you!
 

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