What does [A,1] equal in a commutator?

binbagsss · Nov 9, 2019

I'm just trying to follow the below

Derivation of the commutator [A, e^(λB)] in three steps, labelled Appendix 2 to 4, expanding the exponential as a power series and concluding that it equals λ c e^(λB), with c commuting with A and B.

And I understand all, I think, except what's happened to the term when A hits 1: [A,1] ?
If I'm correct basically we're just hitting on the first operator so reducing the power by one each time of the operator in the right hand bracket

thanks

tnich · Nov 9, 2019

If I understand this correctly, ##1## is the identity operator. So what would be the value of ##[A,1]##?

Abhishek11235 · Nov 9, 2019

[A,1]=0 since 1 is pure number!

binbagsss · Nov 9, 2019

Abhishek11235 said:

[A,1]=0 since 1 is pure number!

omg i think i need some sleeep ! thank you haha :)

Abhishek11235 · Nov 9, 2019

binbagsss said:

omg i think i need some sleeep ! thank you haha :)

A typo! 1 here refers to identity operator as @tnich says. Anyways,final result is same

What does [A,1] equal in a commutator?

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