What Happens When the Ladder Operator Hits the Identity Operator?

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SUMMARY

The discussion centers on the behavior of the ladder operator when it interacts with the identity operator, specifically in the context of the expression [A,1]. Participants clarify that the identity operator, denoted as ##1##, simplifies the expression to [A,1]=0, as 1 is considered a pure number. The consensus is that the ladder operator reduces the power of the operator in the right-hand bracket by one each time it is applied. Despite initial confusion, the final result remains consistent across interpretations.

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binbagsss
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I'm just trying to follow the below
expop.png

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
 
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If I understand this correctly, ##1## is the identity operator. So what would be the value of ##[A,1]##?
 
[A,1]=0 since 1 is pure number!
 
Abhishek11235 said:
[A,1]=0 since 1 is pure number!
omg i think i need some sleeep ! thank you haha :)
 
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
 

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