Quantum Mechanics - Induction Method

[izzmach]

Let a be a lowering operator and a† be a raising operator.

Prove that a((a†)^n) = n (a†)^(n-1)

Professor suggested to use induction method with formula:

((a†)(a) + [a,a†]) (a†)^(n-1)

But before start applying induction method, I would like to know where the given formula comes from. Someone please explain it briefly? Thank you.

 

[Simon Bridge]

If ##a## is a lowering operator and ##a^\dagger##
They are also known as "ladder" operators ... because ##a## means "go down one step" and ##a^\dagger## means "go up one step".
##(a^\dagger)^n## means "go up n steps"... and so on.
##a(a^\dagger)## means "go up one step then go down one" (operators work right-to-left).

Therefore the formula ##a(a^\dagger)^n = n(a^\dagger)^{(n-1)}## means ...

 

[strangerep]

Let a be a lowering operator and a† be a raising operator.
Prove that a((a†)^n) = n (a†)^(n-1)
[...]

You also have a https://www.physicsforums.com/threads/quantum-mechanics-lowering-operator.820823/[/url , so this discussion should probably continue there.

In any case, I think you have not stated the problem question accurately. (In general, it is "commutation by ##A##" which can be interpreted as "differentiation by ##A^\dagger##". Your formula only applies if the expression acts on a vacuum state ##|0\rangle## which is annihilated by ##A##.)

There is a reason why the homework guidelines emphasize that you must state the question exactly as given, and also write out relevant equations (in this case, the commutation relation between ##A## and ##A^\dagger##).