Operators, symbols, are no good if one need to keep thinking about the underlying representation in order to use it. To this end, we need to be able to reason about operators without going back to its underlying representation. I call it operator calculus.
The simplest rule of operator is linearity.
$ (\hat{A + B})\hat{C} = \hat{A}\hat{C} + \hat{B}\hat{C} $.
In general, operators do not commute, i.e. $ \hat{A} \hat{B} \ne \hat{B}\hat{A} $. But they can be compensated by commutator $ [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} $. This definition automatically implies $ [\hat{A}, \hat{B}] = -[\hat{B}, \hat{A}] $.
Commutators are also linear $ [\hat{A} + \hat{B}, \hat{C}] = (\hat{A} + \hat{B})\hat{C} - \hat{C}(\hat{A} + \hat{B}) = \hat{A}\hat{C} + \hat{B}\hat{C} - \hat{C}\hat{A} + \hat{C}\hat{B} = [\hat{A},\hat{C}] + [\hat{B}, \hat{C}] $.
This identity is useful to decompose product of operators inside a commutator.
$ [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C}+\hat{B}[\hat{A}, \hat{C}] $.
To show that, we expands the commutators on the right.
$ [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}] = (\hat{A}\hat{B}-\hat{B}\hat{A})\hat{C} + \hat{B}(\hat{A}\hat{C}-\hat{C}\hat{A}) = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{A}\hat{C} + \hat{B}\hat{A}\hat{C} -\hat{B}\hat{C}\hat{A} = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{C}\hat{A} = [\hat{A},\hat{B}\hat{C}] $
Same principle applies to commutator, once we prove a certain identity in commutator, then let's not expand the operator. That's how one maybe able to handle the complexity.
To apply that, let's see what can we do to deal with an operator product on the left, we have
$ [\hat{A}\hat{B}, \hat{C}] = -[\hat{C}, \hat{A}\hat{B}] = -([\hat{C}, \hat{A}]\hat{B} + \hat{A}[\hat{C}, \hat{B}]) = [\hat{A}, \hat{C}]\hat{B} + \hat{A}[\hat{B}, \hat{C}] $
The simplest rule of operator is linearity.
$ (\hat{A + B})\hat{C} = \hat{A}\hat{C} + \hat{B}\hat{C} $.
In general, operators do not commute, i.e. $ \hat{A} \hat{B} \ne \hat{B}\hat{A} $. But they can be compensated by commutator $ [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} $. This definition automatically implies $ [\hat{A}, \hat{B}] = -[\hat{B}, \hat{A}] $.
Commutators are also linear $ [\hat{A} + \hat{B}, \hat{C}] = (\hat{A} + \hat{B})\hat{C} - \hat{C}(\hat{A} + \hat{B}) = \hat{A}\hat{C} + \hat{B}\hat{C} - \hat{C}\hat{A} + \hat{C}\hat{B} = [\hat{A},\hat{C}] + [\hat{B}, \hat{C}] $.
This identity is useful to decompose product of operators inside a commutator.
$ [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C}+\hat{B}[\hat{A}, \hat{C}] $.
To show that, we expands the commutators on the right.
$ [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}] = (\hat{A}\hat{B}-\hat{B}\hat{A})\hat{C} + \hat{B}(\hat{A}\hat{C}-\hat{C}\hat{A}) = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{A}\hat{C} + \hat{B}\hat{A}\hat{C} -\hat{B}\hat{C}\hat{A} = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{C}\hat{A} = [\hat{A},\hat{B}\hat{C}] $
Same principle applies to commutator, once we prove a certain identity in commutator, then let's not expand the operator. That's how one maybe able to handle the complexity.
To apply that, let's see what can we do to deal with an operator product on the left, we have
$ [\hat{A}\hat{B}, \hat{C}] = -[\hat{C}, \hat{A}\hat{B}] = -([\hat{C}, \hat{A}]\hat{B} + \hat{A}[\hat{C}, \hat{B}]) = [\hat{A}, \hat{C}]\hat{B} + \hat{A}[\hat{B}, \hat{C}] $
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