By Sunil Bhardwaj

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An operator is a symbol for a certain mathematical procedure which transforms one function to another.

In other words,

An operator is an instruction to carry out certain operations.

For every physical measurable or observable quantity like position, velocity, linear momentum, angular momentum, energy of the system, there is a corresponding operator in quantum mechanics.

e.g. square root $$\sqrt { }$$ is an operator, when it is written like this it has no meaning. When some quantity or number is put under it, it transforms that quantity into its square root.

Similarly $$d/dx$$ is an operator, if the function $${ x }^{ n }$$ is placed with the operator .

$$\frac { d }{ dx } \left( { x }^{ n } \right) =n.{ x }^{ n-1 }$$

So in general,

(operator).(function)= (Another function)

Generally, an operator is written with the symbol overhead (?). e.g.

Suppose $$f\left( x \right)$$ is a function and $$\widehat { A }$$ and $$\widehat { B }$$ are two different operators, then $$\boxed { \widehat { \left( A+B \right) } f\left( x \right) = \widehat { A } f\left( x \right) + \widehat { B } f\left( x \right) } \qquad ....(1)$$ e.g. Let $$\widehat { A } = \log _{ e }{ }$$ and $$\widehat { B } = \frac { d }{ dx }$$ , So for a function $$f\left( x \right) = { x }^{ 2 }$$ $$\widehat { \left( A+B \right) } f\left( x \right) = \left\{ \log _{ e }{ } + \frac { d }{ dx } \right\} { x }^{ 2 }$$ $$= \log _{ e }{ { x }^{ 2 } } + \frac { d }{ dx } { x }^{ 2 }$$ $$= 2\log _{ e }{ { x } } + 2x \qquad ....(2)$$ $$\widehat { A } f\left( x \right) + \widehat { B } f\left( x \right) = \left\{ \log _{ e }{ } \right\} { x }^{ 2 } + \left\{ \frac { d }{ dx } \right\} { x }^{ 2 }$$ $$\qquad = \log _{ e }{ { x }^{ 2 } } + \frac { d }{ dx } { x }^{ 2 }$$ $$= 2\log _{ e }{ { x } } + 2x \qquad ....(3)$$ As RHS of equation (2) and (3) are equal, therefore LHS must also be equal. i.e. $$\boxed { \widehat { \left( A+B \right) } f\left( x \right) = \widehat { A } f\left( x \right) + \widehat { B } f\left( x \right) }$$

Substraction:

Suppose $$f\left( x \right)$$ is a function and $$\widehat { A }$$ and $$\widehat { B }$$ are two different operators, then $$\boxed { \widehat { \left( A-B \right) } f\left( x \right) = \widehat { A } f\left( x \right) - \widehat { B } f\left( x \right) } \qquad ....(1)$$ e.g. Let $$\widehat { A } = \log _{ e }{ }$$ and $$\widehat { B } = \frac { d }{ dx }$$ , So for a function $$f\left( x \right) = { x }^{ 2 }$$ $$\widehat { \left( A-B \right) } f\left( x \right) = \left\{ \log _{ e }{ } - \frac { d }{ dx } \right\} { x }^{ 2 }$$ $$= \log _{ e }{ { x }^{ 2 } } - \frac { d }{ dx } { x }^{ 2 }$$ $$= 2\log _{ e }{ { x } } - 2x \qquad ....(2)$$ $$\widehat { A } f\left( x \right) - \widehat { B } f\left( x \right) = \left\{ \log _{ e }{ } \right\} { x }^{ 2 } - \left\{ \frac { d }{ dx } \right\} { x }^{ 2 }$$ $$= \log _{ e }{ { x }^{ 2 } } - \frac { d }{ dx } { x }^{ 2 }$$ $$= 2\log _{ e }{ { x } } - 2x \qquad ....(3)$$ As RHS of equation (2) and (3) are equal, therefore LHS must also be equal. i.e. $$\boxed { \widehat { \left( A-B \right) } f\left( x \right) = \widehat { A } f\left( x \right) - \widehat { B } f\left( x \right) }$$

Multiplication:

If Suppose $$f\left( x \right)$$ is a function and $$\widehat { A }$$ and $$\widehat { B }$$ are two different operators, then $$\widehat { A } \widehat { B } f\left( x \right)$$ means first operator $$\widehat { B }$$ is operated on $$f\left( x \right)$$ to produce new function $$f'\left( x \right)$$ and the operator $$\widehat { A }$$ is operated on function $$f'\left( x \right)$$. In other words we can write, $$\widehat { A } \left\{ \widehat { B } f\left( x \right) \right\}$$.

If the result of two operations is the same irrespective of the sequence in which they are performed, the two operators are said to be Commute.

i.e. If $$\widehat { A } \widehat { B } f\left( x \right) = \widehat { B } \widehat { A } f\left( x \right) \qquad$$ Then $$\widehat { A }$$ and $$\widehat { B }$$ are commutative.

If $$\widehat { A } \widehat { B } f\left( x \right) \neq \widehat { B } \widehat { A } f\left( x \right) \qquad$$ Then $$\widehat { A }$$ and $$\widehat { B }$$ are non-commutative.

e.g. if $$\widehat { A } = 4{ x }^{ 3 }, \widehat { B } = 3{ x }^{ 4 }$$ and $$f\left( x \right) = { x }^{ 2 }$$ Then $$\widehat { A } \widehat { B } f\left( x \right) = \left\{ 4{ x }^{ 3 } \right\} \left\{ 3{ x }^{ 4 } \right\} { x }^{ 2 } = 13{ x }^{ 9 }$$ and $$\widehat { B } \widehat { A } f\left( x \right) = \left\{ 3{ x }^{ 4 } \right\} \left\{ 4{ x }^{ 3 } \right\} { x }^{ 2 } = 13{ x }^{ 9 }$$ $$\therefore \widehat { A } \widehat { B } f\left( x \right) = \widehat { B } \widehat { A } f\left( x \right)$$ Thus, we can say $$\widehat { A }$$ and $$\widehat { B }$$ are commutative. But if $$\widehat { A } = \frac { d }{ dx } , \widehat { B } = 3{ x }^{ 2 }$$ and $$f\left( x \right) = \sin { x }$$ Then $$\widehat { A } \widehat { B } f\left( x \right) = \left\{ \frac { d }{ dx } \right\} \left\{ 3{ x }^{ 2 } \right\} \sin { x }$$ $$= \frac { d }{ dx } \left[ 3{ x }^{ 2 }\sin { x } \right]$$ $$= 6x\sin { x } + 3{ x }^{ 2 }\cos { x }$$ and $$\widehat { B } \widehat { A } f\left( x \right)$$ $$= \left\{ 3{ x }^{ 2 } \right\} \left\{ \frac { d }{ dx } \right\} \sin { x }$$ $$= \left\{ 3{ x }^{ 2 } \right\} \cos { x }$$ $$= 3{ x }^{ 2 }\cos { x }$$ $$\therefore \widehat { A } \widehat { B } f\left( x \right) \neq \widehat { B } \widehat { A } f\left( x \right)$$ Thus, we can say $$\widehat { A }$$ and $$\widehat { B }$$ are non-commutative.

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