Proving U=eiA is Unitary: Exploring -1 as Exponent and Inverse

[dyn]

Hi
If A is a Hermitian operator then U = e^iA is a unitary operator.
To prove this we take the Hermitian conjugate of U

U⁺ = e^-iA = (e^iA)^-1 (1)
⇒ U⁺ = U^-1 (2)

My question is - In line (1) , -1 is used as an exponent or power while in line (2) , -1 is used to refer to the inverse of a matrix. Are these not 2 different uses of -1 ?
Thanks

 

[PeroK]

My question is - In line (1) , -1 is used as an exponent or power while in line (2) , -1 is used to refer to the inverse of a matrix. Are these not 2 different uses of -1 ?

In general, $X^{-1}$ denotes the multiplicative inverse of $X$, where $X$ is some mathematical object. In this case, you have to justify the index rule $e^{-iA} = (e^{iA})^{-1}$ for a (Hermitian) operator $A$.

 

[dyn]

In general, $X^{-1}$ denotes the multiplicative inverse of $X$, where $X$ is some mathematical object. In this case, you have to justify the index rule $e^{-iA} = (e^{iA})^{-1}$ for a (Hermitian) operator $A$.

Thanks. So the -1 in the equation you wrote is not an exponent ; it is the inverse of e^iA ? How do i justify that ?

 

[PeroK]

Thanks. So the -1 in the equation you wrote is not an exponent ; it is the inverse of e^iA ? How do i justify that ?

You prove it, using the definition of $e^A$.

 

[dyn]

I know the definition of e^A as an infinite power series but i don't know how to get the inverse of that

 

[PeroK]

I know the definition of e^A as an infinite power series but i don't know how to get the inverse of that

More generally, if operators/matrices $A$ and $B$ commute, then you can show that:$$e^Ae^B = e^{A+B}$$It shouldn't be hard to find a proof online if you don't want to work it out for yourself.

 

[dyn]

It's not that i don't work it out myself ; i don't know how to !

 

[dyn]

I found it online. Thanks