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

• I
• dyn
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.f

#### dyn

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

U+ =
e-iA = (eiA)-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

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##.

Last edited:
topsquark
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 eiA ? How do i justify that ?

Thanks. So the -1 in the equation you wrote is not an exponent ; it is the inverse of eiA ? How do i justify that ?
You prove it, using the definition of ##e^A##.

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

I know the definition of eA 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.

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

topsquark and PeroK
I found it online. Thanks