Not if x and x0 are real, which I suspect they are. What is it in that case?1. The problem statement, all variables and given/known data
Find the conjugate of
[tex] \varphi[/tex]=[tex]exp(-x^2/x_0^2)[/tex]
2. Relevant equations
3. The attempt at a solution
Isn't the conjugate [tex] \varphi[/tex]^{*}=[tex]exp(x^2/x_0^2)[/tex]
oh ,my solution would only be correct if x/x0 is imaginary.would my expressionNot if x and x0 are real, which I suspect they are. What is it in that case?
Right, sort of. If x is imaginary the conjugate(exp(x))=exp(-x). If x is real then conjugate(exp(x))=exp(x). But your solution is only correct if (x/x0)^2 is purely imaginary.oh ,my solution would only be correct if x/x0 is imaginary.would my expression
[tex]
exp(-x^2/x_0^2)
[/tex] not change when taking its conjugate??
but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugateRight, sort of. If x is imaginary the conjugate(exp(x))=exp(-x). If x is real then conjugate(exp(x))=exp(x). But your solution is only correct if (x/x0)^2 is purely imaginary.
Yesss.but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugate