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Finding the conjugate of psi

  • Thread starter noblegas
  • Start date
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]
 

Dick

Science Advisor
Homework Helper
26,255
618
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]
Not if x and x0 are real, which I suspect they are. What is it in that case?
 
Last edited:
Not if x and x0 are real, which I suspect they are. What is it in that case?
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??
 

Dick

Science Advisor
Homework Helper
26,255
618
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??
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.
 
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.
but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugate
 

Dick

Science Advisor
Homework Helper
26,255
618
but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugate
Yesss.
 

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