Proof that (p + ix) operator is non-hermitian (easy)

theres one line that keeps coming up in proofs that I don't get. How do i get from

[tex]\int[/tex] ([tex]\hat{p}[/tex][tex]\Psi[/tex]1)*[tex]\Psi[/tex]2 + i [tex]\int[/tex] ([tex]\hat{x}[/tex][tex]\Psi[/tex]1)[tex]\Psi[/tex]2

to

[tex]\int[/tex] ( ([tex]\hat{p}[/tex]-i[tex]\hat{x}[/tex]) [tex]\Psi[/tex]1)*[tex]\Psi[/tex]2

using the fact that p and x are Hermitian.

im sure its painfully simple but i cant see it.
 

nrqed

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theres one line that keeps coming up in proofs that I don't get. How do i get from

[tex]\int[/tex] ([tex]\hat{p}[/tex][tex]\Psi[/tex]1)*[tex]\Psi[/tex]2 + i [tex]\int[/tex] ([tex]\hat{x}[/tex][tex]\Psi[/tex]1)[tex]\Psi[/tex]2
I guess you mean that there is a complex conjugate on (x Psi_1) in the second term)
to

[tex]\int[/tex] ( ([tex]\hat{p}[/tex]-i[tex]\hat{x}[/tex]) [tex]\Psi[/tex]1)*[tex]\Psi[/tex]2

using the fact that p and x are Hermitian.

im sure its painfully simple but i cant see it.
Well, for that step you actually do not need at all to use the fact that x and p are hermitian.

All you need to use is that i = (-i)*
 

reilly

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Why the * in the first integral, but not the second? The second integral only makes sense in QM if the wave function is real, when, in fact, most wave functions are complex.

Anyway, both p and x are Hermitian. Thus (P+iX)* =(P-iX), virtually by definition, as i is anti-Hertmitian.
Regards,
Reilly Atkinson
 
okay got it.

I put the operator into the RHS of the hermitian condition, took the complex conjugate and re-arranged it so that it was in the same form as the LHS of the hermitian condition. the inequality obviously doesnt hold because the 'i' put a minus in one, so the operator wasn't hermitian. thankyou so much guys. and sorry for the typo
 

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