For a Hermitian operator Â, ſy°(x)[Â¥(x)]dx = [y(x)[Â¥(x)]*dx . Assume th ƒ(x)= (a +ib)f(x) where a and b are constants. Show that if  is a Hermitian operator, b = 0, so that the eigenvalues of f(x) are real.

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How do I demonstrate a function is a Hermitian operator? That is my biggest question. If I am understanding well, I need to show that A is a Hermitian operator to get started on this problem. Please help.  

For a Hermitian operator Â, ſy'(x)[Â¥(x)]dx= [y(x)[Â¥(x)] dx . Assume th
Af(x) = (a + ib) f(x) where a and b are constants. Show that if  is a Hermitian
operator, b =0, so that the eigenvalues of f(x) are real.
Transcribed Image Text:For a Hermitian operator Â, ſy'(x)[Â¥(x)]dx= [y(x)[Â¥(x)] dx . Assume th Af(x) = (a + ib) f(x) where a and b are constants. Show that if  is a Hermitian operator, b =0, so that the eigenvalues of f(x) are real.
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