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.
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.
Related questions
Question
100%
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F546fe60a-3e0f-4eb3-919d-41650ae193cb%2F78c47eaa-9478-4d3c-809b-e2be0954e985%2F56ttzan_processed.png&w=3840&q=75)
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.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 19 images
