If A, B Hermitian, then <v|AB|v>=<v|BA|v>*. Why?

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The discussion centers on the relationship between two Hermitian operators, A and B, and their effect on a vector v in the context of inner product notation. It establishes that if =x+iy, then =x-iy, leveraging the properties of Hermitian operators and the definition of the inner product. The proof utilizes the adjoint operator's definition, confirming that the inner product's conjugate symmetry leads to the stated equality. This conclusion is essential for understanding operator behavior in quantum mechanics.

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nomadreid
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This is certainly an elementary question, so I would be all the more grateful for the answer. Given: A and B are two Hermitian operators and v is a vector in C. If <v|AB|v>=x+iy (for x and y real), then <v|BA|v> = x-iy.
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Bra-ket notation can be confusing, so I will translate to inner product notation. If I denote the inner product of two arbitrary vectors x and y by (x,y), and denote the ket |v> by v, then <v|AB|v> means (v,ABv). The definition of "inner product" says that ##(x,y)^*=(y,x)## for all x,y. This implies that ##(v,BAv)^*=(BAv,v)##. Now you can use that A and B are hermitian, and the definition of the adjoint operator to evaluate the right-hand side.
$$(v,BAv)^*=(BAv,v)=(B^*A^*v,v)=(A^*v,Bv)=(v,ABv).$$
 
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Many thanks, Fredrik. That clears that step up.
 

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