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**1. The problem statement, all variables and given/known data**

If B is Hermitian, show that B

^{N}and the real, smooth function f(B) is as well.

**2. Relevant equations**

The operator B is Hermitian if [itex]\int { { f }^{ * }(x)Bg(x)dx= } { \left[ \int { { g }^{ * }(x)Bf(x) } \right] }^{ * }[/itex]

**3. The attempt at a solution**

Below is my complete solution. I'm hoping someone can review and let me know if it is okay.

Given the operator B, we can define the operator B

^{N}as B[itex]\cdot[/itex]B[itex]\cdot[/itex]B[itex]\cdot[/itex]...B

_{N}. Using this, we can show:

[itex]

\int { { f }^{ * }(x){ B }^{ N }g(x)dx= } \\ \int { { f }^{ * }(x){ B }BB...{ B }_{ N }g(x)dx= } \\ { \left[ \int { { g }^{ * }(x){ B }BB...{ B }_{ N }f(x) } \right] }^{ * }=\\ \int { { g }(x){ { B }^{ * } }{ B }^{ * }{ B }^{ * }...{ B }^{ * }_{ N }{ f }^{ * }(x) } =\\ \int { { g }(x){ B }^{ { N }^{ * } }{ f }^{ * }(x) } \\

[/itex]

Therefore, B

^{N}is also Hermitian.

Given the operator B, we can define the smooth function f(B) by a Taylor series expansion:

f(B)=[itex]\sum _{ N=0 }^{ \infty }{ \frac { 1 }{ N! } \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } } { B }^{ N }[/itex]

Since B

^{0}= 1 and [itex] \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } [/itex] can be treated as scalar derivatives of a normal function (a number) and B

^{N}was shown to be Hermitian, we're left with a scalar multiplied by a Hermitian, which is itself Hermitian.