- #1
Ikaros
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Homework Statement
If B is Hermitian, show that BN and the real, smooth function f(B) is as well.
Homework Equations
The operator B is Hermitian if [itex]\int { { f }^{ * }(x)Bg(x)dx= } { \left[ \int { { g }^{ * }(x)Bf(x) } \right] }^{ * }[/itex]
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 BN as B[itex]\cdot[/itex]B[itex]\cdot[/itex]B[itex]\cdot[/itex]...BN. 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, BN 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 B0 = 1 and [itex] \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } [/itex] can be treated as scalar derivatives of a normal function (a number) and BN was shown to be Hermitian, we're left with a scalar multiplied by a Hermitian, which is itself Hermitian.