Show a real, smooth function of Hermitian operator is Hermitian

1. Apr 9, 2013

Ikaros

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

If B is Hermitian, show that BN and the real, smooth function f(B) is as well.

2. Relevant equations

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

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 BN as B$\cdot$B$\cdot$B$\cdot$...BN. Using this, we can show:

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

Therefore, BN is also Hermitian.

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

f(B)=$\sum _{ N=0 }^{ \infty }{ \frac { 1 }{ N! } \frac { { \partial }^{ N }f }{ \partial { B }^{ N } } } { B }^{ N }$

Since B0 = 1 and $\frac { { \partial }^{ N }f }{ \partial { B }^{ N } }$ 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.

2. Apr 16, 2013

Ikaros

I've sought assistance from others regarding the above, and everyone believes it's okay . However, I'd love some feedback in this thread on my approach.

3. Apr 16, 2013

Dick

I would say exactly the same thing. If B is hermitian then B^N is hermitian. If B^N is hermitian then any power series in B is also hermitian. Do you see any flaw in that?

4. Apr 17, 2013

Ikaros

Hi Dick and thanks for your reply. I don't see any flaw in the reasoning, but I'm hoping my solution above conveys this clearly.

5. Apr 17, 2013

Dick

It's clear enough to me. But I do think the notation is a little ugly. And are you clear on what something like $B^*$ means? Using bra-ket notation would make this a lot nicer.

6. Apr 17, 2013

micromass

Staff Emeritus
What does the notation $\frac{\partial^N f}{\partial B^N}$ mean? I don't think I've ever encountered that...

7. Apr 18, 2013

Ikaros

Thanks Dick. I should be practicing my Dirac notation more...

$\left< { f }|{ \hat { B } }|{ g } \right> ={ \left< { g }|{ \hat { B } }|{ f } \right> }^{ * }$

As for $B^*$, it is the Hermitian conjugate. In this case, I probably should be using the dagger notation ${ B }^{ \dagger }$.

I may have been a bit lazy with the notation there because my cut and paste from LaTex didn't work. I'll cut and paste the image.

I've been shown that a function of an operator f(A) can be defined by a Taylor series expansion:

where is a scalar calculated in the same way as the derivative of a normal function.

8. Apr 18, 2013

Dick

Sure. So Hermitian means $<f|Bg>=<Bf|g>$ for any f and g. That makes it 'almost' obvious that $<f|B^ng>=<B^nf|g>$. If it's not sufficiently obvious, it's easy to formalize the proof with induction.