# Hermitian calculation question

1. Dec 28, 2005

### maverick6664

Reading back in my book, Greiner's "QM :an introduction" I found a formula I don't understand.

Let $$\alpha$$ be a real number, $$\Delta \hat{A}, \Delta \hat{B}$$ be Hermitian operators. Now I have

$$\int (\alpha \Delta \hat{A} - i \Delta \hat{B})^* \psi^* (\alpha \Delta \hat{A} - i \Delta \hat{B}) \psi dx = \int \psi^* (\alpha \Delta \hat{A} + i \Delta \hat{B}) ( \alpha \Delta \hat{A} - i \Delta \hat{B}) \psi dx$$

This leads to an expected result to prove Heisenberg's Uncertainty Relations, but I think the right-hand side of this formula should be

$$\int \psi^* (\alpha \Delta \hat{A}^* + i \Delta \hat{B}^*) (\alpha \Delta \hat{A} - i \Delta \hat{B}) \psi dx$$

I should be wrong, but I don't know why. Operators $$\Delta \hat{A}, \Delta \hat{B}$$ can be complex... (or are they always real?) So will anyone tell me how or why it's correct?

Thanks in advance!

Last edited: Dec 28, 2005
2. Dec 29, 2005

### Staff: Mentor

Since the operators are Hermitian, you know that:

$$\Delta \hat{A}^* = \Delta \hat{A}$$

$$\Delta \hat{B}^* = \Delta \hat{B}$$

3. Dec 29, 2005

### Staff: Mentor

Just a note for clarity. A Hermitian operator is one that equals its adjoint:
$$\hat{A} = \hat{A}^{\dagger}$$

This implies that the eigenvalues of a Hermitian operator are real (and of course that their mean values are also real). But the operator can certainly be complex and still be Hermitian.

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