I Hermitian operators in quantum mechanics

1. Sep 12, 2016

sams

Hello everyone,

There's something I am not understanding in Hermitian operators.

Could anyone explain why the momentum operator:
px = -iħ∂/∂x
is a Hermitian operator? Knowing that Hermitian operators is equal to their adjoints (A = A), how come the complex conjugate of px (iħ∂/∂x) = px (-iħ∂/∂x) ?

Thank you so much for your support...

2. Sep 12, 2016

Demystifier

3. Sep 12, 2016

vanhees71

A hermitean operator $\hat{p}$ fulfills
$$\langle \psi_1|\hat{p} \psi_2 \rangle=\langle \hat{p} \psi_1|\psi_2 \rangle$$
for all vecotrs $|\psi_1 \rangle$ and $|\psi_2 \rangle$ in the domain of the operator. In the position representation for the momentum operator you have $\hat{p}=-\mathrm{i} \partial_x$ and the scalar product is the usual $\mathrm{L}^2$ product. Now indeed $\hat{p}$ is hermitean since via integration by parts you get
$$\langle \psi_1|\hat{p} \psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_2^*(x)(-\mathrm{i} \partial_x) \psi_1(x) = \int_{\mathbb{R}} \mathrm{d} x +\mathrm{i} \partial_x \psi_1^* \psi_2(x) = \int_{\mathbb{R}} \mathrm{d} x [-\mathrm{i} \partial_r \psi_1(x)]^{*} \psi_2(x)=\langle{\hat{p} \psi_1}|\psi_2 \rangle.$$
So $\hat{p}$ is hermitean.

4. Sep 12, 2016

sams

5. Sep 12, 2016

vanhees71

No, of course I didn't prove an obviously wrong result. Please read more carefully what I posted! The linked document doesn't make sense to me.

6. Sep 12, 2016

DrDu

What makes you think that the adjoint of an operator is the same as the complex conjugate of an operator?

7. Sep 12, 2016

vanhees71

Well, I guess that's due to the misleading notation of the linked paper :-(.

8. Sep 12, 2016

sams

I am sorry Dr. vanhees71 for this confusion. My reply was for response #2.

9. Sep 12, 2016

sams

I think any operator should be inserted in an eigenvalue equation. If this operator is Hermitian, it should leads to a physical observable and gives the same eigenvalue as its complex conjugate.

However, the definition of a Hermitian operator is:
A = A*

If the latter equation is true, can we state that the following:
px = -iħ∂/∂x = iħ∂/∂x = (px)* ?

We can clearly realize that px ≠ (px)*

10. Sep 12, 2016

vanhees71

This is very misleading, as this discussion shows. The hermitean conjugation of an operator is not simply its complex conjugation. It's not even clear, how to define it, but the adjoint operator $\hat{A}^{\dagger}$ to an operator $\hat{A}$ is defined such that
$$\langle \psi_1 |\hat{A} \psi_2 \rangle=\langle \hat{A}^{\dagger} \psi_1|\psi_2 \rangle.$$
Note that I leave out the somewhat subtle questions on domains and codomains of the operators here. This we can discuss as soon as the usual physicists' meaning of hermitean conjugation is understood.

11. Sep 12, 2016

DrDu

Congratulations! You just rediscovered the fact that complex conjugation of an operator depends on the chosen basis.

12. Sep 13, 2016

sams

Dear Sirs,

Thank you so much for this valuable discussion.

Best wishes