# Having trouble showing hermitian-ness.

1. May 19, 2013

### mateomy

(Is that a word? I dunno.)

Anyway,

I'm going through Griffiths QM and I'm also supplementing it with Lifschitz QM. I can't seem to show whether or not an operator is hermitian or not.

For instance, Lifschitz shows the hermitian-ness of the Hamiltonian,
$$\frac{d}{dt}\int \psi \psi^* dq\,=\,\int\psi\frac{\partial \psi^*}{\partial t}dq\,=\,0$$
Substituting...
$$\frac{\partial \psi}{\partial t}=\,-i\hat{H}\psi$$
$$\frac{\partial \psi^*}{\partial t}=\,i\hat{H}^*\psi^*$$

$$\int\psi\left(i\hat{H}^*\right)\psi^* dq\,-\,\int\psi^*\left(-i\hat{H}\right)\psi dq$$

In the next step he does away with the $i$ and I'm not sure how he pulls that off because, say, you're checking if the deriviative $\frac{d}{dx}$ is hermitian or not, it ends up being crucial to the hermitian-ness that it be multiplied by $i$. Moving on with Lifschitz...
$$\int\psi^*\hat{H}^*\psi dq \,-\,\int\psi^*\hat{H}\psi dq$$
$$\int\psi^*\left(\hat{H}^*\,-\,\hat{H}\right)\psi dq=\,0$$
Which shows that (due to the constancy of the norm'd $\psi$'s) $\hat{H^*}-\hat{H}=0$.

Except for the part I mentioned above, I understand how this works. I just don't know how to show it for other operators. Is the method pretty much the same?

2. May 19, 2013

### Thaakisfox

By definition the adjoint operator is:

$$<Ax|y>=<x|A^{*}y>$$

So if it is self adjoint : $$A=A^{*}$$ then $$<Ax|y>=<x|Ay>$$

3. May 19, 2013

### mateomy

Yeah, I understand the definition...its the demonstration I'm having issues with. For instance if I have an operator such as the parity operator ($P$), such that,
$$P f(x) = f(-x)$$
and I use the definition
$$<Pf(x)|g(x)>\,=\,<f(x)|P^*g(x)>$$
$$<f(-x)|g(x)>\,=\,<f(x)|g(-x)>$$
I don't know which step to take after this. I can't justify pulling the negative outside of the functions because that depends on whether or not they are even or odd. So confused.

4. May 19, 2013

### Thaakisfox

Expand f(x) and g(x') in an orthonormal basis (this exists as we have an inner product defined on the space ...).