Homework Help: 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 ...).