Having trouble showing hermitian-ness.

[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,
<br /> \frac{d}{dt}\int \psi \psi^* dq\,=\,\int\psi\frac{\partial \psi^*}{\partial t}dq\,=\,0<br />
Substituting...
<br /> \frac{\partial \psi}{\partial t}=\,-i\hat{H}\psi<br />
<br /> \frac{\partial \psi^*}{\partial t}=\,i\hat{H}^*\psi^*<br />

<br /> \int\psi\left(i\hat{H}^*\right)\psi^* dq\,-\,\int\psi^*\left(-i\hat{H}\right)\psi dq<br />

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...
<br /> \int\psi^*\hat{H}^*\psi dq \,-\,\int\psi^*\hat{H}\psi dq<br />
<br /> \int\psi^*\left(\hat{H}^*\,-\,\hat{H}\right)\psi dq=\,0<br />
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?

 

[Thaakisfox]

By definition the adjoint operator is:

&lt;Ax|y&gt;=&lt;x|A^{*}y&gt;

So if it is self adjoint : A=A^{*} then &lt;Ax|y&gt;=&lt;x|Ay&gt;

 

[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,
<br /> P f(x) = f(-x)<br />
and I use the definition
<br /> &lt;Pf(x)|g(x)&gt;\,=\,&lt;f(x)|P^*g(x)&gt;<br />
<br /> &lt;f(-x)|g(x)&gt;\,=\,&lt;f(x)|g(-x)&gt;<br />
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.

 

[Thaakisfox]

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