Having trouble showing hermitian-ness.

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The discussion focuses on the challenges of demonstrating the hermitian nature of operators in quantum mechanics, specifically referencing Griffiths and Lifschitz texts. The user struggles with the steps involved in showing that the Hamiltonian operator is hermitian, particularly how to handle the imaginary unit 'i' in calculations. They also express confusion about applying the definition of adjoint operators to other operators, such as the parity operator, and how to manipulate functions under the inner product. The conversation highlights the importance of understanding the mathematical framework and definitions involved in proving hermitian-ness. Overall, the user seeks clarity on the demonstration process for various operators.
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(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?
 
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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;
 
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.
 
Expand f(x) and g(x') in an orthonormal basis (this exists as we have an inner product defined on the space ...).
 
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