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,
$$\frac{d}{dt}\int \psi \psi^* dq\,=\,\int\psi\frac{\partial \psi^*}{\partial t}dq\,=\,0$$
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?

 

[Thaakisfox]

By definition the adjoint operator is:

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

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

 

[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.

 

[Thaakisfox]

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

 

[paranormal]

I can understand your frustration in trying to show the hermitian-ness of operators. It is a fundamental concept in quantum mechanics and is essential in understanding the physical significance of operators and their corresponding observables.

Firstly, I would like to clarify that "hermitian-ness" is indeed a word and it refers to the property of an operator being hermitian.

To address your question, the method for showing the hermitian-ness of an operator is generally the same. It involves checking if the operator satisfies the hermitian property, which is defined as the operator being equal to its own adjoint. In mathematical terms, this can be written as A = A^*.

In the case of the Hamiltonian operator, Lifschitz showed its hermitian-ness by using the properties of complex conjugates and integration. The key point to note here is that the operator \hat{H} is self-adjoint, meaning that its adjoint is equal to itself. Therefore, when you substitute the expressions for \frac{\partial \psi}{\partial t} and \frac{\partial \psi^*}{\partial t}, you end up with \int \psi^*\hat{H}^*\psi dq - \int \psi^*\hat{H}\psi dq = 0, which proves the hermitian-ness of the Hamiltonian operator.

For other operators, the method may differ slightly depending on the specific properties of the operator. However, the key idea remains the same - to show that the operator is equal to its own adjoint. This can be done by using mathematical properties and techniques such as integration, complex conjugation, and eigenvalue equations.

I would suggest practicing with different operators and understanding their properties to better grasp the concept of hermitian-ness. Additionally, consulting with your instructor or peers can also be helpful in understanding and solving any difficulties you may encounter. Keep persevering and I am sure you will be able to confidently show the hermitian-ness of operators in no time.