Show that the Hamiltonian is Hermitian for a particle in 1D

[Moolisa]

Homework Statement

(d) Show that $\hat H$ Hermitian for a particle in one dimension.

Relevant Equations

$\hat H$= $\frac{\hat p^2}{2m} +V(x)$
$\hat p^\dagger=\hat p$

I need help with part d of this problem. I believe I completed the rest correctly, but am including them for context

(a)Show that the hermitian conjugate of the hermitian conjugate of any operator $\hat A$ is itself, i.e. $(\hat A^\dagger)^\dagger$
(b)Consider an arbitrary operator $\hat A$. Show that $\hat A^\dagger \hat A$, a Hermitian operator, and its expectation value satisfies $\langle \hat A^\dagger \hat A \rangle \geq 0$
(c) Show that ˆx is Hermitian for a particle in one dimension.
(d) Show that $\hat H$ is Hermitian for a particle in one dimension.

Attempt

Using $\hat H$= $\frac{\hat p^2}{2m} +V(x)$

$\langle f~|\frac{\hat p^2}{2m} +V(x)~g \rangle$

$=\langle f~|\frac{\hat p^2}{2m} ~g \rangle~+\langle f~|V(x)~g \rangle$

$=\frac{1}{2m}~\langle f~|\hat p^2 ~g \rangle~+\langle f~|V(x)~g \rangle$

$=\frac{1}{2m}~\langle (\hat p^2)^\dagger f~|~g \rangle~+\langle\mathbf {V^*(x)} f~|~g \rangle$ Is it wrong to write V(x) like this? Since it isn't an operator, I assumed it might act like a constant

$=\frac{1}{2m}~\langle \hat p^2 f~|~g \rangle~+\langle V^*(x) f~|~g \rangle$ Since $\hat p$ is hermitian $(\hat p^2)^\dagger=\hat p^2$

$=\frac{1}{2m}~\langle g~|~\hat p^2 f \rangle^*~+\langle g~|~V^*(x) f \rangle^*$

$=(\langle g|\frac{\hat p^2}{2m}+ V^*(x) f \rangle^*)^*$

$=\langle \frac{\hat p^2}{2m}+ V^*(x) f|g \rangle$

Is this the correct approach? If it is, what do I do with my $V^*(x)$

 

[PeroK]

In general, you can only show something is Hermitian relative to a given inner product. For $V(x)$ you will need to use the definition of the inner product you are using.

 

[Moolisa]

In general, you can only show something is Hermitian relative to a given inner product. For $V(x)$ you will need to use the definition of the inner product you are using.

Thank you!

Do I need to use the definition of the inner product from the very beginning, or can I start from here?
$+\langle f~|V(x)~g \rangle$

 

[PeroK]

Thank you!

Do I need to use the definition of the inner product from the very beginning, or can I start from here?
$+\langle f~|V(x)~g \rangle$

That's not a definition of an inner product. That could be any inner product.

 

[Moolisa]

That's not a definition of an inner product. That could be any inner product.

I'm sorry, I'm not entirely sure how to word it, but I was wondering if I could do this

$=\langle f~|\frac{\hat p^2}{2m} ~g \rangle~+\langle f~|V(x)~g \rangle$

$=\frac{1}{2m}~\langle (\hat p^2)^\dagger f~|~g \rangle~+\langle f~|V(x)~g \rangle$

$=\frac{1}{2m}~\langle \hat p^2 f~|~g \rangle+\langle f|V(x) g\rangle$

$=\int_{-\infty}^{\infty} (\frac{\hat p^2}{2m} f)^*g~dx +\int_{-\infty}^{\infty} f^*Vg~dx$

$=\int_{-\infty}^{\infty} (\frac{\hat p^2}{2m} f)^*g~dx +\int_{-\infty}^{\infty} (Vf)^*g~dx$

$=\int_{-\infty}^{\infty} (\frac{\hat p^2}{2m} f +Vf)^*~g~dx$

$=\langle \hat H f~|~g \rangle$

So since $\langle \hat H f~|~g \rangle= \langle f|\hat H~g \rangle$, $\hat H$ is hermitian

Is this possible, or do I have to start with the integration method from the beginning, and use integration method for both sides of the "+" sign? ( basically, use integration for both the $\hat p^2$ side and the V(x) side)