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

• Moolisa

#### 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)##

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##

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)

• PeroK and vanhees71