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

• Moolisa
In summary: You can start with the integration method from the beginning, and use integration method for both sides of the "+" sign.
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
PeroK said:
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##

Moolisa said:
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
PeroK said:
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

## 1. What is a Hamiltonian in physics?

The Hamiltonian is a mathematical operator used in quantum mechanics to determine the energy of a system. It is often denoted by the symbol H and contains information about the system's position and momentum.

## 2. How is a Hamiltonian related to a particle in 1D?

A Hamiltonian for a particle in 1D is a specific form of the Hamiltonian operator that describes the energy of a particle moving in one dimension, such as along a straight line. It takes into account the particle's position and momentum in that dimension.

## 3. Why is it important to show that the Hamiltonian is Hermitian for a particle in 1D?

Hermiticity is an important property of the Hamiltonian because it guarantees that the energy eigenvalues (solutions to the Schrödinger equation) are real and the corresponding eigenstates are orthogonal. This ensures that the Hamiltonian is a valid representation of the physical system.

## 4. How can you mathematically prove that the Hamiltonian is Hermitian for a particle in 1D?

To show that the Hamiltonian is Hermitian, we need to prove that it satisfies the Hermitian operator property, which states that the operator is equal to its own conjugate transpose. In other words, H = H†. This can be done using mathematical operations and transformations on the Hamiltonian equation.

## 5. What are the implications if the Hamiltonian is not Hermitian for a particle in 1D?

If the Hamiltonian is not Hermitian, it means that the energy eigenvalues may not be real, and the corresponding eigenstates may not be orthogonal. This would make the Hamiltonian an invalid representation of the physical system and could lead to incorrect predictions and results in calculations.

Replies
10
Views
1K
Replies
3
Views
1K
Replies
0
Views
757
Replies
17
Views
1K
Replies
9
Views
763
Replies
0
Views
659
Replies
2
Views
489
Replies
1
Views
796