# B Hamiltonian vs Momentum

1. Jul 15, 2017

### mieral

To get the dynamics of particles in a box. You are supposed to get the Hamiltonian which is potential energy plus kinetic energy. But does the potential energy take into account the momentum of the particles in the box? What happens if you change the momentum of the particles.. do the potential changes too?

And what would happen if you don't solve for the Hamiltonian (or potential plus kinetic energy) but only momentum of the particles in the box.... from the momentum alone.. can you get the potential?

2. Jul 15, 2017

### Orodruin

Staff Emeritus
The potential is independent of momentum, it only depends on the position.

Momentum is not a well-defined property of a particle in an (infinite potential well), just as position is not.

3. Jul 15, 2017

### mieral

But once you get the potential. You can use it to solve for the momentum?

In the Schrodinger Equation. You need to first get the potential then solve for momentum. Or can you only solve for momentum in the Schrodinger equation without any potential and what would be the result?

4. Jul 15, 2017

### Orodruin

Staff Emeritus
No. Again, the momentum is not well defined.

You cannot solve for momentum.

5. Jul 15, 2017

### mieral

But why are we solving for the momentum operator in quantum mechanics?

6. Jul 15, 2017

### Orodruin

Staff Emeritus
We are not. If you want a more elaborated answer you will have to provide an example of this being done.

There is no momentum operator on the particle in a box Hilbert space.

7. Jul 15, 2017

### mieral

What setup where you need to use the momentum operator in the QM? Maybe it is solving for the standing waves or probabilistic orbitals momentum?

8. Jul 15, 2017

### stevendaryl

Staff Emeritus
I sort of know what you mean, but it's a little misleading to say that, because even for a particle in a box, we write:

$H = \frac{p^2}{2m} + V$

where $p$ is the momentum operator. So it's confusing to students to say that there is no momentum operator in this case. There is a technical sense in which it's true, but it's confusing to say it.

9. Jul 15, 2017

### mieral

stevendaryl.. is there a version of the Schrodinger Equation where it's not the total energy = potential + kinetic but you use the total momentum only? What would be the result of the PSI(x,t) for this?

10. Jul 15, 2017

### stevendaryl

Staff Emeritus
I don't understand what you mean by "use the total momentum only". Use it for what?

11. Jul 15, 2017

### Orodruin

Staff Emeritus
The operator $\hat p^2$ is well defined on the relevant function space unlike $\hat p$. Writing it like that is an advanced form of lying to children. I do not see a problem in mentioning the actual thing if the question so requires, ie, when it is part of the source of confusion.

12. Jul 15, 2017

### stevendaryl

Staff Emeritus
Well, in this particular case, I don't see that it's the source of confusion, but I really don't understand what the OP is asking, either.

13. Jul 15, 2017

### mieral

Just asking what would happen if Schrodinger didn't make use of the Hamiltonian but directly the Lagrangian in formulating the SE or even directly the Momentum only. But never mind.

What I'd like to know now is how much the momentum operator in the SE is so important. What do you accomplish for solving for the momentum in the Schrodinger Equations? Is it not the total energy is the most important variable. For excited and ground states of atoms or molecules.. what situations do you have to use the momentum operator?

14. Jul 15, 2017

### vanhees71

In the Hilbert space $\mathrm{L}^2(\mathbb{R}^3)$ the momentum operator is given as
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}.$$
It represents canonical momentum. That must be so, because canonical momentum in Hamiltonian mechanics is the generator for spatial translations.

The case of a particle in a box with rigid boundary conditions you don't have a momentum operator, because on this space the above given operator is not self-adjoint. Fortunately its square is, i.e., the Hamiltonian
$$\hat{H}=-\frac{\mathrm{\hbar}^2}{2m} \Delta$$
represents a well-defined observable, namely the energy of the particle.

15. Jul 15, 2017

### mieral

But why do you need the canonical momentum in Hamiltonians mechanics in Quantum Mechanics.. why does QM have to have a generator for spatial translation.. what do you really accomplish with it?

16. Jul 15, 2017

### vanhees71

I don't know, how to understand quantum theory if not from symmetries (starting from spacetime symmetries) and the Hamiltonian formulation of classical mechanics. Given the formal structure of QT, i.e., the Hilbert space and the observable algebra, you have to find a way to specify this algebra for a given problem, and that's very clear if you have a classical analogue for this problem and then you can exploit the symmetries of the problem.

There's a hand-waving shortcut, called "canonical quantization" in the textbook literature, but that's a dangerous thing, because only the group-theoretical methods lead to a safe way to get the correct observable algebra.

In non-relativsitic QT you have Galilei symmetry with its 10 generators, which make up the corresponding conserved quantities in the Hamiltonian formulation (with the Poisson brackets as the symplectic realization of the observable algebra and the Lie algebra of the symmetry group): momentum, energy, angular momentum, center of mass. Mass itself also follows, but that's a bit more subtle and requires a deeper understanding of group-representation theory.

17. Jul 15, 2017

### mieral

I was asking what is the purpose of the momentum operator (or momentum) in QM.. I don't understand what the above got to do with it. Please directly connect momentum to symmetries (if it's the purpose of the momentum operator) because I can't understand what you mean. Perhaps the momentum operator is solving for the orbital momentum of the atoms? Right? Thanks.

18. Jul 15, 2017

### vanhees71

In classical mechanics you derive Noether's theorem, according to which each one-parameter Lie symmetry implies the existence of a conserved quantity. Space in Newtonian and special relativistic physics is homogeneous, i.e., the physics is invariant under translations. In other words, the physics doesn't depend on the place, where it is observed. The conserved quantities related to spatial translation invariance is called momentum (because it turned out to be identical with what Newton termed momentum in his more phenomenological approach to mechanics).

The purpose of the momentum operator in QM is to describe an observable, in this case momentum. What else should its purpose be? Since the momentum operator is the generator of spatial translations, i.e., for an infinitesimal translation by $\delta \vec{x}$ you have
$$\psi'(\vec{x})=\psi(\vec{x})-\mathrm{i} \hat{\vec{p}} \cdot \delta \vec{x}/\hbar \psi(\vec{x})=\psi(\vec{x}) - \delta \vec{x} \cdot \vec{\nabla} \psi(\vec{x})=\psi(\vec{x}-\delta \vec{x})+\mathcal{O}(\delta \vec{x}^2)$$
as it should be for the generator of translations.

19. Jul 15, 2017

### mieral

Ok thanks. But when we are solving for the momentum of the orbitals of the atoms.. what would we do with the momentum information? Would it for example gives data about excited or ground states? I'm assuming these are handled by potential or Hamiltonians.. so what good is knowing the momentum?

20. Jul 15, 2017

### stevendaryl

Staff Emeritus
What do you mean by "solving for the momentum"?