# B Momentum in atoms

1. Jul 5, 2017

### mieral

Is momentum in the atoms part of the potential energy of the system? I read the momentum is a Hamiltonian.. and the potential is also a Hamiltonian.. so are the momentum the potential itself? In the atoms, the potential could be the electromagnetic attraction between electrons and nucleus.. how about the momentum? where does momentum come from? Is it separate from the electromagnetic attraction? If the momentum is photonic energy from environment.. couldn't it affect the potential?

2. Jul 5, 2017

### Staff: Mentor

No. Momentum is not energy.

Please give a reference. I strongly suspect you are misinterpreting something, since the Hamiltonian describes energy, not momentum.

Again, please give a reference. You are obviously misinterpreting something. Potential energy is included in the Hamiltonian, but it is not "a Hamiltonian".

3. Jul 5, 2017

### mieral

"That's not what the equation says: energy is not conserved per branch but over all branches; and total energy does not increase during branching. The energy I calculate is not calculated per branch, but as a sum over all branches. Therefore energy per branch is irrelevant, all what counts is total energy over all branches. And this total energy remains visible in each branch, therefore in some sense other branches remain visible in a certain sense - we observe their energy"

When Tomstoer talked about energy. Was he referring to the momentum? or energy as potential? But I thought the potential is not duplicated in Many worlds. He seemed to be suggesting the universal wave function has total energy or potential that is greater than one branch?

4. Jul 5, 2017

### Staff: Mentor

Neither. He was referring to the Hamiltonian operator.

The Hamiltonian is not duplicated--there is only one Hamiltonian, because there is only one overall system. The potential energy is part of the Hamiltonian; it's not something separate.

I think he was saying that "the energy of a branch" is a meaningless concept, because you can't apply the Hamiltonian operator to just one branch; you can only apply it to the state of the system as a whole. That, at any rate, is how I would respond.

5. Jul 5, 2017

### mieral

He used the words "Energy Eigenstates".. what are energy eigenstates composed of? The kinetic or/and potential energy of the system?

Does the kinetic energy contains the so called momentum basis of atoms?

6. Jul 5, 2017

### Staff: Mentor

Energy eigenstates are eigenstates of the energy operator, i.e., of the Hamiltonian operator. They aren't "composed" of anything. They're quantum states of the system.

The Hamiltonian operator, generally speaking, will contain parts that correspond to the kinetic and potential energy of the system. But it is an operator, not a state. QM does not work like classical mechanics.

If you want more, go get a basic QM textbook and learn it. It will amply repay your time, since you appear to be lacking some very basic background in QM, which if you had it would go a long way towards answering the questions you are asking.

This question is not even well-defined.

7. Jul 5, 2017

### mieral

So this is the secret. The Hamiltonian operator of the universal wave function has value that is greater than one branch.. is this correct?

If yes then it makes sense why the Hamiltonian is not duplicated (which is not possible).. because the Hamiltonian already takes into account all the branches.. meaning the Hamiltonian operator of the universal wave function already contains a million you.. is this a way to put it?

Look. I only know how to do basic arithmetic.. I don't even know calculus.. so any QM textbook would be unintelligible to me. I only want to understand the verbal summary of basic stuff for laymen for now. That's all.

8. Jul 5, 2017

### Staff: Mentor

No. Did you read what I posted? You can't even apply the Hamiltonian operator to a single branch.

No. The Hamiltonian is an operator, not a state. The state--the universal wave function--is a highly entangled state of lots of subsystems, including you, me, and everyone else. The Hamiltonian operator, applied to this state, tells you probabilities for the whole universe to have various values for its total energy. It doesn't tell you anything about "copies" of subsystems.

Then unfortunately I don't think you are equipped to ask questions about QM. See below.

You can't, because you are unwilling to stop there. Every time you are given a "verbal summary of basic stuff for laymen", you ask more questions. You can't do that unless you are willing to go further and learn more. There's nothing wrong with that, but it is not compatible with an attitude of "this is too complicated for me". Either you're willing to do the work to learn or you aren't.

9. Jul 5, 2017

### mieral

What is a good basic list of books about QM? Look. all the books suggested by Hobba and others are college level textbooks. Right now we have purely laymen books that don't discuss the Hamiltonian, they only discuss interpretations.. then it suddenly jumps to college or graduate textbook with Neumaier kind of contents that are obviously for the Einstein level folks.. we need books that is in the middle. Actually there is one called Deep Down Things where it mentions the potential and kinetic energy of the system but it doesn't mention about the Hamiltonian. I want a book where it at least gives intro of the Hamiltonian.. anyone has encountered such books? Please give the titles. Thanks.

10. Jul 5, 2017

### Staff: Mentor

I can't recommend any books other than textbooks, which it sounds like you are not interested in. What you are basically asking for are pop science treatments that get the science right. If such things exist, which I doubt, I still would not recommend them, because at the end of them, you will still have unanswered questions, and the only way to answer them is to go further and study textbooks. If you want answers, you might as well cut out the middleman and go straight to the textbooks.

The most basic textbook I can think of would be the QM book from Susskind's "Theoretical Minimum" series. AFAIK it does not require calculus. It does require some linear algebra (it's impossible to teach QM at all without some linear algebra), but AFAIK it develops most of that as part of the presentation.

11. Jul 5, 2017

### mieral

I tried reading it.. but I can't understand any of the math parts. I can only understand the verbal parts. And because the verbal parts are few and disconnected.. I can't understand the whole concepts.

But it's good you at least emphasized the Hamiltonian is not state.. but Hamiltonian operator.. when I mentioned Hamiltonian before.. I assume it's potential energy...

Before I spent a lot of days rereading Wikipedia. What is the momentum in QM? Is it part of the kinetic or potential energy (which one)? And if they are not kinetic or potential energy. What is the momentum in QM?

12. Jul 5, 2017

### Staff: Mentor

Then I'm afraid there won't be any textbook you could follow, unless you are willing to take the time to learn enough math.

Did you read what I wrote before about the Hamiltonian operator? What did I say about potential energy?

An operator. A different operator from energy (the Hamiltonian).

I already said that momentum is not energy. Are you reading my posts? Even if you are unable to follow a lot of math, the least you can do is read the ordinary language I'm writing. Then you won't have to ask questions to which I have already given the answers.

13. Jul 6, 2017

### mieral

Oh. I thought momentum in QM is like the momentum in classical mechanics.. so it's an operator.. ok.. i'll read Wikipedia again and I may understand it this time.

"Anyways, the book is a bit unorthodox. It starts off talking about spin states and vector spaces. This is different than the typical quantum books that start off with talking about solving the Schrodinger equation. Which, now that I think about it, isn't really a great place to start. It's a wave equation that's only once-piece of the bigger picture.

The result is a book that has a very gradual learning curve.

That said, readers still need a bit of math background if they want to get through this book. I'd say at least a decent understanding of complex numbers and matrix algebra are a prerequisite. Both those subjects are thoroughly explained in places online (Khan Academy) for free."

One of the best laymen friendly book is Deep Down Things: The Breathtaking Beauty of Particle Physics".. does anyone know a QM version of it?

14. Jul 6, 2017

### Staff: Mentor

If you are thinking "x in QM is like x in classical mechanics", you're probably wrong, for any x. In QM every observable is an operator.

15. Jul 6, 2017

### mieral

momentum is an operator.. but "angular momentum" in QM is not an operator.. is it.. according to Deep Down Things:

"We’ve seen that quantum mechanics is required to understand and describe the behavior of individual particles. Thus, we might expect that quantum mechanics and, in particular, its attendant scale factor h,might have a role to play in the spin of fundamental particles, especially since the unit of h, as we’ve seen, is just that of angular momentum.
The amount of angular momentum possessed by an electron or a proton is 1⁄2h ¯, where for expediency we have introduced the reduced Planck’s constant h ¯ h/2p (to be read as “h-bar”). An individual photon, on the other hand, possesses an angular momentum of h ¯. We often forget about the h ¯ and just say that electrons and protons are spin-1⁄2 and that photons are spin-1."

16. Jul 6, 2017

### Staff: Mentor

Yes, it is. (More precisely, momentum in a particular direction, and angular momentum about a particular axis, are operators.)

Which just illustrates what I said about pop science sources. What they should have said is that the eigenvalues of the operator "angular momentum about a particular axis" are $\pm \hbar / 2$ for an electron or proton, and $\pm \hbar$ for a photon. In units where $\hbar = 1$, which are common in QM, these eigenvalues are $\pm 1/2$ and $\pm 1$.

17. Jul 6, 2017

### mieral

Oh.. but how about the momentum in Heisenberg's Uncertainly principle which states that "lessening the uncertainty in momentum increases the uncertainty in position"... is it "lessening the uncertainty in momentum operator increases the uncertainty in position operator"? But then.. position is position and not operator at all (?)

18. Jul 6, 2017

### Staff: Mentor

Pop science sources are not valid sources. Stop asking questions about what you read in them. If you want to know how the uncertainty principle actually works, go find a QM textbook and learn it. PF is not here for the purpose of giving you a free complete course in QM.