High School Does Momentum Affect Potential Energy in a Particle Box?

[mieral]

The problem is that you ask questions that don't make any sense. What do you mean by "solving for the momentum of atoms"? If you want to know the momentum of an electron within the atom you have to measure it. As for any other observable, if you know the quantum state of the atom you can calculate the probability for finding the momentum of this electron within a certain range, it's given by the momentum-space wave function via Born's rule.

vanhees71 and stevedaryl.. what i mean by solving for the momentum of atoms is why do we have to calculate the probability for finding the momentum of this electron or quantum system within a certain range... is it to know the position which doesn't commute with it? What would happen if we eliminate the momentum operator and ignore everything about momentum.. what would we miss in quantum mechanics? by knowing the answer, then I'd know the importance of momentum in QM. Thanks a lot!

 

[vanhees71]

Have you ever read a real textbook on QT? To write down the most simple Hamiltonian in atomic physics (the hydrogen atom) you need already momentum:
$$\hat{H}=\frac{\hat{\vec{p}}_1^2}{2m} + \frac{\hat{\vec{p}}_2^2}{2M} -\frac{e^2}{4 \pi |\hat{\vec{x}}_1 - \hat{\vec{x}}_2|}.$$
Here $\vec{p}_1$ and $\vec{p}_2$ are the momenta of the electron and the proton and $\hat{\vec{x}}_1$ and $\hat{\vec{x}}_2$ their position vectors, respectively.

To solve the energy eigenvalue problem, what you indeed only need to know are the commutation relations between these quantities. Pauli solved the hydrogen problem within matrix mechanics by knowing the classical physics of the problem very well, making use of the large dynamical symmetry and the Runge-Lenz vector.

Schrödinger solved the same problem, using his wave mechanics, which was more within the mathematical standard techniques of the physicists at his time, i.e., he wrote down the energy eigenvalue problem in terms of the partial differential equation, which we now call the "time-independent Schrödinger equation". For this he needed to know that in the position representation the momentum operators are $\hat{\vec{p}}_1=-\mathrm{i} \hbar \vec{\nabla}_1$ etc.

 

[mieral]

Have you ever read a real textbook on QT? To write down the most simple Hamiltonian in atomic physics (the hydrogen atom) you need already momentum:
$$\hat{H}=\frac{\hat{\vec{p}}_1^2}{2m} + \frac{\hat{\vec{p}}_2^2}{2M} -\frac{e^2}{4 \pi |\hat{\vec{x}}_1 - \hat{\vec{x}}_2|}.$$
Here $\vec{p}_1$ and $\vec{p}_2$ are the momenta of the electron and the proton and $\hat{\vec{x}}_1$ and $\hat{\vec{x}}_2$ their position vectors, respectively.

Oh. But Orodruin replied in message #2: "The potential is independent of momentum, it only depends on the position." I initially posted this in the General Physics to be familiar with the classical idea about Hamiltonian and momentum. I guess these are simply extended to QM.. hope Orodruin can explain why he said the potential was independent of momentum... when in fact in QT.. it need already momentum as you described above.

To solve the energy eigenvalue problem, what you indeed only need to know are the commutation relations between these quantities. Pauli solved the hydrogen problem within matrix mechanics by knowing the classical physics of the problem very well, making use of the large dynamical symmetry and the Runge-Lenz vector.

Schrödinger solved the same problem, using his wave mechanics, which was more within the mathematical standard techniques of the physicists at his time, i.e., he wrote down the energy eigenvalue problem in terms of the partial differential equation, which we now call the "time-independent Schrödinger equation". For this he needed to know that in the position representation the momentum operators are $\hat{\vec{p}}_1=-\mathrm{i} \hbar \vec{\nabla}_1$ etc.

Ok. I tried to understand that part in Balentine but couldn't before. Now I am getting it. Thanks.

 

[vanhees71]

Yes, in my expression the potential is indeed independent of momentum, it's the Coulomb potential between two point particles. I can only again strongly advice to go back and first learn classical mechanics up to and including the Hamiltonian formulation. The goal must be to have a very good understanding of Poisson brackets and basics of Lie algebras and groups. Then you can attach quantum theory again, e.g., with a textbook like Sakurai.

 

[Orodruin]

Oh. But Orodruin replied in message #2: "The potential is independent of momentum, it only depends on the position." I initially posted this in the General Physics to be familiar with the classical idea about Hamiltonian and momentum. I guess these are simply extended to QM.. hope Orodruin can explain why he said the potential was independent of momentum... when in fact in QT.. it need already momentum as you described above

Beecause it is independent of momentum. I concur with @vanhees71 ...

 

[mieral]

Have you ever read a real textbook on QT? To write down the most simple Hamiltonian in atomic physics (the hydrogen atom) you need already momentum:
$$\hat{H}=\frac{\hat{\vec{p}}_1^2}{2m} + \frac{\hat{\vec{p}}_2^2}{2M} -\frac{e^2}{4 \pi |\hat{\vec{x}}_1 - \hat{\vec{x}}_2|}.$$
Here $\vec{p}_1$ and $\vec{p}_2$ are the momenta of the electron and the proton and $\hat{\vec{x}}_1$ and $\hat{\vec{x}}_2$ their position vectors, respectively.

While it is generally agreed potential is independent of momentum. But how come potential+kinetic is no longer independent to momentum as you showed above?

Yes, in my expression the potential is indeed independent of momentum, it's the Coulomb potential between two point particles. I can only again strongly advice to go back and first learn classical mechanics up to and including the Hamiltonian formulation. The goal must be to have a very good understanding of Poisson brackets and basics of Lie algebras and groups. Then you can attach quantum theory again, e.g., with a textbook like Sakurai.

What you mean it's the Coulomb potential between two point particles? Why did you include Coulomb here.
Yes. I agree I need to first get good handle of classical mechanics up to and including the Hamiltonian formulation. What would be a good book about this.

To solve the energy eigenvalue problem, what you indeed only need to know are the commutation relations between these quantities. Pauli solved the hydrogen problem within matrix mechanics by knowing the classical physics of the problem very well, making use of the large dynamical symmetry and the Runge-Lenz vector.

Schrödinger solved the same problem, using his wave mechanics, which was more within the mathematical standard techniques of the physicists at his time, i.e., he wrote down the energy eigenvalue problem in terms of the partial differential equation, which we now call the "time-independent Schrödinger equation". For this he needed to know that in the position representation the momentum operators are $\hat{\vec{p}}_1=-\mathrm{i} \hbar \vec{\nabla}_1$ etc.

 

[PeterDonis]

how come potential+kinetic is no longer independent to momentum as you showed above?

Um, because kinetic energy is not independent of momentum? Isn't that obvious?

What you mean it's the Coulomb potential between two point particles?

The potential due to the Coulomb interaction (i.e., static electric field) between two point particles.

 

[mieral]

The phrase "solve for momentum of atoms" doesn't mean anything to me. Why do you think anyone solves for the momentum of atoms? What does "solve for the momentum of atoms" mean?

You keep explaining one statement in terms of another statement that doesn't make any sense.

As far as I know, "solving for momentum" and "solving for the momentum observable" and "solving for the momentum of atoms" and "solving for the momentum of observables" are equally mysterious. I don't know what you mean by any of them. You're asking why people do something, and that something is something that, as far as I know, nobody does.

It's as if I asked you: "Why take the square-root of a banana?"

Let me clarify something (this is for Vanhees71 too) . According to Vanhees71. "As for any other observable, if you know the quantum state of the atom you can calculate the probability for finding the momentum of this electron within a certain range, it's given by the momentum-space wave function via Born's rule.". Is this not the same as the statement:

1. solving for momentum
2. solving for the momentum observable
3. solving for the momentum of observables

Isn't the mere calculating the probability for finding the momentum of this electron within a certain range also referred to as "solving for momentum" or "solving for momentum observables?" I've been pondering about this semantic nitpicking.. so I need to know what is wrong with the description or language.

 

[Orodruin]

It is not nitpicking. It is trying to understand what you mean. "Solving for momentum" is generally something you might do in classical physics where it has a particular value. The way you are applying it here is not standard and so you have several experienced physicists trying to figure out what you are actually trying to accomplish.

 

[mieral]

May i know why would anyone want to calculate the probability for finding the momentum of this electron within a certain range as given by the momentum-space wave function via Born's rule. What would you do with the momentum information? What further or other computations where you first need to have the momentum probability information? Thanks.

 

[vanhees71]

I don't know. It was you asking about the momentum, although in a way that we had to guess what you might mean. We cannot teach you basic QT in this forum. So please, first read a good book on QT (e.g., Sakurai).

 

[PeterDonis]

We cannot teach you basic QT in this forum.

And with that, this thread is closed.