Does Momentum Affect Potential Energy in a Particle Box?

[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?

 

[Orodruin]

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.

 

[mieral]

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.

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?

 

[Orodruin]

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

No. Again, the momentum is not well defined.

You cannot solve for momentum.

 

[mieral]

No. Again, the momentum is not well defined.

You cannot solve for momentum.

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

 

[Orodruin]

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

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.

 

[mieral]

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.

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?

 

[stevendaryl]

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.

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.

 

[mieral]

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.

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?

 

[stevendaryl]

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?

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

 

[Orodruin]

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.

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.

 

[stevendaryl]

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.

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.

 

[mieral]

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

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?

 

[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.

 

[mieral]

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.

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?

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.

 

[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.

 

[mieral]

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.

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.

 

[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.

 

[mieral]

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?

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?

 

[stevendaryl]

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?

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

 

[mieral]

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

Getting the momentum operator to solve for the momentum observable of the atoms.. but why do we need to know the momentum.. what are the applications of knowing the momentum? I just need examples. Thanks.

 

[vanhees71]

I don't understand this question. The "orbitals" are the energy eigenstates of the atom. Since the Hamiltonian is not commuting with the momentum operators, it cannot be a momentum eigenstate at the same time, i.e., in an energy eigenstate the momentum will be undetermined.

If you want to know the probability to find the atom in a certain momentum, you have to calculate the corresponding probability. If $|\psi \rangle$ is the state vector of the atom, then the probability density for momentum is given by Born's rule,
$$P(\vec{p})=|\langle \vec{p}|\psi \rangle|^2.$$
If you have, as usual, your state given in terms of a position-space wave function, you just need
$$\langle \vec{p}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \vec{p}|\vec{x} \langle \vec{x}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi \hbar)^3}} \exp(\mathrm{i} \vec{x} \vec{p}/\hbar) \psi(\vec{x}).$$

 

[mieral]

I don't understand this question. The "orbitals" are the energy eigenstates of the atom. Since the Hamiltonian is not commuting with the momentum operators, it cannot be a momentum eigenstate at the same time, i.e., in an energy eigenstate the momentum will be undetermined.

If you want to know the probability to find the atom in a certain momentum, you have to calculate the corresponding probability. If $|\psi \rangle$ is the state vector of the atom, then the probability density for momentum is given by Born's rule,
$$P(\vec{p})=|\langle \vec{p}|\psi \rangle|^2.$$
If you have, as usual, your state given in terms of a position-space wave function, you just need
$$\langle \vec{p}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \vec{p}|\vec{x} \langle \vec{x}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi \hbar)^3}} \exp(\mathrm{i} \vec{x} \vec{p}/\hbar) \psi(\vec{x}).$$

Ok. But what do we accomplish by knowing the probability density for momentum in the atoms. Or let's take a classical example. When our car runs in street.. position of car is important.. but why do we need to know the momentum of the car? to compute for collision energy? So maybe in the atoms, we need to know the momentum to compute for collision energies of the particles?

 

[mieral]

Ok. But what do we accomplish by knowing the probability density for momentum in the atoms. Or let's take a classical example. When our car runs in street.. position of car is important.. but why do we need to know the momentum of the car? to compute for collision energy? So maybe in the atoms, we need to know the momentum to compute for collision energies of the particles?

Or what examples of momentum eigenstates of an atoms or particles or whatever?

 

[vanhees71]

I don't understand your problem obviously. If you prepare your atom in an energy eigenstate it's not in a momentum eigenstate, and thus all you know are probabilities for finding an electron's momentum to have a certain value when measuring it.

 

[mieral]

I don't understand your problem obviously. If you prepare your atom in an energy eigenstate it's not in a momentum eigenstate, and thus all you know are probabilities for finding an electron's momentum to have a certain value when measuring it.

Or maybe a good question is what are some examples of preparing a system in momentum eigenstates?

For example.. in human cells.. we don't solve for the momentum of the particles inside the cell.. so I just want to know practical applications of momentum eigenstates in QM and what good it is.

 

[stevendaryl]

Getting the momentum operator to solve for the momentum observable of the atoms.. but why do we need to know the momentum.. what are the applications of knowing the momentum? I just need examples. Thanks.

What do you mean by "solving for the momentum observable"?

 

[mieral]

What do you mean by "solving for the momentum observable"?

My question is only very simple. Can't understand why people don't understand. I'm simply asking why solve for the momentum of atoms? I'm not talking about angular momentum.. but just momentum.. maybe linear momentum.. why solve for linear momentum of atoms (or other momentum that is not angular)..

 

[vanhees71]

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.

 

[stevendaryl]

My question is only very simple. Can't understand why people don't understand. I'm simply asking why solve for the momentum of atoms?

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?"

 

[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 ...