Quantum superposition of Coulomb potential

[zonde]

As I understand Coulomb potential associated with charged particle is described classically.
My question is if there is a way how to describe Coulomb potential of charged particle that is in quantum superposition of being "here" and "there"?
My motivation for question is that I am trying to imagine amplification process of microscopic event. In avalanche photodiodes single excited electron interacts with population of other electrons in unstable configuration. I am interested if this process can rule out some viewpoints about QM.

 

[DrClaude]

The potential is a single-valued function at every point in space. As the wave function of any charged particle is necessarily spread out over some region of space, for any non-constant Coulomb potential, the particle will interact with different values of the potential at the same time. A simple representation of this is an asymmetric potential well, with a particle in a state such that its wave function covers the two wells.

There is nothing unusual about such a situation. It has been studied numerous time, and the only thing "weird" that comes out it, to my knowledge, is tunneling.

 

[zonde]

The potential is a single-valued function at every point in space. As the wave function of any charged particle is necessarily spread out over some region of space, for any non-constant Coulomb potential, the particle will interact with different values of the potential at the same time. A simple representation of this is an asymmetric potential well, with a particle in a state such that its wave function covers the two wells.

Thanks for replay. I think you misunderstood my question. I do not ask how a particle in superposition will interact with Coulomb potential but how it's own charge will contribute to Coulomb potential. Or maybe in interaction you include that contribution as well?

 

[DrClaude]

I did misunderstand what you had in mind, but it doesn't change the gist of my answer. When you treat atoms, you have delocalized charges interacting with other delocalized charges, nothing unusual.

 

[stevendaryl]

Thanks for replay. I think you misunderstood my question. I do not ask how a particle in superposition will interact with Coulomb potential but how it's own charge will contribute to Coulomb potential. Or maybe in interaction you include that contribution as well?

You're talking about nonrelativistic quantum mechanics, right? Well, in that case, the Coulomb potential is part of the Hamiltonian, not the wave function. There is no principle of superposition for Hamiltonians.

If you have a number of particles interacting via the Coulomb potential, then the many-particle Hamiltonian is something like:

$H = (\sum_j - \frac{\hbar^2}{2 m_j} \nabla^2_j) + (\sum_{i < j} \frac{q_i q_j}{r_{ij}})$

where $\nabla^2_j$ means that we are taking derivatives with respect to the coordinates of the j^th particle, and $r_{ij} = |\vec{r_j} - \vec{r_i}|$, and $q_i$ is the charge of the i^th particle.

The notion of superposition only comes into play in the states, not the Hamiltonian.

 

[zonde]

You're talking about nonrelativistic quantum mechanics, right? Well, in that case, the Coulomb potential is part of the Hamiltonian, not the wave function. There is no principle of superposition for Hamiltonians.

If you have a number of particles interacting via the Coulomb potential, then the many-particle Hamiltonian is something like:

$H = (\sum_j - \frac{\hbar^2}{2 m_j} \nabla^2_j) + (\sum_{i < j} \frac{q_i q_j}{r_{ij}})$

where $\nabla^2_j$ means that we are taking derivatives with respect to the coordinates of the j^th particle, and $r_{ij} = |\vec{r_j} - \vec{r_i}|$, and $q_i$ is the charge of the i^th particle.

The notion of superposition only comes into play in the states, not the Hamiltonian.

So with such Hamiltonian changing kinetic energy (that would change $\nabla^2_j$ as I understand) of k^th electron might give us irreversible process (avalanche) while leaving it as is (plus free photon somewhere) would leave the process stable (except that photon will fly away).
But to plug Hamiltonian in Schrodinger equation we need single Hamiltonian that describes both processes. So my simplified understanding says that we can use Schrodinger equation only when we can ignore particles own feedback on Hamiltonian (particle in a box) or when it's time independent for the whole Hamiltonian.

But probably I am missing something.

 

[stevendaryl]

So with such Hamiltonian changing kinetic energy (that would change $\nabla^2_j$ as I understand) of k^th electron might give us irreversible process (avalanche) while leaving it as is (plus free photon somewhere) would leave the process stable (except that photon will fly away).
But to plug Hamiltonian in Schrodinger equation we need single Hamiltonian that describes both processes. So my simplified understanding says that we can use Schrodinger equation only when we can ignore particles own feedback on Hamiltonian (particle in a box) or when it's time independent for the whole Hamiltonian.

But probably I am missing something.

Well, to take into account the production of photons, you have to go beyond nonrelativistic quantum mechanics and go to quantum field theory. In quantum field theory, the interactions are again found in the Hamiltonian, but now the Hamiltonian is written in terms of field operators, rather than kinetic and potential energy. Superpositions again are about the state, not the hamiltonian. I know that doesn't directly answer your question, but hopefully it gives context for answering followup questions.

 

[A. Neumaier]

now the Hamiltonian is written in terms of field operators, rather than kinetic and potential energy.

On the level of naive QFT, the quadratic part describing free particles is still the kinetic energy, and the interaction part the potential energy. Only renormalization makes this distinction problematic, and with it the whole concept of particles at finite times.