# Explicit non-local form for the vector potential?

• I
• Amentia
In summary, the conversation discusses the use of an explicitly non-local vector potential in light matter interaction Hamiltonians and the introduction of a delta function in the calculation. It is concluded that this is a valid assumption as long as the exact form of the vector potential is known.
Amentia
Hello everyone,

I was looking at the light matter interaction Hamiltonian and I worked out a simple calculation where I was surprised to see that I had to introduce an explicitly non-local vector potential if I want to go further:

$$\langle\psi| \boldsymbol{\hat{A}}(t)\cdot\boldsymbol{\hat{p}}|\phi\rangle = \int\int d^{3}rd^{3}r' \langle\psi|\boldsymbol{r'}\rangle\langle\boldsymbol{r'}|\boldsymbol{\hat{A}}(t)|\boldsymbol{r}\rangle\cdot\langle\boldsymbol{r}|\boldsymbol{\hat{p}}|\phi\rangle$$

Giving:

$$\langle\psi|\boldsymbol{\hat{A}}(t)\cdot\boldsymbol{\hat{p}}|\phi\rangle = \int\int d^{3}rd^{3}r' \psi^{*}(\boldsymbol{r'})\langle \boldsymbol{r'}|\boldsymbol{\hat{A}}(t)|\boldsymbol{r}\rangle\cdot\left(\frac{\hbar}{i}\right)\boldsymbol{\nabla}_{\boldsymbol{r}}\phi(\boldsymbol{r})$$

I would rewrite ##\langle \boldsymbol{r'}|\boldsymbol{\hat{A}}(t)|\boldsymbol{r}\rangle## as ##\boldsymbol{\hat{A}}(\boldsymbol{r'},\boldsymbol{r},t)##. But only ##\boldsymbol{\hat{A}}(\boldsymbol{r},t)## has been considered as a correct form for the vector potential in the literature (usually the dependence with r is a plane-wave). Perhaps I have done something wrong in my calculation although it looks simple? Or is there something in the physics related to vector potentials that I have been missing until now?

Thank you for any thoughts about that!

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Amentia said:
I would rewrite ##\langle r'|\boldsymbol{\hat{A}}(t)|r\rangle## as ##\boldsymbol{\hat{A}}(\boldsymbol{r'},\boldsymbol{r},t)##.
It's actually
$$\langle r'|\boldsymbol{\hat{A}}(t)|r\rangle=\boldsymbol{A}(\boldsymbol{r'},\boldsymbol{r},t)$$
i.e. there is no hat on the right-hand side because the matrix element of an operator is a c-number, not an operator. Furthermore, this c-number function is of the form
$$\boldsymbol{A}(\boldsymbol{r'},\boldsymbol{r},t)=\delta(\boldsymbol{r'}-\boldsymbol{r}) \boldsymbol{A}(\boldsymbol{r},t)$$
which removes non-locality. All this is nothing but a simple generalization of
$$\langle r'|\boldsymbol{\hat{r}}|r\rangle=\delta(\boldsymbol{r'}-\boldsymbol{r}) \boldsymbol{r}$$

vanhees71 and Amentia
Thank you, this is what I was looking for because I was trying to see how a delta function could be introduced here. However, is it always correct in general? If the dependence of A was not a plane-wave or a simple analytical function of r, could we still use this trick? It could happen that the exact form of A is not known in some complex medium, a nanosystem with position-varying dielectric constant, etc.

Edit: Also A is still an operator in quantum mechanics here... I do not like hybrid notations but we usually find it written both with the dependence on r and the creation and annihilation operators. I assume the initial A is supposed to be some kind of tensor product of operators acting on two different vector spaces.

Last edited:
Amentia said:
Thank you, this is what I was looking for because I was trying to see how a delta function could be introduced here. However, is it always correct in general? If the dependence of A was not a plane-wave or a simple analytical function of r, could we still use this trick? It could happen that the exact form of A is not known in some complex medium, a nanosystem with position-varying dielectric constant, etc.
Think of A as
$$\hat{\bf A}(t)={\bf A}(\hat{\bf r},t)$$
where ##{\bf A}({\bf r},t)## is a known function. It doesn't need to be simple or analytic, it only needs to be known. If ##{\bf A}({\bf r},t)## is unknown, then you cannot analyze the system. It's like studying Schrodinger equation with unknown potential ##V(x)##; when it's unknown then you cannot say much about the system.

vanhees71 and Amentia
Thank you, I think it is a reasonable assumption to make in physics and that completely solves my question!

Demystifier

## 1. What is an explicit non-local form for the vector potential?

An explicit non-local form for the vector potential is a mathematical expression that describes the vector potential at a specific point in space, taking into account the contributions from all other points in the surrounding space.

## 2. How is an explicit non-local form for the vector potential different from a local form?

A local form for the vector potential only takes into account the contributions from nearby points in space, while an explicit non-local form considers all points in the surrounding space. This makes the explicit non-local form more accurate, but also more computationally intensive.

## 3. What are some applications of the explicit non-local form for the vector potential?

The explicit non-local form for the vector potential is commonly used in quantum mechanics and electrodynamics, as it allows for a more accurate description of the behavior of particles and electromagnetic fields.

## 4. How is the explicit non-local form for the vector potential derived?

The explicit non-local form for the vector potential is derived from the Maxwell's equations, which describe the behavior of electromagnetic fields. It involves solving a set of differential equations and integrating over all points in space to obtain the complete expression.

## 5. Are there any limitations to using the explicit non-local form for the vector potential?

One limitation of using the explicit non-local form for the vector potential is its computational complexity, which can make it impractical for certain applications. Additionally, it may not be necessary to use the explicit non-local form in situations where the contributions from distant points in space are negligible.

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