I Explicit non-local form for the vector potential?

[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!

 

[Demystifier]

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
$$\langle r'|\boldsymbol{\hat{r}}|r\rangle=\delta(\boldsymbol{r'}-\boldsymbol{r}) \boldsymbol{r}$$

 

[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.

 

[Demystifier]

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.

 

[Amentia]

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