Does electromagnetic radiation make sense in one dimension?

Question

I'm trying to do a simple simulation of a 1D charged quantum particle, which gets irradiated by an electromagnetic wave — in context of non-relativistic QM.

The Schrödinger equation for such a particle would be $$i\hbar\frac{\partial \Psi(x,t)}{\partial t}=\frac1{2m}\left(-i\hbar\frac{\partial}{\partial x}-qA(x,t)\right)^2\Psi(x,t)+q\left(\phi_0(x)+\phi(x,t)\right)\Psi(x,t),$$

where $A(x,t)$ is (what I believe to be) an $x$ component of vector potential $\vec A$ of radiation, $\phi_0(x)$ is static scalar potential of non-irradiated system, and $\phi(x,t)$ is scalar potential of radiation.

What I can't seem to understand is how electromagnetic radiation should be incorporated in this equation. I suppose it shouldn't depend on coordinate, i.e. in non-relativistic context speed of light should be considered infinite, thus wavelength should be infinite.

From basic knowledge of electromagnetism I suppose that such a wave would be represented by:

$$E=E_0 \cos(\omega t)$$ $$B=B_0 \cos(\omega t)$$

But simulation is to be 1D, while electromagnetic wave has three dimensions: one for direction of propagation (which I guess is irrelevant because of infinite wavelength), second for $\vec E$ and third one for $\vec B$.

To add the complexity, I need to determine how vector potential $\vec A$ and scalar potential $\phi$ would look in such case, which I might understand once I realize how to orient the radiation in terms of $\vec B$ and $\vec E$.

So, my question is: does it even make sense to talk about external electromagnetic radiation in 1D case? If yes, how should the field be oriented in this case to make a sensible quantum simulation? What would its vector potential $\vec A$ and scalar potential $\phi$ look like?

Answer 1

If the electromagnetic fields are generated by the charge in 1d, then there is no radiation at all, see SE question Can light exists in 2+1 or 1+1 spacetime dimensions?. A one sentence reason is that there is no magnetic field in 1d.

However, if you are simulating a real problem, then the electromagnetic fields should be in 3d. Think about a classical problem, a charged bead confined on a line. Then only the $x$ component of electric field along the line will influence the classical motion, since the vertical component of the Lorentz force will be cancelled by the normal force of the line confining the bead. Nevertheless, it much harder in my point of view for a 1d quantum problem, because it is the gauge potential, rather than the gauge field that enter into our equation. We need to derive our equation gauge in invariant manner first and then choose an appropriate gauge to simplify out problem.

As long as you want to couple external EM fields, they should be known as a priori, because they are external and can't be figured out within this Schodinger equation.

I would tentatively suggest the following procedures, (please kindly point out the mistake I have made)

1. Couple the EM fields in 3d way, \begin{equation} H = \frac{1}{2m}( {\bf p} - q{\bf A})^2 + q\phi({\bf x},t) \end{equation}

2. Expand out the minimal coupling, \begin{equation} ({\bf p} - q {\bf A} )^2 = {\bf p}^2 + q^2{\bf A}^2 + i \hbar q \nabla \cdot {\bf A} -q {\bf A} \cdot {\bf p} \end{equation}

3. Erase the $y$ and $z$ dependence. Set the operators $\hat{p_y}$ and $\hat{p_z}$ to be constants, and in all functions $y = z= 0$(after taking the derivative). The equation you get is now \begin{equation} i\hbar \frac{\partial}{\partial t} \psi( x, t ) = \frac{1}{2m}[ -\hbar^2 \partial^2_x + i\hbar q \nabla \cdot {\bf A} + i\hbar q A_x \partial_x + p_y^2 + p_z^2 + q^2 {\bf A}^2 - q(A_y p_y + A_z p_z) + \phi( x, t ) ] \psi(x, t ) \end{equation} This process could be understood as a result of the confining boundary condition $\psi(x, \pm \epsilon, \pm\epsilon,t ) = 0$. In the small region $y\in[-\epsilon, +\epsilon], z\in [-\epsilon, +\epsilon]$, as long as the wavelength $\lambda \ll \epsilon$, we can approximate the field by their value at $y=z=0$. Consequently, $[p_y, H] = [p_z, H]=0$ and you can solve the equation in momentum space. Without eletromagnetic field, we can set $p_y=p_z = 0$, because the term $p_y^2 + p_z^2$ only shift the energy level by a constant and will not influence the wavefunction. However that term $A_yp_y + A_zp_z$ will.

4. Perhaps a suitable gauge choice could further simplify the equation. If $\epsilon$ is large such that the ground stage momentum $p_y$ and $p_z \sim 1/\epsilon$ is much smaller compared with $p_x$, then you can neglect the $A_yp_y + A_zp_z$ term. But that requires a very slow varying field.

Answer 2

does it even make sense to talk about external electromagnetic radiation in 1D case?

It depends on what you want to achieve. EM radiation in the common meaning refers to three-dimensional vectors $\mathbf E, \mathbf B$ in three-dimensional space. Sometimes the field can be assumed to depend only on one coordinate $x$ and time - like when we have plane wave - so this can be regarded as 1D wave. But as you mentioned the electric field will necessarily have components perpendicular to direction $x$ in the plane $yz$ so other dimensions are still needed.

If you simply discard everything that is not in the dimension $x$, the electric field along $x$ will be 1D vector with no wave propagation along $x$ (wave components of $\mathbf E, \mathbf B$ are always perpendicular to the direction of propagation).

If you do not require that $A$ give EM field obeying Maxwell's equations, you may take any function of $x,t$ $A(x, t)$, interpret $-\partial_t A_(x,t)$ as "electric field", insert it in the Schr. equation and look for consequences for $\psi$.

Answer 3

The electromagnetic field $A_\mu$, in a $D$ dimensional space-time, has $D$ off-shell degrees of freedom, and $(D-2)$ on-shell degrees of freedom. I let you conclude.