The Klein-Gordon equation with relativistic mass: a relativistic Schrödinger equation

Using the relativistic Lagrangian $L_{R}$ defined by

$$L_{R}=-\frac{m_{0}c^{2}}{\gamma},$$

(2)

it can be shown that the total relativistic energy of Eq. (1) is a Hamiltonian of motion when energy is conserved, since $E=\partial_{\dot{\textbf{r}}}L_{R}\cdot\dot{\textbf{r}}-L_{R}$. Further, since any Lagrangian $L_{R}$ is indefinite with respect to addition of a constant kinetic energy [18], we can define a new energy $U$ simply by adding $m_{0}c^{2}$ to $L_{R}$, i.e. $U=\partial_{\dot{\textbf{r}}}(L_{R}+m_{0}c^{2})\cdot\dot{\textbf{r}}-(L_{R}+m_{0}c^{2})\text{ or }U=E-m_{0}c^{2}$. Using this new energy we get [19]

$$\left(m_{0}c^{2}+U\right)^{2}=\textbf{p}^{2}c^{2}+m_{0}^{2}c^{4}.$$

(3)

When a particle with rest mass $m_{0}$ and charge $q$ is moving inside an electromagnetic field with scalar potential $\phi$ and vector potential A, the Lagrangian $L_{P}$ of the particle is given by [20],

$$L_{P}=L_{R}-q\phi+q\textbf{A}\cdot\dot{\textbf{r}},$$

(4)

When the potentials $\phi$ and A depend explicitly neither on time nor particle velocities, the total energy of the particle,

$$\mathscr{E}_{P}=\partial_{\dot{\textbf{r}}}L_{P}\cdot\dot{\textbf{r}}-L_{P}=mc^{2}+q\phi,$$

is conserved. First, we suppose the particle to be at rest and located at infinity, where $\phi_{0}=0$. In this case, the total energy of the particle $\mathscr{E}_{P}$ is

$$\mathscr{E}_{P}=m_{0}c^{2}$$

Now, for this particle to move inside the potentials $\phi$ and A, we need $q\phi<0$. This requirement can be deduced explicitly from the fact that $m\geqslant m_{0}$ since $\gamma\geqslant 1$. In this case

$$\mathscr{E}_{P}=mc^{2}+q\phi$$

Because the energy of the particle is conserved, we can link the initial (at infinity) and final states, yielding

$$m_{0}c^{2}=mc^{2}+q\phi.$$

(5)

Now, we may want to work with an electromagnetic field that varies in time. In this case, this is not the particle energy $\mathscr{E}_{P}$ is not conserved but this is the total energy $\mathscr{E}_{T}$ of the particle at rest and the electromagnetic field that is conserved. As before, we first compute the total energy at infinity

$\displaystyle\mathscr{E}_{T}=m_{0}c^{2}+\frac{1}{2}\int\left[\varepsilon_{0}\textbf{E}_{0}^{2}+\frac{\textbf{B}_{0}^{2}}{\mu_{0}}\right]dV.$

Here $\textbf{E}_{0}(t,x,y,z)$, $\textbf{B}_{0}(t,x,y,z)$ are the total electric and magnetic fields with the particle at infinity. When the particle moves inside $\phi$ and A we have

$\displaystyle\mathscr{E}_{T}=mc^{2}+q\phi+\frac{1}{2}\int\left[\varepsilon_{0}\textbf{E}^{2}+\frac{\textbf{B}^{2}}{\mu_{0}}\right]dV,$

where $\textbf{E}(t,x,y,z)$ and $\textbf{B}(t,x,y,z)$ are the total electric and magnetic fields when the particle is moving inside the potentials $\phi$ and A. Note that the electric and magnetic fields of the particle are included in all these electric and magnetic fields. We now define an effective mass

$$\tilde{m}=m+\frac{1}{2c^{2}}\int\left[\varepsilon_{0}\left(\textbf{E}^{2}-\textbf{E}_{0}^{2}\right)+\frac{\textbf{B}^{2}-\textbf{B}_{0}^{2}}{\mu_{0}}\right]dV,$$

(6)

which encapsulates these non-conservative processes. If we suppose that there is no outgoing energy flux at infinity (full system isolation hypothesis), the Poynting theorem guarantees the conservation of $\mathscr{E}_{T}$, and we have

$$m_{0}c^{2}=\tilde{m}c^{2}+q\phi.$$

(7)

Despite the existence of non conservative effects, it is interesting to note that the equation above yields

$$\partial_{t}\tilde{m}=-\frac{q}{c^{2}}\partial_{t}\phi.$$

(8)

We can turn any equation with non-conservative effects to an equation with conservative effects simply by setting $\tilde{m}\rightarrow m$. However, conservative effects also require $\phi$ and A to be time independent. To avoid any inconsistencies between time dependent and time-independent gauges, we chose to use the Lorenz gauge [21]

$$c^{-2}\partial_{t}\phi+\nabla\cdot\textbf{A}=0$$

(9)

throughout, which turns into the Coulomb gauge

$$\nabla\cdot\textbf{A}=0$$

for steady state fields. Further, if we use Eqs. (8) and (9) we get

$$\partial_{t}\tilde{m}=q\nabla\cdot\textbf{A},$$

(10)

which again is consistent when we transition to time independent phenomena. Further, we can obtain a non-relativistic equation by setting $\tilde{m}\rightarrow m_{0}$, and a quasi static version by setting $c\rightarrow+\infty$.

Using minimal coupling, we can replace the particle energy with the quantum energy operator (i.e. $U\rightarrow\hat{U}-q\phi$), its momentum with the momentum operator (i.e. $\textbf{p}\rightarrow\hat{\textbf{p}}-q\textbf{A}$), and its rest mass with the effective mass from Eq. (7) inside Eq. (3), and we get

$$\left[\left(\tilde{m}c^{2}+\hat{U}\right)^{2}-\left(\hat{\textbf{p}}-q\textbf{A}\right)^{2}c^{2}\right]\psi=\left[\tilde{m}c^{2}+q\phi\right]^{2}\psi.$$

(11)

Here the quantum energy operator $\hat{U}$ is $i\hbar\partial_{t}$, $\hat{\textbf{p}}$ is the quantum momentum operator $\hat{\textbf{p}}=-i\hbar\nabla$, and $\psi$ is the function that capture the wave behavior of the particle. Since the mass depends on time explicitly, $\hat{U}$ and $\tilde{m}$ do not commute and we must rewrite Eq. (11) as

$$\begin{split}\left[i\tilde{m}\hbar\partial_{t}-\frac{\hbar^{2}}{2c^{2}}\partial_{tt}-\frac{1}{2}(\hat{\textbf{p}}-q\textbf{A})^{2}\right]\psi=\\ \left[\tilde{m}q\phi+\frac{1}{2}\left(\frac{q^{2}\phi^{2}}{c^{2}}-i\hbar\partial_{t}\tilde{m}\right)\right]\psi.\end{split}$$

(12)

We can expand the momentum operator as

$$(\hat{\textbf{p}}-q\textbf{A})^{2}\psi=-\hbar^{2}\nabla^{2}\psi+2i\hbar q\textbf{A}\cdot\nabla\psi+i\hbar q\psi\nabla\cdot\textbf{A}+q^{2}\textbf{A}^{2}\psi$$

since $\nabla\cdot(\psi\textbf{A})=\textbf{A}\cdot\nabla\psi+\psi\nabla\cdot\textbf{A}$. Now, using Eq. (10), we can replace $q\nabla\cdot\textbf{A}$ with $\partial_{t}\tilde{m}$ in the equation above. Once this equation is injected into Eq. (12), we get the Klein-Gordon equation for non-conservative effects

$$\begin{split}i\hbar\tilde{m}\partial_{t}\psi+\frac{\hbar^{2}}{2}\Box\psi-i\hbar q\textbf{A}\cdot\nabla\psi=&\\ \left[\tilde{m}q\phi+\frac{q^{2}}{2}\left(\frac{\phi^{2}}{c^{2}}+\textbf{A}^{2}\right)\right]&\psi,\end{split}$$

(13)

where $\Box=\nabla^{2}-c^{-2}\partial_{tt}$ is the d’Alembertian operator. When only conservative effects are considered, $\tilde{m}\rightarrow m$, and Eq. (13) turns into the Klein-Gordon equation with relativistic mass.

We construct the probability current density using Noether’s framework [22, 23] rather than using the usual method [19] involving $\psi$ and its complex conjugate inside Eq. (13), as the mass is not constant. We start with the first order Euler-Lagrange equation given by

$$\frac{\delta\mathcal{L}}{\delta\psi_{i}}-\partial_{\nu}\frac{\delta\mathcal{L}}{\delta\partial_{\nu}\psi_{i}}=0\,\,\,\forall i\in\{1,\dots,N\},$$

(14)

where $\mathcal{L}$ is the Lagrangian density, and we used summation for repeated Greek index $\nu\in\{t,x,y,z\}$. Here $N=2$, since we will use only the function $\psi$ and its complex conjugate $\psi^{*}$. We can verify that

$$\begin{split}\mathcal{L}=&-i\frac{\hbar\tilde{m}}{2}(\psi^{*}\partial_{t}\psi-\psi\partial_{t}\psi^{*})\\ &+\frac{\hbar^{2}}{2}(\nabla\psi^{*}\cdot\nabla\psi-\frac{1}{c^{2}}\partial_{t}\psi\partial_{t}\psi^{*})\\ &+i\frac{q\hbar}{2}(\psi^{*}\textbf{A}\cdot\nabla\psi-\psi\textbf{A}\cdot\nabla\psi^{*})\\ &+\left[\tilde{m}q\phi+\frac{q^{2}}{2}\left(\frac{\phi^{2}}{c^{2}}+\textbf{A}^{2}\right)\right]\psi^{*}\psi,\end{split}$$

(15)

used in Eq. (14) gives Eq.(13) using the Lorenz gauge. With the first order Euler-Lagrange equation, there is a corresponding Noether’s current density for $\mu,\nu\in\{t,x,y,z\}$

$$j_{\mu}=\sum_{i}\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\psi_{i}}\delta\psi_{i},$$

(16)

where $\delta\psi$ and $\delta\psi^{*}$ correspond to infinitesimal variations under a $U(1)$ phase transformation, i.e. $\psi\rightarrow e^{i\theta}\psi$ and $\psi^{*}\rightarrow e^{-i\theta}\psi^{*}$. In this case, $\delta\psi\rightarrow i\theta\psi$, $\delta\psi^{*}\rightarrow-i\theta\psi^{*}$, $\partial_{\nu}\delta\psi=i\theta\partial_{\nu}\psi$, and $\partial_{\nu}\delta\psi^{*}=-i\theta\partial_{\nu}\psi^{*}$. Computing the four currents we get

$$\left.\begin{split}J_{t}&=\hbar\theta\left(\tilde{m}\psi\psi^{*}-i\hbar\frac{\psi\partial_{t}\psi^{*}-\psi^{*}\partial_{t}\psi}{2c^{2}}\right)\\ J_{\mu}&=\hbar\theta\left(i\hbar\frac{\psi\partial_{\mu}\psi^{*}-\psi^{*}\partial_{\mu}\psi}{2}-qA_{\mu}\psi\psi^{*}\right)\end{split}\right\}$$

(17)

with $\mu\in\{x,y,z\}$. From Eq. (16) we can infer that the term inside the square brackets on the RHS of Eq. (15) does not contribute to the Noether currents. Since the Noether theorem guarantees that

$$\partial_{\nu}J_{\nu}=0,$$

(18)

we can drop $\hbar\theta$ from the four-current density.

We can write this equation in polar form using $\psi=Re^{iS/\hbar}$. Here we take the modulus $R(t,x,y,z):\mathbb{R}^{4}\rightarrow\mathbb{R}^{+}$ and the phase $S(t,x,y,z):\mathbb{R}^{4}\rightarrow\mathbb{R}$, as this approach is compatible with the quantum operators $\hat{\textbf{p}}$ and $\hat{H}$ (e.g. Ref. 24). We can now compute explicitly the current density using the polar form of $\psi$. When $R\neq 0$, we have

$$\forall\nu\in\{t,x,y,z\},\,\frac{\partial_{\nu}\psi}{\psi}=\partial_{\nu}\ln\psi,$$

where $\ln\psi=\ln R+iS/\hbar$. So, $\psi\partial_{\nu}\psi^{*}-\psi^{*}\partial_{\nu}\psi=-2iR^{2}\partial_{\nu}S/\hbar$, even when $R=0$. Using this result, the four-current density becomes

$$\left.\begin{split}J_{t}&=\left[\tilde{m}-\partial_{t}S/c^{2}\right]R^{2}\\ J_{\mu}&=\left[\partial_{\mu}S-qA_{\mu}\right]R^{2}\end{split}\right\}$$

(19)

where $\mu\in\{x,y,z\}$. Using Eqs. (8) and (19), Eq. (18) yields

$$\partial_{t}\rho+\textbf{v}_{\psi}\cdot\nabla\rho=-\frac{\rho c^{2}}{\tilde{m}c^{2}-\partial_{t}S}\Box S.$$

(20)

where

$$\textbf{v}_{\psi}=\frac{(\nabla S-q\textbf{A})c^{2}}{\tilde{m}c^{2}-\partial_{t}S},$$

(21)

and the density $\rho$ is $R^{2}$. Using Eq. (10), we find

$$\begin{split}\nabla\cdot\textbf{v}_{\psi}=\frac{c^{2}\nabla^{2}S-\partial_{t}\tilde{m}c^{2}-\textbf{v}_{\psi}\cdot\nabla\left(\tilde{m}c^{2}-\partial_{t}S\right)}{\tilde{m}c^{2}-\partial_{t}S}.\end{split}$$

(22)

Adding $\rho\nabla\cdot\textbf{v}_{\psi}$ on both sides of Eq. (20) yields

$$\begin{split}\partial_{t}\rho+\nabla\cdot(\rho\textbf{v}_{\psi})=\\ -\frac{\partial_{t}\left(\tilde{m}c^{2}-\partial_{t}S\right)+\textbf{v}_{\psi}\cdot\nabla\left(\tilde{m}c^{2}-\partial_{t}S\right)}{\tilde{m}c^{2}-\partial_{t}S}\rho.\end{split}$$

(23)

At this point, we can easily identify the momentum of $\psi$ as $\textbf{p}_{\psi}=\hbar\textbf{k}$, and its energy as $E_{\psi}=\hbar\omega$, where the angular wave vector[25] is given by

$$\textbf{k}=\nabla S/\hbar,$$

(24)

and the angular frequency[25] is

$$\omega=-\partial_{t}S/\hbar.$$

(25)

So, we can write Eq. (21) as

$$\textbf{v}_{\psi}=\frac{(\textbf{p}_{\psi}-q\textbf{A})c^{2}}{\tilde{m}c^{2}+E_{\psi}}.$$

(26)

As this equation shows, $\textbf{v}_{\psi}$ conforms to the definition of a velocity according to special relativity. Using this insight, we see that the LHS of Eq. (23) is, in fact, a continuity equation for the density $\rho$ inside a velocity field $\textbf{v}_{\psi}$. The RHS of Eq. (23) is an advection equation (i.e. material derivative) for the energy $\tilde{m}c^{2}+E_{\psi}$, and $\textbf{v}_{\psi}$ is the speed at which this energy is advected. While both equations are conservation equations, there is a fundamental difference. The continuity equation implies conservation inside a static volume while the advection equation implies conservation inside a volume following along the streamlines of the vector field $\textbf{v}_{\psi}$.

The necessary and sufficient condition required to conserve the density $\rho$ can be found in Eq. (23), namely

$$\begin{split}\partial_{t}\left(\tilde{m}c^{2}+E_{\psi}\right)+\textbf{v}_{\psi}\cdot\nabla\left(\tilde{m}c^{2}+E_{\psi}\right)=0.\end{split}$$

(27)

Using Eq. (7) we find that the equation above is equivalent to,

$$\begin{split}\partial_{t}\left(E_{\psi}-q\phi\right)+\textbf{v}_{\psi}\cdot\nabla\left(E_{\psi}-q\phi\right)=0.\end{split}$$

(28)

So, if the effective wave energy $E_{\psi}-q\phi$ is conserved then

$$\partial_{t}\rho+\nabla\cdot(\rho\textbf{v}_{\psi})=0.$$

(29)

Hence the density $\rho$ is conserved in the sense of the continuity equation, and it is always positive when the energy in conserved. As a result, $\rho$ can be interpreted as a probability density, and $\psi$ is a probability amplitude.

When $\Box R=0$, Eq. (12) has a simple dispersion relation (see the appendix)

$$\left(\tilde{m}c^{2}-\partial_{t}S\right)^{2}=m_{0}^{2}c^{4}+(\hbar\textbf{k}-q\textbf{A})^{2}c^{2}.$$

(30)

In this case, Eq. (30) guarantees that $\tilde{m}c^{2}-\partial_{t}S\neq 0$ for massive particles. We can use Eqs. (24) and (25) to write the dispersion relation explicitly as a function of the wave vector and the angular frequency,

$$\omega_{\pm}=-\frac{\tilde{m}c^{2}}{\hbar}\pm\sqrt{\left(\frac{m_{0}c^{2}}{\hbar}\right)^{2}+\left(\textbf{k}-\frac{q}{\hbar}\textbf{A}\right)^{2}c^{2}}.$$

(31)

This is the de Broglie dispersion relation for matter-wave [6], where the bias term $\tilde{m}c^{2}/\hbar$ comes from using $U$ rather than $E$. Under these conditions, we can now compute the group velocity of the probability amplitude of the particle

$$\frac{\partial\omega_{+}}{\partial\textbf{k}}=\frac{(\hbar\textbf{k}-q\textbf{A})c^{2}}{\sqrt{m_{0}^{2}c^{4}+\left(\hbar\textbf{k}-q\textbf{A}\right)^{2}c^{2}}}.$$

(32)

and

$$\frac{\partial\omega_{-}}{\partial\textbf{k}}=\frac{(-\hbar\textbf{k}+q\textbf{A})c^{2}}{\sqrt{m_{0}^{2}c^{4}+\left(-\hbar\textbf{k}+q\textbf{A}\right)^{2}c^{2}}}.$$

(33)

The positive group velocity is the velocity of the particle with charge $q$ and the negative group velocity is the velocity of the dual particle with charge $-q$. We see here that the particle and its dual particle travel in opposite directions. Note that we are not using the term “antiparticle” for the dual particle here because we have made no assumption on the actual value of the charge $q$. It can be positive or negative. The only requirement is this paper is that $q\phi<0$. As a result, we see that the particle has an energy $\tilde{m}c^{2}+E_{\psi}>0$ and its dual particle has an energy $\tilde{m}c^{2}+E_{\psi}<0$. Since $q$ must satisfy $q\phi<0$ everywhere, then dual particles cannot propagate in this potential since their charge is $-q$ as it would violate Eq. (5). As a result, this work only applies to the particle only.

Using the dispersion relation above we also find that the velocity $\textbf{v}_{\psi}$ defined in Eq. (21) is the group velocity given by Eq. (32) if $\tilde{m}c^{2}+E_{\psi}>0$ and Eq. (33) if $\tilde{m}c^{2}+E_{\psi}<0$. Note that the amplitude $R$ has a phase velocity $c$ since $\Box R=0$.

So, the particle is formed by a group of waves and its most probable location is where these waves interfere constructively, leading to a match between the particle velocity and the wave group velocity. Since $\textbf{v}_{\psi}$ matches exactly the velocity of a relativistic classical particle in this section, we also conclude that quantum effects are inexistent when $\Box R=0$.

In reality, we know that particles and antiparticles can propagate in electrical potentials of any sign, the restriction herein comes from the fact that we only consider a single particle in this work. So the restriction of the sign of $q\phi$ are stringent. A many-body version of Eq. (34) would be required to capture the time evolution of the probability amplitude for a particle and its antiparticle inside a potential of arbitrary sign, a topic beyond the scope of this paper.

In relativity, mass and energy are exchangeable, as a result, we expect Eq. (5) to be violated when mass is converted to energy or vice versa. However, when the wave gains or loses energy from $\tilde{m}$ (i.e. conservative mass acceleration/deceleration or non-conservative electromagnetic effects, see Eq. (27)), or from the potential energy $q\phi$ (i.e. conservative time-independent or non-conservative time-dependent effects, see Eq. (28)), then the energy is conserved because there is there is really no other energies to tap on here, especially because interactions with other particles (e.g. pair creation/annihilation) cannot be captured by Eq. (13).