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External Potentials and Ehrenfest Relations in
Lagrangian Field Theories

Rayn Samson [email protected] Department of Physics, Grinnell College, Grinnell, Iowa 50112, USA
Abstract

This paper develops a general method to construct Ehrenfest-like relations for Lagrangian field theories when an external, coordinate-dependent scalar potential is applied. To do so, we derive continuity equations in which the spatial and temporal derivatives of the potential can be interpreted as a source of field momentum and field energy, respectively. For a non-relativistic Schrödinger field theory, these continuity equations yield Ehrenfest’s theorem for energy, linear momentum, and angular momentum. We then derive a relativistic counterpart for these relations using complex Klein-Gordon fields coupled with an electric potential.

I Introduction

It is well understood in classical mechanics that the conservation of linear momentum in a system only holds when no external force is acting on it. Put in the language of Noether’s theorem, a system is not symmetric under translation when a spatially dependent potential acts on it—in which case the conservation of linear momentum is broken. However, much of the richness in classical mechanics comes from understanding how mechanical momentum changes in time when such a potential is applied. This is captured by Newton’s second law,

ddt𝐩=V.\frac{d}{dt}\mathbf{p}=-\nabla V. (1)

Of course, this expression reduces to a statement of momentum conservation when the potential VV is not spatially dependent. While this result can be easily derived as the equation of motion in a classical Lagrangian of the form L(𝐪,𝐪˙,t)L(\mathbf{q},\dot{\mathbf{q}},t), there has so far been no systematic way to produce analogous results for the field momentum of a field Lagrangian (φ1,,φn,μφ1,,μφn)\mathcal{L}(\varphi_{1},\dots,\varphi_{n},\partial_{\mu}\varphi_{1},\dots,\partial_{\mu}\varphi_{n}). This paper develops such a method by first considering a non-relativistic field theory that describes the Schrödinger equation with an external, coordinate-dependent potential in Section III. Conservation of energy and momentum does not hold in general for such a system, but a close examination of the potential’s response to coordinate transformations yields Ehrenfest’s theorem for linear and angular momentum as well as a similar result for energy. We then use a generalized version of the same method in Section IV to derive analogous results in a complex Klein-Gordon field theory with minimal electric potential coupling. In this paper, such results are referred to as “Ehrenfest” or “Ehrenfest-like” relations owing to their structural similarities with the original results obtained from the Schrödinger field theory. However, the content of relativistic Ehrenfest relations has an entirely different interpretation from the non-relativistic case.

I.1 Notation and Conventions

We will be working in flat Minkowski spacetime with Cartesian coordinates. The mostly positive signature (+++)(-+++) is used. Coordinate derivatives are abbreviated with a comma before the appropriate index, e.g. M,μμMM_{,\mu}\equiv\partial_{\mu}M. Greek indices run through all four spacetime coordinates, while latin indices run through the three spatial coordinates. We will switch freely between spatial index notation and vector notation. For example, the four-divergence is given by

μ(0),\partial_{\mu}\equiv\begin{pmatrix}\partial_{0}\\ \nabla\end{pmatrix}, (2)

where =(1,2,3)\nabla=(\partial_{1},\partial_{2},\partial_{3}) may be represented with the open spatial index operator i\partial_{i}.

II Noether Currents

In the context of classical field theory, Noether’s theorem describes a fascinating equivalence between continuous symmetries of a Lagrangian system and its conserved quantities. Starting with an action of the form

𝒮=(φ,μφ)d4x,\mathcal{S}=\int\mathcal{L}(\varphi,\partial_{\mu}\varphi)\,d^{4}x, (3)

where \mathcal{L} is a Lagrangian density (henceforth: “Lagrangian”) that depends on a field φ\varphi and its derivatives. Using the principle of stationary action, the field φ\varphi must satisfy the Euler-Lagrange field equation of motion

μ(φ,μ)+φ=0.-\partial_{\mu}\left(\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\right)+\frac{\partial{\mathcal{L}}}{\partial{\varphi}}=0. (4)

A “symmetry” of the system is a transformation that yields the same physical predictions. A transformation of the field,

φφ+δφ,\varphi\to\varphi+\delta\varphi, (5)

is a symmetry if it leaves Eq. (4) invariant. This holds most generally when the transformation deforms the Lagrangian by a divergence of some function KμK^{\mu}:

(φ+δφ,μ(φ+δφ))(φ,μφ)=μKμ.\mathcal{L}\Big{(}\varphi+\delta\varphi,\partial_{\mu}(\varphi+\delta\varphi)\Big{)}-\mathcal{L}(\varphi,\partial_{\mu}\varphi)=\partial_{\mu}K^{\mu}. (6)

(We will later argue that the reverse is not true: invariance of Eq. (4) does not imply symmetry under a given transformation.) Calculating the left-hand side of Eq. (6) with a series expansion and simplifying with the equation of motion yields

(φ+δφ,μ(φ+δφ))(φ,μφ)=μ(φ,μδφ).\mathcal{L}\Big{(}\varphi+\delta\varphi,\partial_{\mu}(\varphi+\delta\varphi)\Big{)}-\mathcal{L}(\varphi,\partial_{\mu}\varphi)=\partial_{\mu}\left(\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\delta\varphi\right). (7)

Equating this to the surface term μKμ\partial_{\mu}K^{\mu} yields the vanishing divergence

μ(φ,μδφKμ)=0,\partial_{\mu}\left(\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\delta\varphi-K^{\mu}\right)=0, (8)

which defines a conservation law in the form of a continuity equation. The term in parentheses defines a Noether current. Importantly, if the action depends on more than one field, we would need to add the response of each field to the transformation when defining the Noether current. In this paper, we will examine Lagrangian systems that depend on two fields which are complex conjugates of each other, so Noether currents will take the form

Jμφ,μδφ+φ¯,μδφ¯Kμ.J^{\mu}\equiv\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\delta\varphi+\frac{\partial{\mathcal{L}}}{\partial{\overline{\varphi}_{,\mu}}}\delta\overline{\varphi}-K^{\mu}. (9)

We now proceed to discuss how the conservation of these Noether currents is affected by an external scalar potential.

III Schrödinger Field Theory

Suppose we are given a physical system that is influenced by an external, coordinate-dependent potential. One such example is a system described by the Schrödinger equation

22m2Ψ+VΨ=iΨt.-\frac{\hbar^{2}}{2m}\nabla^{2}\Psi+V\Psi=i\hbar\frac{\partial{\Psi}}{\partial{t}}. (10)

Crucially, Eq. (10) does not specify the source or dynamics of the potential VV, which is why we say that it is external to the theory. To formulate this equation in terms of field theory, consider a Schrödinger LagrangianFranklin2017 TakahashiRop1986

=22mjΨ¯jΨ+VΨ¯Ψi2(Ψ¯Ψ˙ΨΨ¯˙),\mathcal{L}=\frac{\hbar^{2}}{2m}\partial_{j}\overline{\Psi}\partial^{j}\Psi+V\overline{\Psi}\Psi-\frac{i\hbar}{2}\left(\overline{\Psi}\dot{\Psi}-\Psi\dot{\overline{\Psi}}\right), (11)

where the fields are Ψ\Psi and Ψ¯\overline{\Psi}. The equation of motion associated with the latter,

t(Ψ¯˙)xj(Ψ¯,j)+Ψ¯=0,-\frac{\partial{}}{\partial{t}}\left(\frac{\partial{\mathcal{L}}}{\partial{\dot{\overline{\Psi}}}}\right)-\frac{\partial{}}{\partial{x^{j}}}\left(\frac{\partial{\mathcal{L}}}{\partial{\overline{\Psi}_{,j}}}\right)+\frac{\partial{\mathcal{L}}}{\partial{\overline{\Psi}}}=0, (12)

reduces to Eq. (10). To account for the potential’s response to a coordinate transformation, we treat the Lagrangian as an explicit function of VV in addition to the fields and their derivatives:

=(Ψ,Ψ¯,μΨ,μΨ¯,V).\mathcal{L}=\mathcal{L}(\Psi,\overline{\Psi},\partial_{\mu}\Psi,\partial_{\mu}\overline{\Psi},V). (13)

Despite the awkward dependence on the potential, this Lagrangian still admits a U(1)U(1) symmetry. Consider the complex rotation ΨeiθΨ\Psi\to e^{i\theta}\Psi (and cc). For small θ\theta we have eiθ1+iθe^{i\theta}\approx 1+i\theta, and so the fields respond with

ΨΨ+iθΨandΨ¯Ψ¯iθΨ¯.\Psi\to\Psi+i\theta\Psi\quad\text{and}\quad\overline{\Psi}\to\overline{\Psi}-i\theta\overline{\Psi}. (14)

The associated conservation law becomes a statement of probability conservation,

t|Ψ|2+𝐣=0,\frac{\partial{}}{\partial{t}}|\Psi|^{2}+\nabla\cdot\mathbf{j}=0, (15)

where |Ψ|2Ψ¯Ψ|\Psi|^{2}\equiv\overline{\Psi}\Psi is interpreted as a probability density and 𝐣\mathbf{j} is the probability current defined byGriffiths2018-vs

𝐣2mi(Ψ¯ΨΨΨ¯).\mathbf{j}\equiv\frac{\hbar}{2mi}\left(\overline{\Psi}\nabla\Psi-\Psi\nabla\overline{\Psi}\right). (16)

This result is unaffected by the presence of VV in the Lagrangian, but the same is not true for coordinate transformations. Consider a spacetime translation

xμxμ+aμ,x^{\mu}\to x^{\mu}+a^{\mu}, (17)

for some (small) constant aμa^{\mu}. The fields are scalars, so they transform as

ΨΨ+aμμΨandΨ¯Ψ¯+aμμΨ¯,\Psi\to\Psi+a^{\mu}\partial_{\mu}\Psi\quad\text{and}\quad\overline{\Psi}\to\overline{\Psi}+a^{\mu}\partial_{\mu}\overline{\Psi}, (18)

and crucially, so does the potential:

VV+aμμV.V\to V+a^{\mu}\partial_{\mu}V. (19)

As a scalar, the Lagrangian also transforms as +aμμ\mathcal{L}\to\mathcal{L}+a^{\mu}\partial_{\mu}\mathcal{L}. Thus the change in the Lagrangian (which we will denote as δ\delta\mathcal{L}) calculated by treating \mathcal{L} as a scalar is given by

δ=μ(aμ),\delta\mathcal{L}=\partial_{\mu}(a^{\mu}\mathcal{L}), (20)

which gives the surface term. Care must be taken to understand what is meant here: as a scalar, the Lagrangian does changes by an overall divergence, and so the equations of motion are invariant even in the presence of VV. However, computing δ\delta\mathcal{L} from a first-order series expansion in all arguments of the transformed Lagrangian does not yield an overall divergence, and dependence on VV is the culprit:

δ=aν[μ(Ψ,μΨ,ν+Ψ¯,μΨ¯,ν)+VνV].\delta\mathcal{L}=a^{\nu}\left[\partial_{\mu}\left(\frac{\partial{\mathcal{L}}}{\partial{\Psi_{,\mu}}}\Psi_{,\nu}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\Psi}_{,\mu}}}\overline{\Psi}_{,\nu}\right)+\frac{\partial{\mathcal{L}}}{\partial{V}}\partial_{\nu}V\right]. (21)

We can thus see that \mathcal{L} is not translationally symmetric, and hence there is no conservation law to be obtained from a vanishing divergence. That is, invariance of a potential-dependent Lagrangian under a coordinate transformation does not imply a symmetry under that transformation. However, we can still set Eqs. (20) and (21) equal to each other and see what is obtained instead of a conservation law. After discarding the overall constant aνa^{\nu}, we define the energy-momentum tensor in a standard way:

TνμΨ,μΨ,ν+Ψ¯,μΨ¯,νδνμ.T^{\mu}_{\,\,\,\nu}\equiv\frac{\partial{\mathcal{L}}}{\partial{\Psi_{,\mu}}}\Psi_{,\nu}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\Psi}_{,\mu}}}\overline{\Psi}_{,\nu}-\delta^{\mu}_{\,\,\,\nu}\mathcal{L}. (22)

We still say that TνμT^{\mu}_{\,\,\,\nu} represents four Noether currents containing the energy (ν=0\nu=0) and momentum (ν=1,2,3\nu=1,2,3) densities stored in the fields. However, the divergence of TνμT^{\mu}_{\,\,\,\nu} is not in general zero. Instead, the derivatives of the potential act as a “source” in the continuity equations

μTνμ=VνV.\partial_{\mu}T^{\mu}_{\,\,\,\nu}=-\frac{\partial{\mathcal{L}}}{\partial{V}}\partial_{\nu}V. (23)

The conservation of these currents depends on the presence of an external potential and its derivatives. In analogy with classical mechanics, the momentum of a system is conserved only when there is no coordinate-dependent external potential acting on it. When an external potential is applied, momentum is no longer conserved and the change in momentum is summarized by Newton’s second law: d𝐩/dt=Vd\mathbf{p}/dt=-\nabla V. The equivalent result in non-relativistic quantum mechanics would be Ehrenfest’s theorem for momentum, d𝐩^/dt=Vd\langle\hat{\mathbf{p}}\rangle/dt=\langle-\nabla V\rangle. Section III.1 will show that this is one of two statements contained in Eq. (23).

III.1 Ehrenfest Relations

Having defined the energy-momentum tensor associated with the Schrödinger Lagrangian, it is helpful to write out the components explicitly:

T   00\displaystyle T^{0}_{\,\,\,0} =22miΨ¯iΨVΨ¯Ψ\displaystyle=-\frac{\hbar^{2}}{2m}\partial^{i}\overline{\Psi}\partial_{i}\Psi-V\overline{\Psi}\Psi (24)
T   0i\displaystyle T^{i}_{\,\,\,0} =22mc(Ψ¯˙iΨ+Ψ˙iΨ¯)\displaystyle=\frac{\hbar^{2}}{2mc}\left(\dot{\overline{\Psi}}\partial^{i}\Psi+\dot{\Psi}\partial^{i}\overline{\Psi}\right) (25)
Ti0\displaystyle T^{0}_{\,\,\,i} =ic2(Ψ¯iΨ+ΨiΨ¯)\displaystyle=-\frac{i\hbar c}{2}\left(\overline{\Psi}\partial_{i}\Psi+\Psi\partial_{i}\overline{\Psi}\right) (26)
Tji\displaystyle T^{i}_{\,\,\,j} =22m(iΨ¯jΨ+iΨjΨ¯)δji.\displaystyle=\frac{\hbar^{2}}{2m}\left(\partial^{i}\overline{\Psi}\partial_{j}\Psi+\partial^{i}\Psi\partial_{j}\overline{\Psi}\right)-\delta^{i}_{\,j}\mathcal{L}. (27)

The momentum density is given by the Tj0T^{0}_{\,\,\,j} components, where j=1,2,3j=1,2,3 represents components along each spatial coordinate. By noting that /V=Ψ¯Ψ\partial\mathcal{L}/\partial V=\overline{\Psi}\Psi is the probability density, the sourced continuity equation says

μTjμ=Ψ¯(jV)Ψ,\partial_{\mu}T^{\mu}_{\,\,\,j}=\overline{\Psi}(-\partial_{j}V)\Psi, (28)

where it can be seen that the right-hand side represents a force expectation upon integration. For comparison, we put the left-hand side into integral form, simplify using the divergence theorem, and assume boundary terms vanish. This yields

μTjμd3x\displaystyle\int\!\!\partial_{\mu}T^{\mu}_{\,\,\,j}\,d^{3}x =\displaystyle= (1cTj0t+Tjixi)d3x\displaystyle\int\!\!\left(\frac{1}{c}\frac{\partial{T^{0}_{\,\,\,j}}}{\partial{t}}+\frac{\partial{T^{i}_{\,\,\,j}}}{\partial{x^{i}}}\right)d^{3}x (29)
=\displaystyle= ddt1cTj0d3x\displaystyle\frac{d}{dt}\int\!\frac{1}{c}T^{0}_{\,\,\,j}\,d^{3}x
=\displaystyle= i2ddt(Ψ¯jΨΨjΨ¯)d3x\displaystyle-\frac{i\hbar}{2}\frac{d}{dt}\int\!\!\left(\overline{\Psi}\partial_{j}\Psi-\Psi\partial_{j}\overline{\Psi}\right)\,d^{3}x
=\displaystyle= i2ddt(2Ψ¯jΨj(Ψ¯Ψ))d3x\displaystyle-\frac{i\hbar}{2}\frac{d}{dt}\int\!\!\big{(}2\overline{\Psi}\partial_{j}\Psi-\partial_{j}(\overline{\Psi}\Psi)\big{)}\,d^{3}x
=\displaystyle= ddtΨ¯(ij)Ψd3x\displaystyle\frac{d}{dt}\int\!\overline{\Psi}\!\left(\frac{\hbar}{i}\partial_{j}\right)\!\Psi\,d^{3}x
=\displaystyle= dp^jdt,\displaystyle\frac{d\langle\hat{p}_{j}\rangle}{dt},

where the momentum operator p^jij\hat{p}_{j}\equiv\frac{\hbar}{i}\partial_{j} has been identified. We thus conclude that Eq. (28) in integral form is Ehrenfest’s theorem for momentum,Takahashi1986

d𝐩^dt=V.\frac{d\langle\hat{\mathbf{p}}\rangle}{dt}=\langle-\nabla V\rangle. (30)

There are several points to make regarding this familiar result obtained by unfamiliar means. Locally, the negative gradient of an external potential—the external force—acts as a source of field momentum, which is captured by Eq. (28). Globally, this results in Ehrenfest’s theorem for momentum. Moreover, this result does not require a probabilistic interpretation of the wavefunction, nor does it reference the field equations associated with Eq. (11). This indicates that Eq. (23), which makes no reference to a specific choice of Lagrangian, holds more generally for field theories with an external potential.

There is now a clear motivation to ask about the remaining Noether current representing energy density. The sourced continuity equation for this component is

μT   0μ=1cΨ¯VtΨ.\partial_{\mu}T^{\mu}_{\,\,\,0}=-\frac{1}{c}\overline{\Psi}\frac{\partial{V}}{\partial{t}}\Psi. (31)

Putting the left-hand side in integral form as before, we now have an expression in terms of the Hamiltonian operator:

μT   0μd3x\displaystyle\int\!\!\partial_{\mu}T^{\mu}_{\,\,\,0}\,d^{3}x =\displaystyle= (1cT   00t+T   0ixi)d3x\displaystyle\int\!\!\left(\frac{1}{c}\frac{\partial{T^{0}_{\,\,\,0}}}{\partial{t}}+\frac{\partial{T^{i}_{\,\,\,0}}}{\partial{x^{i}}}\right)d^{3}x (32)
=\displaystyle= 1cddtT   00d3x\displaystyle\frac{1}{c}\frac{d}{dt}\int\!T^{0}_{\,\,\,0}\,d^{3}x
=\displaystyle= 1cddt(22miΨ¯iΨVΨ¯Ψ)d3x\displaystyle\frac{1}{c}\frac{d}{dt}\int\!\!\left(-\frac{\hbar^{2}}{2m}\partial^{i}\overline{\Psi}\partial_{i}\Psi-V\overline{\Psi}\Psi\right)d^{3}x
=\displaystyle= 1cddt(22mΨ¯iiΨVΨ¯Ψ)d3x\displaystyle\frac{1}{c}\frac{d}{dt}\int\!\left(\frac{\hbar^{2}}{2m}\overline{\Psi}\partial^{i}\partial_{i}\Psi-V\overline{\Psi}\Psi\right)d^{3}x
=\displaystyle= 1cddtΨ¯(22m2+V)Ψd3x\displaystyle-\frac{1}{c}\frac{d}{dt}\int\!\!\overline{\Psi}\!\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V\right)\!\Psi\,d^{3}x
=\displaystyle= 1cdH^dt.\displaystyle-\frac{1}{c}\frac{d\langle\hat{H}\rangle}{dt}.

Thus integrating both sides of Eq. (31) and dividing out the factor of 1/c-1/c, we have an Ehrenfest-like relation for energy:

dH^dt=Vt.\frac{d\langle\hat{H}\rangle}{dt}=\left\langle\frac{\partial{V}}{\partial{t}}\right\rangle. (33)

Just as the spatial derivatives of an external potential are sources of linear momentum, the time derivative of an external potential acts as a source of energy. Thus the expected energy is not a constant in time when a time-dependent potential is applied, but we still have a formal expression to track its rate of change. In Section III.2, we demonstrate one example of this result holding true for an infinite square well with a moving floor.

We now briefly cover the case of angular momentum, which yields a similar Ehrenfest relation. Consider a coordinate rotation

xμxμ+Rνμxν,x^{\mu}\to x^{\mu}+R^{\mu}_{\,\,\,\nu}x^{\nu}, (34)

where RνμR^{\mu}_{\,\,\,\nu} represents an infinitesimal rotation of the spatial coordinates. We can show that there are six associated Noether currents

μνσxνTμσxσTμν,\mathcal{M}^{\mu\nu\sigma}\equiv x^{\nu}T^{\mu\sigma}-x^{\sigma}T^{\mu\nu}, (35)

where TμνT^{\mu\nu} is the energy-momentum tensor as defined before with raised indices. A detailed derivation is provided in Appendix A.2. We can use the Levi-Civita symbol to pick out the components representing the angular momentum density along each spatial coordinate,

i1cεijkxjT0k=12cεijk0jk.\ell_{i}\equiv\frac{1}{c}\varepsilon_{ijk}x^{j}T^{0k}=\frac{1}{2c}\varepsilon_{ijk}\mathcal{M}^{0jk}. (36)

On one hand, we can now directly compute the divergence of Eq. (35) with the product rule and pick out the terms of interest,

12εijkμ0jk=Ψ¯τiΨ,\frac{1}{2}\varepsilon_{ijk}\partial_{\mu}\mathcal{M}^{0jk}=\overline{\Psi}\tau_{i}\Psi, (37)

where τiεijkxi(jV)\tau_{i}\equiv\varepsilon_{ijk}x^{i}(-\partial^{j}V) is the iith component of the torque associated with VV. It is thus clear that integration along all components of Eq. (37) yields the expected torque, 𝝉\langle\boldsymbol{\tau}\rangle. On the other hand, computing the integral in a similar fashion to Eqs. (29) and (32) yields

12εijkμ0jkd3x\displaystyle\frac{1}{2}\varepsilon_{ijk}\!\!\int\!\!\partial_{\mu}\mathcal{M}^{0jk}\,d^{3}x =\displaystyle= 12εijk(1c0jkt+jkx)d3x\displaystyle\frac{1}{2}\varepsilon_{ijk}\!\!\int\!\!\left(\frac{1}{c}\frac{\partial{\mathcal{M}^{0jk}}}{\partial{t}}+\frac{\partial{\mathcal{M}^{\ell jk}}}{\partial{x^{\ell}}}\right)d^{3}x (38)
=\displaystyle= 12cεijkddt0jkd3x\displaystyle\frac{1}{2c}\varepsilon_{ijk}\frac{d}{dt}\!\int\!\!\mathcal{M}^{0jk}\,d^{3}x
=\displaystyle= 12cεijkddt(xjT0kxkT0j)d3x\displaystyle\frac{1}{2c}\varepsilon_{ijk}\frac{d}{dt}\!\int\!\!\left(x^{j}T^{0k}-x^{k}T^{0j}\right)d^{3}x
=\displaystyle= ddtΨ¯(εijkx^jp^k)Ψd3x\displaystyle\frac{d}{dt}\!\int\!\!\overline{\Psi}\left(\varepsilon_{ijk}\hat{x}^{j}\hat{p}^{k}\right)\Psi\,d^{3}x
=\displaystyle= dL^idt,\displaystyle\frac{d\langle\hat{L}_{i}\rangle}{dt},

where L^iεijkx^jp^k\hat{L}_{i}\equiv\varepsilon_{ijk}\hat{x}^{j}\hat{p}^{k} is the angular momentum operator. Comparing this with the integral of Eq. (37), we have Ehrenfest’s theorem for angular momentum:Griffiths2018-vs

d𝐋^dt=𝝉.\frac{d\langle\hat{\mathbf{L}}\rangle}{dt}=\langle\boldsymbol{\tau}\rangle. (39)

This derivation serves as an important template for deriving analogous results in the relativistic Klein-Gordon setting, which is the subject of Section IV.

III.2 Moving Floor Infinite Square Well

We demonstrate one instance of the result in Eq. (33) using the time-varying potential

V(x,t)={V~(t)if 0<x<LotherwiseV(x,t)=\begin{cases}\tilde{V}(t)&\text{if }0<x<L\\ \infty&\text{otherwise}\end{cases} (40)

for some differentiable function V~(t)\tilde{V}(t). This reduces to the more familiar stationary floor infinite square well when V~(t)=0\tilde{V}(t)=0. With boundary condition Ψ(0,t)=Ψ(L,t)=0\Psi(0,t)=\Psi(L,t)=0, the full solution to the time-dependent Schrödinger equation can be expanded into a complete orthonormal basis with time-varying coefficients

Ψ(x,t)=k=1ak(t)sin(kπxL).\Psi(x,t)=\sum_{k=1}^{\infty}a_{k}(t)\sin\left(\frac{k\pi x}{L}\right). (41)

The coefficients ak(t)a_{k}(t) absorb the overall normalization and satisfy the residual requirementFranklin2013

0=2k2π22mL2ak(t)ia˙k(t)+ak(t)V~(t).0=\frac{\hbar^{2}k^{2}\pi^{2}}{2mL^{2}}a_{k}(t)-i\hbar\dot{a}_{k}(t)+a_{k}(t)\tilde{V}(t). (42)

The solution to this set of differential equations is

ak(t)=AkeEkt/eiw(t)/,a_{k}(t)=A_{k}e^{-E_{k}t/\hbar}e^{-iw(t)/\hbar}, (43)

where Ek=2k2π22mL2E_{k}=\frac{\hbar^{2}k^{2}\pi^{2}}{2mL^{2}} are the energy states of the stationary square well and

w(t)t=0tV~(t)𝑑tw(t)\equiv\int_{t^{\prime}=0}^{t}\tilde{V}(t^{\prime})\,dt^{\prime} (44)

is the additional phase picked up due to the time-dependent potential that does not depend on kk. We can thus write the solution

Ψ(x,t)=eiw(t)/k=1AkeiEkt/sin(kπxL),\Psi(x,t)=e^{-iw(t)/\hbar}\sum_{k=1}^{\infty}A_{k}e^{-iE_{k}t/\hbar}\sin\left(\frac{k\pi x}{L}\right), (45)

and the additional phase will disappear in any expression containing |Ψ|2|\Psi|^{2}. With this, we can calculate

dH^dt\displaystyle\frac{d\langle\hat{H}\rangle}{dt} =ddt0LΨ¯(22m2x2+V~(t))Ψ𝑑x\displaystyle=\frac{d}{dt}\!\int_{0}^{L}\!\overline{\Psi}\left(\frac{-\hbar^{2}}{2m}\frac{\partial{{}^{2}}}{\partial{x^{2}}}+\tilde{V}(t)\right)\Psi\,dx
=22mddt0LΨ¯Ψ2x2𝑑x+ddt0LV~(t)|Ψ|2𝑑x.\displaystyle=\frac{-\hbar^{2}}{2m}\frac{d}{dt}\!\!\int_{0}^{L}\!\!\!\overline{\Psi}\frac{\partial{{}^{2}\Psi}}{\partial{x^{2}}}\,dx+\frac{d}{dt}\!\!\int_{0}^{L}\!\!\!\tilde{V}(t)|\Psi|^{2}dx. (46)

To handle the first integral, we use the orthonormal decomposition

0LΨ¯Ψ2x2𝑑x\displaystyle\int_{0}^{L}\!\!\overline{\Psi}\frac{\partial{{}^{2}\Psi}}{\partial{x^{2}}}\,dx =\displaystyle= 0Lδjk(kπL)2A¯jAkei(EjEk)t/\displaystyle\!\int_{0}^{L}\!\!\!-\delta_{jk}\!\left(\frac{k\pi}{L}\right)^{2}\overline{A}_{j}A_{k}e^{i(E_{j}-E_{k})t/\hbar} (47)
×sin(jπxL)sin(kπxL)dx\displaystyle\quad\quad\times\sin\left(\frac{j\pi x}{L}\right)\sin\left(\frac{k\pi x}{L}\right)dx
=\displaystyle= (jπL)2|Aj|20Lsin2(jπxL)𝑑x\displaystyle-\left(\frac{j\pi}{L}\right)^{2}|A_{j}|^{2}\!\!\int_{0}^{L}\!\!\sin^{2}\left(\frac{j\pi x}{L}\right)dx
=\displaystyle= (jπL)2.\displaystyle-\left(\frac{j\pi}{L}\right)^{2}.

Since there is no time dependence here, the first term in Eq. (46) vanishes. To handle the second term, we pass the time derivative into the integral and use the product rule:

ddt0LV~(t)|Ψ|2𝑑x\displaystyle\frac{d}{dt}\!\int_{0}^{L}\!\!\tilde{V}(t)|\Psi|^{2}dx =\displaystyle= 0LV~(t)t|Ψ|2𝑑x+V~(t)0Lt|Ψ|2𝑑x\displaystyle\int_{0}^{L}\!\frac{\partial{\tilde{V}(t)}}{\partial{t}}|\Psi|^{2}dx+\tilde{V}(t)\!\!\int_{0}^{L}\!\!\frac{\partial{}}{\partial{t}}|\Psi|^{2}dx (48)
=\displaystyle= 0LΨ¯V~(t)tΨ𝑑x\displaystyle\int_{0}^{L}\!\overline{\Psi}\frac{\partial{\tilde{V}(t)}}{\partial{t}}\Psi\,dx
=\displaystyle= Ψ¯V(x,t)tΨ𝑑x\displaystyle\int_{-\infty}^{\infty}\!\overline{\Psi}\frac{\partial{V(x,t)}}{\partial{t}}\Psi\,dx
=\displaystyle= V(x,t)t.\displaystyle\left\langle\frac{\partial{V(x,t)}}{\partial{t}}\right\rangle.

We have thus verified that Eq. (33) holds for this system for any differentiable motion of the floor V~(t)\tilde{V}(t).

IV Complex Klein-Gordon Field Theory with Minimal Electric Potential Coupling

The general outline of the previous section appears to be a way to generate Ehrenfest-like relations directly from a Lagrangian, without reference to the probabilistic interpretation of the wavefunction. We use this as motivation to develop analogous results in a relativistic setting using complex Klein-Gordon fields coupled with an electric potential. We use a minimal coupling given by the covariant derivative

DμμiqAμ,D_{\mu}\equiv\partial_{\mu}-i\frac{q}{\hbar}A_{\mu}, (49)

where qq determines the coupling strength and

Aμ(V/c𝐀)A^{\mu}\equiv\begin{pmatrix}V/c\\ \mathbf{A}\end{pmatrix} (50)

is the electromagnetic four-potential (note that VV now has a dimension of energy per unit charge). Since we are only interested in the scalar potential term, we set 𝐀=0\mathbf{A}=0. In order to treat the potential as external to the theory, we will further omit the 14μ0FαβFαβ\frac{1}{4\mu_{0}}F^{\alpha\beta}F_{\alpha\beta} term used in a scalar QED action. With these modifications, the Lagrangian we use is

\displaystyle\mathcal{L} =(Dμϕ)¯(Dμϕ)+m2c22ϕ¯ϕ\displaystyle=\overline{\left(D_{\mu}\phi\right)}\left(D^{\mu}\phi\right)+\frac{m^{2}c^{2}}{\hbar^{2}}\overline{\phi}\phi
=ϕ¯,μϕ,μ+iqVc2(ϕ¯ϕ˙ϕϕ¯˙)+(m2c22q2V22c2)ϕ¯ϕ.\displaystyle=\overline{\phi}_{,\mu}\phi_{,}^{\,\,\mu}+i\frac{qV}{\hbar c^{2}}\left(\overline{\phi}\dot{\phi}-\phi\dot{\overline{\phi}}\right)+\left(\frac{m^{2}c^{2}}{\hbar^{2}}-\frac{q^{2}V^{2}}{\hbar^{2}c^{2}}\right)\,\overline{\phi}\phi. (51)

Using the d’Alembertian operator μμ\square\equiv\partial_{\mu}\partial^{\mu}, the field equations are

ϕ¯+(m2c22q2V22c2)ϕ¯+2iqVc2ϕ¯˙+iqV˙c2ϕ¯=0\displaystyle-\square\overline{\phi}+\left(\frac{m^{2}c^{2}}{\hbar^{2}}-\frac{q^{2}V^{2}}{\hbar^{2}c^{2}}\right)\overline{\phi}+2i\frac{qV}{\hbar c^{2}}\dot{\overline{\phi}}+\frac{iq\dot{V}}{\hbar c^{2}}\overline{\phi}=0 (52)
ϕ+(m2c22q2V22c2)ϕ2iqVc2ϕ˙iqV˙c2ϕ=0.\displaystyle-\square\phi+\left(\frac{m^{2}c^{2}}{\hbar^{2}}-\frac{q^{2}V^{2}}{\hbar^{2}c^{2}}\right)\phi-2i\frac{qV}{\hbar c^{2}}\dot{\phi}-\frac{iq\dot{V}}{\hbar c^{2}}\phi=0. (53)

As before, we will first find the conservation law associated with a U(1)U(1) symmetry, ϕeiθϕ\phi\to e^{i\theta}\phi. The conserved current is

Jμ=iθ(ϕ¯μϕϕμϕ¯2iqAμϕ¯ϕ).J^{\mu}=i\theta\left(\overline{\phi}\partial^{\mu}\phi-\phi\partial^{\mu}\overline{\phi}-2\frac{iq}{\hbar}A^{\mu}\overline{\phi}\phi\right). (54)

The 0-component can be simplified by separating the fields into their spatial and time-dependent components

ϕ=ψ(𝐫)eiEt/,\phi=\psi(\mathbf{r})e^{-iEt/\hbar}, (55)

and the complex conjugate. This allows us to write the time derivatives of the fields as

ϕ˙=Eiϕandϕ¯˙=Eiϕ¯.\dot{\phi}=\frac{E}{i\hbar}\phi\quad\text{and}\quad\dot{\overline{\phi}}=-\frac{E}{i\hbar}\overline{\phi}. (56)

After setting units, the conservation law reads

tρ+𝐣=0,\frac{\partial{}}{\partial{t}}\rho+\nabla\cdot\mathbf{j}=0, (57)

where

ρ2q2c2(EqV)ϕ¯ϕ\rho\equiv\frac{2q}{\hbar^{2}c^{2}}(E-qV)\overline{\phi}\phi (58)

is interpreted as an electric charge density,Winter1959 and

𝐣iq(ϕϕ¯ϕ¯ϕ)\mathbf{j}\equiv i\frac{q}{\hbar}(\phi\nabla\overline{\phi}-\overline{\phi}\nabla\phi) (59)

is the corresponding electric current. Since electric charge conservation appears precisely where probability conservation appeared in the Schrödinger field theory, we anticipate the same replacement to appear in the continuity equations associated with the energy-momentum tensor. Section IV.1 shows that this is indeed the case.

IV.1 Relativistic Ehrenfest Relations

Given our Lagrangian that depends on ϕ\phi, its complex conjugate, their derivatives, and the electric potential VV

=(ϕ,ϕ¯,μϕ,μϕ¯,V),\mathcal{L}=\mathcal{L}(\phi,\overline{\phi},\partial_{\mu}\phi,\partial_{\mu}\overline{\phi},V), (60)

we can follow the exact same procedure outlined from Eq. (17) to Eq. (23). Again defining the energy-momentum tensor as

Tνμϕ,μϕ,ν+ϕ¯,μϕ¯,νδνμ,T^{\mu}_{\,\,\,\nu}\equiv\frac{\partial{\mathcal{L}}}{\partial{\phi_{,\mu}}}\phi_{,\nu}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\phi}_{,\mu}}}\overline{\phi}_{,\nu}-\delta^{\mu}_{\,\,\,\nu}\mathcal{L}, (61)

the continuity equation still reads as Eq. (23). However, the factor of /V\partial\mathcal{L}/\partial V in the source term is not as straightforward to calculate. Computing it explicitly gives

V\displaystyle\frac{\partial{\mathcal{L}}}{\partial{V}} =\displaystyle= iqc2(ϕ¯ϕ˙ϕϕ¯˙)2q2V2c2ϕ¯ϕ\displaystyle\frac{iq}{\hbar c^{2}}\left(\overline{\phi}\dot{\phi}-\phi\dot{\overline{\phi}}\right)-2\frac{q^{2}V}{\hbar^{2}c^{2}}\overline{\phi}\phi (62)
=\displaystyle= iqc2(2Eiϕ¯ϕ)2q2V2c2ϕ¯ϕ\displaystyle\frac{iq}{\hbar c^{2}}\left(2\frac{E}{i\hbar}\overline{\phi}\phi\right)-2\frac{q^{2}V}{\hbar^{2}c^{2}}\overline{\phi}\phi
=\displaystyle= 2q2c2(EqV)ϕ¯ϕ\displaystyle\frac{2q}{\hbar^{2}c^{2}}(E-qV)\overline{\phi}\phi
=\displaystyle= ρ,\displaystyle\rho,

where we used Eq. (56) to simplify the time derivatives as before. Thus the derivatives of the potential are weighted by the electric charge density in the source term of Eq. (23), exactly taking the place of probability density in the Schrödinger theory. The continuity equation is thus

μTνμ=ρνV,\partial_{\mu}T^{\mu}_{\,\,\,\nu}=-\rho\partial_{\nu}V, (63)

where the energy-momentum tensor components are

T   00\displaystyle T^{0}_{\,\,\,0} =1c2ϕ¯˙ϕ˙ϕ¯ϕ(m2c22q2V22c2)ϕ¯ϕ\displaystyle=-\frac{1}{c^{2}}\dot{\overline{\phi}}\dot{\phi}-\nabla\overline{\phi}\cdot\nabla\phi-\left(\frac{m^{2}c^{2}}{\hbar^{2}}-\frac{q^{2}V^{2}}{\hbar^{2}c^{2}}\right)\overline{\phi}\phi (64)
T   0i\displaystyle T^{i}_{\,\,\,0} =1c(ϕ¯˙iϕ+ϕ˙iϕ¯)\displaystyle=\frac{1}{c}\left(\dot{\overline{\phi}}\partial^{i}\phi+\dot{\phi}\partial^{i}\overline{\phi}\right) (65)
Ti0\displaystyle T^{0}_{\,\,\,i} =1c(ϕ¯˙iϕ+ϕ˙iϕ¯)+iqVc(ϕ¯iϕϕiϕ¯)\displaystyle=-\frac{1}{c}\left(\dot{\overline{\phi}}\partial_{i}\phi+\dot{\phi}\partial_{i}\overline{\phi}\right)+i\frac{qV}{\hbar c}\Big{(}\overline{\phi}\partial_{i}\phi-\phi\partial_{i}\overline{\phi}\Big{)} (66)
Tji\displaystyle T^{i}_{\,\,\,j} =iϕ¯jϕ+iϕjϕ¯δji.\displaystyle=\partial^{i}\overline{\phi}\partial_{j}\phi+\partial^{i}\phi\partial_{j}\overline{\phi}-\delta^{i}_{\,\,\,j}\mathcal{L}. (67)

The component T   00T^{0}_{\,\,\,0} is the Hamiltonian density or energy density (up to a sign) stored in the fields. Upon integration, there is no longer a familiar expression like the expected Hamiltonian in Eq. (32). Instead, we write the field Hamiltonian directly in terms of T   00T^{0}_{\,\,\,0}:

HT   00d3x.H\equiv\int-T^{0}_{\,\,\,0}\,d^{3}x. (68)

With this definition and assuming boundary terms vanish as before, the ν=0\nu=0 continuity equation in integral form reads

ddtH=ρVtd3x.\frac{d}{dt}H=\int\!\!\rho\frac{\partial{V}}{\partial{t}}\,d^{3}x. (69)

Structurally, this is the same as Eq. (33) with a charge density replacing probability density. In a similar fashion, we directly use the field momentum components

Pi1cTi0d3x,P_{i}\equiv\int\frac{1}{c}T^{0}_{\,\,\,i}\,d^{3}x, (70)

and 𝐏=(P1,P2,P3)\mathbf{P}=(P_{1},P_{2},P_{3}). Noting that 𝐄V\mathbf{E}\equiv-\nabla V defines an electric field, the ν=1,2,3\nu=1,2,3 continuity equations yield

ddt𝐏=ρ𝐄d3x.\frac{d}{dt}\mathbf{P}=\int\!\!\rho\mathbf{E}\,d^{3}x. (71)

Stated in words, the rate of change of the field momentum is equal to the external Coulomb force acting on a charge distribution. This is a fitting counterpart to Ehrenfest’s theorem, which relates the rate of change of expected momentum to the external force acting on a probability distribution. Of course, Ehrenfest’s theorem for momentum is itself a quantum mechanical counterpart of Newton’s second law, which relates the rate of change of mechanical momentum to the external force acting on a mass distribution.

A similar result can be shown for angular momentum. We still have the six Noether currents associated with a spatial coordinate rotation

μνσxνTμσxσTμν,\mathcal{M}^{\mu\nu\sigma}\equiv x^{\nu}T^{\mu\sigma}-x^{\sigma}T^{\mu\nu}, (72)

from which the angular momentum density components are

i=12cεijk0jk.\ell_{i}=\frac{1}{2c}\varepsilon_{ijk}\mathcal{M}^{0jk}. (73)

Now defining the field angular momentum

Liid3x,L_{i}\equiv\int\ell_{i}\,d^{3}x, (74)

it is straightforward to show that

12εijkμμjkd3x=ddtLi,\frac{1}{2}\varepsilon_{ijk}\int\partial_{\mu}\mathcal{M}^{\mu jk}\,d^{3}x=\frac{d}{dt}L_{i}, (75)

assuming boundary terms vanish so that only the μ=0\mu=0 term survives integration. Compare this to an explicit calculation of the divergence of the four-current using the product rule, which after simplifying with Eq. (63) yields

12εijkμμjk=εijkxj(kV)ρ,\frac{1}{2}\varepsilon_{ijk}\partial_{\mu}\mathcal{M}^{\mu jk}=\varepsilon_{ijk}x^{j}(-\partial^{k}V)\rho, (76)

where the right-hand side is a torque per unit charge multiplied by the charge density ρ\rho. The right-hand side is a torque per unit charge multiplied by the charge density ρ\rho. Integrating and comparing with Eq. (76) we have a relativistic Ehrenfest relation for angular momentum,

ddt𝐋=(ρ𝐫×𝐄)d3x.\frac{d}{dt}\mathbf{L}=\int(\rho\mathbf{r}\times\mathbf{E})\,d^{3}x. (77)

Here ρ𝐫=ρ(x1,x2,x3)\rho\mathbf{r}=\rho(x^{1},x^{2},x^{3}) represents a dipole moment vector, and the electric field is defined as before. Since the integrand is weighted by a charge density, we say that the change in field angular momentum is equal to the external torque acting on a charge distribution.

V Conclusion

This paper examined the effect of external scalar potentials on Lagrangian systems. When the potential is coordinate-dependent, such systems are not symmetric under a coordinate transformation and hence conservation of energy and momentum do not hold. By accounting for the response of the potential to spacetime translation, however, we find a continuity equation wherein the derivatives of the potential act as a source of field energy and field momentum. In integral form, these continuity equations express familiar Ehrenfest relations directly from a Schrödinger Lagrangian. This approach makes no appeal to the probabilistic interpretation of the wavefunction, nor does it require explicit reference to the Schrödinger equation. We thus postulate that Eq. (23) is a general way to derive Ehrenfest-like relations for a Lagrangian acted on by an external, coordinate-dependent scalar potential.

We then demonstrated one example of Ehrenfest-like relations for a complex Klein-Gordon Lagrangian coupled to an electric potential. These results for relativistic spin 0 particles have an electric charge density ρ\rho taking the place of the probability density |Ψ|2|\Psi|^{2} from the Schrödinger Lagrangian. A natural progression of this work would be to replicate these results for a Dirac Lagrangian describing spin 1/2 particles.

It should be noted that the derivation leading to Eq. (23) is not the only way to derive the Ehrenfest relations shown this paper, although other methods are far less elegant. One could instead explicitly calculate μTνμ\partial_{\mu}T^{\mu}_{\,\,\,\nu} using Eqs. (2427) for the Schrödinger tensor and Eqs. (6467) for the Klein-Gordon tensor. This process is much more involved and requires extensive use of the field equations in order to arrive at the desired continuity equations. However, the correspondence of both methods lends support to the notion of treating a Lagrangian as an explicit function of an external scalar potential when one is applied.

*

Appendix A Noether Currents From Coordinate Transformations

We discuss more generally the standard Noether currents spacetime translations and spatial rotations for a Lagrangian of two fields φ\varphi and φ¯\overline{\varphi}. Generalizing to a Lagrangian with any number of fields is straightforward.

A.1 Spacetime Translation

The infinitesimal spacetime translation xμxμ+aμx^{\mu}\to x^{\mu}+a^{\mu} has the linearized field response

φ(xμ+aμ)φ(xμ)+aμμφ,\varphi(x^{\mu}+a^{\mu})\approx\varphi(x^{\mu})+a^{\mu}\partial_{\mu}\varphi, (78)

and the complex conjugate. After calculating the response of the Lagrangian (I) as a scalar function of the coordinates and (II) as a function of the fields and their derivatives (and simplifying with field equations), we see that the associated Noether current is

Jμ=aν(φ,μφ,ν+φ¯,μφ¯,νδνμ).J^{\mu}=a^{\nu}\left(\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\varphi_{,\nu}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\varphi}_{,\mu}}}\overline{\varphi}_{,\nu}-\delta^{\mu}_{\,\,\nu}\mathcal{L}\right). (79)

In taking the divergence of this current, μ\partial_{\mu} only acts on the expression inside the parentheses. This allows us to write the term inside as four Noether currents expressed as a mixed second-rank stress tensor,

Tνμφ,μφ,ν+φ¯,μφ¯,νδνμ.T^{\mu}_{\,\,\,\nu}\equiv\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\varphi_{,\nu}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\varphi}_{,\mu}}}\overline{\varphi}_{,\nu}-\delta^{\mu}_{\,\,\nu}\mathcal{L}. (80)

We say that the four currents are associated with the energy and linear momentum densities stored in the fields. To see this explicitly, first note that

T   00=φ˙φ˙+φ¯˙φ¯˙,T^{0}_{\,\,\,0}=\frac{\partial{\mathcal{L}}}{\partial{\dot{\varphi}}}\dot{\varphi}+\frac{\partial{\mathcal{L}}}{\partial{\dot{\overline{\varphi}}}}\dot{\overline{\varphi}}-\mathcal{L}, (81)

and we define

π(xμ)φ˙π¯(xμ)φ¯˙\pi(x^{\mu})\equiv\frac{\partial{\mathcal{L}}}{\partial{\dot{\varphi}}}\quad\overline{\pi}(x^{\mu})\equiv\frac{\partial{\mathcal{L}}}{\partial{\dot{\overline{\varphi}}}} (82)

to be the canonical momentum densities conjugate to the fields φ\varphi and φ¯\overline{\varphi} respectively.Peskin:1995ev We then see that this component of the stress tensor defines the Hamiltonian density \mathcal{H} of the system,

π(xμ)φ˙+π¯(xμ)φ¯˙=T   00.-\mathcal{H}\equiv\pi(x^{\mu})\dot{\varphi}+\overline{\pi}(x^{\mu})\dot{\overline{\varphi}}-\mathcal{L}=T^{0}_{\,\,\,0}. (83)

Thus the Hamiltonian, or total energy of the system, is found by integrating the above expression over all three-space,

Hd3x=T   00d3x.H\equiv\int\!\!\mathcal{H}\,d^{3}x=-\int\!\!T^{0}_{\,\,\,0}\,d^{3}x. (84)

Similarly, the spatial components give the physical linear momentum stored in the field,

𝒫i[π(xμ)iφ+π¯(xμ)iφ¯]=1cTi0,\mathcal{P}_{i}\equiv[\pi(x^{\mu})\partial_{i}\varphi+\overline{\pi}(x^{\mu})\partial_{i}\overline{\varphi}]=\frac{1}{c}T^{0}_{\,\,\,i}, (85)

and

Pi𝒫id3x=1cTi0d3x.P_{i}\equiv\int\!\!\mathcal{P}_{i}\,d^{3}x=\int\!\!\frac{1}{c}T^{0}_{\,\,\,i}\,d^{3}x. (86)

A.2 Spatial Rotation

A rotation of spatial coordinates in classical mechanics is associated with the conservation of angular momentum. A rotation of spatial coordinates can be written in terms of a 4×44\times 4 rotation matrix AA which transforms our position vector with xμAxμx^{\mu}\to Ax^{\mu} but leaves the time component unchanged. In the case of an infinitesimal transformation, the rotation matrix can be decomposed into

AI+R,A\approx I+R, (87)

where II is the identity and RR is an antisymmetric matrix with an infinitesimal parameter (omitted here for simplicity). We can represent this rotation as a mixed second-rank tensor AνμA^{\mu}_{\,\,\,\nu} applied to our position vector, and using the decomposition (87) we have

xμxμ+Rνμxν.x^{\mu}\to x^{\mu}+R^{\mu}_{\,\,\,\nu}x^{\nu}. (88)

The field response to this transformation is approximated by Taylor expansion about xμx^{\mu}:

φ(xμ+Rνμxν)φ(xμ)+(μφ)Rνμxν,\varphi(x^{\mu}+R^{\mu}_{\,\,\,\nu}x^{\nu})\approx\varphi(x^{\mu})+(\partial_{\mu}\varphi)R^{\mu}_{\,\,\,\nu}x^{\nu}, (89)

and complex conjugate. The Lagrangian, as a scalar function of coordinates, responds with

(xμ+Rνμxν)(xμ)+(μ)Rνμxν.\mathcal{L}(x^{\mu}+R^{\mu}_{\,\,\,\nu}x^{\nu})\approx\mathcal{L}(x^{\mu})+(\partial_{\mu}\mathcal{L})R^{\mu}_{\,\,\,\nu}x^{\nu}. (90)

To write the last term as an overall divergence, note that

μ(Rνμxν)\displaystyle\partial_{\mu}(\mathcal{L}R^{\mu}_{\,\,\,\nu}x^{\nu}) =(μ)Rνμxν+Rνμ(μxν)\displaystyle=(\partial_{\mu}\mathcal{L})R^{\mu}_{\,\,\,\nu}x^{\nu}+\mathcal{L}R^{\mu}_{\,\,\,\nu}(\partial_{\mu}x^{\nu})
=(μ)Rνμxν+Rνμδμν\displaystyle=(\partial_{\mu}\mathcal{L})R^{\mu}_{\,\,\,\nu}x^{\nu}+\mathcal{L}R^{\mu}_{\,\,\,\nu}\delta^{\nu}_{\,\,\,\mu}
=(μ)Rνμxν+Rμμ\displaystyle=(\partial_{\mu}\mathcal{L})R^{\mu}_{\,\,\,\nu}x^{\nu}+\mathcal{L}R^{\mu}_{\,\,\,\mu}
=(μ)Rνμxν,\displaystyle=(\partial_{\mu}\mathcal{L})R^{\mu}_{\,\,\,\nu}x^{\nu}, (91)

where the last step holds since RνμR^{\mu}_{\,\,\,\nu} is antisymmetric and hence the diagonal components vanish. Thus we can write the change in the Lagrangian given in Eq. (90) as an overall divergence,

(xμ+Rνμxν)(xμ)=μ(Rνμxν).\mathcal{L}(x^{\mu}+R^{\mu}_{\,\,\,\nu}x^{\nu})-\mathcal{L}(x^{\mu})=\partial_{\mu}(\mathcal{L}R^{\mu}_{\,\,\,\nu}x^{\nu}). (92)

A similar calculation can be made for Eq. (89). The Noether current for rotations is then

Jμ=Rνσxν(φ,μφ,σ+φ¯,μφ¯,σδσμ).J^{\mu}=R^{\sigma}_{\,\,\,\nu}x^{\nu}\left(\frac{\partial{\mathcal{L}}}{\partial{\varphi_{,\mu}}}\varphi_{,\sigma}+\frac{\partial{\mathcal{L}}}{\partial{\overline{\varphi}_{,\mu}}}\overline{\varphi}_{,\sigma}-\delta^{\mu}_{\,\,\,\sigma}\mathcal{L}\right). (93)

The term in parentheses is just the energy-momentum tensor TσμT^{\mu}_{\,\,\,\sigma} found in (80). After moving the σ\sigma index, this reduces to

Jμ=RσνxνTμσ.J^{\mu}=R_{\sigma\nu}x^{\nu}T^{\mu\sigma}. (94)

We want to exploit the fact that RσνR_{\sigma\nu} is an antisymmetric tensor being multiplied into a third-rank contravariant tensor. First note that

xνTμσ=12(xνTμσ+xσTμν)+12(xνTμσxσTμν),x^{\nu}T^{\mu\sigma}=\frac{1}{2}(x^{\nu}T^{\mu\sigma}+x^{\sigma}T^{\mu\nu})+\frac{1}{2}(x^{\nu}T^{\mu\sigma}-x^{\sigma}T^{\mu\nu}), (95)

where the first (second) term is symmetric (antisymmetric) under νσ\nu\leftrightarrow\sigma interchange. When we multiply the expression above by RσνR_{\sigma\nu} only the antisymmetric components survive the summation, so we have

Jμ=12Rσν(xνTμσxσTμν).J^{\mu}=\frac{1}{2}R_{\sigma\nu}(x^{\nu}T^{\mu\sigma}-x^{\sigma}T^{\mu\nu}). (96)

Discarding the constant term out front as before, we are left with six Noether currents

μνσxνTμσxσTμν.\mathcal{M}^{\mu\nu\sigma}\equiv x^{\nu}T^{\mu\sigma}-x^{\sigma}T^{\mu\nu}. (97)

To calculate the divergence, we use the product rule and find

,μμνσ\displaystyle\mathcal{M}_{,\mu}^{\,\,\,\mu\nu\sigma} =xνμTμσxσμTμν+TνσTσν\displaystyle=x^{\nu}\partial_{\mu}T^{\mu\sigma}-x^{\sigma}\partial_{\mu}T^{\mu\nu}+T^{\nu\sigma}-T^{\sigma\nu}
=xν(VσV)xσ(VνV)+TνσTσν,\displaystyle=x^{\nu}\!\!\left(-\frac{\partial{\mathcal{L}}}{\partial{V}}\partial^{\sigma}V\right)-x^{\sigma}\!\!\left(-\frac{\partial{\mathcal{L}}}{\partial{V}}\partial^{\nu}V\right)+T^{\nu\sigma}-T^{\sigma\nu}, (98)

where we have in general assumed that the Lagrangian depends on an external potential so that Eq. (23) holds. We have also not assumed a symmetric energy-momentum tensor, since that was the case for both systems studied in this paper. However, note that the linear momentum density is stored in the T0kT^{0k} components, so the μjk\mathcal{M}^{\mu jk} components will be relevant for studying angular momentum. The divergence of these terms is then

,μμjk=xj(VkV)xk(VjV),\mathcal{M}_{,\mu}^{\,\,\,\mu jk}=x^{j}\!\!\left(-\frac{\partial{\mathcal{L}}}{\partial{V}}\partial^{k}V\right)-x^{k}\!\!\left(-\frac{\partial{\mathcal{L}}}{\partial{V}}\partial^{j}V\right), (99)

where now TjkTkj=0T^{jk}-T^{kj}=0 since the spatial components of the energy-momentum tensor are in general symmetric. This result is used to write Eqs. (37) and (76) using the Levi-Civita symbol. Thus conservation of angular momentum only holds when a spherically symmetric (if any) potential is applied. In general, however, the torque density in Eq. (99) is a source of angular momentum from which Ehrenfest-like relations can be derived.

Acknowledgements.
I would first like to thank Joel Franklin, my primary thesis advisor for this project at Reed College. Joel discovered a way to derive Ehrenfest’s theorem for momentum by explicitly calculating the divergence of the energy-momentum tensor associated with the Schrödinger Lagrangian, which inspired me to formulate the general expression in Eq. (23). Joel also came up with the choice of the relativistic Lagrangian Eq. (51) among many other key suggestions in this paper. I would also like to thank Alexander Moll, my secondary thesis advisor, for providing extensive comments that led to this work.

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