Noether's Theorem and Conserved Currents for the Klein-Gordon Field

1. Noether's Theorem for Fields

In classical mechanics, you know that if a system's Lagrangian is invariant under a continuous transformation (like spatial translation or rotation), a corresponding physical quantity (like momentum or angular momentum) is conserved. Noether's theorem is the beautiful mathematical statement of this principle. In field theory, it takes on an even more powerful role, giving rise to locally conserved currents.

The theorem states: For every continuous symmetry of the action, there exists a corresponding conserved current $j^{\mu }$ such that ${\partial }_{\mu }{j}^{\mu }=0$.

A conserved current implies a conserved charge, $Q=\int d^{3}x{j}^0$, whose total amount in space is constant over time, since $dQ/dt=0$.

For a detailed, step-by-step proof of the theorem in the context of field theory, the following video is an excellent resource. It demonstrates how a symmetry, defined as a transformation that changes the Lagrangian by at most a total derivative, leads directly to a conserved current.

As derived in the video and summarized succinctly in David Tong's QFT notes, if a transformation $\delta {\varphi }_a$ is a symmetry, such that the Lagrangian changes by a total derivative, $\delta \mathcal{L}={\partial }_{\mu }{F}^{\mu }$, the resulting conserved current is given by:

$j^{\mu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }_a)}\delta {\varphi }_a-F^{\mu }$

This formula is our primary tool for the rest of this lesson.

2. Spacetime Translations and the Energy-Momentum Tensor

Our first application of Noether's theorem will be to the symmetry of spacetime itself. Physical laws should not depend on where or when you perform an experiment. This corresponds to invariance under spacetime translations.

Consider an infinitesimal translation: $x^{\mu }\to x^{\prime \mu }=x^{\mu }-{ϵ}^{\mu }$, where ${ϵ}^{\mu }$ is a small, constant four-vector. We are performing an active transformation, which means we are shifting the field configuration itself. A field value that was at $x^{\mu }-{ϵ}^{\mu }$ is now at $x^{\mu }$. Equivalently, the field at $x^{\mu }$ changes as:
$\varphi (x)\to {\varphi }^{\prime }(x)=\varphi (x+ϵ)\approx \varphi (x)+{ϵ}^{\nu }{\partial }_{\nu }\varphi (x)$
The infinitesimal change in the field is $\delta \varphi ={ϵ}^{\nu }{\partial }_{\nu }\varphi (x)$.

Since the Lagrangian density $\mathcal{L}$ depends only on the fields and their derivatives at a point, it transforms in the same way:
$\mathcal{L}(x)\to \mathcal{L}(x+ϵ)\approx \mathcal{L}(x)+{ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}(x)$
The change in the Lagrangian is $\delta \mathcal{L}={ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}={\partial }_{\mu }({ϵ}^{\nu }{\delta }_{\nu }^{\mu }\mathcal{L})$. This is a total derivative, so we have a symmetry. Comparing with the general form $\delta \mathcal{L}={\partial }_{\mu }{F}^{\mu }$, we identify $F^{\mu }={ϵ}^{\nu }{\delta }_{\nu }^{\mu }\mathcal{L}$.

We have four independent symmetries, one for each component of ${ϵ}^{\nu }$ ($\nu =0,1,2,3$). This will give us four conserved currents. Let's construct the current associated with the $\nu$-th transformation. Plugging $\delta \varphi ={ϵ}^{\nu }{\partial }_{\nu }\varphi$ and $F^{\mu }={ϵ}^{\nu }{\delta }_{\nu }^{\mu }\mathcal{L}$ into the Noether current formula:

$j^{\mu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}({ϵ}^{\nu }{\partial }_{\nu }\varphi )-{ϵ}^{\nu }{\delta }_{\nu }^{\mu }\mathcal{L}$

Since ${ϵ}^{\nu }$ is an arbitrary constant vector, we can factor it out to define a tensor quantity, the energy-momentum tensor $T^{\mu \nu }$:

$T^{\mu \nu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}{\partial }^{\nu }\varphi -g^{\mu \nu }\mathcal{L}$
Note: We have raised the index $\nu$ on ${\partial }_{\nu }\varphi$ and ${\delta }_{\nu }^{\mu }$ to factor out ${ϵ}_{\nu }$, using the metric $g^{\mu \nu }$. The resulting conservation law is ${\partial }_{\mu }{T}^{\mu \nu }=0$.

Energy-Momentum Tensor for the Klein-Gordon Field

Now, let's compute this for our real scalar field Lagrangian: $\mathcal{L}=\frac{1}{2}({\partial }_{\alpha }\varphi {)}^2-\frac{1}{2}{m}^2{\varphi }^2$.
We already know $\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}={\partial }^{\mu }\varphi$. Substituting this into our expression for $T^{\mu \nu }$:

$T^{\mu \nu }=({\partial }^{\mu }\varphi )({\partial }^{\nu }\varphi )-g^{\mu \nu }\left(\frac{1}{2}({\partial }_{\alpha }\varphi )({\partial }^{\alpha }\varphi )-\frac{1}{2}{m}^2{\varphi }^2\right)$

This tensor contains all the information about the energy and momentum of the field. Let's examine its components:

• Conserved Energy ($\nu =0$):
  The conserved charge associated with time translation is the total energy, or the Hamiltonian $H$.
  $H=\int d^{3}x{T}^{00}$
  The component $T^{00}$ is the Hamiltonian density, $\mathcal{H}$. Let's calculate it:
  $T^{00}=({\partial }^0\varphi )({\partial }^0\varphi )-g^{00}\mathcal{L}=(\dot{\varphi }{)}^2-\left(\frac{1}{2}({\dot{\varphi }}^2-(\nabla \varphi {)}^2)-\frac{1}{2}{m}^2{\varphi }^2\right)$
  $\mathcal{H}=T^{00}=\frac{1}{2}{\dot{\varphi }}^2+\frac{1}{2}(\nabla \varphi {)}^2+\frac{1}{2}{m}^2{\varphi }^2$
  This result should be reassuring. It's the sum of the "kinetic", "gradient", and "potential" energy densities we intuitively identified in the last lesson.

• Conserved Momentum ($\nu =i$):
  The conserved charge associated with spatial translation in the $i$-direction is the total momentum $P^i$.
  $P^i=\int d^{3}x{T}^{0i}$
  The component $T^{0i}$ is the momentum density.
  $T^{0i}=({\partial }^0\varphi )({\partial }^i\varphi )-g^{0i}\mathcal{L}=\dot{\varphi }({\partial }^i\varphi )=-\dot{\varphi }({\partial }_i\varphi )$
  The total momentum of the field is $\mathbf{P}=-\int d^{3}x\dot{\varphi }(\nabla \varphi )$.

The same video you started watching covers this derivation clearly.

3. Internal Symmetry and Conserved Charge Current

Spacetime symmetries are not the only kind. Fields can also have internal symmetries, which are transformations that act on the field values themselves but not on the spacetime coordinates.

To discuss a conserved charge, like electric charge, we need a field that can carry it. A real scalar field isn't sufficient. We must introduce a complex scalar field, $\varphi (x)$, which can be thought of as a combination of two real fields. The Lagrangian density for a free complex scalar field is:
$\mathcal{L}=({\partial }_{\mu }{\varphi }^{\ast })({\partial }^{\mu }\varphi )-m^2{\varphi }^{\ast }\varphi$
Note that this Lagrangian is constructed to be real (${\mathcal{L}}^{\ast }=\mathcal{L}$). Applying the Euler-Lagrange equations to $\varphi$ and ${\varphi }^{\ast }$ as if they are independent fields yields the Klein-Gordon equation for both: $(\square +m^2)\varphi =0$ and $(\square +m^2){\varphi }^{\ast }=0$.

This Lagrangian has a new, continuous symmetry. It is invariant under a global phase rotation (a U(1) transformation):
$\varphi (x)\to e^{i\alpha }\varphi (x)\text{and}{\varphi }^{\ast }(x)\to e^{-i\alpha }{\varphi }^{\ast }(x)$
where $\alpha$ is a real, constant parameter. The Lagrangian is unchanged because the $e^{i\alpha }$ and $e^{-i\alpha }$ factors cancel in the ${\varphi }^{\ast }\varphi$ terms.

To find the conserved current, we consider an infinitesimal transformation, where $\alpha \ll 1$:

• $\delta \varphi =i\alpha \varphi$
• $\delta {\varphi }^{\ast }=-i\alpha {\varphi }^{\ast }$

The Lagrangian is perfectly invariant, so $\delta \mathcal{L}=0$, which means $F^{\mu }=0$ in our Noether formula. The current is given by summing over the fields $\varphi$ and ${\varphi }^{\ast }$:
$j^{\mu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}\delta \varphi +\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }^{\ast })}\delta {\varphi }^{\ast }$
The required derivatives of the Lagrangian are:

• $\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}={\partial }^{\mu }{\varphi }^{\ast }$
• $\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{\varphi }^{\ast })}={\partial }^{\mu }\varphi$

Substituting these in:
$j^{\mu }=({\partial }^{\mu }{\varphi }^{\ast })(i\alpha \varphi )+({\partial }^{\mu }\varphi )(-i\alpha {\varphi }^{\ast })=i\alpha (\varphi {\partial }^{\mu }{\varphi }^{\ast }-{\varphi }^{\ast }{\partial }^{\mu }\varphi )$
Factoring out the arbitrary parameter $\alpha$, we identify the conserved current:
$j^{\mu }=i(\varphi {\partial }^{\mu }{\varphi }^{\ast }-{\varphi }^{\ast }{\partial }^{\mu }\varphi )$
This is the conserved charge current. The conserved charge is $Q=\int d^{3}x{j}^0$. After quantization, this $Q$ will be interpreted as the total charge of the system (e.g., number of particles minus number of antiparticles).

For a concise, written derivation of both the energy-momentum tensor and this charge current, the following resource is excellent.

For those with a strong interest in the mathematical formalism, the paper "Pedagogical Systematic Derivation of Noether Point..." (LINK) offers an exhaustive method to find all point symmetries of a given Lagrangian, including Lorentz boosts and rotations which we haven't discussed. It is a much more in-depth treatment than required for this lesson, but you may find its systematic approach satisfying given your background.

Conclusion

Today, we've unlocked a deeper layer of the Lagrangian formalism, connecting abstract symmetries to tangible physical laws.

Key Takeaways:

• Noether's Theorem: A continuous symmetry of the Lagrangian implies the existence of a locally conserved current ${\partial }_{\mu }{j}^{\mu }=0$.
• Energy-Momentum Conservation: The invariance of the Klein-Gordon Lagrangian under spacetime translations leads to the conservation of the energy-momentum tensor, $T^{\mu \nu }=({\partial }^{\mu }\varphi )({\partial }^{\nu }\varphi )-g^{\mu \nu }\mathcal{L}$. Its components give the energy density ($T^{00}$) and momentum density ($T^{0i}$).
• Charge Conservation: The invariance of the complex scalar field Lagrangian under a global U(1) phase rotation leads to the conservation of a charge current, $j^{\mu }=i(\varphi {\partial }^{\mu }{\varphi }^{\ast }-{\varphi }^{\ast }{\partial }^{\mu }\varphi )$.

We have now established the classical field theory of the simplest relativistic field, including its dynamics and fundamental conservation laws. We are fully prepared to take the leap into the quantum world.

Preview of the Next Lesson:

In the next lesson, we will finally quantize the theory. We will treat the field $\varphi (x)$ and its conjugate momentum as operators and impose canonical commutation relations. This is the crucial step that introduces quantum mechanics into our field theory, turning fields into operators that create and destroy the particles that are the subject of QFT.