Klein-Gordon Equation: Derivation and Lagrangian Density

1. From Particles to Fields: The Need for QFT

In your study of quantum mechanics, you became familiar with the Schrödinger equation, which describes the evolution of a particle's wavefunction, $\psi (\mathbf{x},t)$. While incredibly successful, this is an inherently non-relativistic theory. A natural first step towards a relativistic quantum theory is to find a relativistic counterpart to the Schrödinger equation.

However, as you may have encountered, this approach is plagued by fundamental problems like negative probability densities or negative energy states. More profoundly, any relativistic theory must respect causality. Information cannot travel faster than the speed of light, $c$. This principle of locality is difficult to reconcile with a single-particle quantum theory but is naturally incorporated in a field theory.

To get a clear intuition for why fields are necessary to enforce locality, please watch the beginning of the following video.

As the video explains, a field, $\varphi (x,t)$, which has a value at every point in spacetime, provides a local description of physical reality. Excitations or disturbances in this field propagate outwards at a finite speed, respecting causality. In QFT, we identify these very excitations with particles.

2. The Klein-Gordon Equation from Relativistic Principles

Let's derive the simplest possible relativistic wave equation. We'll start with Einstein's energy-momentum relation for a particle of mass $m$:

$E^2={\mathbf{p}}^2{c}^2+m^2{c}^4$

In non-relativistic quantum mechanics, you used the correspondence principle to turn classical observables into quantum operators acting on a wavefunction:

• Energy: $E\to i\hslash \frac{\partial }{\partial t}$
• Momentum: $\mathbf{p}\to -i\hslash \nabla$

Let's postulate that these same operator substitutions hold in the relativistic case, but now they act on a scalar field, denoted $\varphi (x,t)$, which we assume for now is a simple complex-valued function of spacetime coordinates. Applying these operators to our field $\varphi$ in the energy-momentum relation gives:

${\left(i\hslash \frac{\partial }{\partial t}\right)}^2\varphi =(-i\hslash \nabla {)}^2{c}^2\varphi +m^2{c}^4\varphi$

$-{\hslash }^2\frac{{\partial }^2\varphi }{\partial t^2}=-{\hslash }^2{c}^2{\nabla }^2\varphi +m^2{c}^4\varphi$

Rearranging the terms, we get:

${\hslash }^2\frac{{\partial }^2\varphi }{\partial t^2}-{\hslash }^2{c}^2{\nabla }^2\varphi +m^2{c}^4\varphi =0$

Dividing by ${\hslash }^2{c}^2$ yields:

$\frac{1}{c^2}\frac{{\partial }^2\varphi }{\partial t^2}-{\nabla }^2\varphi +\frac{m^2{c}^2}{{\hslash }^2}\varphi =0$

This is the Klein-Gordon equation. It is the simplest wave equation consistent with the principles of special relativity.

Covariant Notation

For the rest of this course, it will be immensely helpful to use a more compact, manifestly Lorentz-covariant notation. We will adopt the natural units system where $\hslash =c=1$. This simplifies many equations and is standard practice in QFT. In these units, the Klein-Gordon equation becomes:

$\frac{{\partial }^2\varphi }{\partial t^2}-{\nabla }^2\varphi +m^2\varphi =0$

We can make this even more compact. We define the four-vector for position as $x^{\mu }=(t,\mathbf{x})$ and the four-gradient as ${\partial }_{\mu }=\frac{\partial }{\partial x^{\mu }}=(\frac{\partial }{\partial t},\nabla )$. The invariant interval in spacetime is described by the Minkowski metric, $g_{\mu \nu }$ (or ${\eta }_{\mu \nu }$), which we'll take with the $(+,-,-,-)$ signature. The Lorentz-invariant "dot product" of two four-vectors $A^{\mu }$ and $B^{\mu }$ is $A\cdot B=g_{\mu \nu }{A}^{\mu }{B}^{\nu }=A^0{B}^0-\mathbf{A}\cdot \mathbf{B}$.

Using this, we can define the D'Alembertian operator, denoted $\square$ (or ${\partial }^2$):

$\square \equiv {\partial }_{\mu }{\partial }^{\mu }=g_{\mu \nu }{\partial }^{\mu }{\partial }^{\nu }=\frac{{\partial }^2}{\partial t^2}-{\nabla }^2$

With this operator, the Klein-Gordon equation takes its most elegant form:

$(\square +m^2)\varphi (x)=0$

This form makes it clear that the equation treats space and time on an equal footing, a requirement of any relativistic theory.

3. The Lagrangian Formalism for Fields

While deriving the equation of motion from the energy-momentum relation is intuitive, a more powerful and fundamental approach in QFT is to start with a Lagrangian density, $\mathcal{L}$. As you know from classical mechanics, the Lagrangian formalism, based on the principle of least action, provides a systematic way to derive equations of motion and, crucially, to identify conserved quantities associated with symmetries (which we will cover in the next lesson).

In field theory, the Lagrangian $L$ is the spatial integral of the Lagrangian density $\mathcal{L}$:
$L=\int d^{3}x\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )$
The action $S$ is the time integral of the Lagrangian, which becomes an integral over all of spacetime:
$S=\int dtL=\int d^{4}x\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )$

For the action $S$ to be Lorentz-invariant (a scalar), the Lagrangian density $\mathcal{L}$ must also be a Lorentz scalar.

The Euler-Lagrange Equation for Fields

The equation of motion for the field $\varphi$ is found by applying the principle of least action, which states that the field configuration that occurs in nature is the one that extremizes the action, i.e., $\delta S=0$. This requirement leads to the Euler-Lagrange equation for fields:

$\frac{\partial \mathcal{L}}{\partial \varphi }-{\partial }_{\mu }\left(\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}\right)=0$

Given your background, you've likely seen the derivation for a system of particles. The extension to fields is a key step. For a very clear and rigorous walk-through of this derivation, I highly recommend the following video.

4. The Klein-Gordon Lagrangian

Now we can address the central part of our learning outcome: finding the Lagrangian density $\mathcal{L}$ that produces the Klein-Gordon equation when plugged into the Euler-Lagrange equation.

Our task is to construct the simplest, non-trivial Lagrangian density $\mathcal{L}$ that is a Lorentz scalar built from the scalar field $\varphi$ and its derivatives ${\partial }_{\mu }\varphi$. The simplest Lorentz scalars we can form are:

1. ${\varphi }^2$
2. $({\partial }_{\mu }\varphi )({\partial }^{\mu }\varphi )$

Any other terms, like ${\varphi }^3$ or $({\partial }_{\mu }\varphi {\partial }^{\mu }\varphi {)}^2$, would be more complicated. A general Lagrangian density composed of the simplest terms would be a linear combination:
$\mathcal{L}=A({\partial }_{\mu }\varphi )({\partial }^{\mu }\varphi )+B{\varphi }^2$
where A and B are constants. Let's find what A and B must be.

To use the Euler-Lagrange equation, we need two derivatives of $\mathcal{L}$:

1. Derivative with respect to $\varphi$:
   $\frac{\partial \mathcal{L}}{\partial \varphi }=2B\varphi$
2. Derivative with respect to ${\partial }_{\nu }\varphi$:
   $\frac{\partial \mathcal{L}}{\partial ({\partial }_{\nu }\varphi )}=\frac{\partial }{\partial ({\partial }_{\nu }\varphi )}[A(g^{\mu \lambda }{\partial }_{\mu }\varphi {\partial }_{\lambda }\varphi )]=A{g}^{\mu \lambda }({\delta }_{\mu }^{\nu }{\partial }_{\lambda }\varphi +{\partial }_{\mu }\varphi {\delta }_{\lambda }^{\nu })=2A{g}^{\nu \lambda }{\partial }_{\lambda }\varphi =2A{\partial }^{\nu }\varphi$

Now, we plug these into the Euler-Lagrange equation:
$2B\varphi -{\partial }_{\nu }(2A{\partial }^{\nu }\varphi )=0$
$2B\varphi -2A{\partial }_{\nu }{\partial }^{\nu }\varphi =0$
$-A\square \varphi +B\varphi =0$

Comparing this to the Klein-Gordon equation $(\square +m^2)\varphi =0$, or $\square \varphi =-m^2\varphi$, we can identify the coefficients:
$-A=1⟹A=-1$
$B=-m^2$
Wait, something is not quite right. The convention is to have a positive kinetic term. Let's redefine our starting point: $\mathcal{L}=A({\partial }_{\mu }\varphi )({\partial }^{\mu }\varphi )-B{\varphi }^2$.
Then $\frac{\partial \mathcal{L}}{\partial \varphi }=-2B\varphi$ and the derivative term is the same, $2A{\partial }^{\nu }\varphi$. The Euler-Lagrange equation becomes:
$-2B\varphi -{\partial }_{\nu }(2A{\partial }^{\nu }\varphi )=0⟹-2A\square \varphi -2B\varphi =0⟹A\square \varphi +B\varphi =0$
Comparing to $\square \varphi +m^2\varphi =0$, we get $A=1$ and $B=m^2$. The conventional choice includes factors of 1/2, for reasons that will become clear when we discuss the Hamiltonian. So, we propose:

$\mathcal{L}=\frac{1}{2}({\partial }_{\mu }\varphi )({\partial }^{\mu }\varphi )-\frac{1}{2}{m}^2{\varphi }^2$

This is the Lagrangian density for the free, real scalar field.

Let's quickly verify this.
$\frac{\partial \mathcal{L}}{\partial \varphi }=-m^2\varphi$
$\frac{\partial \mathcal{L}}{\partial ({\partial }_{\nu }\varphi )}={\partial }^{\nu }\varphi$
Plugging into the Euler-Lagrange equation:
$-m^2\varphi -{\partial }_{\nu }({\partial }^{\nu }\varphi )=0⟹-m^2\varphi -\square \varphi =0⟹(\square +m^2)\varphi =0$
It works perfectly.

For a final consolidation of these ideas, you can refer to the following textbook section.

Physical Interpretation of the Lagrangian Terms

It's useful to see how the terms in $\mathcal{L}$ relate to concepts from classical mechanics. Let's expand the kinetic term:
$\mathcal{L}=\frac{1}{2}({\partial }_0\varphi {)}^2-\frac{1}{2}(\nabla \varphi )\cdot (\nabla \varphi )-\frac{1}{2}{m}^2{\varphi }^2$

• $\frac{1}{2}({\partial }_0\varphi {)}^2=\frac{1}{2}{\dot{\varphi }}^2$: This is analogous to the kinetic energy density ($\sim \frac{1}{2}m{v}^2$). It depends on the rate of change of the field in time.
• $-\frac{1}{2}(\nabla \varphi {)}^2$: This is often called the gradient energy density. It penalizes field configurations that vary sharply in space. You can think of it as a kind of potential energy stored in the "stiffness" of the field.
• $-\frac{1}{2}{m}^2{\varphi }^2$: This is a potential energy density term, analogous to the potential for a harmonic oscillator ($\sim \frac{1}{2}k{x}^2$). The parameter $m$ determines the "restoring force" pulling the field value back to $\varphi =0$. As we will see, this parameter m is precisely the mass of the particles that arise when we quantize the field.

Conclusion

In this first lesson, we have taken a significant step into the world of Quantum Field Theory.

Key Takeaways:

• A consistent relativistic quantum theory requires fields to maintain locality and account for particle creation and annihilation.
• The simplest relativistic equation for a scalar field is the Klein-Gordon equation: $(\square +m^2)\varphi =0$.
• The dynamics of fields are most fundamentally described by a Lagrangian density $\mathcal{L}$ through the principle of least action.
• The Lagrangian density for a free scalar field of mass $m$ is $\mathcal{L}=\frac{1}{2}({\partial }_{\mu }\varphi )({\partial }^{\mu }\varphi )-\frac{1}{2}{m}^2{\varphi }^2$. This Lorentz-scalar object elegantly contains all the information about the field's free propagation.

You have now seen how to derive the fundamental equation of motion for a free scalar field from two different perspectives. This Lagrangian density will be our starting point for the next few lessons.

Preview of the Next Lesson:

In the next lesson, we will explore one of the most beautiful aspects of the Lagrangian formalism: its connection to symmetries and conservation laws. We will apply Noether's theorem to the Klein-Gordon Lagrangian to derive the conserved energy-momentum tensor and a conserved charge current, which are essential physical quantities for any field theory.