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The Lagrangian density for the real Klein-Gordon field, which describes a real scalar field $\phi$ with mass $m$ is given by

\begin{equation} \mathcal{L} = \frac{1}{2} \partial^\mu \phi \, \partial_\mu \phi - \frac{1}{2} m^2 \phi^2 \end{equation}

This is directly introduced by using the action is invariant under Lorentz transformation in ad hoc manner in few books and then they show it satisfies the Euler-Lagrange equation

\begin{equation} \frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \right) = 0 \end{equation}

But what is the importance of $\frac{1}{2}$ here and why $m^2$, why not $m$ ?

The Hamiltonian density of this field,

\begin{equation} \mathcal{H} = \frac{1}{2} \pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2 \end{equation}

which also satisfies Hamiltonian equations of motions $\dot{\phi} = \frac{\delta \mathcal{H}}{\delta \pi}$ and $\dot{\pi} = -\frac{\delta \mathcal{H}}{\delta \phi}$, if we don't take $\frac{1}{2}$.

Give a physical importance of $\frac{1}{2}$ and $m^2$ instead of $m$. Give the mathematical intution also.

The related questions are physics.stackexchange.com/q/314426/2451 , physics.stackexchange.com/q/732564/2451

AVOY JANA
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2 Answers2

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The answer is clear if focusing on the the stress energy tensor obtained either by the Noether theorem or the functional derivative with respect to the metric. It turns out to be the correct one only if one includes that factor $1/2$ in the lagrangian.

That is because, when quantizing, the integral of the normal order stress energy tensor operator $:\hat{T}_{00}:$ reproduces the expected total Hamiltonian in the Fock space. $$\int_{\Sigma} :\hat{T}_{00}(0, {\bf x}): d^3x = H = H_1 \oplus (H_1 \otimes I + I \otimes H_1) \oplus \cdots$$ where $H_1$ is the known Hamiltonian in the one-particle Hilbert space $H_1 = \sqrt{{\bf p}^2 + m^2}$.

(Differently, if considering a charged KG field, that factor $1/2$ has to be omitted because we have here two types of particles...)

Valter Moretti
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The particular normalization is just a mathematical convenience. Let's say you normalize the field differently so that it is $\frac{K}{2}(\partial\phi)^2$, then the conjugate momentum is $\pi = \frac{1}{K} \partial_t \phi$, the two-point correlation has an $1/K$ prefactor, and so on. Now you get loads of unnecessary $K$ and $1/K$ to keep track of, and will make a mistake sooner or later.

But sometimes you are not allowed to rescale. For instance, if $\phi$ represents an angle, rescaling is blasphemous. This is often the case in sigma model, Luttinger liquid, bosonization and related topics in field theory. And yes, I mess up my $K$ all the time.

Sean
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T.P. Ho
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