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According to Mankowski flat space dimensions We can write, $$L= \int \text{dt} \text d^d{x} \left[ \frac{1}{2} \dot\phi^2 - \frac{1}{2} \left(\frac{\partial \phi}{\partial r} \right)^2 -V(\phi)\right] \tag{1}$$ Where $V$ can be written as $$V = \frac{1}{8} \phi^2 (\phi-2)^2$$

But the author wrote in his article in the equation (1) including dimensions. $$V= \frac{1}{2} m^2 \phi^2+ \frac{\lambda_3}{3!}m^\frac{5-d}{2} \phi^3+ \frac{\lambda_4}{4!}m^{3-d} \phi^4$$

My question is how the dimensions incorporated with the potential?

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When expanded, $(1/8)\phi^2(\phi-2)^2$ contains quadratic, cubic, quartic (2nd, 3rd, 4th degree) terms in $\phi$. So it's a polynomial with these monomials. The final form of the potential places general coefficients in front of these terms.

The quadratic term is universally written as $(1/2)m^2\phi^2$ because it contributes the usual mass term $m^2\psi$ to the Klein-Gordon equation of motion. Note that the potential has units of $m^{d+1}$ where $d+1$ is the spacetime dimension in your conventions because when integrated over $length^{d+1}$ spacetime, we should get a dimensionless action.

It follows from the $m^{d+1}$ dimension of $m^2 \phi^2$ that $\phi$ has the dimension of $m^{d/2-1/2}$. In the cubic term, $\phi^3$ therefore has dimension $m^{3d/2-3/2}$ and we have to multiply it by a coefficient with units $m^{-d/2+5/2}$ to obtain another $m^{d+1}$ term. This $m^{-d/2+5/2}$ coefficient is written as a product of the same power of $m$, the mass from the quadratic term, and a $\lambda_3$ which is may be kept dimensionless.

In the same way, the quartic term contains $\phi^4$ whose dimension is $m^{2d-2}$ but we need the dimension of the whole $V$ to be $m^{d+1}$ so we need to add $m^{d-3}$, dimensionally speaking, which the formula does, and it adds a new dimensionless coefficient $\lambda_4$ to this term.

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  • $\begingroup$ How $\phi$ has the dimension of $m^{d/2-1/2}$? $\endgroup$ – Raisa Jul 1 '13 at 10:32
  • $\begingroup$ See my answer. The Langrangian density must have dimension $m^D = m^{d+1}.$ Derivatives have dimension $m^1$, so the dimension of $\phi$ follows from the fact that $(\partial \phi)^2 \sim m^{d+1}.$ $\endgroup$ – Vibert Jul 1 '13 at 12:01
  • $\begingroup$ @Vibert, I assure you that I wrote the same thing and at least minutes before you. ;-) $\endgroup$ – Luboš Motl Jul 7 '13 at 5:48
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You have probably already seen this in your QFT class, but anyway... Note that your reference works in $D \equiv d+1$-dimensional spacetime. I will use brackets $[\dotsm]$ to denote the mass dimension. In particular, the Lagrangian density will have dimension $[\mathcal{L}] = D.$

Now since $[(\partial_\mu \phi)^2] = D$, we must have $[\phi] = \tfrac{1}{2} (D-2).$ Now consider an interaction term of the form $~ y \phi^4.$Then $[y \cdot \phi^4] = D,$ so $[y] = D - 2(D-2) = 4-D.$ Similarly, if $[u \cdot \phi^3] = D$, then $[u] = \tfrac{1}{2}(6-D).$

But often we want to work with dimensionless coupling constants - they are more fundamental than dimensionful ones. So you multiply interaction terms by the 'right' power of $m$ to make the coupling constant $[\lambda_n] = 0.$ It is a miniscule exercise to find the right power of $m$ in terms of $d$, given the above calculations.

The potential in the form $V =\tfrac{1}{8} \phi^2(\phi^2-2)$ is fully dimensionless. They have scaled out the dimension of the field $\phi$ and $V$ itself is dimensionless there, too.

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  • $\begingroup$ Now since $[(\partial_\mu \phi)^2] = D$, we must have $[\phi] = \tfrac{1}{2} (D-2).$ can you please tell me that why $\phi= \frac{1}{2} (D-2)$, I'm actually QFT beginner. $\endgroup$ – Raisa Jul 1 '13 at 21:06
  • $\begingroup$ @Raisa : Any differentiable operator has mass dimension $1$ or length dimension $-1$. This because $i\partial_\mu = P_\mu$, where $P_\mu$ is momentum/energy. And momentum-energy has mass dimension $1$. Remember that in QFT, we take $\hbar = c = 1$, so speeds $v$ for instance are considered as $\frac{v}{c}$, so speeds have zero mass or length dimension. $\endgroup$ – Trimok Jul 2 '13 at 8:06
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This comes from the idea that the action $S=\int d^4x\,\mathcal{L}$ is a scalar. From planck's law it is easy to see that the energy $E=\hbar \omega$ has the dimension of inverse time. From de-broglie relation you can see that momentum has the inverse dimension of space. Hence $\mathcal{L}$ should have a dimension of energy ${[\mathcal{L}]=E^{4}}$. This will help us to identify the dimension of the field $[\phi]=E$. Hence the maximum possible exponent in field is $\phi^4$ and the coupling constant is dimensionless. All the other higher order exponent has the negative energy dimension of coupling constant in 4d. In the Dirac spinor field $\Psi$, Lagrangian density $\mathcal{L}$ contain only one derivative $\partial_{\mu}$ hence dimensional analysis will give us $[\Psi]=E^{3/2}$ which is little bit higher than the scalar field $[\Phi]=E^1$. Once we know this, we can write any Lagrangian density !!!

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