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?


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 !!!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.