# Qualitative explanation of scale dependence

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1. Sep 23, 2015

### metroplex021

Hi all -- can anyone offer a qualitative explanation of why it is that couplings run with the energy in *relativistic* quantum theory, and not in non-relativistic? Some insight here would be much appreciated. Thanks.

2. Sep 23, 2015

### RGevo

Couplings run in renormalised field theory.

Relativistic quantum theory, what are you referring to here? What couplings are there.

3. Sep 25, 2015

### fzero

There are a few ideas that probably need explaining here, so this is going to be long. First is the difference between relativistic and nonrelativistic quantum theory. For nonrelativistic quantum mechanics to be valid, we must be dealing with energy scales that are small compared to the masses of the particles that being described. So, for example, in studying the energy levels of the hydrogen atom, the electron mass is $\sim 0.5~\text{MeV}$, while the spacing between energy levels is of order tens of $\text{eV}$. Therefore the energy scales are around five orders of magnitude smaller than the electron mass. This ensures that the electron will never be excited to relativistic speeds and a nonrelativistic quantum mechanics where the number of particles in the system is fixed can be used to describe the physics. However, the relativistic theory remains useful in order to describe the properties due to spin, as well as very small corrections to the nonrelativistic approximation.

At large enough energies, it is possible to have processes in which new particles are created, so one has to expand quantum mechanics to quantum field theory, where the number of particles can change, so long as certain conservation laws are obeyed. For example, if we collide an electron and a positron at a center of mass energy of $1~GeV$, then it is possible to create a muon-anti-muon pair (since $m_\mu \sim 106~\text{MeV}$.) It is natural to develop the theory of quantum fields in a relativistic-covariant way, because of this contact with processes in particle physics. However, nonrelativistic quantum field theories are very important in condensed matter physics, where the energy might not be relativistically large, but particle numbers are not necessarily conserved.

So quantum field theories, in a sense, replace the wavefunctions of quantum mechanics with operators known as quantum fields that act on quantum states to give new quantum states. The new states consist of adding or removing a particle from the original state, so processes that change the number and type of particles are naturally described. Since we often want to use momentum to describe a particle state, it is common to decompose a quantum field as a linear combination of operators that create or destroy a particle with a fixed value of the momentum. For a real scalar field we would have something like
$$\phi(\mathbf{x}) \sim \int d\mathbf{k} \left( a(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} +a^\dagger(\mathbf{k}) e^{-i \mathbf{k} \cdot \mathbf{x}} \right),$$
where in the relativistic case $\mathbf{k}$ is a 4-vector, while in a nonrelativistic case, it might be a 3-vector. Here $a^\dagger(\mathbf{k})$ creates the particle with momentum $\mathbf{k}$, while $a(\mathbf{k})$ destroys it.

With this background, we can give a basic description of renormalization and explain why coupling constants run with the energy scale used to define them. We will see that it is a property of quantum field theory and does not explicitly depend on whether the particular theory is relativistic or not.

Let's consider the example of electron-positron scattering from above. From energy conservation, we know that the center-of-mass energy of the initial system is the same as that of the final state, no matter what particular particles appear after the collision. By allowing each of the initial or final state particles to have a fraction of this energy, we can put a bound on the possible momentum of any external state, which will be less than the center of mass energy. Let us call the energy bound $\Lambda$, which we can also refer to as the ultraviolet (UV) cutoff. Then only modes from a quantum field with momentum $|\mathbf{k}| \leq \Lambda$ can appear in the external states of the process. Particles in the theory with a mass $> \Lambda$ can't appear at all in the external states either. We can call these allowed modes the low energy modes or low energy degrees of freedom, since they contain an energy which is lower than the bound. We can schematically denote them by $\phi_L$ and ignore the specific momentum or particle-type labels.

However, since this is quantum theory, the high energy modes, $\phi_H$ can still appear in whatever intermediate states describe the process. You might already be familiar from this in ordinary quantum mechanics, where in 2nd order perturbation theory, one sums over all possible states, even those that themselves have an energy much larger than those of the initial state. In QFT, this situation arises when we have loops in Feynman diagrams, where we sum over all momenta, including infinitely large ones.

These considerations naturally lead to the idea of effective field theory. Since only the low energy modes appear as external states, we can imagine that there is some effective description that only includes low energy modes. The full quantum theory might be described by some action $S(\phi_L,\phi_H)$ that depends on all of the degrees of freedom, but the effective theory will be described by a theory with action $S_\Lambda(\phi_L)$. The process of deriving the effective action is referred to as "integrating out" the high energy degrees of freedom $\phi_H$, since in the path integral description, we essentially have
$$e^{-i S_\Lambda(\phi_L)} = \int \mathcal{D}\phi_H e^{-i S(\phi_L,\phi_H)}.$$

We can see that the coupling constants in the effective theory must depend on the cutoff scale $\Lambda$ (hence they "run" with the energy) by considering processes at a slightly lower center of mass energy so that there is a new cutoff $\Lambda'< \Lambda$. We haven't changed the processes that are allowed to happen in a scattering event, so the new effective action $S_{\Lambda'}(\phi_{L'})$ should have the same functional form as $S_\Lambda(\phi_L)$, but the coupling constants must be different since the new theory is obtained from the old one by integrating out the modes with momenta $\Lambda' \leq |\mathbf{k}| \leq \Lambda$. A careful analysis in perturbation theory would allow us to derive the renormalization group equations that describe the coupling constants $g_i(\Lambda')$ in terms of the coupling constants $g_i(\Lambda)$.

This is what is called the Wilsonian approach to renormalization, which gives a detailed qualitative picture of renormalization in QFT, which had previously been thought of as a mere mathematical trick. It also makes clear that renormalization is a basic property of quantum field theories. In particular, it is not especially tied to Lorentz invariance.

Last edited: Sep 25, 2015
4. Sep 25, 2015

### metroplex021

that's a really wonderful answer. Thank you! Going to try and digest it a bit, and get back...

But thanks so much!