# Nonrelativistic limit of scalar field theory

• I
• stevendaryl
In summary, the Klein-Gordon equation has the Schrodinger equation as a nonrelativistic limit when a new function is defined and the equation is approximated. However, when considering the field theory, a relativistic Lagrangian density that yields the Klein-Gordon equation does not have the appropriate nonrelativistic limit. By correctly applying the field redefinition and Legendre transform, the correct nonrelativistic Hamiltonian density can be obtained.

#### stevendaryl

Staff Emeritus
Science Advisor
The Klein-Gordon equation has the Schrodinger equation as a nonrelativistic limit, in the following sense:

Start with the Klein-Gordon equation (for a complex function ##\phi##)

## \partial_\mu \partial^\mu \phi + m^2 \phi = 0##

Now, define a new function ##\psi## via: ##\psi = e^{i m t} \phi##. Then the equation for ##\psi## is:

## \ddot{\psi} -2 i m \dot{\psi} -\nabla^2 \psi = 0##

Now, if we assume that ##\ddot{\psi}## is small compared with the other terms, then we have approximately:

## -2 i m \dot{\psi} -\nabla^2 \psi = 0 \Rightarrow i \dot{\psi} = - \frac{1}{2m} \nabla^2 \psi##

That's Schrodinger's equation. So that seems to work. But now, instead of looking at equations of motion, let's look at the field theory.

The Schrodinger equation follows from a field-theoretic lagrangian density:

##\mathcal{L} = - i (\psi^* \dot{\psi} - \dot{\psi^*} \psi) - \frac{1}{2m} (\nabla \psi^*) \cdot (\nabla \psi)##

which corresponds to the hamilton density:

##\mathcal{H} = \frac{1}{2m} (\nabla \psi^*) \cdot (\nabla \psi)##

Now, what I would expect is that just as the Klein Gordon equation has the Schrodinger equation as a nonrelativistic limit, the relativistic hamiltonian density should have the appropriate nonrelativistic limit, as well. But it doesn't quite work.

A relativistic lagrangian density that yields the Klein-Gordon equation is:

##\mathcal{L} = \frac{1}{2} \dot{\phi^*} \dot{\phi} - \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) - \frac{m^2}{2} \phi^* \phi##

This corresponds to a Hamiltonian density (I think).

##\mathcal{H} = \frac{1}{2} \dot{\phi^*} \dot{\phi} + \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) + \frac{m^2}{2} \phi^* \phi##

Now, let's try the same trick: Let ##\phi = e^{-imt}\psi##. Then in terms of ##\psi##:

##\mathcal{H} = \frac{1}{2} \dot{\psi^*}\dot{\psi} + \frac{im}{2} (\psi^* \dot{\psi} - \dot{\psi^*} \dot{\psi}) + \frac{1}{2} (\nabla \psi^*) \cdot (\nabla \psi) + m^2 \psi^* \psi##

Assuming once again that the first term is negligible compared to the others gives you:

##\mathcal{H} = \frac{im}{2} (\psi^* \dot{\psi} - \dot{\psi^*} \psi) + \frac{1}{2} (\nabla \psi^*) \cdot (\nabla \psi) + m^2 \psi^* \psi##

That doesn't look anything like the nonrelativistic Hamiltonian density. It doesn't even have the right units (although I guess you could fix that by rescaling ##\psi##).

stevendaryl said:
A relativistic lagrangian density that yields the Klein-Gordon equation is:

##\mathcal{L} = \frac{1}{2} \dot{\phi^*} \dot{\phi} - \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) - \frac{m^2}{2} \phi^* \phi##
Here you made one mistake by including the $1/2$-factor in the complex scalar Lagrangian which has the form $$\mathcal{L}(\varphi , \partial \varphi) = \dot{\varphi} \dot{\varphi}^{\ast} - \nabla \varphi \cdot \nabla \varphi^{\ast} - m^{2} \varphi \varphi^{\ast} .$$ Now, if you redefine the field according to $$\varphi (t , \vec{x}) = e^{- i mt} \psi (t , \vec{x}) ,$$ the new Lagrangian becomes $$\mathcal{L}(\psi , \dot{\psi} , \nabla \psi) = \dot{\psi}\dot{\psi}^{\ast} + I am (\dot{\psi}\psi^{\ast} - \psi \dot{\psi}^{\ast}) - \nabla \psi \cdot \nabla \psi^{\ast} . \ \ \ \ \ (1)$$ Now, you can, if you wish, take the non-relativistic limit by ignoring the first term and dividing by $2m$: $$\mathcal{L}_{Sch} = \frac{i}{2} (\dot{\psi}\psi^{\ast} - \psi \dot{\psi}^{\ast}) - \frac{1}{2m} \nabla \psi \cdot \nabla \psi^{\ast} .$$ This gives you the Schrödinger field equations as well as the correct non-relativistic Hamiltonian.
##\mathcal{H} = \frac{1}{2} \dot{\phi^*} \dot{\phi} + \frac{1}{2} (\nabla \phi^*) \cdot (\nabla \phi) + \frac{m^2}{2} \phi^* \phi##

Now, let's try the same trick: Let ##\phi = e^{-imt}\psi##. Then in terms of ##\psi##:

##\mathcal{H} = \frac{1}{2} \dot{\psi^*}\dot{\psi} + \frac{im}{2} (\psi^* \dot{\psi} - \dot{\psi^*} \dot{\psi}) + \frac{1}{2} (\nabla \psi^*) \cdot (\nabla \psi) + m^2 \psi^* \psi##
Here, you made another mistake by substituting the field redefinition in the old Hamiltonian. This is not correct because field redefinition changes the conjugate momentum. So, to obtain $\mathcal{H}(\psi)$, you need to apply the Legendre transform to $\mathcal{L}(\psi)$ of eq(1): $$\frac{\partial \mathcal{L}}{\partial \dot{\psi}} = \dot{\psi}^{\ast} + I am \psi^{\ast}, \ \ \frac{\partial \mathcal{L}}{\partial \dot{\psi}^{\ast}} = \dot{\psi} - I am \psi ,$$ $$\mathcal{H}(\psi) = ( \dot{\psi}^{\ast} + I am \psi^{\ast}) \dot{\psi} + ( \dot{\psi} - I am \psi ) \dot{\psi}^{\ast} - \mathcal{L}(\psi) .$$ Or $$\mathcal{H}(\psi) = \dot{\psi} \dot{\psi}^{\ast} + \nabla \psi \cdot \nabla \psi^{\ast} .$$ Now, go to the non-relativistic limit and introduce the correct scaling dimension to obtain $$\mathcal{H}_{Sch} = \frac{1}{2m} \nabla \psi \cdot \nabla \psi^{\ast} .$$

Spinnor, dextercioby, kith and 3 others

## 1. What is the nonrelativistic limit of scalar field theory?

The nonrelativistic limit of scalar field theory is a mathematical approximation used to describe physical systems in which particles move at speeds much slower than the speed of light. This limit is often used in quantum mechanics to simplify calculations and make them more manageable.

## 2. How is the nonrelativistic limit different from the full scalar field theory?

The nonrelativistic limit is a simplified version of the full scalar field theory, in which the effects of relativity are neglected. This means that the equations used in the nonrelativistic limit only apply to systems with slow-moving particles, while the full scalar field theory applies to all systems.

## 3. What are the assumptions made in the nonrelativistic limit of scalar field theory?

The nonrelativistic limit makes several assumptions, including that the particles are non-relativistic, that the interactions between particles are weak, and that the particles are not affected by external electromagnetic fields. These assumptions allow for the simplification of the equations used in the limit.

## 4. When is the nonrelativistic limit of scalar field theory useful?

The nonrelativistic limit is useful in situations where the particles involved are moving at speeds much slower than the speed of light. This includes many common physical systems, such as atoms, molecules, and condensed matter systems. It is also used in quantum mechanics to simplify calculations and make them more manageable.

## 5. Are there any limitations to the nonrelativistic limit of scalar field theory?

While the nonrelativistic limit is a useful approximation in many situations, it does have limitations. It cannot be used to accurately describe systems with particles moving at speeds close to the speed of light, and it does not take into account the effects of relativistic phenomena such as time dilation and length contraction. In these cases, the full scalar field theory must be used.