Nonrelativistic limit of scalar field theory

[stevendaryl]

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 $\psi$).

 

[Demystifier]

Yes, you have to rescale it. See e.g. my http://lanl.arxiv.org/abs/1205.1992 , Eq. (8.71) and the analysis around it.

 

[samalkhaiat]

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 $$\mathcal{H}_{Sch} = \frac{1}{2m} \nabla \psi \cdot \nabla \psi^{\ast} .$$