Nonrelativistic limit of scalar field theory

  • #1
stevendaryl
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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##).
 

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  • #3
samalkhaiat
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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 [itex]1/2[/itex]-factor in the complex scalar Lagrangian which has the form [tex]\mathcal{L}(\varphi , \partial \varphi) = \dot{\varphi} \dot{\varphi}^{\ast} - \nabla \varphi \cdot \nabla \varphi^{\ast} - m^{2} \varphi \varphi^{\ast} .[/tex] Now, if you redefine the field according to [tex]\varphi (t , \vec{x}) = e^{- i mt} \psi (t , \vec{x}) ,[/tex] the new Lagrangian becomes [tex]\mathcal{L}(\psi , \dot{\psi} , \nabla \psi) = \dot{\psi}\dot{\psi}^{\ast} + im (\dot{\psi}\psi^{\ast} - \psi \dot{\psi}^{\ast}) - \nabla \psi \cdot \nabla \psi^{\ast} . \ \ \ \ \ (1)[/tex] Now, you can, if you wish, take the non-relativistic limit by ignoring the first term and dividing by [itex]2m[/itex]: [tex]\mathcal{L}_{Sch} = \frac{i}{2} (\dot{\psi}\psi^{\ast} - \psi \dot{\psi}^{\ast}) - \frac{1}{2m} \nabla \psi \cdot \nabla \psi^{\ast} .[/tex] 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 [itex]\mathcal{H}(\psi)[/itex], you need to apply the Legendre transform to [itex]\mathcal{L}(\psi)[/itex] of eq(1): [tex]\frac{\partial \mathcal{L}}{\partial \dot{\psi}} = \dot{\psi}^{\ast} + im \psi^{\ast}, \ \ \frac{\partial \mathcal{L}}{\partial \dot{\psi}^{\ast}} = \dot{\psi} - im \psi ,[/tex] [tex]\mathcal{H}(\psi) = ( \dot{\psi}^{\ast} + im \psi^{\ast}) \dot{\psi} + ( \dot{\psi} - im \psi ) \dot{\psi}^{\ast} - \mathcal{L}(\psi) .[/tex] Or [tex]\mathcal{H}(\psi) = \dot{\psi} \dot{\psi}^{\ast} + \nabla \psi \cdot \nabla \psi^{\ast} .[/tex] Now, go to the non-relativistic limit and introduce the correct scaling dimension to obtain [tex]\mathcal{H}_{Sch} = \frac{1}{2m} \nabla \psi \cdot \nabla \psi^{\ast} .[/tex]
 
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