Equations of QFT

1. Aug 22, 2008

ghery

Hello:

In quantum field theory, there is the Klein-Gordon Equation that describes particles with Spin 0, this equation reduces to the SchÖdinger equation when the non relativistic limit is taken, Does the Dirac equation that describes particles with spin ½, reduce to Pauli's equation when the non-relativistic limit is taken? and if it does reduce, could you explain me how?

Regards

2. Aug 23, 2008

malawi_glenn

I know that Dirac reduces to Pauli equation, can give you a pdf I have if you send me a PM.

But for KG, which is non-linear, I don't know.

3. Aug 23, 2008

humanino

KG is linear but 2nd order. A wave satisfying Dirac does satisfy KG as well.

4. Aug 24, 2008

malawi_glenn

Ah, well, yes :-) I think I meant 2nd order.. hehe

5. Aug 24, 2008

jostpuur

The Klein-Gordon equation alone does not reduce to the Schrödinger equation in the non-relativistic limit. The Klein-Gordon equation has both positive and negative frequency solutions, and the Schrödinger's equation has only positive frequency solutions, but the negative frequency solutions of the Klein-Gordon equation don't vanish in the non-relativistic limit.

In other words, the non-relativistic solutions of the Klein-Gordon equation are linear combinations of solutions of the equations

$$\pm i\hbar\partial_t\psi = H\psi.$$

6. Aug 25, 2008

Orion1

Relativistically covariant quantized Schrödinger equation...

I note that in the Schrödinger equation, when momentum is zero, the energy is also zero. This equation does not account for the particle rest energy. This is why this equation is not relativistically covariant and is not invariant under a Lorentz transformation.

Schrödinger equation for a free particle is:
$$- \frac{\hbar^2}{2m} \nabla^2 \ \psi = i \hbar\frac{\partial}{\partial t} \psi$$

$$\boxed{\nabla = 0 \; \; \; i \hbar\frac{\partial}{\partial t} \psi = 0}$$

However, for the Klein–Gordon equation, when momentum is zero, energy is equivalent to the particle rest energy.

Klein–Gordon equation is:
$$- \hbar^2 c^2 \mathbf{\nabla}^2 \psi + m^2 c^4 \psi = - \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi$$

$$\nabla = 0 \; \; \; \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = m^2 c^4 \psi$$
$$\boxed{\nabla = 0 \; \; \; i \hbar\frac{\partial}{\partial t} \psi = m c^2 \psi}$$

$$m = 0 \; \; \; - \hbar^2 c^2 \mathbf{\nabla}^2 \psi = - \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi$$

$$\boxed{m = 0 \; \; \; - i \hbar c \nabla \psi = i \hbar\frac{\partial}{\partial t} \psi}$$

Therefore, the equation I derived for a relativistically covariant Schrödinger equation:
$$- \frac{\hbar c}{\overline{\lambda}} \nabla \psi + mc^2 \psi = i \hbar\frac{\partial}{\partial t} \psi$$

$$\boxed{\nabla = 0 \; \; \; i \hbar\frac{\partial}{\partial t} \psi = m c^2 \psi}$$

Relativistically covariant quantized Schrödinger equation:
$$\boxed{- i \hbar c \nabla \psi + mc^2 \psi = i \hbar\frac{\partial}{\partial t} \psi}$$

$$\boxed{m = 0 \; \; \; - i \hbar c \nabla \psi = i \hbar\frac{\partial}{\partial t} \psi}$$

Reference:
http://en.wikipedia.org/wiki/Free_particle#Non-Relativistic_Quantum_Free_Particle"
http://en.wikipedia.org/wiki/Klein-Gordon_equation#Derivation"

Last edited by a moderator: Apr 23, 2017
7. Aug 25, 2008

malawi_glenn

Orion, all this was done in the 1920's, we all know how the story goes.

Besides, is $$p + m$$
An apropriate hamiltonian? I would say that its not.

8. Aug 25, 2008

jostpuur

Re: Relativistically covariant quantized Schrödinger equation...

Not agreed. The solutions of the Klein-Gordon equation

$$-\hbar^2\partial_t^2\psi = m^2c^4\psi - \hbar^2 c^2\nabla^2\psi$$

are linear combinations of the solutions of the equations

$$\pm i\hbar\partial_t\psi = \sqrt{m^2c^4 - \hbar^2 c^2\nabla^2}\psi.$$

So if we are only interested in the positive frequency solutions with no relativistic frequencies, this equation can be approximated to be

$$i\hbar\partial_t\psi = \Big(mc^2 - \frac{\hbar^2}{2m}\nabla^2\Big)\psi.$$

So when we arrive at the Schrödinger's equation in this way, the rest energy term is there. The Schrödinger's equation in any case is not Lorentz invariant, for the simple reason that it has some non-trivial transforming in boosts. The rest energy term is not key issue with Lorentz invariance. The reason why the rest term can be dropped is that it has no relevance, since only energy differences matter.

In this equation the right side is a complex number (a member of $$\mathbb{C}$$), and the left side is a three component object (a member of $$\mathbb{C}^3$$). The equation doesn't mean anything.

9. Aug 25, 2008

malawi_glenn

man why didn't I see that before you did Jostpuur? :-(

he seems to think that one can square-root operators as if they would been ordinary numbers..

10. Aug 25, 2008

Count Iblis

These are not " equations of QFT" at all! In QFT the Dirac equation, Klein Gordon equation, etc. are to be interpreted as classical field equations that have yet to be quantized.

So, this is all classical physics.

11. Aug 25, 2008

humanino

Good point

12. Aug 25, 2008

samalkhaiat

Re: Relativistically covariant quantized Schrödinger equation...

13. Aug 25, 2008

samalkhaiat

14. Aug 25, 2008

jostpuur

I don't want to put this.

The wave packet solutions of Klein-Gordon equation can be in general written in the form

$$\phi(t,\boldsymbol{x}) = \int \frac{d^3p}{(2\pi\hbar)^3}\Big(\phi^+_{\boldsymbol{p}} e^{i(\boldsymbol{x}\cdot\boldsymbol{p} \;-\; \sqrt{|\boldsymbol{p}|^2 c^2 + m^2c^4}t)/\hbar} \;+\; \phi^-_{\boldsymbol{p}} e^{i(\boldsymbol{x}\cdot\boldsymbol{p} \;+\; \sqrt{|\boldsymbol{p}|^2 c^2 + m^2c^4}t)/\hbar}\Big)$$

The wave packet solutions of the Schrödinger's equation instead have the form

$$\psi(t,\boldsymbol{x}) = \int\frac{d^3p}{(2\pi\hbar)^3} \phi_{\boldsymbol{p}} e^{i(\boldsymbol{x}\cdot\boldsymbol{p} \;-\; (|\boldsymbol{p}|^2/(2m))t)/\hbar}$$

If we substitute the non-relativistic approximation

$$\sqrt{|\boldsymbol{p}|^2c^2 + m^2c^4} \approx mc^2 + \frac{|\boldsymbol{p}|^2}{2m}$$

into the solutions of the Klein-Gordon equation, and absorb the rest mass somewhere (like into the definition of the phi as its background phase oscillation), we get

$$\phi(t,\boldsymbol{x}) \approx \int \frac{d^3p}{(2\pi\hbar)^3}\Big(\phi^+_{\boldsymbol{p}} e^{i(\boldsymbol{x}\cdot\boldsymbol{p} \;-\; (|\boldsymbol{p}|^2/(2m))t)/\hbar} \;+\; \phi^-_{\boldsymbol{p}} e^{i(\boldsymbol{x}\cdot\boldsymbol{p} \;+\; (|\boldsymbol{p}|^2/(2m))t)/\hbar}\Big)$$

and this is not necessarily a solution of the Schrödinger's equation.

For me, for now, these are merely equations.