Talk:Schrödinger field

Reason For Being

I added this page because this sort of thing is discussed in dozens of places, each place with a different name. Theres the "Coherent state path integral" in Negele and Orland, there is the "Gross Pitaevski equation" in BEC theory. There is the "Bogoliubov DeGennes Equation" in superconductivity and the "nonlinear Schrodinger equation" in fiber-optics. It's all the same thing. It confuses people, because there's very little conceptual relation to the equation for the quantum mechanical wavefunction, other than obeying the same equation.Likebox 19:06, 8 October 2007 (UTC)

On Pair potentials

In the section on Pair potentials there's written that the non relativistic limit of for example QED is the following action:

$S=\int _{xt}\psi ^{\dagger }\left(i{\partial \over \partial t}+{\nabla ^{2} \over 2m}\right)\psi -\int _{xy}\psi ^{\dagger }(x)\psi (x)V(x,y)\psi ^{\dagger }(y)\psi (y)$

with

$V(x,y)={q^{2} \over |x-y|}$

and that this describes nearly all of condensed matter physics.

That's ok, but in condesed matter I'm used the following expression for the pair potential in the hamiltonian:

$H_{int}=\int _{xy}\psi ^{\dagger }(x)\psi ^{\dagger }(y)V(x,y)\psi (y)\psi (x)\,$

This is different from the previous one...could someone comment about the difference. Dave (talk) 17:54, 28 March 2009 (UTC)

It's the same expression.Likebox (talk) 21:16, 28 March 2009 (UTC)

Up to some normal ordering stupidities. There might be some difference for singular potentials, like the one in the example. Drat. You should think of the potential as smoothed out in some way near r=0. But you're right, the article should be more careful about operator ordering.Likebox (talk) 18:49, 30 March 2009 (UTC)

Yep, I was exactly thinking to this small difference for divergent potentials. I was wandering how this difference could be linked to the self-interaction renormalization and the Lamb shift. In fact the expression you wrote seems to be the right non relativistic limit of QED (isn't it), and has to be renormilized, while the expression I wrote is what one obtains from second quantization of Shroedinger equation, and does not present divergences. Am I missing something? Dave (talk) 20:55, 30 March 2009 (UTC)

This article is all about second quantization of the Schrodinger equation, because this case is free from the conceptual difficulties of particles going backwards in time. The Lamb shift in QED is caused by loops involving virtual electrons and photons, and will be absent in this approximation.

The ordering ambiguity you noticed is real: use the commutation relation for psi and psi-dagger to sort it out. The different between the two orders is a local term psi*(x)psi(x)V(x,x), which, when V is regular at x=0 is just a constant potential everywhere in space. This can be removed by a time-dependent redefinition of the phase of psi, it's not physically important because the zero of energy is only relative to a reference. But it is annoying that the particular V of physical interest is divergent when the electrons are sitting right on top of each other.Likebox (talk) 01:19, 2 April 2009 (UTC)

Tiding up form and treatment similar to Klein-Gordon page

I think this article page could provide from a closer alignment to the discussion of its relativistic counterpart the Klein-Gordon Lagrangian and drawing more on conventions introduced in the Lagrangian field theory article.

This would include for example a proper distinction between the Lagrangian and Hamiltonian and its densities. Partial Integration are often done without highlighting them and the conjugate field is not properly discussed for my taste.

I think another example that could be included is a discussion of the Probability current conservation from Noether's theorem in a similiar fashion to the discussion of the Klein-Gordon field conserved current:

Using the asymmetric density a scetch of the calculation would be:

$${\begin{aligned}{\mathcal {L}}&=i\psi ^{\dagger }{\dot {\psi }}{\frac {\nabla \psi ^{\dagger }\nabla \psi }{2m}}\\\psi &\mapsto e^{i\theta }\psi =\psi +\theta \delta \psi +\dots \implies \delta \psi =i\psi \\j^{\mu }&=-{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\psi )}}\delta \psi :\\{\frac {\mathcal {L}}{\partial (\partial _{t}\psi )}}&=i\psi ^{\dagger }\quad {\frac {\mathcal {L}}{\partial (\nabla \psi )}}={\frac {1}{2m}}\nabla \psi ^{\dagger }\\j^{0}=\rho &=-(i\psi ^{\dagger })(i\psi )=\psi ^{\dagger }\psi \\{\vec {j}}&={\frac {\mathcal {L}}{\partial (\nabla \psi )}}\delta \psi +{\frac {\mathcal {L}}{\partial (\nabla \psi ^{\dagger })}}\delta \psi ^{\dagger }=i\psi 1/2m\nabla \psi ^{\dagger }-i\psi ^{\dagger }1/2m\nabla \psi ={\frac {i}{2m}}\left(\psi \nabla \psi ^{\dagger }-\psi ^{\dagger }\nabla \psi \right)={\frac {1}{2mi}}\left(\psi ^{\dagger }\nabla \psi -\psi \nabla \psi ^{\dagger }\right)\\\partial _{\mu }j^{\mu }&=\partial _{t}\rho +\nabla {\vec {j}}=0\end{aligned}}$$ Conscious 'n' curious (talk) 16:05, 18 April 2025 (UTC)