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Introduction to GrossPitaevskii equation 
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#1
May812, 12:49 PM

P: 49

Hi to everyone :)
Like the topic's title says, I am starting my journey to BoseEinstein condensate. What I am looking for are some basic introductory papers or materials that deal with derivation of the GrossPitaevskii equation on the basic, most elementary level that is possible. I am a undergraduate student and finishing last year of Bachelor degree, so I have only basic knowledge of quantum mechanics, statistical physics and their application in solid state physics, nanoelectronics and photonics. So any advice from all of you is really welcome, and if you know some material that could be useful and if you btw recommend them to me I would be really grateful. PS I was looking for material on derivation of GP equation, but what I have found were only comprehensive and advance theories that deal with second quantisation and are filled with highly rigorous mathematics, which I think is not a starting point for someone like me :(. 


#2
May812, 01:23 PM

P: 3,014

I have used the following textbook in my Atomic Gasses graduate level class:
BoseEinstein Condensation in Dilute Gases, C.J. Pethick and H. Smith (Cambridge University Press, Second Edition, 2008) Chapter 6 deals with the GrossPitaevskii Equation. I will list the references at the end of that chapter:



#3
May2612, 04:08 AM

P: 49

I have found another good view of the derivation of GP equation, which is I think suitable for beginners like me. There is one more thing that is confusing to me in pdf that is attached, for which I am asking for help.
How do they construct the functional for minimization that is different from the usual one? [itex]E[ψ]=[/itex][itex]\frac{<ψHψ>}{<ψψ>}[/itex] is fine for me, but the next one is troublesome. [itex]F[ψ]=[/itex][itex]<ψHψ>[/itex][itex]μ<ψψ>[/itex]. I understand that we need to find extreme value of the functional under constraint that number of particles is conserved, but it isn't obvious why [itex]E[ψ][/itex] is replaced exactly with [itex]F[ψ][/itex]. And also I don't see how [itex]<ψψ>[/itex] represents condition of conservation of particles in formalism of Lagrange multipliers where constraint condition is simply restriction of the domain given by [itex]g(x,y)=c[/itex], for function [itex]z=f(x,y)[/itex]. I want to ask how [itex]<ψψ>[/itex] restricts the domain of the functional [itex]E[ψ][/itex]. 


#4
May2612, 07:58 AM

P: 3,014

Introduction to GrossPitaevskii equation
The method of Lagrange multipliers is used to exactly avoid restrictions on the domain, because it is often hard to solve w.r.t. any one of the variables.



#5
May2612, 08:15 AM

P: 3,014

Notice that :
[tex] \begin{array}{rcl} \delta E & = & \frac{\delta \langle \psi \vert H \vert \psi \rangle}{\langle \psi \vert \psi\rangle} \\ &  & \frac{\langle \psi \vert H \vert \psi \rangle}{\langle \psi \vert \psi\rangle^{2}} \, \langle \delta \psi \vert \psi\rangle \\ & = & \frac{1}{\langle \psi \vert \psi\rangle} \, \langle \delta \psi \left\vert H  \frac{\langle \psi \vert H \vert \psi \rangle}{\langle \psi \vert \psi \rangle} \right \vert \psi \rangle \end{array} [/tex] where we only varied the bra. Equating this variation to zero, we get Schroedinger equation: [tex] H \, \vert \psi \rangle = E \, \vert \psi \rangle [/tex] If you vary F, you will get the same equation, except that [itex]E = \mu[/itex]. This is understandable, since when all the particles are in the ground state, the average energy E is equal to this lowest energy state. But, a level becomes macroscopically occupied when the chemical potential [itex]\mu[/itex] equals the energy of that level. Indeed, then the exponential [itex]e^{\frac{\epsilon_{i}  \mu}{T}} = 1[/itex] becomes 1, and the denominator of the Bose Einstein distribution function vanishes. 


#6
May2612, 09:59 AM

P: 49

Thank you very much for reply :).. Maybe I said it in a wrong way about Lagrange multipliers, but it surely gives a chance to find an extreme value of some function on a restricted domain which is implicitly contained in a new defined function..
It might be that I misunderstood something in your previous post, but I still can't see how they found that they should minimize functional F rather than E? Is there the answer to this, without calling on some results that are on the end of the calculation, and is more likely based on some energy reasons? Here is something that I am thinking on: If your rearrange functional [itex]F[/itex] into: [itex]\frac{F[ψ]}{<ψψ>}=E[ψ]μ[/itex] you get the expression which represents some kind of available energy which I don't know how to interpret... 


#7
May2612, 10:10 AM

P: 49

And one more thing, why is only bra varied? It is the first time for me to use variation technique.



#8
May2612, 10:16 AM

P: 3,014

I'm afraid you have problem with the mathematical technique, not with the Physics. I suggest you open another thread in a section on Calculus, or, go through a textbook on Mathematical Methods for Physicists.



#9
May2612, 11:41 AM

P: 49

Well, could you help me with the problem about defining the function that should be minimized? I think that its selection is about physics mainly.



#10
May2612, 12:31 PM

P: 3,014

The functional that is minimized F corresponds to the so called Grand Thermodynamic Potential [itex]\Omega = E  T S  \mu N[/itex] at absolute zero. It may be shown that when a system is in a contact with a heat and particle reservoir that it can exchange energy and particles with, the grand thermodynamic potential reaches a minimum in equilibrium.



#11
May2612, 12:36 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

The expression
$$E[\psi] = \frac{ \langle \psi  H  \psi \rangle }{ \langle \psi  \psi \rangle }$$ is appropriate when the wavefunction has not been normalized. To work in the space of normalized wavefunctions, we must enforce the constraint ##\langle \psi  \psi \rangle=1##. This can be accomplished by extremizing the functional $$F[\psi] = \langle \psi  H  \psi \rangle + \mu ( \langle \psi  \psi \rangle 1 ).$$ The author of those notes has omitted the term involving ##(1) \mu## since it is just a constant, so its variation is zero. We get the same GP equation whether we include it or not. 


#12
May2612, 12:45 PM

P: 49

I feel a little bit stupid right now :)... thanks you all!



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