1
$\begingroup$

I wanted to derive time-dependent Gross-Pitaevskii equation in an alternative way, but I don't know if something presented below is allowed. Hamiltonian is the following (I do not assume translational invariance) $$\hat{\mathcal{H}} = \int d^3x\ \hat{\Psi}^{\dagger}(\mathbf{x})\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x}) \right]\hat{\Psi}(\mathbf{x}) + g\hat{\Psi}^{\dagger}(\mathbf{x})\hat{\Psi}^{\dagger}(\mathbf{x})\hat{\Psi}(\mathbf{x})\hat{\Psi}(\mathbf{x})$$ Motivated by the BCS theory I will linearize the Hamiltonian in the following way: $$\hat{\mathcal{H}} = \int d^3 x\ \hat{\Psi}^{\dagger}(\mathbf{x})\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x}) \right]\hat{\Psi}(\mathbf{x}) + g\hat{\Psi}^{\dagger}(\mathbf{x})\hat{\Psi}(\mathbf{x})\xi(\mathbf{x},t)$$ with some position and time dependent factor $\xi(\mathbf{x},t) = \langle \hat{\Psi}^{\dagger}(\mathbf{x})\hat{\Psi}(\mathbf{x})\rangle$. Now Hamiltonian is quadratic so we can write the solution for field oprator: $$\hat \Psi(\mathbf{x},t) = T\exp\left\{-\frac{i}{\hbar}\left[\left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x}) \right)t + g\int\limits_{0}^{t} dt'\ \xi(\mathbf{x},t') \right]\right\}\hat{\Psi}(\mathbf{x})$$ If we assume that initially all particles were in the condensate: $\hat{\Psi}(\mathbf{x}) = \hat{a}\phi_{0}(\mathbf{x})$, then $$\phi(\mathbf{x},t) = T\exp\left\{-\frac{i}{\hbar}\left[\left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x}) \right)t + g\int\limits_{0}^{t} dt'\ \xi(\mathbf{x},t') \right]\right\}\phi_{0}(\mathbf{x})$$ and $$i\hbar \partial_t \phi(\mathbf{x},t) = \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x}) \right)\phi(\mathbf{x},t) + g \xi(\mathbf{x},t)\phi(\mathbf{x},t)$$ but now $\xi(\mathbf{x},t)$ should be equal to $N|\phi(\mathbf{x},t)|^2$, if the wavefunction is normalized to unity. I don't know how to justify this. What do you think?

$\endgroup$
  • $\begingroup$ I think the only way you can identify those two quantities is to assume that your $\xi(\mathbf{x},t) = \langle \hat{\Psi}^{\dagger}(\mathbf{x})\hat{\Psi}(\mathbf{x})\rangle$ actually factorizes, which is the standard BEC approximation. I think this might even be necessary, because it corresponds to the requisite symmetry breaking (I think?). $\endgroup$ – march Feb 29 '16 at 18:30
  • $\begingroup$ @Nex_Friedrich How does $\frac{d\xi}{dt}$ arise from the integral term in the generator? $\endgroup$ – udrv Mar 1 '16 at 5:01
  • $\begingroup$ @udrv You are right! Of course Liebniz integral rule. My mistake. I am correcting this right now. $\endgroup$ – WoofDoggy Mar 1 '16 at 9:57
  • $\begingroup$ Now the question arises what is the justification of such simplification? $\endgroup$ – WoofDoggy Mar 1 '16 at 17:39
  • $\begingroup$ When you write $\xi({\vec x}, t) = \langle \hat\Psi^\dagger({\vec x})\hat\Psi({\vec x})\rangle$, I guess the time dependence comes from the time dependent state $\rho(t) = e^{-iHt}\rho(0)e^{iHt}$ ($\hbar =1$) in Schroedinger representation, whereas your $\hat\Psi({\vec x}, t)$ is in Heisenberg representation, right? Then why not switch to Heisenberg rep in $\xi({\vec x}, t)$ and write instead $\xi({\vec x}, t) = \langle \hat\Psi^\dagger({\vec x},t)\hat\Psi({\vec x},t)\rangle$ as average on the initial state. Doesn't this then relate trivially to your $|\phi({\vec x}, t)|^2$? $\endgroup$ – udrv Mar 1 '16 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.