I guess this is a straightforward question but I was wondering if I can get an explicit steps toward the answer.

Using the Gross-Pitaevskii equation: $$ \tag{1} i \hbar\frac{\partial\psi\left(x,t\right)}{\partial t} =\left(-\frac{\hbar^{2}}{2m} \nabla^{2} + V_{ext} + g \left|\psi\left(x,t\right)\right|^{2} \right) \psi\left(x,t\right)$$ With the variational relation: $$\tag{2} i \hbar\frac{\partial\psi}{\partial t} = \frac{\delta\epsilon}{\delta \psi^{*}}$$ We can find the energy density by equating the right hand side of equation (1) and equation (2).

$$ \tag{3} \frac{\delta\epsilon}{\delta \psi^{*}} = \left(-\frac{\hbar^{2}}{2m} \nabla^{2} + V_{ext} + g \left|\psi\left(x,t\right)\right|^{2} \right) \psi\left(x,t\right)$$

By integrating both sides of equation (3) over $\psi^{*}$ we get:

$$ \tag{4} \epsilon \left[\psi \right] = \left(\frac{\hbar^{2}}{2m} \left|\nabla \psi\right|^{2} + V_{ext} \left|\psi\right|^{2} + \frac{g}{2} \left|\psi\right|^{4} \right)$$

My question is:

What are the explicit steps to get equation (4) from equation (3) ?

In my calculations I have a problem only in getting the factor $\frac{g}{2}$ in the last term in equation (4) and also getting $\epsilon[\psi]$ from $\frac{\delta\epsilon}{\delta \psi^{*}}$.

Thank you!


1 Answer 1


First of all, the form of eq-n (4) is usually just guessed from eq-n (3). It's almost like integration. Let's try to check term-by-term backwards (which will allow us to put the right coefficients).

First term: $$ \frac{\delta}{\delta \psi^\dagger}\left|\nabla \psi\right|^2=\frac{\delta}{\delta \psi^\dagger}\nabla\psi\cdot\nabla\psi^\dagger=\nabla\psi^\dagger\cdot\frac{\delta}{\delta \psi^\dagger}\nabla\psi+\nabla\psi\cdot\frac{\delta}{\delta \psi^\dagger}\nabla\psi^\dagger=0+\nabla\psi\cdot\frac{\delta}{\delta \psi^\dagger}\nabla\psi^\dagger=\\=\left|\text{ integrating by parts & setting surface part to zero }\right|=-\nabla^2\psi $$

Second term: $$ \frac{\delta}{\delta \psi^\dagger}\left|\psi\right|^2=\frac{\delta}{\delta \psi^\dagger}\psi\psi^\dagger=0+\psi=\psi $$

Third term: $$ \frac{\delta}{\delta \psi^\dagger}\left|\psi\right|^4=\frac{\delta}{\delta \psi^\dagger}\psi\psi^\dagger\psi\psi^\dagger=0+\psi\psi\psi^\dagger+0+\psi\psi^\dagger\psi=2\left|\psi\right|^2\psi $$

I used the fact that $\delta\psi/\delta\psi^\dagger=0$. Remember, that $\psi$ and $\psi^\dagger$ are thought independent here.

So the method is pretty much by guessing, but, as you see, it's not difficult to guess.

UPD How's the integration by parts done?

A functional derivative is defined from the variation, so let's try to write the functional variation of $E=\int \epsilon[\psi]\mathrm{d}^3 x$ (first term).

$$ \delta E^{(1)}=\int_\Omega\mathrm{d}^3x\left[\nabla\psi\cdot \delta \nabla\psi^\dagger\right]=\left|\text{ $\delta$ and $\nabla$ can be swapped }\right|=\\ =\int_\Omega\mathrm{d}^3x\left[\nabla\psi\cdot \nabla \delta\psi^\dagger\right]=\left|\text{ integrating by parts }\right|=\\ =\int_S\delta\psi^\dagger\nabla\psi\cdot\mathbf{\mathrm{d}\Gamma}-\int_\Omega\mathrm{d}^3x\left[\nabla^2\psi \delta\psi^\dagger\right] $$

In $\delta E^{(1)}$ the first term is zero, as $\delta\psi^\dagger\biggr\rvert_S=0$. So $\delta E^{(1)}/\delta\psi^\dagger=\int_\Omega \mathrm{d}^3x\left[-\nabla^2\psi\right]$.

  • $\begingroup$ I though we can integrate both sides of equation (3) over $\psi^{*}$ to get equation (4). But nonetheless, this answer my question. Thank you very much. I just have one comment. Can you please describe the integration by parts in the first term? It seems that I have a problem when we have a functional derivative multiplies by ordinary derivative.(i.e. $\frac{\delta}{\delta \psi^{*}} \nabla \psi^{*}$). $\endgroup$
    – MA13
    Commented Aug 21, 2016 at 6:15
  • 1
    $\begingroup$ Added in the UPD. $\endgroup$
    – hayk
    Commented Aug 21, 2016 at 15:36

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