I am reading through the derivation of the Gross-Pitaevskii equation in the Heisenberg picture, and I am having some trouble interpreting the following identificiation. In the derivation the field operator $\psi(r)$ is defined as the position space equivalent of the creation/annihilation operator $\hat{a}_k$. $$\hat{\psi}(r,t) = \sum_k \phi_k(r, t) \hat{a}_k.$$ Here $\phi_k(r,t)$ is a single particle state of momentum mode $k$. The GP equation is now defined in terms of the following c-number. $$\psi_0(r,t) = \langle \hat{\psi}(r,t)\rangle.$$ The derivation I understand, but what confused me is that this complex function $\psi_0(r,t)$ is identified as the "wavefunction of the condensate". In other words the absolute value squared of this function gives the local particle density, and can thus be used to examine the spatial profile of the BEC.

I cannot seem to find a satisfactory explanation for this identification. Surely the density is actually related to the expectation value of the number operator, e.g. $\langle\psi^{\dagger}\psi\rangle$? Also I don't understand how this is even a wavefunction at all, what does this function have to do with probability of position?

Any help would be greatly appreciated!


I think that the answer relies on the fact that to get the Gross Pitaevski equation (GPE) you treat the field operator $\hat{\Psi}(\bf{r},t)$ in a mean field way. The mean field approximation consists in writing the field operator as a non operatorial part plus operatorial fluctuations on top of it: $$ \hat{\Psi}(\bf{r},t) = \left\langle \hat{\Psi}(\bf{r},t) \right\rangle + \delta \hat{\Psi}(\bf{r},t) \equiv \Phi(\bf{r},t) + \delta \hat{\Psi}(\bf{r},t). $$ Now to get GPE what you have to do is neglect the fluctuating operatorial part and keep the mean field $\Phi$: $$ \hat{\Psi}(\bf{r},t) \simeq \Phi(\bf{r},t). $$

As you said the natural density of condensate would be $n(\bf{r},t) = \left\langle \hat{\Psi}^{\dagger}\hat{\Psi } \right\rangle $, but what is this quantity in our approximation? Well, substituting $\hat{\Psi}$ with the function $\Phi$, this is $$ n(\bf{r},t) = \left\langle \Phi^{*}(\bf{r},t)\Phi(\bf{r},t) \right\rangle = \Phi^{*}(\bf{r},t)\Phi(\bf{r},t) = |\Phi(\bf{r},t)|^2, $$ where we get rid of the expectation value since the argument is a function, not an operator. From this equation you clearly see that $\Phi(\bf{r},t)$ can be interpreted as the wave function of the condensate, since its square modulus naturally gives the number density.

Hopefully this answers your question. If not please let me know! Sorry for changing notation $\psi_0 \to \Phi$, but too many $\psi$s around confuse me!


You don't seem to be considering interactions here, so you don't even need the GPE. This is just about Bose-Einstein condensation (non-interacting phenomenon).

Without interactions, number operators are defined and the state is a Fock state:

$$ |\Psi\rangle = \left ( \prod_{j=1}^\nu \frac{1}{\sqrt{N_j!} }(a_j^\dagger)^{N_j} \right ) |0\rangle, $$ where $|0\rangle$ is the vacuum.

For BECs, you generally look at the ground state.
The expectation value of the number operator in the ground state is: $$ \langle n \rangle = \langle \psi^\dagger\psi\rangle = \langle 0| \psi^\dagger \psi |0\rangle \propto \sum_k \langle 0|a^\dagger_k a_k|0\rangle = 0, $$

because $a_k |0\rangle =0$. Just like in quantum mechanics (first quantisation). So the number operator is quite useless for the ground state, i.e. for a BEC.


You know that a BEC breaks the $U(1)$ symmetry corresponding to particle conservation. Hence,

  • for $T>T_{\mathrm{c}}$, the number operator makes sense and you can build a Fock state like above. Hence, $\langle \hat\psi \rangle = \langle 0 | \hat\psi|0 \rangle \propto a_0 |0\rangle = 0.$

  • for $T<T_{\mathrm{c}}$, the number operator is not useful any longer as you've broken $U(1)$ and hence particle conservation -- you went from a complex field to a real field, $\hat\psi \in \mathbb{C} \rightarrow \hat\psi \in \mathbb{R}$.
    Hence: $$\langle \hat\psi \rangle = \langle 0 | \hat\psi|0 \rangle \neq 0,$$ which is one of the definitions for the BEC phase.

In fact, the BEC is actually a coherent state of the $|\mathbf{k}=0\rangle$ mode: $$ |BEC\rangle = \mathrm{e}^{-N/2} \mathrm{e}^{\sqrt{N}a_0^\dagger} |0\rangle, $$ where $|0\rangle$ is, again, the vacuum, and $N$ is the expectation value of the number operator, since the state is not a Fock state and hence does not have a fixed eigenvalue for $N$. In fact, you know that a coherent state has a fixed phase $\theta$, and the coherent state $\leftrightarrow$ Fock state correspond to the extremes of the commutation relation $[\hat N, \hat \theta] \propto \mathrm{i} \hbar$ (with an associated uncertainty principle).

Ok, your ground state is now $|BEC\rangle$. So let's use this for the averages from earlier:

  • $\langle BEC | \hat \psi \psi | BEC \rangle = \langle n \rangle = N$ (mean number of atoms),

  • $\sigma_n = ... = \sqrt{N}$ (standard deviation of atom number: non-zero because it is not a Fock state,

  • $\langle BEC | \hat \psi(\mathbf{r},t) | BEC \rangle = \langle BEC| \sum_k \phi_k(\mathbf{r},t) \hat{a}_k |BEC\rangle = \phi_0(\mathbf{r},t)\langle BEC | BEC \rangle = \phi_0(\mathbf{r},t)$, because a coherent state $|BEC\rangle$ is an eigenstate of the annihilation operator $a_k$, and the BEC is only built of $a_0$.

Now since $\psi^\dagger$ creates an excitation of the field $\psi$ at some position and some time, $\phi(\mathbf{r},t)$ is basically the spatial profile of the excitation.

This whole thing was done in second quantisation, so it's a bit weird to jump to first quantisation and call this "a wavefunction". It still has a "normalisation" interpretation in that the state is a 100% pure BEC, so you have a 100% probability of finding anything at $|\mathbf{k}=0\rangle$. However, the concept of "local particle density" is a bit ill-defined since this is not a Fock state and particle number $N$ has an uncertainty $\sqrt{N}$. Of course, for large number of particles $\sqrt{N}/N$ becomes negligible so you can just treat it as a "wavefunction" and not care about the nuanced meaning of particle number.

  • $\begingroup$ This answer actually cleared up a lot of other small points of confusion I was having, thanks a lot! $\endgroup$ – Jasper Jun 16 '20 at 8:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.