2
$\begingroup$

Background

Consider a lattice system described by the Hamiltonian $$ H = - J \sum_{\left\langle ij \right\rangle} \left( \Delta^{\dagger}_i \Delta_j + \mathrm{h.c.} \right) $$ where $\Delta_i$, $\Delta^{\dagger}_i$ destroy and create a particle on site $i$ respectively, and $J$ is the hopping amplitude, here involving nearest neighbours only. Now if $\Delta_i$, $\Delta^{\dagger}_i$ are elementary bosonic operators this hamiltonian describes the Bose-Hubbard model in the non interacting limit, and the ground state if there is no superfluid symmetry breaking is a Bose-Einstein condensate which is obtained by writing the single particle energy levels in momentum space $\varepsilon_k = -2Jcos(k)$ and by massively occupying the lowest state at $k=0$ with all the $N$ particles: $\left|\mathrm{G.S.}\right\rangle = \left( \Delta^{\dagger}_{k=0} \right)^N \left| 0 \right\rangle$, where $\left| 0 \right\rangle$ is the vacuum. To write the ground state in terms of real space Fock states we can use the fact that $\Delta^{\dagger}_{k=0}= \sum_i \Delta^{\dagger}_i$, and we get a linear combination involving many states, including Fock states where more than a particle per site is present.

First question

You can just give a very quick answer here: is it true that the symmetry is not broken, or is it?

Second question

This is the real question: no matter if the symmetry is broken or not, the ground state will always contain Fock states with more than one particle on each lattice site. So, what if now the particles are composite bosons made of two fermions in a singlet spin state? More concretely, if $\Delta^{\dagger}_i = c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}$, where $c^{\dagger}_{i\sigma}$ satisfy fermionic anticommuation rules, you can't have more than one boson per site, due to Pauli exclusion on the fermions! So can you write explicitly the ground state in this case?

My thoughts

I think that the Pauli principle introduces a constraint on the Hilbert space: the constrained Hilbert space only contains Fock states with $0$ or $1$ boson per site. But still I am confused by the fact that essentially these composite particles apparently don't condense at all...

$\endgroup$
1

1 Answer 1

4
$\begingroup$

First question Is the symmetry broken in a BEC state?

This is a subtle point, and probably not really what you want to know. The thing is that the phase and number operators are canonically conjugate. In the usual sense, the symmetry is broken when there is a definite expectation value of the phase (with respect to a standard reference phase), i.e. the phase is the order parameter. However, if we know exactly how many bosons there are in the system, the system is in an eigenstate of the number operator, which means the phase is maximally undetermined.

A more useful question is: is there long-range order in the BEC state? And the answer is yes. (In this context it is usually called off-diagonal long-range order.)

This absence of broken symmetry in the strict sense was and is emphasized particularly by Leggett, see for instance https://doi.org/10.1007/BF01883640. Another good treatment is in the recent book by Tasaki https://doi.org/10.1007/978-3-030-41265-4.

Second question What is the BEC state of composite fermions?

This is the BCS wavefunction

$|BCS\rangle = \prod_k (u_k + v_k \hat{c}^\dagger_{k} \hat{c}^\dagger_{-k}) |\rangle$

where $u_k$ and $v_k$ are complex coefficients satisfying $|u_k|^2 + |v_k|^2 =1$ for each $k$.

At first sight, this seems rather different than just putting all bosons (Cooper pairs) in the same $k=0$ state. But there is another, more useful point of view:

The free bosons of the non-interacting Bose-Hubbard model are maximally delocalized. The Cooper pairs in the BCS state are also as maximally delocalized as possible for fermions, which suffer the Pauli exclusion principle. They all cuddle $k=0$ as much as possible (which is perhaps not so much, the action all happens at the Fermi level which is of the order of 1eV or 10000K).

But the specific $k$-values of the occupied states are not so important. In fact, for many calculations we can regard the BCS state as being an eigenstate of the Cooper pair annihilation operator, i.e.

$\hat{c}_{k} \hat{c}_{-k} |BCS\rangle = |BCS\rangle$ for any $k$.

This implies that one can remove (and therefore add) pairs at will from the condensate. How can this be? Because the number of states with momentum exactly $k$ becomes measure zero. This is just a very good, and useful, approximation. In this approximation, the number of pairs in the BCS state is undetermined, while the phase gets a definite expectation value. This is the broken symmetry state. In superconductivity (as well as superfluids) it makes a lot of sense, since we can pull bosons/pairs from the normal state in the two-fluid model.

Another insight is that the BCS state is a BEC state, macroscopically occupied by bosons $\hat{b}^\dagger$, where

$\hat{b}^\dagger = \sum_k (v_k/u_k) \hat{c}^\dagger_{k} \hat{c}^\dagger_{-k}$

So these bosons do not have a definite momentum eigenvalue, still the state is well specified. See for instance the book by Piers Coleman: https://doi.org/10.1017/CBO9781139020916

Another good reference for this point of view is the book by Annett: https://global.oup.com/academic/product/superconductivity-superfluids-and-condensates-9780198507550

$\endgroup$
5
  • 1
    $\begingroup$ To be slightly picky, I think the BCS wavefunction is only valid for very weak interactions between fermions, which don't bind them into composite bosons. However, it still does show the basic structure of how you can have something that condenses without violating Pauli exclusion. $\endgroup$
    – Rococo
    Feb 24, 2021 at 23:44
  • $\begingroup$ I know this is an old post, but where can I learn more about this approximation of "the BCS state as being an eigenstate of the Cooper pair"? I've asked a question about this some months ago (here), but haven't gotten an answer yet. I would appreciate if you could give your shot on answering it. $\endgroup$ Dec 27, 2021 at 18:43
  • 1
    $\begingroup$ @LucasBaldo Did you read the answers linked by the comment to the question above? You can't prove that the BCS state is an eigenstate of the annihilation operator because it isn't in the strict sense. In the end it comes down to the approximation N ~ N + 1 for very large N. But the approximation of the BCS state as a coherent state is very good and gives you access to all the symmtry-breaking machinery. $\endgroup$ Jan 4, 2022 at 19:27
  • $\begingroup$ Thanks. Yes, I have read those answers, but none of them seem to mention the BCS being an eigenstate. I do understand that that the BCS is an eigenstate only in some approximate sense, not in a strict sense. However, I don't really understand the approximation that needs to be made. In your comment you say this is related to N being very large. However, for the BCS state to have a well defined phase it cannot have a well deifned N, so it doesn't make sense to me to say N is very large. On the other hand, if we consider BCS state of well defined N, then the phase is not well defined. $\endgroup$ Jan 5, 2022 at 1:19
  • $\begingroup$ In other words, I would like to derive this approximate equation mathematically, but I'm currently not able to do so. Do you know sources where I can find this derivation? $\endgroup$ Jan 5, 2022 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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