1
$\begingroup$

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
3
$\begingroup$

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

$\endgroup$
1
  • 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 at 23:44

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.