What is the Löwenheim number of $L(Q^c)$ Let $L(Q^c)$ be an extended FOL with Chang's quantifier $Q^c$. What is the Löwenheim number of $L(Q^c)$?
 A: I believe I have a partial answer to your question. I had to look up the meanings of "Chang's quantifier" and "Löwenheim number" so before I begin let me state the definitions that I am using. If you are using different definitions, please let me know.
Definition: Chang's quantifier, $Q^c$, is the quantifier where $Q^c x\, \varphi(x)$ holds in a model $\mathcal{M}$ if and only if $\{x \in \mathcal{M} \mid \varphi(x)\}$ has size equal to the size of $\mathcal{M}$.
Definition: The Löwenheim number of a language $L$ is the least $\kappa$ such that for any sentence $\varphi$ in $L$, if $\varphi$ has a model then $\varphi$ has a model of size at most $\kappa$.
Furthermore, I am assuming that $L(Q^c)$ has countably many relation symbols of every arity.

First, we prove a couple easy claims about what can be expressed in $L(Q^c)$.
Claim 1: There is a sentence $\varphi$ in $L(Q^c)$ such that for any cardinal $\kappa$, $\varphi$ has a model of size $\kappa$ if and only if $\kappa$ is a singular cardinal.
Proof: Let $U$ be a unary relation symbol and $R$ be a binary relation symbol.
$$
\varphi := (\lnot\, Q^c x\, U(x)) \land (\forall x\, \lnot\, Q^c y\, R(x, y)) \land (Q^c y\, \exists x\, U(x) \land R(x, y))
$$
Essentially $\varphi$ says that $U$ picks out a set of small size and for each $x$, $R(x, -)$ picks out a set of small size, but the union over $x$ in $U$ of the sets $R(x, -)$ is not small.
Claim 2: There is a sentence $\psi$ in $L(Q^c)$ such that for any cardinal $\kappa$, $\psi$ has a model of size $\kappa$ if and only if for some $\lambda < \kappa$, $2^\lambda \geq \kappa$.
Proof: Let $V$ be a unary relation symbol and $S$ a binary relation symbol.
$$
\psi := (\lnot\, Q^c x\, V(x)) \land (\forall y, z\, (y \neq z) \rightarrow \exists x\, (V(x) \land \lnot\,(S(y,x) \leftrightarrow S(z, x))))
$$
Essentially $\psi$ says that $V$ picks out a set of small size but $S$ gives an injection from the entire model into subsets of the set picked out by $V$. 

Now that we know some things $L(Q^c)$ can express, we can make a few observations. The first claim implies that the Löwenheim number of $L(Q^c)$ is at least $\aleph_\omega$ (the first singular cardinal). The two claims combined imply that the Löwenheim number is at least the first singular cardinal $\kappa$ such that for some $\lambda < \kappa$, $2^\lambda \geq \kappa$, if such a cardinal exists (it won't if GCH holds, for instance).
This second observation means that it is consistent with ZFC that the Löwenheim number is arbitrarily large. The point is that we can use forcing to find a model of ZFC in which GCH holds up until a very large cardinal and then fails dramatically. For instance, if we force over $L$ (where GCH holds) we can make GCH hold below $\aleph_\omega$ but also make sure that $2^{\aleph_{\omega + 1}} > \aleph_{\omega\times 2}$. This gives a model of ZFC in which the Löwenheim number is at least $\aleph_{\omega\times 2}$. But you can use the same idea to do much more dramatic sounding things as well. For instance, if ZFC + "there is an inaccessible cardinal" is consistent then it is consistent that the Löwenheim number is at least the first inaccessible.
So the Löwenheim number can be arbitrarily large and its value seems to depend partly on the behavior of the continuum function. I think the following conjecture is at least superficially plausible:
Conjecture: If GCH holds then the Löwenheim number of $L(Q^c)$ is $\aleph_\omega$.
I don't have much idea how this could be proved, if indeed it's true at all.
Here are some other observations one can make. If $\varphi$ is any sentence in $L(Q^c)$ and it is consistent with ZFC that $\varphi$ has a model of uncountable regular cardinality then it is consistent with ZFC that $\varphi$ has a model of size $\aleph_1$. The idea is just that when $\kappa$ is a regular uncountable cardinal there is a forcing extension in which $\kappa$ is $\aleph_1$ (i.e. all cardinals smaller than $\kappa$ were collapsed to be countable). This forcing preserves truth of sentences in $L(Q^c)$ for models of size $\kappa$. On the other hand, there is a sentence which has a model of size $\aleph_1$ but not $\aleph_0$. This is because you can already express in first order logic that something is infinite (use a function symbol to say that there is a bijection from it to a proper subset) and so in $L(Q^c)$ you can write a sentence expressing that "some subset of this model is infinite but has cardinality smaller than the model."
After having thought briefly about models of regular cardinality, I am ready to make the following irresponsible conjecture:
Conjecture: If a sentence $\varphi$ in the language $L(Q^c)$ has a model of uncountable regular cardinality then for all uncountable regular cardinals $\kappa$, $\varphi$ has a model of size $\kappa$.
Once again, I don't have any great ideas of how to prove this, or even much confidence that it's really true.
