The answer is no.

Neither the number of bosons nor the number of fermions notr their difference is conserved in nature. This means that these number change all the time.

Should they be equal by coincidence for some time, this would last only for a moment. Even such a coincidence would be extremely unlikely, given the huge number of particles in the universe.

Edit: I had added in the discussion to other answers that due to the infrared problem, the number of each kind of massless particles (in particular that of photons, but possibly also that of one species of neutrinos) is infinite.

This makes the answer to the question somewhat trivial: If there is a massless species of neutrinos (still a possibilty - we now only know that not all neutrinos are massless), the answer would be yes, though not for an interesting reason, as both numbers are countably infinite and hence equal. On the other hand, if all neutrinos are massive, the answer would be no, as there are then infinitely many bosons but only finitely many Fermions.

The real answer to the question is is that counting particles makes no sense; particle number is not a physically relevant observable, and tells nothing interesting about the universe. (Except perhaps an entry in the Guinness Book of Records - but isn't the universe extremal in every respect?)

My intuition tells me that the ground state energy of bosons should
always be lower than the ground state of fermions -- no matter what
kind of interactions or other external properties we've chosen.

I think your intuition is usually correct; but it's possible to define systems where the ground state fermions would have less energy than the ground state bosons. First, a note on why fermions tend to have higher energies, than a note on how one can arrange for the bosons to have more energy than the fermions.

**(1) Assume no particle interactions.**

Under the assumption that the Hamiltonian for the boson and fermion are identical, the energies for single particles (i.e. $N=1$) will also be identical. This follows from the fact that the Schroedinger wave equation applies equally to a boson or fermion. In particular, under this assumption, the ground state energies are identical, call that energy $E_1$.

The simplest assumption for particle interactions is that there is none. In this case, the ground state energy for the boson is easy, it's just $N\; E_1$ as all the bosons are in the same state.

The ground state for the fermion will be a higher energy (due to the Pauli exclusion principle) except in the case that the ground state is $N$-fold degenerate in which case the bosons and fermions will have the same energy.

To some people, the above might be obvious in itself. For others, they might want a little less hand waving and a little more mathematics. So let the $N$ lowest energy eigenstates be $\psi_n(x)$ for $n=1,2,3,...,N$ with no degeneracy so that $H\psi_n = E_n\psi_n$. In the boson case the ground state has all the particles in this state so the wave function is a symmetrization of $\psi(x_1,x_2,...,x_n) = \psi_1(x_1)\psi_1(x_2)...\psi_1(x_n)$. But no matter how the positions are permuted, the energy of this state is $E_1+E_1+...+E_1 = N\;E_1$. Similarly, the energy for any permutation of the fermi ground state $\psi_1(x_1)\psi_2(x_2)...\psi_n(x_n)$ is $E_1+E_2+...E_N > N\;E_1$.

**(1) Assume arbitrary particle interactions.**

With arbitrary particle interactions, it's easy to create a situation where bosons have the same ground state energy as fermions. One simply adds energy to the bose wave functions without adding energy to the fermi states. Let's do this explicitly for a 2-particle wave function. To do this, we need to define the wave functions for $\psi_1 \times \psi_2$, that is, we need to define the tensor product. Let's use the simplest possible wave functions, spinors, and define the combined wave function $\psi_1\otimes \psi_2$ by:

$$|1\rangle=\left(\begin{array}{c}a_1\\b_1\end{array}\right)$$

$$|2\rangle=\left(\begin{array}{c}a_2\\b_2\end{array}\right)$$

$$|1\rangle\otimes |2\rangle=\left(\begin{array}{c}a_1a_2\\b_1a_2\\a_1b_2\\b_1b_2\end{array}\right).$$

Now (ignoring an unimportant factor of $\sqrt{1/2}$ from here on) a fermi symmetrization looks like this:

$$|1\rangle\otimes|2\rangle-|2\rangle\otimes|1\rangle=\left(\begin{array}{c}a_1a_2-a_2a_1\\b_1a_2-b_2a_1\\a_1b_2-a_2b_1\\b_1b_2-b_2b_1\end{array}\right)= \left(\begin{array}{c}0\\b_1a_2-b_2a_1\\a_1b_2-a_2b_1\\0\end{array}\right),$$

while a bose symmetrization looks like this:
$$|1\rangle\otimes|2\rangle+|2\rangle\otimes|1\rangle=\left(\begin{array}{c}a_1a_2+a_2a_1\\b_1a_2+b_2a_1\\a_1b_2+a_2b_1\\b_1b_2+b_2b_1\end{array}\right)= \left(\begin{array}{c}2a_1a_2\\b_1a_2+b_2a_1\\a_1b_2+a_2b_1\\2b_1b_2\end{array}\right).$$

So to add energy only to the bose symmetry case, all we have to do is to include a potential term that looks like:

$$V = \left(\begin{array}{cccc}E&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&E\end{array}\right).$$

Note that the above matrix has eigenvalues of 0 and $E$, but the $E$ eigenvalues are only accessible to a boson, the fermions automatically have an energy 0 for this potential. For a bose state to avoid the energy addition, it needs to have $a_1a_2=b_1b_2=0$. There are two ways to do this; either have $a_1=0$ or $a_2=0$. Since you can't have both $a_1$ and $a_2$ zero, or have both $b_1$ and $b_2$ zero, a little algebra will show that there is only one bose state that avoids the $E$:

$$\left(\begin{array}{c}0\\1\\1\\0\end{array}\right)$$

The above is the bose symmetrization of (1,0) x (0,1). On the other hand, the fermi symmetrization of these two states is:
$$\left(\begin{array}{c}0\\+1\\-1\\0\end{array}\right).$$

So to add an energy to the bose symmetrization but not the fermi symmetrization, use a potential of:

$$V = \left(\begin{array}{cccc}E&0&0&0\\0&E/2&E/2&0\\0&E/2&E/2&0\\0&0&0&E\end{array}\right).$$
The above has three eigenvectors with eigenvalue $E$ and only a single eigenvector with eigenvalue 0; this eigenvalue is accessible only to fermions.

## Best Answer

The real question is "What's the matter?". "Things with mass" is clearly problematic with regards to photons vs Z-bosons. If by matter you mean "stuff that is conserved" [proton decay notwithstanding], then some bosons are matter, and some aren't. If you mean "stuff you can hold", consider a ballon full of Helium (4, the cheap stuff..not 3He)--it's full of bosons.