# A confusion: Why are composite bosons possible?

I am not a physicist, but trying to understand the standard model to some extent. My understanding is that the essential property of Bosons and Fermions is that two distinct Bosons can occupy the same quantum state, while two distinct fermions cannot.

I take this to mean the following: Assume for simplicity we have two non-entangled particles (either two bosons or two fermions). It is not an allowable state of the composite quantum system where the two bosons have the same state as separate systems, whereas this is allowed for fermions.

Given this definition, I am confused why it is possible for a composite particle made of fermions to be a boson. (e.g. consider a meson, or a helium 4 atom) Here is my argument why this shouldn't be possible:

1. The quantum state of a composite particle equals the product state of its constituent particles (assuming the constituent particles are non-entangled). i.e. To fully describe the composite particle's quantum state we need to simply describe the states of all its constituent particles.

2. Consider two instances A and B of the same composite boson (e.g. two helium-4 atoms). Since the constituent particles of both A and B are all fermions, they cannot have the same quantum state as each other.

3. But if A and B have the same quantum state, then by 1 the constituent particles must have the same quantum state. By 2 this is not allowed, so A and B cannot be bosons.

I know this is just an informal argument, but where is the misunderstanding in my argument?

• Are you comfortable with the language of second quantization? Commented Apr 15 at 17:34
• @NandagopalManoj, no I haven't heard of it. Commented Apr 16 at 4:30

Consider a boson comprised of two fermions A and B. The boson has spin $$0$$.

One way this can happen is one fermion is spin up and the other is spin down. Another way is A and B are mixtures of spin up and down. It isn't true that constituent A's and B's have to be in the same state for composite boson to be in the same state.

Note that if A and B are identical particles, you can't say that A in one state and B is the other. Just that they are opposite.

• Ok so the problem is in my statement 1? I am a little confused by what you/physicists mean by "state" of a particle then. I was taking state to mean the entire state of the composite particle as a system. Do you only mean the spin state? Commented Apr 16 at 4:33
• Also, presumably there are only a finite number of ways for the boson to have spin 0, so even though it is a boson, there is not the possibility for there to be an arbitrary number of particles that occupy the same state, but instead only a handful? I am assuming it IS possible for an arbitrary number of fundamental bosons such as photons to occupy the same state (e.g. 10^20 of them)? Commented Apr 16 at 4:35
• Two particles are in the same state if all their quantum numbers are the same. Two electrons can occupy the s orbital if their spins are different. More electrons can occupy higher orbitals because they have different orbital angular momenta. If you have $10^{20}$ composite bosons in the same state. that state is presumably spread out over space. This gives lots of room for a pair (or larger group) of constituent fermions to have different states with the same mutual center of mass. Commented Apr 16 at 13:11
• "if all their quantum numbers are the same". What are the quantum numbers of a composite particle, e.g. of a meson? I'm a little confused because I had assumed that e.g. for a free particle, the "state" refers to the entire wave function of the particle, so for an electron it would be the wave function + the spin and so on. Commented Apr 16 at 16:06

This is a great question and something that confused me while studying superconductivity, where we often think of bound states of two electrons as a boson. I think the other answer and comments by mmesser314 captures the gist of the idea, but let me elaborate.

Your flow of logic in points 1,2,3 are perfect. The idea works when your assumption in point 1 is not valid, i. e. the two fermions are not in a product state and are instead entangled.

To explain this, I will introduce the occupation number formalism and the language of second quantization (but very simplified, I will sacrifice precision to get the point across easily).

## Occupation number formalism

Instead of thinking in terms of particles and their wavefunctions, let us think of quantum states in terms of the "orbitals" and whether they are occupied or not. This is useful and natural when we are dealing with identical particles. For simplicity, let us think of an electronic system with $$n$$ orbitals, where each orbital can have both spin-$$\uparrow$$ and spin-$$\downarrow$$ electrons. So the empty quantum state with no electrons would be written as $$\vert \rangle$$ and a state with one spin-$$\downarrow$$ electron in site 3 and one spin-$$\uparrow$$ electron in site 5 would be represented as $$\vert \downarrow_3 \uparrow_5 \rangle$$ and so on. This is the occupation number representation of quantum states with many particles. We say two electrons are entangled if the state is in some superposition $$\vert \downarrow_3 \uparrow_5 \rangle + \vert \downarrow_4 \uparrow_7 \rangle$$ since if we measure a spin-down electron at orbital 4, we gain information about the location and spin of the other electron.

## Creation and annihilation operators

A useful object in this language (sometimes called second quantization) are operations that act on quantum states to create ($$c$$) or remove ($$a$$) particles. Let us define $$c_{\downarrow_7}\vert \downarrow_3 \uparrow_5 \rangle = \vert \downarrow_3 \uparrow_5 \downarrow_7 \rangle \qquad a_{\downarrow_3}\vert \downarrow_3 \uparrow_5 \rangle = \vert \uparrow_5 \rangle$$ Because of Pauli's exclusion principle for fermions, we want them to satisfy $$c_i c_i \vert \psi \rangle = 0 \qquad a_i a_i \vert \psi \rangle = 0$$ for any state $$\vert \psi \rangle$$, as we cannot put two electrons on the same single-particle state $$i$$ (which now includes both orbital and spin), which also means that we cannot remove two electrons from a single-particle state. The right-hand-side being zero (not to be confused with the empty state $$\vert \rangle$$) tells us that this is an invalid operation.

If I put your intuitive idea into this language, this is how you would say it. I want to think of a state with two electrons, say $$\vert \uparrow_1 \downarrow_1\rangle$$ as a boson. In other words, we want to define the boson creation operator $$c_b = c_{\uparrow_1} c_{\downarrow_1}$$ But this does not work! If I try to put two bosons in this state, I end up getting $$0$$ because of Pauli's exclusion principle. $$c_b c_b \vert \rangle = c_{\uparrow_1} c_{\downarrow_1} c_{\uparrow_1} c_{\downarrow_1} \vert \rangle = c_{\uparrow_1} c_{\downarrow_1} \vert \uparrow_1 \downarrow_1\rangle = 0$$ So that is not allowed, and we cannot think of this two-particle state as a boson.

The trick is to put the two electrons in a large superposition. Let us define the two particle state to be $$\vert \uparrow_1 \downarrow_1 \rangle + \vert \uparrow_2 \downarrow_2 \rangle + \vert \uparrow_3 \downarrow_3 \rangle + \dots +\vert \uparrow_n \downarrow_n \rangle,$$ Equivalently, let us define the boson creation operator as $$c_b = c_{\uparrow_1} c_{\downarrow_1} + c_{\uparrow_2} c_{\downarrow_2} + c_{\uparrow_3} c_{\downarrow_3} + \dots + c_{\uparrow_n} c_{\downarrow_n}$$ Now we can think of this as a boson (approximately), because \begin{align} c_b c_b \vert \rangle &= c_b (\vert \uparrow_1 \downarrow_1 \rangle + \vert \uparrow_2 \downarrow_2 \rangle + \vert \uparrow_3 \downarrow_3 \rangle + \dots +\vert \uparrow_n \downarrow_n \rangle ) \\ &= \sum_{i \neq j} \vert \uparrow_{i} \downarrow_{i} \uparrow_{j} \downarrow_{j} \rangle \end{align} In fact, you can convince yourself that you can add up to $$n$$ bosons in this fashion without violating Pauli's exclusion principle! So, as long as your number of "bosons" is much less than $$n$$, it is okay to think of this two particle bound superposition state as approximately a boson!

Summary: The misunderstanding in your argument is that, when the constituent fermions are entangled, the statement "if A and B have the same quantum state, then by 1 the constituent particles must have the same quantum state" is meaningless. You cannot think of the constituent particles being in a definite quantum state if they are entangled.

• Please verify your $c_b = c_{\uparrow_1} c_{\downarrow_1} + c_{\uparrow_2} c_{\downarrow_2} + c_{\uparrow_3} c_{\downarrow_3} + \dots + c_{\uparrow_n} c_{\downarrow_n}$ satisfy the boson relation $[c_b^\dagger, c_b] = 1$. I bet you can't, 'cause it's only "approximately a boson". Commented Apr 16 at 19:28
• Indeed, it will not satisfy that relation since is not a true boson which can have arbitrarily many particles in the same "orbital". But that relation is easily recovered approximately using some commutator identities (you will find it is $= 1 - \sum_{i,\sigma} c^\dagger_{i\sigma} c_{i\sigma} / n$ in standard notation) in the limit where the number of bosons is much smaller than $n$. I did not include that because of OP's background. Commented Apr 16 at 20:59