Two Particle System

Question

I've read that if the wavefunction of a combined state can be represented as the tensor product of the individual wavefunctions, the two particles involved are non-entangled.

Now taking a scenario in which the two particles can only occupy two state, i.e. : $$ \rvert\psi_a\rangle=a_1\rvert0\rangle+a_2\rvert1\rangle$$ $$ \rvert\psi_b\rangle=b_1\rvert0\rangle+b_2\rvert1\rangle$$ The combined two particle wavefunction is: $$\begin{align} \rvert\Psi_{ab}\rangle &=\rvert\psi_a\rangle\otimes\rvert\psi_b\rangle \\ &= a_1b_1\rvert 0\rangle\otimes\rvert 0\rangle+a_1b_2\rvert 0\rangle\otimes\rvert 1\rangle+a_2b_1\rvert 1\rangle\otimes\rvert 0\rangle+a_2b_2\rvert 1\rangle\otimes\rvert 1\rangle\\ &=a_1b_2\rvert 0\rangle\otimes\rvert 1\rangle+a_2b_1\rvert 1\rangle\otimes\rvert 0\rangle \\ &= (a_1b_2-a_2b_1)\rvert 0\rangle\otimes\rvert 1\rangle \\ &= (a_1b_2-a_2b_1)\rvert 01\rangle \\ \end{align}$$

So here, evidently the two particles are not allowed to occupy the same state (among other restrictions).

1. Is there a name for this restriction?
2. Is there more physically intuitive content behind this picture?

EDIT: The mathematics done above is missing steps, as elucidated below.

Answer 1

This is precisely the Pauli exclusion principle: two indistinguishable fermions are forbidden from occupying the exact same quantum state. The formulation you give is the more rigorous, grown-up version. The effect is fully quantum mechanical and it cannot really be explained in a particularly intuitive fashion. Fermions cannot occupy the same state because that wavefunction is symmetric under exchange, and fermions - by definition - need an exchange- antisymmetric wavefunction.

Note that this effect, while counterintuitive, is perfectly physical. Among many other things, it is directly responsible for electron degeneracy pressure, which is what enables white dwarf stars to survive.

Answer 2

Your algebra is wrong (or at least very hand-wavy) and I don't think we can directly fix it. Here's how I look at the whole situation, by example.

Fermion antisymmetry, and the effects it has.

Let's just start with Pauli exclusion.

Supposing you have two single-particle states which measure some observable $U$ as $0$ or $1$, and you label these states $|0\rangle$ and $|1\rangle,$ and you try to prepare two electrons in these states: you are constrained by the fact that they are fermions and must be indistinguishable under particle exchange, with a $-$ sign appearing between them. This means that the $|00\rangle$ and $|11\rangle$ states are forbidden as indeed is the $|01\rangle + |10\rangle$ state, leaving only the singlet state $|01\rangle - |10\rangle$. This is "Pauli exclusion;" the fact that they are fermions means that they cannot be in the same state and reduces the 4-dimensional Hilbert space to 1-dimensional.

So this can be viewed as what's happening in the $s$ orbital of an atom; an unpaired electron can either have spin "up" or "down" in this orbital, but two electrons can only enter if they assume opposite spins.

Spin singlet/triplets and the physics of phosphorescence.

Now let's add another $s$ orbital there with some extra energy: the single-particle picture is that there are four orbital states $|g_\downarrow\rangle, |g_\uparrow\rangle, |e_\downarrow\rangle, |e_\uparrow\rangle.$ Now there is some interesting structure between these "ground" and "excited" states. You still have the singlet state $|g_\downarrow g_\uparrow\rangle - |g_\uparrow g_\downarrow\rangle$ in the ground state, and four singly-excited states:$$|g_\downarrow e_\uparrow\rangle - |e_\uparrow g_\downarrow\rangle,\\ |g_\uparrow e_\downarrow\rangle - |e_\downarrow g_\uparrow\rangle,\\ |g_\downarrow e_\downarrow\rangle - |e_\downarrow g_\downarrow\rangle,\\ |g_\uparrow e_\uparrow\rangle - |e_\uparrow g_\uparrow\rangle.\\$$ Now the third state for example can be written as an outer product, $\big(~|ge\rangle - |eg\rangle ~\big)\otimes|\downarrow\downarrow\rangle,$ where we just distribute the first down-arrow over the first symbol and the second over the second symbol. And if your Hamiltonian only depends on position, not on spin, then this will be degenerate with the fourth state, $\big(~|ge\rangle - |eg\rangle ~\big)\otimes|\uparrow\uparrow\rangle.$ But there is one more degenerate state with this particular spatial configuration, and it is given by adding the first two states: $$|g_\downarrow e_\uparrow\rangle - |e_\uparrow g_\downarrow\rangle + |g_\uparrow e_\downarrow\rangle - |e_\downarrow g_\uparrow\rangle = \big(~|ge\rangle - |eg\rangle~\big)~\big(~|\downarrow\uparrow\rangle + |\uparrow\downarrow\rangle~\big)$$Please expand out the right hand side as a sum and then use the above "distribute arrows over symbols" rules to see that these are equivalent! Your equations contain a lot of "magic" but I want to emphasize that notationally there should be no magic, things should just make direct algebraic sense from applying a few rules.

Anyway, these three are called the "spin triplet" of states with that particular spatial distribution; one has spin $+1,$ another has spin $0$, and the last has spin $-1.$ Since there are 4 excited states, you can quickly realize that there is a fourth state lurking around here, and it looks like this: $$ \big(~|ge\rangle + |eg\rangle~\big)~\big(~|\downarrow\uparrow\rangle - |\uparrow\downarrow\rangle~\big).$$ Coincidentally, this is the only electron which has the same spin state as our ground state $|gg\rangle~\big(~|\downarrow\uparrow\rangle - |\uparrow\downarrow\rangle~\big).$ And this is where phosphorescence gets interesting: phosphorescent molecules have this general structure. Usually you excite the phosphorescent molecule into one of the triplet states, and it does not have the right spin configuration to spatially relax to the ground state, so you have to wait a longer time for effects like spin-orbit coupling to kick in, and those can eventually get the phosphorescent molecule into the spin-singlet state, from which it relaxes into the ground state almost immediately by emitting a photon.

There is also a singlet doubly-excited state which is not remarkable; it's just worth mentioning that the state I gave is only allowed, by fermion antisymmetry, to live in the 6-dimensional Hilbert space, not some $4^2 = 16$-dimensional one.

How to reconcile this with the quantum outer-product.

Here's the deal: when you have a 2-qubit system, you're not just adding two electrons into a two-level system: rather, each electron also brings its own two-level system to occupy. Then the positional information of each of these systems is also distinct.

So we might speak of our first qubit as being "1" and our second qubit as being "0". But if these are spin qubits ("up" = 1), those are really spins up or down at two lattice sites, $a$ and $b$. Presumably we've built our lattice so that there is vanishing probability that both electrons occupy the same lattice sites, so we are only interested in studying the case where the first electron is in lattice site $a$ with spin up and the second is in lattice site $b$ with spin down, and the particle-reverse of that: $|10\rangle$ in the computational basis is $|a_\uparrow b_\downarrow\rangle - |b_\downarrow a_\uparrow\rangle$ if we actually look at the electrons.

And the key here is to realize that the arbitrary labeling $a, b$ of lattice sites is the same as the labeling $e,g$ of excited and ground states above: there are 4 such states because the lattice sites are forced to be different while the spin states don't have to be. With $n$ sites there will be $2^n$ basis states; just start by writing $|001\dots\rangle$ as $|a_\downarrow b_\downarrow c_\uparrow\dots\rangle$ and then complete the antisymmetric permutations of that state. But the added detail has no real point.

Then one can effortlessly write in the computational basis that $|001\rangle + |010\rangle + |101\rangle + |110\rangle$ is only entangled between the second and third qubit because it can be written as the outer product $\big(~|0\rangle + |1\rangle~\big)~\big(~|01\rangle + |10\rangle~\big).$ The "qubits" do not depend on which electron precisely happens to be occupying them, only that some electron is, so fermion antisymmetry can be largely neglected.

In a tiny bit of depth: we saw above that the $|01\rangle-|10\rangle$ spin state had an symmetric spatial wavefunction and the $|01\rangle + |10\rangle$ spin state had an antisymmetric one; does this matter? No: suppose we want to transfer between them, we will probably apply a magnetic field to the second qubit mapping $|\downarrow\rangle \mapsto -i|\downarrow\rangle$ and $|\uparrow\rangle \mapsto i|\uparrow\rangle$. In the actual microscopic state $$|01\rangle + |10\rangle = |a_\downarrow b_\uparrow\rangle - |b_\uparrow a_\downarrow\rangle + |a_\uparrow b_\downarrow\rangle - |b_\downarrow a_\uparrow\rangle $$ the resulting state is $$i |a_\downarrow b_\uparrow\rangle - i |b_\uparrow a_\downarrow\rangle - i |a_\uparrow b_\downarrow\rangle + i |b_\downarrow a_\uparrow\rangle = i \big(~|01\rangle - |10\rangle~\big).$$ Since a global phase difference (the prefactor of $i$) doesn't matter in quantum mechanics, this is 100% valid. We didn't have to specially do anything to change spatial symmetry into spatial antisymmetry; the wavefunction took care of that for us.

Put another way, we care about the symmetry/antisymmetry in the phosphorescence case because there's a law of conservation of angular momentum which needed to be violated to change overall spin-states; we don't care about it here precisely because there's no analogous law entering into play, and there never will be.