I need some clarification of what is meant when someone says "fermions cannot occupy the same quantum state". Consider two bosons:
$$\psi(\vec{r_1}, s_1, \vec{r_2}, s_2) = \frac{1}{\sqrt{2}} \left( \phi_A(\vec{r_1}, s_1)\phi_B(\vec{r_2}, s_2) + \phi_A(\vec{r_2}, s_2)\phi_B(\vec{r_1}, s_1) \right)$$
This is one wavefunction of two particles. A wavefunction directly corresponds to a state, and since this is only one wavefunction, it seems there is only one state — that two bosons occupy.
But now consider two fermions:
$$\psi(\vec{r_1}, s_1, \vec{r_2}, s_2) = \frac{1}{\sqrt{2}} \left( \phi_A(\vec{r_1}, s_1)\phi_B(\vec{r_2}, s_2) – \phi_A(\vec{r_2}, s_2)\phi_B(\vec{r_1}, s_1) \right)$$
Again, one wavefunction (=> one state) and two particles that occupy it.
Yeah, $\psi(\vec{r_1}, s_1, \vec{r_2}, s_2) = -\psi(\vec{r_2}, s_2, \vec{r_1}, s_1)$, but it's still just one state — occupied by two fermions.
Could someone clarify?
Best Answer
Great question that exposes some really confusing terminology. This is a rather long answer, and the punchline is basically in the second-to-last paragraph, but I think (hope) it's worthwhile to read the whole answer because I tried to give a somewhat systematic description of fermionic states using a specific, simple example along the way.
Firstly, let's use Dirac notation; it makes things a bit more clear in my opinion. Let's also restrict the initial discussion to the spin states of two spin-$1/2$ particles (which are therefore fermions) so that the Hilbert space for the state of each particle is two-dimensional.
The Hilbert space $\mathcal H_{1/2}$ for a single spin-$1/2$ particle is spanned by the vectors $|+\rangle, |-\rangle$ corresponding to the spin being "up" and "down" respectively. The Hilbert space for the composite system of two distinguishable spin $1/2$ particles is the tensor product $\mathcal H=\mathcal H_{1/2}\otimes\mathcal H_{1/2}$ of the the single spin $1/2$ Hilbert space with istelf. This Hilbert space is four-dimensional and is spanned by the four states \begin{align} |+\rangle|+\rangle, \qquad |+\rangle|-\rangle, \qquad |-\rangle|+\rangle,\qquad |-\rangle|-\rangle \end{align} Every state of the system is some linear combination of these four. Now suppose, instead that the spins are identical, then it turns out that the physical Hilbert space of the system is no longer the tensor product; it is a subspace of the tensor product called the "antisymmetric subspace" which is defined as follows. We define the exchange operator $P$ on $\mathcal H$ as the unique linear operator with the following action on any tensor product basis state \begin{align} P|i\rangle|j\rangle = |j\rangle|i\rangle \end{align} In other words, the exchange operator just exchanges the two factors of any product state. We say that a state $|\psi\rangle$ in the tensor product space is antisymmetric provided \begin{align} P|\psi\rangle = -|\psi\rangle \end{align} The antisymmetric subspace of $\mathcal H$ is then defined as the set of all vectors that are antisymmetric. We then have the following physical fact:
Now let's go back to the spin example to see what this means concretely. An arbitrary state $|\psi\rangle$ of the two spin $1/2$ system can be written as \begin{align} |\psi\rangle = c_{++}|+\rangle|+\rangle + c_{+-}|+\rangle|-\rangle + c_{-+}|-\rangle|+\rangle + c_{--}|-\rangle|-\rangle \end{align} The exchance operator acting on this state gives \begin{align} P|\psi\rangle = c_{++}|+\rangle|+\rangle + c_{+-}|-\rangle|+\rangle + c_{-+}|+\rangle|-\rangle + c_{--}|-\rangle|-\rangle \end{align} but for identical fermions, the state must be antisymmetric, and this implies constraints on the coefficients \begin{align} c_{++} = 0, \qquad c_{--} = 0, \qquad c_{-+} = -c_{+-} \end{align} so the most general (normalized) fermionic state for the system is \begin{align} |\psi\rangle = \frac{1}{\sqrt{2}}(|+\rangle|-\rangle - |-\rangle|+\rangle) \end{align} When we say that the particles cannot occupy the same state, this is just another way of pointing out in this case that the coefficients of the states $|+\rangle|+\rangle$ and $|-\rangle|-\rangle$ must vanish; these are states in which either both spins are "up" or both are "down".
In particular, you say
Well certainly that's true since the (pure) state of any quantum mechanical system must be some vector in some Hilbert space. The above example shows, however, that the "same state" terminology can be thought of in terms of the two tensor factors in the Hilbert space; namely the product basis vectors in which the single-particle states of both particles are the same should be excluded from the Hilbert space.
Note: I have concentrated on low dimensional examples, the analysis goes through analogously for Hilbert spaces of any dimension; the Fermionic state are always just those in the antisymmetric subspace, so any product basis vectors in which both factors are the same should be excluded from the Hilbert space basis; such vectors do not live in the antisymmetric subspace.