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?


2 Answers 2


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:

For a system consisting of two identical fermions, the state of the system must reside in the antisymmetric subspace of the tensor product of the single-particle Hilbert spaces.

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

but it's still just one state -- occupied by two fermions.

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.

  • $\begingroup$ One important thing is missing: the exchange operator is applicable if and only if particles are identical. $\endgroup$ Aug 21, 2014 at 16:41
  • $\begingroup$ @IncnisMrsi I'm not sure what you mean by "is applicable," but it is not true that the exchange operator is "only defined if the particles are identical." For example, if the two particles are both of spin $1/2$, then the spin Hilbert space will be (a subpspace of) the tensor product $\mathcal H_{1/2}\otimes\mathcal H_{1/2}$, and the exchange operator can be defined on that Hilbert space regardless of whether or not the particles are identical. The particles being identical means that the spin state must be an eigenstate of the exchange operator with appropriate eigenvalue $(\pm 1)$. $\endgroup$ Aug 21, 2014 at 18:38
  • $\begingroup$ Sure, I understand the difference between “can be defined” and “should be used to extract the −1 eigenspace”. That’s why I said what I said. $\endgroup$ Aug 21, 2014 at 18:45
  • $\begingroup$ @IncnisMrsi In that case, I'm not sure what part of the response is missing. I explicitly address the role of the exchange operator in determining appropriate states of identical particles. $\endgroup$ Aug 21, 2014 at 18:47
  • $\begingroup$ Only the small silly thing that if two fermions are not identical, then nothing hinders them to have the same wavefunction. In other words “the same quantum state” is not only about wavefunction, but necessarily about identity. $\endgroup$ Aug 21, 2014 at 18:56

The idea is actually simple. However, most book usually use sloppy terms, or they have not give explicit discussion on this issue, so it usually confused students. The correct phrase should be:

Individual fermion in a system cannot have the same single particle wavefunction

It is clear that the whole system itself always described by a total wavefunction $\Psi$. However, if the particles are not interacting, we can solve each individual particle wavefunction $\psi$ separately and construct the total wavefunction as:

$$\Psi(r_1,r_2,...,r_n) \propto \prod_{\sigma_i} \sigma_i \psi(r_i)$$

where $\sigma_i$ are all possible sign permutation for fermion and boson. The symmetrization and anti-symmetrization is the direct results from the indistinguishably of particles.

So why we usually discuss about the single particle wavefunction $\psi$ rather than the total wavefunction $\Psi$? Though it is possible to measure the total wavefunction, however, each individual particle is actually the smallest measurable subsystem (corresponding to partial trace). When we treat each particle separately, interesting phenomenon appears such as entanglement.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.