How does Pauli's Exclusion Principle relate to a quantum superposition of states?

Question

Pauli's Exclusion principle states 2 fermions can not occupy the same quantum state. However, a particle can occupy a superposition of quantum states. Does this mean you can have an infinite amount of particles occupying a slightly different superposition of states where the superposition of states all have the same two basis states? See comment for an example. This has been answered before here but I don't understand the mathematical notation. I tried searching up bra-ket notation, anti-symmetrizing function, and whatnot, but found it confusing.

Additionally, many answers express the system wavefunction as a linear combination of products of individual wavefunctions. But this neglects particle particle interaction. Does Pauli's Exclusion Principle still apply if you consider particle particle interaction?

Answer 1

The Pauli principle states that the full many-body fermionic state must be antisymmetric (i.e. pick up a minus sign) under permutation of any two fermions. If you have 2 fermions occupying any two states $\psi$ and $\phi$, then the 2-fermion state will be (up to an overall phase and normalization) $$ \psi(1)\phi(2)-\psi(2)\phi(1)\, . $$ This generalizes to a determinant if you have $n$ particles.

There is no infinite number of particles. Usually the states $\phi,\psi$ are orthogonal so it’s not clear what you mean by “slightly different superpositions”. The coefficients of each term in the superposition cannot be varied continuously since $$ a\psi(1)\phi(2)-b\psi(2)\phi(1) $$ is only fully antisymmetric if $a=b$.

Note that the non-interacting wavefunctions form a complete set so that the “true” wavefunction which includes the interaction terms can be expressed as a linear combo of (possibly very many) determinants, each individually fully antisymmetric.

To include interaction term one would start with a set of single particle states $\psi_m$ and construct (in the case of 2 particles) the antisymmetric combinations \begin{align} \psi_{mn}(1,2)=\psi_m(1)\psi_n(2)-\psi_n(1)\psi_m(2) \end{align} All antisymmetric states are of this form so that an 2-fermion state including interaction would be of the type \begin{align} \psi_k(1,2)=\sum_{m,n} c^k_{m,n}\psi_{mn}(1,2) \end{align} with the $c^k_{m,n}$ expansion coefficient of the eigenstate number $k$ of the Hamiltonian with interaction on the set $\psi_{mn}(1,2)$ of non-interacting antisymmetric states.

Note that $$ P_{12}\psi_k(1,2)=\sum_{m,n} c^k_{m,n}P_{12}\psi_{mn}(1,2) =\sum_{m,n} c^k_{m,n}\left(-\psi_{mn}(1,2)\right)=-\psi_k(1,2) $$ as required.

Answer 2

The space of spin states for a single electron is two-dimensional (spanned, say, by UP and DOWN in whatever direction you care to choose).

Therefore (by simple algebra) the space of antisymmetric spin states for a pair of electrons is one-dimensional (spanned by the single vector $\hbox{UP}|\hbox{DOWN}-\hbox{DOWN}|\hbox{UP}$ ).

Here is the simple algebra: The states $UU$, $DD$ and $UD+DU$ are clearly all symmetric and mutually linearly independent. That leaves at most one dimension for the orthogonal complement of the symmetric states (namely the antisymmetric states). Also $UD-DU$ is clearly antisymmetric, so we get at least one dimension.]

Therefore when you projectivize the state space (and account for the Pauli requirement that the state of the ensemble must be antisymmetric), there is only one possible spin state for a pair of electrons.

Further simple algebra shows that the space of antisymmetric spin states for a triple (or more) of electons is zero-dimensional, leaving no possible states at all when you projetivize.

Answer 3

Pauli's Exclusion principle states 2 identical fermions can not occupy the same quantum state. In essence when we have identical particles, the probabilities will be invariant under exchange of particles. In other words, if we have 2 particles one in state $|{\phi}\rangle$ and another in $|{\psi}\rangle$ then we can’t distinguish between this and the state where the first one is in $|{\psi}\rangle$ and the other in $|{\phi}\rangle$. There are two ways we can make this happen: $$|{\phi}\rangle_1\otimes |{\psi}\rangle_2 \pm |{\psi}\rangle_1\otimes |{\phi}\rangle_2$$

Pauli says that fermions are of the negative kind. And this is extendible to many particles, by swapping places two at a time and putting a negative sign. This is easily applicable using Slater determinant.

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Coming to Aaron’s comment, let us consider $N$ particles that are in up eigenstate along slightly different directions.

Let the first spin be along $z$ and say we rotate about $x$ by an angle $-jd\theta$ (clockwise), then the next would be up along $n(j)$ where $j$ is an integer. Then one of the states (base state) would be $|n(0)n(1)n(2)...n(N)\rangle$ which for simplicity we denote by: $$|0123...N\rangle$$ Note that here we have used a place value system.

But since they are fermions, we need to antisymmetrise the state which we do by the means of a slater determinant that we symbolically represent by: $\hat{\mathscr{S}}|0123...N\rangle$

Now if we want to express our base state in terms of up and down along $z$, then we have to relate up along $n(j)$ to up along $z$. This is done simply by the rotation operator $\mathscr{D}(jd\theta,x)$ acting on $|{\uparrow}\rangle$:

$$\mathscr{D}(jd\theta,x)|{\uparrow}\rangle = \cos{\left(\frac{jd\theta}{2}\right)}|{\uparrow}\rangle+ \sin{\left(\frac{jd\theta}{2}\right)} |{\downarrow}\rangle$$

But this representation is complicated in that when we exchange a particular particle, we must ensure to change both the corresponding ups and downs.