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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?

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    $\begingroup$ Do you mean can you have something like $11$ electrons in the ground state such that we have spin states $$\sqrt a\mid\uparrow\rangle+\sqrt{1-a}\mid\downarrow\rangle$$ for $a=0,0.1,0,2,\dots,1$? $\endgroup$ – BioPhysicist Apr 5 at 21:54
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    $\begingroup$ Please be much more specific what confuses me. Until now, your question seams to be a duplicate of the link you posted, physics.stackexchange.com/questions/109437/…. $\endgroup$ – Semoi Apr 5 at 22:05
  • $\begingroup$ @AaronStevens yes. $\endgroup$ – roobee Apr 5 at 22:26
  • $\begingroup$ @Semoi I don't know what the ⊗ means. What ∣↑↓⟩ means. What Alt() is. $\endgroup$ – roobee Apr 5 at 22:26
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    $\begingroup$ You should probably not try to understand the notations, but the idea. Pauli's exclusion principle can be stated as: If you take two identical fermions, their composite wave function must be anti-symmetric. This means, if the two particles swap their labels ($1 \leftrightarrow 2$) you obtain the same state, but with a minus sign. E.g. check out $∣↑↓⟩−∣↓↑⟩$. $\endgroup$ – Semoi Apr 5 at 22:41
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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.

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  • $\begingroup$ @PM2Ring : okay....added. $\endgroup$ – WillO Apr 6 at 4:29
  • $\begingroup$ What is projectivize? And how is this related to quantum superposition of states? $\endgroup$ – roobee Apr 6 at 5:04
  • $\begingroup$ @roobee: "Projectivize" means "take account of the fact that if the vector $v$ is a non-zero scalar multiple of the vector $w$, then $v$ and $w$ represent the same quantum state". The vector space addition is quantum superposition. $\endgroup$ – WillO Apr 6 at 5:08
  • $\begingroup$ so the state UD+DU would represent that the system has a 50% chance that particle 1 is spin up and particle 2 is spin down, and a 50% chance that particle 1 is spin down and particle 2 is spin up? $\endgroup$ – roobee Apr 6 at 5:19
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    $\begingroup$ @roobee: I'm not sure what you mean by "represent". Yes, if that is the state then those are the probabilities (assuming you measure in the right directions). But if those are the probabilities, it does not follow that that is the state. UD+iDU is a different state that gives the same probabilities. $\endgroup$ – WillO Apr 6 at 5:37
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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.

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  • $\begingroup$ See my comment on the question for what the OP means by "slightly different superpositions" $\endgroup$ – BioPhysicist Apr 6 at 2:42
  • $\begingroup$ Can you explain or link a source explaining why a linear combinations of products of individual wavefunctions forms a complete set/general solution to the multielectron wavefunction problem? Most sources I find say that you can not express a multielectron wavefunction as a product of wavefunctions when considering electron electron interaction $\endgroup$ – roobee Apr 7 at 17:08
  • $\begingroup$ @roobee see additional bits to my previous answer. $\endgroup$ – ZeroTheHero Apr 7 at 17:16
  • $\begingroup$ What is 𝑃12? It looks like you are just stating, without justifying, that you can express the system wavefunction as a linear combination of products of individual wavefunctions, even when there are electron electron interactions. If you assume that, I believe that the system wavefunction is antisymmetric. But I don't see how you justify that first step. Or where you show that that form represents a general solution to a system with electron electron interactions. $\endgroup$ – roobee Apr 7 at 17:55
  • $\begingroup$ $P_{12}$ is the permutation of particles 1 and 2. Since the single particle states are a complete set - call it ${\cal H}$, the set of two particles states ${\cal H}_1\otimes{\cal H}_2$ is also complete, and the subset of antisymmetric states is complete for the set of all antisymmetric states. This follows from the completeness properties of states: any state can be written as a combination of a complete set of basis states. $\endgroup$ – ZeroTheHero Apr 7 at 18:33
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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.


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

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.

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  • $\begingroup$ Like always, I’d be grateful for any feedback on the downvote. $\endgroup$ – Superfast Jellyfish Apr 6 at 20:13

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