Question about the exclusion principle

I understand the Pauli exclusion principle like this:

1. For two electrons to occupy the same state their spins must be opposite.
2. If the two electrons are in different states (different spatial wave-functions) their spins are allowed to be parallel if this is otherwise energetically favorable.

The question is this: How different must their wave-functions be for them to allow parallel spins?

Or posing the question differently: If we start to "deform" one of the wave-functions so that it becomes increasingly similar to the other wave-function, at which point will the spins be forced to become anti-parallel? At the point where the wave-functions becomes identical?

I hope this makes sense.. Anyway, thanks in advance!

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this is a discrete symmetry, so "continous parts of the wavefunction" are identical and are not deformed into one-another, it is a discrete symmetry under permutations of particles – Nikos M. Aug 21 '14 at 23:41
The whole state of the $2$ electrons has to be antisymmetric. If the spin part is symmetric, then the spatial part has to be antisymmetric, that is $\psi(x_1,x_2) \sim \psi_1(x_1) \psi_2(x_2) - \psi_2(x_1) \psi_1(x_2)$. – Trimok Aug 22 '14 at 10:20

3 Answers

The wave function isn't something that can be "deformed" in the way that you are thinking. The possible states of an electron in the vicinity of a proton can be found by solving Schrodinger's equation. This gives a discrete set of bound state solutions (energy < 0), labelled by the quantum numbers n and l (and also s, j, etc. once various spin effects have been included that split all the degeneracies in the naive solution). There is also a continuum of positive energy solutions, which correspond to an unbound electron scattered by a nucleus. An actual state of an electron is some superposition of these different energy eigenstates. So, the meaning of a "deformation" of the wave function is actually that the electron is in a superposition of two different states (say two bound states, or a bound state and a free state).

The states of the two electrons in this (presumably helium or a hydrogen ion) atom must be entangled. So, if one electron is in a superposition of two states (i.e. with a "deformed" wave function), its state is something like

$$\alpha \left| n_1, l_1, s_1 \right> + \beta \left| n_2, l_2, s_2 \right>$$

where $\alpha \gg \beta$, so that the electron is predominantly in the first state. This is an incomplete description of the state of the atom, however. Accounting for the other electron, the state is something like

$$\alpha \left| n_1, l_1, s_1 \right> \otimes \left| \text{other electron state 1} \right> + \beta \left| n_2, l_2, s_2 \right> \otimes \left| \text{other electron state 2} \right>$$

Here's the thing to remember: Pauli's exclusion principle applies separately to each entangled term in this superposition. The situation pertinent to your question is where the other electron, in the first state, shares the same orbital as the first electron. By the exclusion principle, it must have opposite spin:

$$\alpha \left| n_1, l_1, \text{up} \right> \otimes \left| n_1, l_1, \text{down} \right> + \beta \left| n_2, l_2, s_2 \right> \otimes \left| \text{other electron state 2} \right>$$

Now, Pauli's exclusion principle would say that if "other electron state 2" is in the orbital n2, l2, then its spin must be the opposite of s2. I don't think this is the scenario you have in mind. I believe you are imagining that the "other electron state 2" is the same as "other electron state 1", i.e. that only the first electron's wave function is in a superposition (i.e. deformed). So, the state of the system is

$$\alpha \left| n_1, l_1, \text{up} \right> \otimes \left| n_1, l_1, \text{down} \right> + \beta \left| n_2, l_2, s_2 \right> \otimes \left| n_1, l_1, \text{down} \right>$$

What does the exclusion principle say in this case? Nothing extra. By definition n1/l1 and n2/l2 are in different orbitals, so the exclusion principle does not apply to the second component in the superposition. s2 could be spin up or spin down.

There is no "threshold" at which the spins are forced to be parallel. There is only the probability

$$\frac{\alpha^2}{\alpha^2 + \beta^2}$$

that, when measured, the system is in the first state and the spins are opposite.

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Think about the two fermions as two traveling wave-packets, more or less overlapping. At which point (where the packets are very similar (and very "overlapping") are the spins forced to become anti-parallel? – Carl Nilsson Aug 21 '14 at 19:40
If the fermions are charged, then they repel/attract and the above applies. If they are electrically neutral (e.g. neutrons) there should still be interactions (strong or weak). I don't think these can be ignored in reality. In theory, what about two non-interacting fermions, travelling as plane waves? Plane wave states represent a continuum, making them tricky. Still, exclusion only applies between exactly equal momentum vectors. The two particle wavefunction must satisfy f(k1,k2) = 0 for k1 = k2. There are no real requirements regarding k1 approaching k2, except continuity. – jwimberley Aug 21 '14 at 22:02

You might want to look into the bonding and anti bonding states of the hydrogen molecule. (scroll down to page 14 here, (just my first hit on a google search) http://www4.ncsu.edu/~franzen/public_html/CH431/lecture/lec_10.pdf

The anti bonding state is with the spins parallel. You can see the energy difference as you bring the two atoms together. At zero temperature the electrons are always going to want to go into the lower energy bonding state. (Even if the two atoms are separated by (say) 10 times the atomic radius.) But at some finite temperature you'll get the atoms sometimes in the anti bonding state.

Does that help?

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Thanks George but I think not after having a look. Let me rephrase the question: Two fermions (lets also imagine they are electrically neutral) are emitted practically simultaneously from an apparatus with virtually the same velocity, direction etc so that their wave-packet "description" are almost the same. How similar can the wave-packets be while still allowing parallel spins? At which "point" of similarity are the spins forced to become anti-parallel? Does this clarify? – Carl Nilsson Aug 21 '14 at 20:13
Not at all. To be in a localized state the particles need to be some how localized. Wrapped up in an atom or quantum well. In free space you can use plane waves states.. they go on forever. – George Herold Aug 22 '14 at 1:44
George, I think traveling wave-packets can be fairly localized? At least for some time.. Sharply localized wave-packets will disperse quickly though. But this will be the same for both particles and should not matter. Actually for this discussion it should not matter whether the particles are in plane wave states or in localized packets(?) According to jwimberleys comment above (if correct) in plane wave states the fermions can have parallel spins unless the wave-vectors are exactly the same.. From this I guess the fermions can have parallel spins even if their states are "nearly" the same. – Carl Nilsson Aug 22 '14 at 13:54
@CarlNilsson, OK I like the hydrogen molecule picture for thinking about this. I think the short answer is wavefunction overlap. If the two electron wavefunctions overlap there will be some energy associated with the parallel spin case. You could also do the case of two 1-D square well potentials with a barrier in between and look at what happens as the barrier is made lower (or thinner). – George Herold Aug 22 '14 at 17:24

The exclusion principle is only a feature of fermions' statistics, not something that can be dynamically forced over a system.

Fermion statistics is mathematically taken into account considering antisymmetric wavefunctions for fermions. The Hilbert space of fermion particles is the antisymmetric Fock space. Let $\mathscr{H}_1$ be the one-particle space and $\mathscr{H}_0=\mathbb{C}$ the vacuum. Then define $$\mathscr{H}_n=\underbrace{\mathscr{H}_1\otimes_a\dotsc\otimes_a \mathscr{H}_1}_n\; ;$$ where $\otimes_a$ is the antisymmetric tensor product. The (antisymmetric) Fock space $\Gamma_a(\mathscr{H}_1)$ over $\mathscr{H}_1$ is defined as $$\Gamma_a(\mathscr{H}_1)=\bigoplus_{n=0}^\infty \mathscr{H}_n\; ,$$ where $\oplus$ stands for the direct sum of Hilbert spaces. $\mathscr{H}_n$ is the $n$-particles subspace.

As an example, here it is the formula for the tensor product of two vectors $\psi$ and $\phi$: $$\psi\otimes_a \phi = (\psi\otimes\phi-\phi\otimes\psi)/2\; .$$ It is then clear that an antisymmetric tensor product of $n$ nonzero vectors $\{\psi_i\}_{i=1}^n$ is zero if and only if $\psi_i=\alpha\psi_j$ for some $i,j\in\{1,\dotsc,n\}$ and $\mathbb{C}\ni\alpha\neq 0$.

The condition $\psi_i=\alpha\psi_j$ is the mathematical equivalent of saying that there are two particles in the same state, and yields zero for the total fermionic wavefunction. Thus, we can say (naïvely) that as long as the state of any fermion particle is different from the others, the wavefunction is not zero.

But the quantum dynamics is implemented by a unitary operator on the Hilbert space, in this case $\Gamma_a(\mathscr{H}_1)$. Let's call this group $(U(t))_{t\in\mathbb{R}}$. Unitary operators preserve the Hilbert space norm, so given any nonzero vector $\Gamma_a(\mathscr{H}_1)\ni \Phi\neq 0$, then $U(t)\Phi\neq 0$ for all $t\in\mathbb{R}$. Therefore there will be no physical way of "deforming" the wavefunction to become zero, or to "force" the spins to move in order to obey the exclusion principle. It is simply a feature of the antisymmetric Fock space, and any meaningful (non-zero) vector will remain non-zero as the dynamics evolve it.

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Thanks! As I understand your answer: As long as there is an infinitesimal difference between the two states the spins can remain parallel? – Carl Nilsson Aug 21 '14 at 23:31
@CarlNilsson It depends what you mean by "infinitesimal difference". Think at finite dimensional vector spaces (for simplicity): let $(v_i)_{i\in\mathbb{N}}$ be a suite of vectors converging to $v$ as $n\to\infty$. You may have, for all $i,j\in\mathbb{N}$, $v_i/\lVert v_i\rVert=v_i/\lVert v_i\rVert=v/\lVert v\rVert$; or $v_i/\lVert v_i\rVert\neq v_i/\lVert v_i\rVert\neq v/\lVert v\rVert$ for all $i,j\in\mathbb{N}$. But in the first case the wavefunction $v_i\otimes_a v=0$ for all $i\in\mathbb{N}$; in the second $v_i\otimes_a v\neq 0$ for all $i\in\mathbb{N}$. – yuggib Aug 22 '14 at 8:44