Is every electron in the Universe in a different quantum state?

Question

Is every electron in the Universe in a different quantum state? Is this what the Pauli Exclusion Principle tells us?

Answer 1

The simple answer is yes, but as usual things are rather more complicated than a simple answer would suggest.

For example, if you take a free electron then it isn't quantised but has a continuous energy spectrum. This means you can't usefully talk about two free electrons being in the same quantum state. In principle two free electrons described by infinite plane waves (so perfectly defined momentum but completely delocalised) cannot have exactly the same momentum. However this is a physically unrealistic situation, and in practice two free electrons can have arbitrarily similar momenta.

Alternatively, consider the electrons in two separated hydrogen atoms. In principle the two atoms are described by a single wavefunction regardless of their separation, and the two electrons cannot be in the same state. In practice once the atoms are more than a handful of Bohr radii apart they can be treated as non-interacting and you now cannot usefully compare the states of the two electrons.

Answer 2

Note that usually electrons are somewhat localised, and that localisation is part of the quantum state. The Pauli principle is a consequence of the symmetrization postulate, that states that, for fermionic particles, the wavefunction must be antysymmetric if you exchange the coordinates of any two particles. A common example is that of two electrons in the same atom, with quantum numbers $n, \ell, m_\ell, s$ and $n', \ell', m_\ell', s'$. The antysymmetrized wavefunction is (assuming that the electrons don't interact): $$\Psi(\vec{x}_1, \vec{x}_2) = \frac{1}{\sqrt{2}}\left[\psi_{n\ell m_\ell s}(\vec{x}_1)\psi_{n'\ell' m_\ell' s'}(\vec{x}_2)-\psi_{n\ell m_\ell s}(\vec{x}_2)\psi_{n'\ell' m_\ell' s'}(\vec{x}_1)\right]$$ where $\psi_{n\ell m_\ell s}(\vec{x})$ is the wavefunction for one single electron in the atom. It's easy to check that if you set $n=n'$, $\ell = \ell'$, $m_\ell = m_\ell'$ and $s=s'$, you get $\Psi =0$, so you can't have a wave function for two electrons with the same quantum numbers.

The situation is different if you have electrons in two different atoms, one of them centered at $\vec{x}=0$ and the other centered at $\vec{x} = \vec{x}_0$. The wavefunction for this system is $$\Psi(x_1, x_2) = \frac{1}{\sqrt{2}}\left[\psi_{n\ell m_\ell s}(\vec{x}_1)\psi_{n'\ell' m_\ell' s'}(\vec{x}_2-\vec{x}_0)-\psi_{n\ell m_\ell s}(\vec{x}_2)\psi_{n'\ell' m_\ell' s'}(\vec{x}_1-\vec{x}_0)\right]$$ In general, this wavefunction won't be zero even when all the quantum numbers are the same for both electrons.

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In principle, the Pauli principle applies to all electrons, as they are excitations of the same fermionic field. But in practice, it has little effect (or not effect at all) when you consider two particles that are far away.

Imagine that you have an electron with wavefunction $\psi(x)$ and other with wavefunction $\phi(x)$, and that both electrons aren't interacting. If you could distinguish both electrons, the colective wavefunction would be $$\chi(x_1, x_2) = \psi(x_1) \phi(x_2)$$ but since the electrons are indistinguishable, and the state must be antisymmetric in the coordinates of both particles, the correct wavefunction is $$\Phi(x_1, x_2) = \frac{1}{\sqrt{2}}\left[ \psi(x_1)\phi(x_2) - \psi(x_2)\phi(x_1)\right]$$ The wavefunctions of both electrons are localised in different regions of space $D_\psi$ and $D_\phi$: $$\psi(x) = 0 \textrm{ if }x\not\in D_\psi \qquad\qquad \phi(x)=0 \textrm{ if } x \not\in D_\phi$$ If the distance between the electrons is large, $D_\psi \cap D_\phi = \emptyset$, and therefore their wavefunctions don't overlap.

We have a measurement device $A$ which operates only in region $D_\psi$ and another measurement device $B$ that operates in $D_\phi$. They have eienstates $a(x)$ and $b(x)$, respectively. As the operators are localised, the eigenstates will verify $$a(x) = 0 \textrm{ if }x\not\in D_\psi \qquad\qquad b(x)=0 \textrm{ if } x \not\in D_\phi$$ The simultenous measurement must be symmetricla under the exchange of both particles, so the eigenstate will be $$\frac{1}{\sqrt{2}}\left[a(x_1) b(x_2) - a(x_2) b(x_1) \right]$$ We can calculate the expectation value of this measurement $$\langle AB \rangle = \int \textrm{d}x_1 \textrm{d}x_2 \left[\frac{1}{\sqrt{2}}a^*(x_1) b^*(x_2) - \frac{1}{\sqrt{2}}a^*(x_2) b^*(x_1) \right]\left[\frac{1}{\sqrt{2}}\psi(x_1) \phi(x_2) - \frac{1}{\sqrt{2}}\psi(x_2) \phi(x_1) \right] = \int \textrm{d}x_1 a^*(x_1)\psi(x_1) \int\textrm{d}x_2 b^*(x_2) \phi(x_2)- \int \textrm{d}x_1 a^*(x_1)\phi(x_1)\int \textrm{d}x_2 b^*(x_2)\psi(x_2)$$

The first integral is called "direct" integral, and the second one "exchange" integral. In our setup of electrons and devices, the exchange integral vanishes because $a(x)$ is zero in all the places where $\phi(x)$ is not zero. $$\langle AB\rangle = \int \textrm{d}x_1 a^*(x_1)\psi(x_1) \int\textrm{d}x_2 b^*(x_2) \phi(x_2) $$ This is exactly what we'd obtain if the electrons were distinguishable and we measured $A$ in the $\psi$ electron and $B$ in the $\phi$ electron (i.e., the $\chi$ state). This is true for all observables $A$ and $B$, so if you want to measure an electron in your lab, you only have to "do nothing", $A = id$, with that electron in Pluto. You can simply forget about the electrons whose wavefunctions don't overlap with the electron that you are describing.

That's the important point: is the extent of the wavefunction what determines if the Pauli exclusion principle is important and whether two elecetrons can share their quantum numbers.

References: Chapter XIV (Systems of identical particles) of Quantum Mechanics by Cohen-Tannoudji, Diu & Lalöe

Answer 3

Is every electron in the Universe in a different quantum state? Is this what the Pauli Exclusion Principle tells us?

No, it doesn't tell us that. Read the Wikipedia article:

"The Pauli exclusion principle is the quantum mechanical principle that states that two identical fermions (particles with half-integer spin) cannot occupy the same quantum state simultaneously. In the case of electrons, it can be stated as follows: it is impossible for two electrons of a poly-electron atom to have the same values of the four quantum numbers (n, ℓ, m_ℓ and m_s). For two electrons residing in the same orbital, n, ℓ, and m_ℓ are the same, so m_s must be different and the electrons have opposite spins".

Also read the history. The Pauli exclusion principle tells us that two electrons can't be in the same state in one atom. It's just the subatomic equivalent of two waves can ride over one another but two whirlpools cannot overlap.