# Do electrons (in a shell) have to be necessarily entangled?

1. Aug 13, 2013

### San K

Are electrons in the, say lowest energy level, shell, of an atom, always entangled?

i.e. Do electrons (in a shell) have to be necessarily entangled?

Is there any maths (law of conservation of spin/momentum etc) that says so?

Any relation with Pauli's exclusion principle?

2. Aug 13, 2013

### sugeet

Fermions obey the paulis exclusion principle and this causes the electrons in a given shell, say the lowest energy level to have an definite dependence on each other while in the shell or in other words entangled.

If the question is as to why the exclusion principle?? Is it an assumption that we have to make or is there any mathematical proof in relation to say the conservation principles?? Then I understand that the answer to that lies in the Pauli-Luders theorem or the CPT invariance principle.

I do await response, comments and criticisms

3. Aug 13, 2013

### vanhees71

You can classify the hydrogen-atom single-electron energy eigenstates by $|n,l,m,\sigma \rangle$.

No consider, e.g., helium, neglecting the interaction of the electrons among each other. Then the lowest single-electron energy eigenstate is given by $n=1$, $l=0$. Then necessarily $m=0$. Since the electrons are fermions, they must be antisymmetric under exchange. The only way to achieve this is to antisymmetrize the state where one electron has spin up the other spin down, i.e.,
$$|\psi_0 \rangle=\frac{1}{\sqrt{2}} (|n=0,l=0,m=0,\sigma=+1/2 \rangle \otimes |n=0,l=0,m=0,\sigma=-1/2 \rangle - |n=0,l=0,m=0,\sigma=-1/2 \rangle \otimes |n=0,l=0,m=0,\sigma=+1/2 \rangle.$$
Since this is necessarily not a product state, the two electrons are entangled. The only single-particle quantum numbers in this two-particle state are the spins, and thus the spins are entangled. I.e., the single-electron spins are indetermined but have a 100% correlation, i.e., if one measures one electron with $\sigma=+1/2$ the other necessarily must have $\sigma=-1/2$ and vice versa.

The other question concerning the quantum statistics is a bit complicated. One can show from pretty general arguments that in 3 space dimensions there can only be bosons or fermions, i.e., the many-body Hilbert space are spanned by either the completely symmetrized (bosons) or antisymmetrized (fermions) products of an arbitrary complete single-particle-basis.

http://en.wikipedia.org/wiki/Indistinguishable_particles#The_homotopy_class

In systems in 2 spatial dimensions, one can have more general symmetries under exchange of identical particles in a many-body system. Such "exotic statistics states (anyons)" have been observed in condensed-matter physics (e.g., in connection with the fractional quantum Hall effect):

http://en.wikipedia.org/wiki/Anyon

Considering relativistic quantum theory, there is the spin-statistics theorem (Pauli-Lüders theorem), according to which in any relativistic, local, microcausal QFT with a Hamiltonian bounded from below particles with integer spin necessarily are bosons and those with half-integer spin are fermions. Under the same conditions the Hamiltonian is necessarily invariant under the "grand reflection", PCT, i.e., the combined space reflection (parity transformation), exchange of particles with their antiparticles (charge conjugation transformation), and reversal of motion ("time reversal" transformation).

4. Aug 13, 2013

### San K

Thanks Vanhees and sugeet. Great replies.

Interesting information, need time to digest, if I can....:)

Also interesting how dimensions come into the picture.....QM is mind boggling, at times (or most of the time...;))