As per Pauli exclusion followed by fermions, the state of the two fermions in their Cooper pair has to be anti-symmetric for their exchange. If electrons that form Cooper pairs come from a single orbital state of L=0 (i.e. S state) then the electrons in the pair have to have anti-parallel spin. Naturally, the net spin would be nothing but s=0. However, if two fermions forming Copper pair have non-zero orbital angular momentum say L=1 (i.e. P state) as we find in case of Helium-3 atoms in superfluid state of liquid Helium-3, they can have different values of $m_L$ quantum number and the spin of two fermions can be identical rendering net spin of the pair to be s =1. As concluded by Leggett, there can be three possible magnetic state of such a system which is found to be true in case of superfluid Helium-3.

Yes, it's is possible.

For example, you can look at references below for an "historical" paper and a recent review :

http://www.nature.com/nature/journal/v396/n6712/abs/396658a0.html

http://arxiv.org/abs/1112.1620

Please tell us if you need more details.

Regards.

Standard cooper parirs have S== and do not couple to S=!

Yes, Yatendra's answer is correct!

Thank your very much for your answers, I also checked the provided references. I see now, that these triplett state exist in reality.

However, I feel, that the three question aren't fully answered. Therefore, I try to sum up from your answers:

Q1: Is there anything preventing them from forming a triplet state S=1 (the spins are parallel)?

--> I guess it is the symmetry as Yatendra Jain said (see next question). But why aren't there L=1 states in standard coopers to allow S=1?

Q2: Is it related to symmetry considerations?

--> Is it parity? P = P_S * P_L = (-1)^S * (-1)^L = 1, if both L,S = 0 or both L,S = 1. If L and S aren't the same, then the parity = -1.

Q3: What would be the consequences of a S=1 state?

--> Since magnetism is microscopically related to spins, isn't there any interaction of the perfect diamagnetism of superconductor and the possible magnetism from S=1 state?

To the sub question of Q1 of Magnus Schlosser:

For l_1 = 1 and l_2 =1, possible m_l are 1, 0 and -1.

Since there are two electrons; their states with total L can correspond to L = 2, 1, 0. (which means the states of the pair can be D, P or S states.

Since the net spin of the pair can be S=1 or S=0, hence if Pauli exclusion was not be followed, the pssible states could have been triplet D, triplet P triplet S, singlet D, singlet P, and singlet S. But the Pauli exclusion allows only singlet D, singlet S and triplet P. This can be followed from the following points.

L=2 is related to a state where both electrons have m_l = 1 (with total sub states with M_L = 2, 1, 0, -1 and -2. Naturally since such a pair of two electrons has identical L, identical m_l they have to have opposite spin 1/2 and -1/2 for Pauli exclusion or for the total state to be anti-symmetric. Thus one possibility is that they have siglet D state as a pair. They can not have triplet D state because it will violate exclusion principle.

Similarly if two electrons have identical m_l = 0, they can correspond to L = 0 and once again the two electrons have identical l_1 = l_2 = 1, and m_l = 0, they have to have opposite spin rendering S=0; the possible allowed state is singlet S.

However, when the two electrons have different m_l e.g m_l=1 and m_l = 0 rendering total M_L = 1 which can come from total L = 1; now two electrons can have parallel spin which gives us a state of L = 1 and S=1 i.e. triplet P. we can not choose anti-parallel spin because orbital state of the electrons have different sets of quantum numers indicating that the orbital state is antisymmetric which can be allowed to combine antisymmetric spin state as this makes total state symmetric.

This is the reason that if pair has anti-parallel spin its net state can be either singlet D and singlet S (but not singlet P)

If they have parallel spin, the net state of the pair can be only triplet P (not triplet D or triplet S).

To Q.2

I think Pauli exclusion explains the situation.

To Q.3

Such a pair can be only have triplet P state.

Its consequences have been worked out by Leggett and it has been discussed elegantly in C. Enss and S. Hunklinger, Low Temperature Physics,

Springer Verlag, Berlin, (2005) in relation to the possible phases of superfluid Helium-3.

yes Cooper pairs can have S=1 for heavy Fermion superconductors. The superconductors with S=1 are called triplet superconductors as their energy gap has p-wave nature

The case orf Cooper pairs with parallel spin is very nicely discussed in the book G. Rickayzen, Theory of Superconductivity, Monographs and Texts in Physics and Astronomy, Volume XIV, Interscience Publishers (John Wiley), 1965 (New York). See Chapter 4, Section 10.

Sr2RuO4 is also a p-wave superconductor, I believe.

If you are looking for a simple explanation as to why nature would prefer an S=0 pairing state, then I was always partial to an arguement Tinkham gave in his Intro to Superconductivity book, "Anticipating an attractive interaction, we expect the singlet coupling to have a lower energy because the cosinudial dependence of its orbital wavefuntion on (r1 - r2) gives a probility amplitude for the electrons to be near each other."

This is very similar to the argument given for why two electrons sharing a covalent bond occupy the singlet state. ie: the expectation distance of two electrons with a symmetric spacial wavefunction (which requires a singlet pairing of spins) is less than that of an antisymetric spacial wavefunction (which requires a triplet pairing.)

Two electrons can not occupy a common pont in the position space (i.e., two electrons can not have r1 = r2) because; (i) their electrostatic energy becomes +ve infinity at r1=r2 and also because (ii) they have hard core structure like any other material particle.

If I am right in having this understanding (if I am wrong, could any abody let me know the easons), then I have serious difficiluty in accepting cosinudial dependennce of the oribital wave function for a pair of electrons because this function assumes its maximum possible value at r1=r2, while above mentioned physical reality demands that the pair wave function should vanish at r1=r2. In what follows one can use sin[|(k2-k1)(r2-r1)|/2] = sin[|q(r2-r1)|/2] (with k=2q) as the orbital function of two electrons because it not only has +ve symmetry but also vanishes at r2=r1.

I believe that nature does not play dice. Two electrons are, therefore, expected to satisfy r2=r1 even when they make a covelent bond. I would appreciate if some body can list any established reason(s) for their occupying common point (r1=r2) in a covalent bond. This will greatly help persons like me in having a correct understanding of a state of pair of electrons.

In respect to the above mentioned observation I may also add that the inconsistency of the Cosinudial nature of orbital wave function (+ve symmetry) of a pair of electrons with physical reality that the electrons can not have r1=r2 rebders unnecessary complications of energy divergence etc which we can easily avoid. There have been numerous theories of superconductivity (perhaps as many as the authors who tried to work them out) and it is obvious that only one of these could be correct. We find that theories such as BCS theory have intrisic problems and they can not reveal complete, clear and physically consistent understanding of the phenomenon as demonstrated in. http://www.scribd.com/doc/110441679/Intrinsic-Problems-Superfluid-Theories

I would appreciate if some scientists can help me understand why should they not consider an alternative theory developed by using first quantization approach.

http://www.scribd.com/doc/110681115/First-Quantization-Theory-of-Superconductivity

Please come forward in the interest of establishing the truth.

Wave function of a Cooper pair with a node at r=0 (r1=r2) would have a higher kinetic energy compared to a smoother wave function with a maximum at r=0. It is true that the latter (say, with an orbital symmetry S) would have a higher expectation value for the (repulsive) potential energy compared to the former (say, with an orbital symmetry P). If the increase in the expectation value of the potential energy is not sufficiently high (especially, since the repulsion is only a soft screened Coulomb), wave function without a node at r=0 will still win. If this increase in the potential energy is too high (as it happens with He3 superfluid because of a hard core repulsion re-12), the p-wave pairing state with a node at r=0 is stabilized. Once the pairing is P-wave, it is necessarily in the spin triplet channel because of the requirement of overall antisymmetry of the total wavefunction (including the orbital and spin parts).

Once we ascribe a two particle wavefunction to a Cooper pair it is difficult to discuss whether two electrons constituting the Cooper pair are at the same spatial point (r1=r2) etc; we can only discuss the expectation values of potential energy and kinetic energy operators in the wavefunctions with a orbital symmetry S (nodeless) or orbital symmetry P (with a node at r=0).

See article PHYSICAL REVIEW B 98, 104513 (2018)

In addition to all interesting answers in this thread Prof. Magnus Schlösser .

I will add to this interesting thread, 2 classical references for triplet (S=1) superconductivity, also called unconventional superconductivity:

• L. Gorkov: Superconductivity in heavy fermion systems. Sov. Phys. Rev. A Phys 9, pp. 1-116, 1987.
• M. Sigrist & K. Ueda: Phenomenological theory of unconventional superconductivity. Rev. Mod. Phys. 63, pp. 239, 1991.

Some heavy fermions crystals & A-phase of 3He isotope are triplet paired. Also the strontium ruthenate crystal as already was mentioned in a previous post.