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An electron can be either spin-up $|\!\uparrow\,\rangle$ or spin-down $|\!\downarrow\,\rangle$.

Moreover, it can be in a superposition of the two states, i.e. $\alpha \,|\!\uparrow\,\rangle + \beta \,|\!\downarrow\,\rangle$, where $\alpha, \beta \in \mathbb C$ and $\alpha\alpha^*=|\alpha|^2$ is the probability of measuring the electron in the spin-up state, while $\beta\beta^* = |\beta|^2$ is the probability of measuring the electron in the spin-down state.

The state space is a two-dimensional, complex vector space with basis vectors $|\!\uparrow\,\rangle$ and $|\!\downarrow\,\rangle$.

Given two electrons, the state space is given by $H_1 \otimes H_2$ where $H_1$ is the state space of the first electron and $H_2$ is the state space of the second electron. This is a four dimensional, complex vector space with basis vectors $|\!\uparrow\,\rangle \otimes |\!\uparrow\,\rangle$, $|\!\downarrow\,\rangle \otimes |\!\uparrow\,\rangle$, $|\!\uparrow\,\rangle\otimes |\!\downarrow\,\rangle$ and $|\!\downarrow\,\rangle \otimes |\!\downarrow\,\rangle$.

These basis vectors are usually written as $|\!\uparrow\uparrow\rangle$, $|\!\downarrow\uparrow\rangle$, $|\!\uparrow\downarrow\rangle$ and $|\!\downarrow\downarrow\rangle$ respectively. Meaning $$H_1 \otimes H_2 = \{\alpha|\!\uparrow\uparrow\rangle +\beta|\!\downarrow\uparrow\rangle +\gamma |\!\uparrow\downarrow\rangle +\delta|\!\downarrow\downarrow\rangle:\alpha,\beta,\gamma,\delta \in \mathbb C\}$$

Why is the basis usually taken to be $$|\!\uparrow\uparrow\rangle, \ \ \tfrac{1}{\sqrt 2}\left(|\!\uparrow\downarrow\rangle+|\!\downarrow\uparrow\rangle\right) , \ \ \tfrac{1}{\sqrt 2}\left(|\!\uparrow\downarrow\rangle-|\!\downarrow\uparrow\rangle\right), \ \ |\!\downarrow\downarrow\rangle \tag{1}$$ I see that these are orthonormal with repsect to the Hermitian inner product, but aren't $$|\!\uparrow\uparrow\rangle, \ \ |\!\downarrow\uparrow\rangle, \ \ |\!\uparrow\downarrow\rangle, \ \ |\!\downarrow\downarrow\rangle \tag{2}$$

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    $\begingroup$ These last four basis functions are not eigenfunctions of the Hamiltonian since $\hat{H}$ commutes with $\hat{P}$ where $\hat{P}$ is the permutation operator. Hence eigenfunctions of $\hat{H}$ should be symmetric or antisymmetric under particle exchange. $\endgroup$
    – Paul
    Mar 21, 2017 at 21:45
  • $\begingroup$ @Paul Thanks for getting back to me so quickly. I was just looking at it from a linear algebra point of view, so it seems I'm missing a lot. Must the basis always comprise eigenvectors of the Hamiltonian? Why? How does one compute the Hamiltonian? Can you recommend a (relatively) straight-forward, basic introduction of where the Hamiltonians come from and how they are used? (I realise that this may be impossible). $\endgroup$
    – Fly by Night
    Mar 21, 2017 at 21:49
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    $\begingroup$ As long as your basis is complete, it does not have to consist of the true eigenfunctions of your Hamiltonian. Typically, using the eigenfunctions as basis makes it easier to diagonalize your matrix (that is, to solve the Schrödinger equation). Concerning your request for literature, it is difficult to recommend a good text as this is very personal and depends a lot on what you already know. Griffiths is a standard introductory textbook or from a more chemical perspective the text by Pauling (which is available for free under archive.org/details/introductiontoqu031712mbp). $\endgroup$
    – Paul
    Mar 21, 2017 at 21:58
  • $\begingroup$ @Paul Thanks again for being so helpful. I'm a mathematician by trade, with very little knowledge of modern Physics. I'm interested in cryptography, and wanted to understand quantum computers - qubits, etc. I'll have a look at the reference, and give it my best shot. Thanks again! $\endgroup$
    – Fly by Night
    Mar 22, 2017 at 17:20

1 Answer 1

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The first, second and last states in (1) are actually a basis for the triplet of $S=1$ states, while the third in (1) is the singlet $S=0$ state.

As noted by @Paul, the triplet states are also symmetric under permutation of the spins, while the singlet is antisymmetric.

In general, Schur-Weyl duality is a deep mathematical result that allows the construction of states in the $n$-fold tensor product $H_1\otimes H_2\ldots \otimes H_n$ so they also transform by irreducible representations of the permutation group $S_n$.

In the case of two spin-$1/2$ particle as you have here, the appropriate permutation group is $S_2$. The two irreducible representations of $S_2$ are the fully symmetric and fully antisymmetric (or alternating) representations.

If you had $3$ spin-$1/2$ particles, you could arrange them in irreps of $S_3$: one fully symmetric irrep containing the $S=3/2$ states, and two irreps of mixed symmetric each containing $S=1/2$ states.

This decomposition does NOT depend on the Hamiltonian.

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  • $\begingroup$ Thanks for your answer ZtH. I'm a mathematician and so I know a little about representations of finite groups, but I have very little knowledge of modern physics. Please could you explain more about how symmetry groups and their representations appear in QM? Does it always have to be symmetry groups? Can the representation of, say, a diherdral group turn up? $\endgroup$
    – Fly by Night
    Mar 22, 2017 at 17:23
  • $\begingroup$ @FlybyNight This is a very broad request. There must be other similar questions on Physics SE. Are you comfortable with annotated links? $\endgroup$
    – ZeroTheHero
    Mar 22, 2017 at 17:43
  • $\begingroup$ Sorry for asking poor questions. Fire away with whatever links you think would help. Thanks again for your patience. $\endgroup$
    – Fly by Night
    Mar 22, 2017 at 19:28
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    $\begingroup$ @FlybyNight see physics.stackexchange.com/questions/6108/… for some ideas. Finite groups like the dihedral group often occur in solid state physics because of the connection with crystals: they also occur because of the symmetries if the weight lattice in Lie theory. Otherwise continuous groups occur a lot because physics is often phrased in terms of differential equations so some kind of continuity of the solutions is usually assumed. $\endgroup$
    – ZeroTheHero
    Mar 23, 2017 at 3:25

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