There are 3 orthogonal bases that describe a spin-1/2 system:

$|S_x \rangle = \pm \hbar/2,|S_y \rangle = \pm \hbar/2,|S_z \rangle = \pm \hbar/2 $.

If any basis vector is expressed as a linear combination of the vectors in another basis, then the each coefficient in that linear combination will have its magnitude equal to $1/\sqrt{2}$.

I read that there are no more orthogonal bases that can be added to the above 3 bases such that all 4 of them mutually satisfy the above property. No proof was given and despite my best efforts I can't work it out. Does the proof involve a physical argument, or is it purely mathematical and the result is due to the fact that field underlying the vector space is a complex field?

  • $\begingroup$ There's an infinite number of orthogonal bases. What you're looking for is known as a set of mutually unbiased bases. $\endgroup$ Commented Sep 10, 2017 at 16:58
  • $\begingroup$ @EmilioPisanty: Thanks! I'll edit the question title accordingly $\endgroup$ Commented Sep 10, 2017 at 17:19
  • $\begingroup$ I believe the proof is purely mathematical but depends on the dimension of the space being $3$, which is the only place that the physics comes in. However trying to work this out in, say, 4 dimensions gives the problem that the angular momentum is now a [0,2]-tensor $\epsilon_{abcd}~r^c~p^d$ and I don't know off the top of my head what that does to all of our arguments regarding the spin operators... ironically the first result I found about this was worldbuilding. $\endgroup$
    – CR Drost
    Commented Sep 10, 2017 at 17:42

1 Answer 1


Up to an overall unitary transformation, there are only three. They are the eigenstates of $\sigma_x$, $\sigma_y$ and $\sigma_z$.

When the dimension $d$ of the space is a prime number (as is $d=2$ here, or if $d=3$) or a power of a prime (v.g. $d=8=2^3$ or $d=9=3^2$), it is known that one can construct complete sets of mutually unbiased bases. In a complete set, there are $d+1$ subsets, each themselves containing $d-1$ commuting operators. Together with the identity, these $d^2-1$ operators can serve as measurements to reconstruct the $d^2-1$ independent entries of the $d\times d$ density matrix.

In your case $d=2$ so there are three sets each containing a single operator. Each member of a set trivially commutes with itself of course.

When the dimension of the space is a composite integer (v.g. $d=6=2\times 3$), the number of MUBs is unknown.

You can read the work of Bill Wootters et al (for instance Wootters, William K., and Brian D. Fields. "Optimal state-determination by mutually unbiased measurements." Annals of Physics 191.2 (1989): 363-381). I believe Wootters actually gave a proof of the statement you're looking for (although maybe in a different paper than the once cited above).

The result is constructive, and all constructions rely on finite fields (in the algebraic sense). The problem is there is no finite field of dimension $6$ (or any other compositive dimension).

Edit: see

  1. Bandyopadhyay, Somshubhro, et al. "A new proof for the existence of mutually unbiased bases." Algorithmica 34.4 (2002): 512-528. (arXiv version here )
  2. Klimov, Andrei B., Luis L. Sánchez-Soto, and Hubert de Guise. "Multicomplementary operators via finite Fourier transform." Journal of Physics A: Mathematical and General 38.12 (2005): 2747. (arXiv version here)

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