# Find a set of vectors on the Bloch sphere such that $\langle \psi_i | \psi_j \rangle = \frac{1}{\sqrt{n}}$

How can I find a set of multiple vectors on the block sphere which satisfies

$$\langle \psi_i | \psi_j \rangle = \frac{1}{\sqrt{n}}$$

where $$n$$ is any natural number greater than $$2$$?

I think I have to do rotations on the Bloch sphere. But how exactly

• Is this in the context of en.wikipedia.org/wiki/Mutually_unbiased_bases? en.wikipedia.org/wiki/SIC-POVM? Commented Apr 27, 2022 at 13:58
• I think you'll have trouble going higher than $n=3$ if you're constrained to the Bloch sphere (dimension 2) Commented Apr 27, 2022 at 14:04
• No. But I think it might be helpful. Thanks Commented Apr 27, 2022 at 14:05

You do this iteratively. Start with $$|\psi_1\rangle=|0\rangle.$$ Then, one component of $$|\psi_2\rangle$$ is fixed to get the correct inner product, so add one more component with an amplitude set to get the length of the vector correct: $$|\psi_2\rangle=\frac{1}{\sqrt{n}}|0\rangle+\sqrt{\frac{n-1}{n}}|1\rangle.$$ Now for $$|\psi_3\rangle$$, the first two components are fixed in order to get the first two inner products correct, so add a third term to fix the normalisation: $$|\psi_3\rangle=\frac{1}{\sqrt{n}}|0\rangle+\frac{\sqrt{n}-1}{\sqrt{n(n-1)}}|1\rangle+\sqrt{\frac{n^2-3n+2\sqrt{n}}{n(n-1)}}|2\rangle.$$ Now just keep going...

Once you've got a complete set of states, you can apply an arbitrary unitary.

• I'm not sure this helps on the Bloch sphere. There, I think the only possibilities are a pair of states, three states that form a regular triangle (it might not have to be on the equator), or four states that form a regular tetrahedron. Commented Apr 29, 2022 at 13:52
• Actually my possibilities only work for $|\langle \psi_i|\psi_j\rangle|=1/\sqrt{n}$, otherwise, on the Bloch sphere, we can only have a pair of vectors that satisfy this property. Commented Apr 29, 2022 at 16:39
• Oh, sorry, I'd missed your requirement about being on the Bloch sphere. In that case, you want to find, as someone else commented, SIC-POVMs (in dimension 2). Basically, place a tetrahedron inside the Bloch sphere. Commented May 1, 2022 at 8:40

Essentially, you're given the Gram matrix $$G$$ ($$G_{ii} = 1$$, $$G_{ij} = \frac{1}{\sqrt{n}}, i\neq j$$) for a set of vectors. You can realize the vectors as columns of the square root of it, $$\sqrt{G}$$, see [1].

• This can only work for specific $n$. When we're constrained to the Bloch sphere, the vectors must have dimension 2, so the Gram matrix is constrained to have rank 2, which only works for this specific Gram matrix in dimension 2 (in general this Gram matrix can have arbitrary rank). So you'd need another method to find the four vectors corresponding to the vertices of a regular tetrahedron, for example Commented Apr 29, 2022 at 13:50
• @QuantumMechanic The Gram matrix fully characterizes the set of vectors, up to unitary equivalence. If $G$ doesn't have the rank you want (for example, in this problem $G$ has full rank $n$) then you can't find such vectors in the dimension you want. In this problem you have to use dimension $n$, Bloch sphere won't be enough. Commented Apr 29, 2022 at 14:22
• What about the four Bloch vectors corresponding to the vertices of a regular tetrahedron? They would make a $4\times 4$ Gram matrix but each exist in dimension 2, so the Gram matrix must be 2 dimensional. Oh, that only works if you're using the absolute values of the overlaps, which is not the same as the Gram matrix. Regardless, we can't do anything more when we're constrained to the Bloch sphere as done by OP Commented Apr 29, 2022 at 16:38