# Is there a simple way to express the 2ⁿ+1 mutually unbiased bases for n qubits?

The title says it. An explanation for only 2 qubits would already be interesting, since I already have difficulties to find the 5 MUBs in this simple case.

• This question might be more appropriate on theoretical computer science. It doesn't really relate to physics directly... I'm happy to leave it open here however. Nov 15, 2010 at 21:50
• Qubits are quantum systems, hence physical systems. Should I have said 'quantum 2-level systems' instead of 'qubits'? Nov 15, 2010 at 23:09
• Quantum computation and quantum information theory is an interesting crossover area, we should think about the extent to which we want to allow those questions. I tend to agree with Frédéric that they could fit in here. Nov 16, 2010 at 2:09
• According to wikipedia, this question was first asked by Schwinger, a physicist, in 1960. So, to my opinion, it is a physics question, which happens to have application in quantum information theory, which has its place here even if it would be decided to reject some quantum information questions. Nov 16, 2010 at 10:08
• – glS
Oct 1, 2018 at 10:27

Edit: Sorry I misread 2n+1 as the number of dimensions. It becomes more systematic then. The case N=4 can be found in http://en.wikipedia.org/wiki/Mutually_unbiased_bases#Example_for_d_.3D_4

The general construction can be found in [4]. Let's consider the case n=2. Firstly, use the Galois ring $GR(4,n=2) = \mathbb Z_4[x] / \langle h(x)\rangle$ where $h(x)=x^2+x+1$ is an irreducible polynomial of the field (i.e. its root cannot be represented in $\mathbb Z_4$. Let $\xi$ be that root. We define the Teichmüller set $T_n = \{0,1,\xi,\xi^2,\dotsc\}$, where \begin{aligned} 1 &= 1 \\ \xi &= \xi \\ \xi^2 &= -\xi-1 = 3\xi+3 \\ \xi^3 &= 3\xi^2 + 3\xi = -3 = 1 \text{ (cycle formed)} \end{aligned} so $T_2 = \{0,1,\xi,3\xi+3\}$. Now, the basis are given by $$v_{a,b} = \frac1{\sqrt{2^n}} \exp\left(\frac{2\pi i}4 \mathrm{tr}((a+2b)x)\right)_{x\in T_n}$$ where $a,b\in T_n$, the trace is defined as $\mathrm{tr}(x)=\sum_{k=0}^{n-1}\sigma^k(x)$ where $\sigma(a+2b)=a^2+2b^2$. In our case, $\mathrm{tr}(a+2b)=a+2b+a^2+2b^2$.

For instance, \begin{aligned} v_{1,\xi} &= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( (1+2\xi) \{ 0, 1, \xi, 3\xi+3 \} ) \right) \\ &= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, \xi+2\xi^2, 9\xi+3+6\xi^2 \} ) \right) \\ &= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, 3\xi+2, 3\xi+1 \} ) \right) \\ &= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, \xi+2(3\xi+3), 3\xi+3+2 \} ) \right) \\ &= \frac12 \exp \left(\frac{2\pi i}4 \{0, 1+2\xi+1+2\xi^2, \xi+2(3\xi+3)+\xi^2+2(3\xi+3)^2, (3\xi+3)+2+(3\xi+3)^2+2 \} \right) \\ &= \frac12 \exp \left(\frac{2\pi i}4 \{0, 0, 1, 3 \} \right) \\ &= \frac12 \{1, 1, i, -i \} \end{aligned} You could then compute the other 15 combinations to get the MUBs (after including the standard basis). (Please ask on math.SE or MathOverflow for the details of Galois ring.)

The N=5 case is kept below for those who are interested.

As far as I can check, the MUBs for N not a power of prime is in general unknown. In fact, even for N = 6 only 3 MUBs are found, and it's still a conjecture mathematically if there the 4th set doesn't exist[1,2].

Since not all 2n+1 are primes, I doubt if it is possible to analytically find MUBs for all n. But for N=5, we could use the "Unitary operators method using Galois fields" as explained in Wikipedia[1,3].

$$X = \sum_0^4 |k+1\rangle\langle k| = \begin{pmatrix} 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 1&0&0&0&0 \end{pmatrix}$$ $$Z = \sum_0^4 \omega^k|k\rangle\langle k| = \begin{pmatrix} 1&0&0&0&0\\ 0&\omega&0&0&0\\ 0&0&\omega^2&0&0\\ 0&0&0&\omega^3&0\\ 0&0&0&0&\omega^4 \end{pmatrix}$$ where $\omega = e^{2\pi i/5}$. Then we compute the eigenvectors of the 6 matrices $$X, Z, XZ, XZ^2, XZ^3, XZ^4$$ to get the MUBs (each row represents a base):

\begin{aligned} Z &\to \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{bmatrix} \\ X &\to \frac1{\sqrt5}\begin{bmatrix} \omega ^2 & \omega ^3 & \omega ^4 & 1 & \omega \\ \omega ^3 & \omega ^2 & \omega & 1 & \omega ^4 \\ -\omega ^2 & -\omega ^4 & -\omega & -\omega ^3 & -1 \\ -\omega ^3 & -\omega & -\omega ^4 & -\omega ^2 & -1 \\ -1 & -1 & -1 & -1 & -1 \end{bmatrix}\\ XZ &\to \frac1{\sqrt5}\begin{bmatrix} \omega ^4 & 1 & 1 & \omega ^4 & \omega ^2 \\ 1 & 1 & \omega ^4 & \omega ^2 & \omega ^4 \\ \omega ^2 & \omega ^4 & 1 & 1 & \omega ^4 \\ 1 & \omega ^3 & 1 & \omega & \omega \\ \omega & 1 & \omega ^3 & 1 & \omega \end{bmatrix} \\ XZ^2 &\to \frac1{\sqrt5}\begin{bmatrix} \omega ^4 & \omega & \omega & \omega ^4 & 1 \\ 1 & \omega ^3 & \omega ^4 & \omega ^3 & 1 \\ \omega ^4 & 1 & \omega ^4 & \omega & \omega \\ 1 & 1 & \omega ^3 & \omega ^4 & \omega ^3 \\ \omega ^3 & 1 & 1 & \omega ^3 & \omega ^4 \\ \omega & 1 & \omega ^2 & \omega ^2 & 1 \end{bmatrix}\\ XZ^3 &\to \frac1{\sqrt5}\begin{bmatrix} 1 & \omega ^2 & \omega & \omega ^2 & 1 \\ \omega & 1 & \omega & \omega ^4 & \omega ^4 \\ 1 & 1 & \omega ^2 & \omega & \omega ^2 \\ \omega ^2 & 1 & 1 & \omega ^2 & \omega \\ \omega ^4 & 1 & \omega ^3 & \omega ^3 & 1 \end{bmatrix}\\ XZ^4 &\to \frac1{\sqrt5}\begin{bmatrix} \omega & 1 & 1 & \omega & \omega ^3 \\ 1 & 1 & \omega & \omega ^3 & \omega \\ \omega ^3 & \omega & 1 & 1 & \omega \\ 1 & \omega ^2 & 1 & \omega ^4 & \omega ^4 \\ \omega ^4 & 1 & \omega ^2 & 1 & \omega ^4 \end{bmatrix} \end{aligned}

Ref:

• $2^n+1$ is the number of MUB, as Marek has pointed out, this is exactly what you expect for $n$ qubits. But the dimension of the Hilbert space of $n$ qubits is $2^n$, which is a power of a prime. Nov 15, 2010 at 19:18

I don't know anything about the topic but just by searching the internet I found this paper which claims to prove that $2^n+1$ MUBs always exist for $n$ qubits and moreover it says in the abstract that the proof is constructive. Check it out.