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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.

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  • $\begingroup$ 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. $\endgroup$
    – Noldorin
    Nov 15, 2010 at 21:50
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    $\begingroup$ Qubits are quantum systems, hence physical systems. Should I have said 'quantum 2-level systems' instead of 'qubits'? $\endgroup$ Nov 15, 2010 at 23:09
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    $\begingroup$ 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. $\endgroup$
    – David Z
    Nov 16, 2010 at 2:09
  • $\begingroup$ 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. $\endgroup$ Nov 16, 2010 at 10:08
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    $\begingroup$ related: physics.stackexchange.com/q/391234/58382 $\endgroup$
    – glS
    Oct 1, 2018 at 10:27

2 Answers 2

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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:

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    $\begingroup$ $2^n+1$ 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. $\endgroup$ Nov 15, 2010 at 19:18
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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.

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