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In trying to understand the mathematical details of high-dimensional QKD with few mutually unbiased bases. First, prepare a state $|\psi\rangle \in H^d$ as a linear combination of a first orthonormal basis $A$. Then use a second mutually unbiased basis B as the Hamiltonian POVM operator.

What is the probability of observing the value associated to $|b_i \rangle$? The probability is $\frac{1}{d}$.

$\langle \psi | b_i \rangle \langle b_i | \psi \rangle = (\sum_s \alpha_s^* \langle a_s|b_i\rangle) (\sum_t \alpha_t \langle b_i |a_t\rangle) = \sum_j \alpha_j^* \alpha_j \langle a_j|b_i\rangle \langle b_i |a_j\rangle + \sum_{s \neq t} \alpha_s^* \alpha_t \langle a_s|b_i\rangle \langle b_i |a_t\rangle = \frac{1}{d}$.

Since $|\langle a_s|b_i\rangle|^2 = \frac{1}{d}$ (mutually unbiased bases), the first summation with the index $j$ is simply $\frac{1}{d}$.

But the problem, I do not see why the second summation with the indices $s \neq t$ is equal to $0$. The phases of each element should cancel out ($\langle a_s|b_i\rangle = e^{i\theta} \cdot 1/ \sqrt{d}$).Furthermore, can I derive this result without having to deal explicitly with the phases?

JMark
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2 Answers2

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There should be additional restrictions on $|\psi\rangle$ for this to hold. The counter example that comes to mind is $|a_i\rangle = |i\rangle$ and $|s_i\rangle = Z^i |+\rangle$. We can write $|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$ which is written with respect to basis $A$ but of course if we measure in basis $B$ the outcome is always $0$.

This isn't too surprising $A$ is a basis, so the property that any state $|\psi\rangle$ can be written as a linear combination of $|a_i\rangle $ is always true and the result of measuring $|\psi\rangle$ in the B basis should be independent of the fact other Bases exist.

I'm guessing the original question was about measuring any state $|a_i\rangle$ in the B basis which as you showed, will yield a uniformly random result.

Ethan Davies
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Effectively, I arrived overnight with the same kind of answer. I was looking too much to the problem of the Mutually Unbiased Bases (MUB) and I was forgiving the original problem (the QKD). In the original BB84 protocol, the two MUB are $(|0\rangle , |1\rangle)$ and $(|+\rangle , |-\rangle)$. So, Alice (the source) choses a basis, and then sends any vector of that basis. If Bob (the legitimate destination) measures the received quantum state in the same basis, he will retrieve the exact vector with probability 1. If he choses the other basis, he will retrieve any vector of that basis with probability 0.5. So the protocol sends a vector of one of the bases and not a linear combination of these vectors. This is the mistake that I did.

I was trying to forge a state of the form $\alpha |0\rangle + \beta |1\rangle$ which measurement with the basis $(|+\rangle , |-\rangle)$ gives either vector with probability 0.5. This is rarely possible. Simply consider the example $0.8 |0\rangle + 0.6 |1\rangle$ - the sum of the squared coefficients is equal to 1, as expected. In such a case, measuring that state with the basis $(|+\rangle , |-\rangle)$ gives $|+\rangle$ with probability 0.98 and $|-\rangle$ with probability 0.02. The error was abstracting the choice of the MUB, and forgiving the classic ones.

JMark
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