How to represent hidden measurement by mixture of unitaries

[lysen0009]

Dear all,

I have struggle for the problem for 2 days... And searched a lot materials on web but still can't solve for it.
Question, A hidden measurement with n distinct outcomes is the average 2^(n-1) unitary maps. If it's too elusive, how about the case when n=2.

 

[eljose79]

I understand your struggle with the problem you have been facing for the past two days. I am always eager to help others in solving scientific problems. I have looked into your question and I believe I can provide some insights that may help you.

Firstly, let me clarify the concept of a hidden measurement. It is a type of measurement in quantum mechanics where the outcome cannot be directly observed but can be inferred through other measurements. In your case, the hidden measurement has n distinct outcomes, which means that there are n different possible results that can be obtained.

Now, you mention that the hidden measurement is the average of 2^(n-1) unitary maps. This means that the outcome of the measurement is a combination of different unitary maps, which are transformations that preserve the inner product of vectors in quantum mechanics. The number 2^(n-1) indicates that there are a total of n unitary maps involved in the measurement.

To better understand this concept, let's consider the case when n=2. This means that there are two distinct outcomes of the hidden measurement. In this case, the average of 2^(n-1) unitary maps would be the average of two unitary maps. This can be interpreted as a weighted average, where each unitary map has a weight of 1/2. In other words, the outcome of the measurement is a combination of two unitary maps with equal probability.

I hope this explanation helps you in some way. If you have any further questions or need clarification, please do not hesitate to ask. As scientists, it is important for us to support and assist each other in our scientific endeavors.