# Question about discarding the second register in the standard approach of hidden subgroup algorithm

My questions:

1. What does discarding the second register mean for the standard approach of hidden subgroup algorithm?
2. Why does discarding let the first register end up in a mixed state?

My understanding so far:

I am trying to understand the effect of discarding the second register in the standard approach of hidden subgroup algorithm using this paper as resource. I will be using $S_5$ as my group of interest.

It is said in the paper that,

I assume $f$ maps $g$ to it's irrep. So, if we take the case of $S_5$, assuming an automorphism group of a $5$-node graph is hidden inside $S_5$, $|G| = |S_5| = 120$. Both $|g\rangle$ and $|f(g)\rangle$ are $120$ element column vectors.

So, when it is said that the second register is discarded what it actually means? Let's take the case of trapped ion qubits. When we say we discard the second register, does it mean that we will not do any unitary transformation or measurement on them anymore but they will always be there? In the Wikipedia page of Simon's problem the second register is measured.

I also don't understand why discarding (whatever it means) will let the first register end up in a mixed state. Shouldn't it be in a pure quantum state?

• I have started with a pure state $$\frac{1}{\sqrt{|G|}} \sum_{g \in G} |g, f(g)\rangle$$. Why or how would I take a partial trace on it? Dec 19, 2015 at 14:30
• that's correct. Moreover it is also true that we can treat a pure state as a mixed state with probability 1. Are you suggesting that I would consider $\frac{1}{\sqrt{|G|}} \sum_{g \in G} |g, f(g)\rangle$ as a mixed state i.e. $\frac{1}{|G|} \sum_{g \in G} |g, f(g)\rangle \langle, f(g)I$ and take the partial trace on the second register? Dec 19, 2015 at 16:07