<p>Discarding in this context can be carried out by performing a <a href="https://en.wikipedia.org/wiki/Partial_trace" rel="nofollow">Partial Trace</a>. The system you're interested in is that of the first register. If you want to know only its state, as opposed to the combined state of the two registers, then you take the density matrix of the whole system and <em>trace out</em> the second register.  The resultant state will, in this case, be mixed.  What you're seeing at work is more or less <a href="https://en.wikipedia.org/wiki/Purification_of_quantum_state" rel="nofollow">purification</a> in reverse. The resultant mixed state is the reduced state of some other pure system, namely the system comprised of the two registers.  </p>

CC BY-SA 3.0
2015-12-19T14:30:21.373Z
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?
Omar Shehab
CC BY-SA 3.0
2015-12-19T16:00:05.133Z
Partial traces are covered in all the standard texts on Quantum computation and information. I'm not even going to attempt to explain them better than they are there.
Logan Mayfield
CC BY-SA 3.0
2015-12-19T16:07:05.983Z
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?
Omar Shehab
CC BY-SA 3.0
2015-12-19T16:28:52.267Z
yes. that's the idea.
Logan Mayfield
CC BY-SA 3.0
2015-12-19T16:47:24.857Z
is that the only way? What could be other alternatives?
Omar Shehab
CC BY-SA 3.0
2015-12-19T14:21:59.463Z
2015-12-19T14:21:59.463Z
Logan Mayfield
<p><strong>My questions:</strong></p>

<ol>
<li>What does discarding the second register mean for the standard approach of hidden subgroup algorithm?</li>
<li>Why does discarding  let the first register end up in a mixed state?</li>
</ol>

<p><strong>My understanding so far:</strong></p>

<p>I am trying to understand the effect of discarding the second register in the standard approach of hidden subgroup algorithm using <a href="http://arxiv.org/pdf/quant-ph/0504083v2.pdf" rel="nofollow noreferrer">this paper</a> as resource. I will be using $S_5$ as my group of interest.</p>

<p>It is said in the paper that,</p>

<p><a href="https://i.sstatic.net/BQ1TQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BQ1TQ.png" alt="enter image description here"></a></p>

<p>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.</p>

<p>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 <a href="https://en.wikipedia.org/wiki/Simon%27s_problem" rel="nofollow noreferrer">Wikipedia page</a> of Simon's problem the second register is measured.</p>

<p>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?</p>

<quantum-computing><quantum-information><gr.group-theory><fourier-analysis><measure-theory>
Question about discarding the second register in the standard approach of hidden subgroup algorithm
Omar Shehab
# Question about discarding the second register in the standard approach of hidden subgroup algorithm

(Asked by: username_0 on 2015-12-19)

**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](http://arxiv.org/pdf/quant-ph/0504083v2.pdf) as resource. I will be using $S\_5$ as my group of interest.

It is said in the paper that,

[![enter image description here](https://i.sstatic.net/BQ1TQ.png)](https://i.sstatic.net/BQ1TQ.png)

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](https://en.wikipedia.org/wiki/Simon%27s_problem) 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?

**Tags:** quantum-computing, quantum-information, gr.group-theory, fourier-analysis, measure-theory

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## 1 answers

### Answer by username_1 on 2015-12-19. Score: 4 | Accepted answer

Discarding in this context can be carried out by performing a [Partial Trace](https://en.wikipedia.org/wiki/Partial_trace). The system you're interested in is that of the first register. If you want to know only its state, as opposed to the combined state of the two registers, then you take the density matrix of the whole system and *trace out* the second register. The resultant state will, in this case, be mixed. What you're seeing at work is more or less [purification](https://en.wikipedia.org/wiki/Purification_of_quantum_state) in reverse. The resultant mixed state is the reduced state of some other pure system, namely the system comprised of the two registers.

> **Comments:**
> **username_0**: 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?  
> **username_1**: Partial traces are covered in all the standard texts on Quantum computation and information. I'm not even going to attempt to explain them better than they are there.  
> **username_0**: 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?  
> **username_1**: yes. that's the idea.  
> **username_0**: is that the only way? What could be other alternatives?  

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