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I am reading some quantum papers (In particular, this one page 34) . One of the theorem statement reads, 

 "For every CPTP operator M, we have that .... "

I know that we usually apply unitary operations on quantum states. Is completely positive trace preserving (CPTP) operator same as unitary operator? What's the difference between them? What's the advantage of proving theorems for CPTP operators rather than unitary operators?

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Unitary operations are a subset of CPTP operations. You can think of a CPTP operation as the description of a unitary over a larger system.

The advantage of using CPTP maps is that you increase the generality of your statement. Think about, for example, a proof of the no-cloning theorem. People usually start talking about the input state, the target state, and an ancilla, so you want to achieve some rotation $$ |0\rangle|0\rangle|0\rangle\mapsto |0\rangle|0\rangle|a\rangle,\qquad|1\rangle|0\rangle|0\rangle\mapsto |1\rangle|1\rangle|b\rangle $$ where the third system is any dimension. We then assume that the process is unitary over that whole space. This is equivalent to allowing a CPTP map on the first two systems. Why don't we just completely ignore the third system and only assume a unitary on the first two? Well, what use would our no-cloning theorem be if it turned out that by adding some extra ancillas, our proof was completely invalidated and that cloning were possible? You want to perform your proof using the full generality of what's available, which means not making any assumptions about availability of ancillas, for example.

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  • $\begingroup$ Is it safe to say the following? Every CPTP operator on $n$ qubits is equivalent to applying a unitary operation on $n+k$ qubits (for some $k \geq 0$ ancilla bits) and then tracing out the $k$ ancilla bits? $\endgroup$
    – satya
    Commented Jul 31, 2019 at 19:30
  • $\begingroup$ I am asking this because, I know every unitary operator can be realized in a quantum system. I wanted to know if every CPTP operator can actually be realized on a quantum system. $\endgroup$
    – satya
    Commented Jul 31, 2019 at 19:37
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    $\begingroup$ @satya yes, what you're describing is called Stinespring dilation, and it's always possible with a CPTP map $\endgroup$
    – user2723984
    Commented Jul 31, 2019 at 20:03
  • $\begingroup$ @satya Moreover, there is always a $k\leq n$. $\endgroup$
    – DaftWullie
    Commented Aug 1, 2019 at 6:38

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