# Why should we purify a mixed state?

I'm new to quantum computing so I know this question may sound stupid. But I still want to know why purification is necessary sometimes. Please help me with it:)

• "Necessary" is a strong word. It's a technique to treat mixed states using the formalism of pure states, but the reason why you'd want to do it will strongly depend on the context. In which context did you see it mentioned as "necessary"?
– glS
Commented May 19, 2022 at 14:26

Here is a short list of things that I know 'require' the notion of a pure state, or also at least make a point in their relevance.

1. Suppose that you are considering a protocol for learning how a parameter $$\theta$$ governs the changes in your system at a given state. The generic goal of this kind of learning problem, a quantum metrology problem, is to maximize the Fisher information of a given quantity. Mixed states represent that you already have some part of your information about the system that 'leaked out' so in general people assume that the initial state being considered is 'pure' or at least ideally should be close to pure.
2. Another possible answer might be that if it is always possible to describe states as pure states with respect to a larger Hilbert space - which I think is what you meant by purification - then states of this kind are more fundamental than mixed states (mixed states are convex combinations of pure states, by definition).
3. In Bell experiments this kind of dilation is used and it is invoked the so-called 'Church of the larger Hilbert space'. This is crucial for the theoretical background in experimental implementations of violations of Bell's notion of local causality in quantum theory.

All that said, it is kind of debatable if pure states are necessary since they may be regarded somewhat as an idealization.

It's a useful tool for certain proofs, as the idea of purification allows you to equate the entropy of the purifying system to that of the rest of the system via the schmidt decomposition ie:

If $$ABC$$ is a mixed system, and I purify it with system $$R$$, then $$S(ABC)=S(R)$$

So for example, using $$S(A)+S(B)\le S(A,C)+S(B,C)$$, and then taking the pure $$ABCR$$ system, you can use $$S(R)+S(B)\le S(R,C)+S(B,C)$$

$$S(R)=S(ABC)$$, due to purification and schmidt decomposition, $$S(R,C)=S(A,B)$$ for the same reason, so subbing those in, you get $$S(ABC)+S(B) \le S(AB)+S(BC)$$

This is just one example of how purification lends itself to proofs.