Consider a quantum system with Hilbert space $\mathscr{H}$ and suppose the quantum state is specified by a density operator $\rho$. Since it is Hermitian, it has a spectral decomposition: $$\rho = \sum p_i |\phi_i\rangle \langle \phi_i |.$$

Now take another quantum system with Hilbert space $\mathscr{H}'$ with dimension at least equal to the first. Take any basis $|\psi_i\rangle$ and consider the state $$|\Psi\rangle = \sum \sqrt{p_i} |\phi_i\rangle \otimes \lvert\psi_i\rangle.$$

A partial trace over the second system yields the first state. This is the purification. A mixed state is always a partial trace of some pure state in a composite system.

There are issues, however: (1) the purification is highly non-unique, any Hilbert space of dimension equal or higher to the first will work, and we can pick any basis we want yielding distinct pure states. (2) this is a mathematical construction. The purifying system seems to have no true meaning physically, this seems to be further implied by the non-uniqueness described in (1).

So is purification a purely mathematical construction with no physical meaning, or it indeed has some physical meaning ? If so, what is the physical meaning of the purification?

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    $\begingroup$ It is mostly just a mathematical trick, however it is also often argued that purifications can represent non-local information, i.e. entanglement in your system. In AdS/CFT for example, the thermofield double (TFD) state is dual to an eternal black hole, and the TFD state is merely a purification of a thermal state. One can show that the two CFT's are entangled even though there are no local interactions between them. This observation essentially lead to the famous ER = EPR. $\endgroup$
    – Akoben
    Aug 21, 2018 at 15:47
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    $\begingroup$ @Akoben With a bit more elaboration that could be a very good answer! $\endgroup$
    – ACuriousMind
    Aug 21, 2018 at 15:48

1 Answer 1


This really depends whether you believe in the "church of the larger Hilbert space". If you feel that pure states are more fundamental than mixed states, then you might argue that any mixed state is just a lack of knowledge, and somewhere out there is the missing piece of the system which will give you full information (i.e., a pure state). Even though you don't know what it is, you know it is out there. (As you can see, this is really more a matter of interpretation of quantum mechanics, since mathematically, the two perspectives are equivalent.)

There are many cases where this is a very reasonable perspective on mixed states regardless of what you believe, e.g. when you have a pure state which becomes mixed by coupling to the environment: While the state of the system looks mixed to you, it has just unitarily interacted with the environment, so the overall state system + environment will be pure, and the environment will hold the purification of your system.

Clearly, this is true more generally as long as your initial global state is pure and you consider part of an isolated sytem (i.e. unitary dynamics).

Beyond that, purifications are of course also a powerful mathematical tool.

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    $\begingroup$ Is the "CotLHS" really about an interpretation of quantum mechanics, or is it the physical assumption that the state of the universe is pure? $\endgroup$ Aug 21, 2018 at 18:36
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    $\begingroup$ @RyanThorngren Well, my impression was always that the CotLHS is to subscribe to the "Whatever can be pure will be pure" perspective. I'd call this an interpretation. In any case, I'm not sure about how to deal with the "the whole universe is pure" perspective since it makes the measurement problem even worse (so we again end up with interpretations). $\endgroup$ Aug 21, 2018 at 21:14
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    $\begingroup$ How does it make the measurement problem worse? $\endgroup$ Aug 22, 2018 at 16:27
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    $\begingroup$ @RyanThorngren Well, it is easier if you as the observer are not part of the system - i.e. QM is a theory about what you see when you measure systems - but if you insist of assigning a WF to the whole universe, it is hard to argue that you are not part of it. $\endgroup$ Aug 22, 2018 at 20:51

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