If we treat our entire system (say an electron and a bunch of atoms) quantum mechanically then all possible interactions will be unitary transformations. Thus any state that I describe will always be a pure state.

But if I observe only a subspace of my system (just the spin of the electron, say) I need to trace out rest of the space and I end up with a density matrix.

If my states were separable to begin with then my density matrix will correspond to a pure state. The only way to get a mixed state would be if the spin of my electron was entangled with the rest of the system. Right?

In other words, is a mixed state always an entangled state in a higher dimension?

Edit: My question is not about purification. I do not care if I can find a state in my complete Hilbert space by purification. Rather, is entanglement the only way to go from a pure state to a mixed state. Thus it isn't a duplicate of this.

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    $\begingroup$ Every density matrix admits a purification; is that your question? $\endgroup$ Commented Jan 26, 2023 at 17:31
  • $\begingroup$ While that question is related (thank you for the link) my question is approaching the mixed state from the other direction. Purification is going from mixed to pure while my query is about going from pure to mixed. $\endgroup$ Commented Jan 26, 2023 at 17:57
  • $\begingroup$ Sorry, I don't understand: In which sense are you "going from pure to mixed"? BTW: You can have a mixed reduce density matrix for a non-entangled density matrix (even a product state) of a larger system. So in this sense, no, a mixed state can arise as a partial trace from a non-entangled mixed state as well. $\endgroup$ Commented Jan 26, 2023 at 17:59
  • $\begingroup$ Can you give me an example where a partial trace over product states results in a mixed density matrix? Because that would answer my question. $\endgroup$ Commented Jan 26, 2023 at 18:04
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    $\begingroup$ This is really a philosophical question, i.e. about interpretations. You can always attribute any uncertainty/randomness to entanglement with a system you don't have control of. (The most radical of this interpretations maybe being many worlds.) $\endgroup$ Commented Jan 26, 2023 at 18:45

1 Answer 1

  1. Purely mathematically, this is certainly true, since any mixed state in a Hilbert space $H$ can be "purified", i.e. we can exhibit a pure state in $H\otimes H$ whose partial trace is the mixed state.

  2. In "physical" terms, a mixed state doesn't need to always arise as the partial trace of a larger entangled system: If I have an electron source and I tell you that half the electrons it spits out are spin-up (with respect to some spin operator) and the other half is spin down, then you will likely model what you know about an electron in the beam by assigning it a mixed state of 50% spin-up and 50% spin-down. So this mixed state models incomplete knowledge about the pure state of the individual electrons, but it didn't arise from any sort of entanglement or larger pure state - whether or not my source internally uses entanglement to achieve this outcome is completely irrelevant for your model.

  • $\begingroup$ True, I do not need entanglement to arrive at a mixed state when I model only the electrons’ spin. But surely entanglement is the only way to arrive at that mixed state if for whatever reason I chose to model the entire system. $\endgroup$ Commented Jan 26, 2023 at 17:51
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    $\begingroup$ @SuperfastJellyfish Only if you have enough information about "the entire system" to determine a unique pure state. Whether or not this is "surely possible" depends on your interpretation, $\psi$-epistemic models (c.f. physics.stackexchange.com/a/290592/50583 for terminology) do not commit themselves to the idea that there is an "actual" quantum state of a physical system, so it is not necessarily the case that we must be able to determine enough information to arrive at one. $\endgroup$
    – ACuriousMind
    Commented Jan 26, 2023 at 17:55
  • $\begingroup$ Upon revisiting this after a few months, I get what you meant. If the wavefunction is just epistemic then there’s no unambiguous way to purify. This, as you and @Norbert Schuch pointed out is indeed a matter of interpretation. $\endgroup$ Commented May 18, 2023 at 6:00

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