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If I want do describe two particles #1 and #2 within Quantum Mechancis - let them be different in some way (Spin for example) - how do "I" decide if the total state of the system should be an entangled state, a product state or some density matrix?

I am concerned about this because I read that for some reason the universe is technically in an entangled state with everything and if we ignore it we should write every state as a density matrix (because we end up with a statistical mixture, if we ignore the universe). But this would mean that writing states as Dirac Brakets would be always wrong.

Edit: I can build up a two particle system ether as product-state or as an entangled state. Say I want to mix an electron with the rest of the universe.

If I choose a product state I can ignore the universe and just look at the electron without losing any information. My electron description is complete without the universe.

If I choose an entangled state for the electron and the universe and I again want to focus only on the electron I end up with a statistical mixture. That I can ONLY write as a density matrix because I choose to ignore the universe I "forget" some information about the state. I add to the quantum probability the probability due to my incomplete knowledge of the actual electron state. This is not possible with Diracs Braket.

My question is how do I choose with what kind of state I start?

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  • $\begingroup$ Any state, pure or mixed can always be written as a density matrix. Given a density matrix, you can easily see if the state is a pure state, because then if you diagonalize it it has one diagonal entry equal to 1, while in the mixed case there will be probabilities smaller than 1 on the diagonal. This means that the square of the trace is equal to 1 in the pure case while it is strictly smaller than 1 in the mixed case. To see if pure state can be factored, you could consider the reduced density matrix and then apply the above test to see if it is a pure state. $\endgroup$ – Count Iblis Jun 24 '15 at 18:46
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Extending David Z's answer:

If you want to answer your question, you can ask the problem differently: your particle is independent or it is part of the system, ergo it is interacting with the system.

In entangled state, although you loose concrete information about your particle, but you get something much more greater: relational information.

I suppose you understand what an entangled state is, so I won't go into details, but if you want I will explain it with pleasure.

So if your particles are interacting with each other, and one's action has a direct effect on the other, then you start as an entangled state, otherwise as a product state. If you have a complex system then entanglement is often broken and remade ( like atom ionization ).

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  • $\begingroup$ Sounds like there are no product states - I guess everything interacts with all the stuff in some way since for example the electromagnetic force reaches infinitely far. Is this reasoning correct? If so, this seems to imply that there is nothing that is not entangled with "everything" else (within its lightcone). -> There are no product states. $\endgroup$ – Thomas Elliot Jun 26 '15 at 6:43
  • $\begingroup$ Yes and no. It all depends on how large your system is, at witch you are looking at. As you said, if you look at the universe, than you get that everything is entangled, and indeed there are no product states. BUT, if you look at a much smaller system, e.g. an atom, you find that the electrons of it have very little ( almost none ) interaction with a particle 1km away from him, so the change of the state of the latter particle has no effect on our atom's state. $\endgroup$ – Geseft Jun 26 '15 at 8:44
  • $\begingroup$ Since all particle interacted "at" the big bang I thought they were in "close contact" and must be entangled. But I think we can kill the entanglement by our state preperation. We prepare states in a defenite state in order to have a pure state. Am I correct? (I just thought of that) $\endgroup$ – Thomas Elliot Jun 26 '15 at 9:24
  • $\begingroup$ If you are preparing for example the spin of a particle, you actually create a superposition of two states ( u and d in the respective basis vector), and when you observe it, the state will collapse to a single state, this is when you collapse the entanglement too. $\endgroup$ – Geseft Jun 26 '15 at 10:00
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Density matrices and linear combinations of brakets are two different but equivalent ways of representing quantum states. On the other hand, entangled states and product states (if we share the same understanding of the terminology) are two different kinds of quantum states. So you're mixing up two questions: first whether a given state is entangled or not, and second, whether it should be represented as a density matrix or as Dirac brakets. If you keep them separate, I think that resolves your issue.

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  • $\begingroup$ You can only with the density matrix formalism take your own ignoranz into account. If you use the Braket you kind of assume that you know everything about the state. So, my question is how one can be bold enough to make such a claim. $\endgroup$ – Thomas Elliot Jun 24 '15 at 20:39
  • $\begingroup$ @ThomasElliot You can always represent a mixed state as a pure state by considering a larger system, see here. $\endgroup$ – Count Iblis Jun 24 '15 at 21:04

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