Quantum entanglement is the norm, is it not? All that exists in reality is the wave function of the whole universe, true? So how come we can blithely talk about the quantum state of subsystems if everything is entangled? How is it even possible to consider subsystems in isolation? Anything less than the quantum state of the whole universe at once. Enlighten me.
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2$\begingroup$ "How is it even possible to consider subsystems in isolation?" Because everything that is physically meaningful may be considered. One may talk about the well-defined eigenstates of a quantity which may be associated with a local region. Whenever we learn about the value of an observable, we know that the state of the system is an eigenstate of this quantity. When we measure a complete set of observables describing a subsystem or a region, we know that the subsystem is in a pure state - the common eigenstate of the set. Any measurement makes a previous correlation/entanglement irrelevant. $\endgroup$– Luboš MotlCommented May 9, 2012 at 11:24
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8 Answers
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The normal state of subsystem is entangled, as follows from the Schroedinger equation, which turns a separable state immediately into an entangled state if there is some interaction.
The state of a subsystem is obtained from the state of the universe by tracing out all other degrees of freedom. This leaves a density matrix, which is a perfectly good state for the subsystem. But this state is usually a mixed state. If the system is sufficiently shielded from the environment, its density matrix will follow a nonunitary evolution of Lindblad type. http://en.wikipedia.org/wiki/Lindblad_equation
This is the equation that must be used in place of the Schroedinger equation whenever quantum optical phenomena must be predicted quantitatively. The usual arguments with pure states are idealizations that ignore imperfections that always exist in real experiments. The quantum Liouville equation (which is equivalent to the Schroedinger equation) arises as a limit of Lindblad equations in which the dissipative terms are dropped. But this limit is an idealization that (in quantum optics) is usually not good enough for quantitative predictions.
Experimental techniques can be used to put (at most) a few of the degrees of freedom of the subsystem into an (approximately) pure state, which are then used as embodiments of pure states in quantum experiments. For example, in a silver atom (which has many degrees of fredom) a magnet influences different spins in a different way; this is the reason why one can make the spin degree of freedom pure. And a photon has a mixed state with respect to the momentum degrees of freedom, but its polarization degree of freedom can be made pure by a polarizer.
Thus talking about pure states of a physical system smaller than the universe as a whole, and using the Schroedinger equation for its dynamics, is an idealization of the same kind as ignoring friction in classical Hamiltonian mechanics. It is appropriate in many cases, and inappropriate in others, and the user must decide whether a particular description used is adequate
answered May 9, 2012 at 10:53
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$\begingroup$ Best answer here. Deserves more points. $\endgroup$ Commented May 27, 2012 at 11:05
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I think there are two sides to this question:
1) Why can we often treat systems with which our measuring devices are entangled as classical systems? This is explained by decoherence as pointed out in John Rennie's answer.
2) Why can we describe some systems in a lab as undergoing unitary evolution of their wavefunctions as prescribed by Schrodinger's equation in a manner that is independent of some of our measuring devices? This is because great care is taken to isloate the system under investigation from premature decoherence with the measuring devices. This does not mean that there is no entanglement in the global wavefunction, it just means that the Hamiltonian of the experiment has been designed very carefully to allow the subspaces of the global wavefunction to evolve with minimal coupling. This is often very hard to do, and it is one of the reasons that scalable quantum computers are so difficult to create.
answered May 9, 2012 at 9:02
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Because of decoherence.
Every time I think I understand decoherence I discover I'm wrong, but basically the phase information between your subsystem and its environment gets scrambled, and they start behaving independently.
answered May 9, 2012 at 8:43
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$\begingroup$ This depends a lot on the subsystem! Making subsystems consisting of more than a few quantum degrees of freedom and behaving as if a pure state is an art that requires enormous experimental diligence. $\endgroup$ Commented May 9, 2012 at 10:36
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You state:
Quantum entanglement is the norm, is it not?
Norm in what framework? Quantum mechanics is the description of nature in the microcosm and even there, entanglement in the sense of one wave function describing a whole system of particles happens only sometimes. Decoherence , see John Rennie's answer, exists even in the microcosm
All that exists in reality is the wave function of the whole universe, true?
Not true, if by wavefunction we mean the solution of the quantum mechanical equation that would govern the zillions of particles contained in the Universe all phases known and calculated. By the time we reach dimensions of centimeters decoherence is completely inevitable. The framework of quantum mechanical fields morphs into the framework of classical fields and the phases of the quantum mechanical solutions are lost ( except in superfuids and supeconductors, but that is another story and is still limited to earth sizes).
So we can talk about the quantum state of subsystems since everything is not entangled. Let alone the universe.
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1$\begingroup$ To be super-pedantic, might there be some truth to the statement that "everything is entangled" as long as you acknowledge that the measurable consequences of the entanglement are often vanishingly small? $\endgroup$ Commented May 9, 2012 at 9:36
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$\begingroup$ Well, I guess yes . But for the size of the universe they are truly infinitesimally small and infinitely complicated in phase space parameters. It is decoherence which allows classical physics to appear. $\endgroup$– anna vCommented May 9, 2012 at 9:40
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$\begingroup$ The superconductor example is nice in this context, as this is a special case where a comparatively large system is entangled compared to the electrons in a normal metal which do not posses the same phase. $\endgroup$ Commented May 9, 2012 at 9:57
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1$\begingroup$ -1: The OP was correct. The normal state of subsystem is entangled, as follows from the Schoedinger equation, which turns a separable state immediately into an entangled state if there is some interaction. Decoherence only applies to a subsystem relative to the environemnt, not to two small quatum systems relative to each other. $\endgroup$ Commented May 9, 2012 at 10:39
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1$\begingroup$ See my answer to the question. As I understand him, the OP wants to understand why ''Anything less than the quantum state of the whole universe at once'' can be assigned a state, and this includes small systems such as a calcium atom in an ion trap. The latter is intinsically entangled with the trap, and to speak of its state needs some justification. Decoherence is not a valid justification in this case (and many other cases). $\endgroup$ Commented May 9, 2012 at 10:58
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Interesting question. There are quantum fluctuations during the inflationary epoch and the CMBR anisotropies are collapsed relics of these fluctuations. Decoherence aside, the universe is still in an entangled state on a cosmic scale.
I think Tegmark at http://arxiv.org/abs/1108.3080 has a nice tripartite framework to answer your question. You have to include an observer. Relative to some knowledge state of the observer, what do you get? If the observer knows about the WMAP data, then relative to that observer, there is a collapse into an unentangled state. The observer need not even need to know what happens far away. Knowing what happens here is enough to get rid of the entanglement.
Otherwise, according to our best inflationary models, tracing over cosmic regions far away will leave our local region in a highly mixed state at a superposition of CMBR temperatures varying by as much as 10^-5 in ratio, and the the structure formation of matter in an even greater superposition.
Decoherence explains why the different terms stop interference, but not why one term is selected out of many.
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Don't you worry about cosmic entanglement coming from the inflationary era. They've collapsed away by now. So don't worry. The wave function has already collapsed. Ordinary Schroedinger time evolution tends to produce more entanglement, but collapses tend to undo entanglement. Are we entangled? No, because the collapse of the wave function destroys the entanglement. As for who or what collapses the wave function, or if it can be reversed, that is the million dollar question.
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It is true everything is entangled. However, in the many worlds interpretation, the wave function of the universe splits into many worlds. Typically, each decohered world taken by itself will be far less entangled than the wave function as a whole. While taking the partial trace to a subsystem can lead to a very mixed state when done over the complete wave function, the partial trace taken only over one of the worlds will most likely lead to a less mixed state.
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If you have two subsystems that are entangled and forming an (ideal) pure state, it is still meaningful to consider a subsystem experimentally. If I entangle two atoms and give one to you, nothing prevents you from making measurements on "your" atom only.
By repeating this same experiment many times, with atoms prepared in the same way, you would be able to characterise the density matrix describing your subsystem. You wouldn't be able to characterise the joint wave function of the two atoms though.
