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1) Can a state be entangled without also being a superposition? (Please give an example.)

2) Can a state be a superposition without being entangled? (Again, an example please.)

3) And what about a cat state?

I am royally confused, and a little bit of googling didn't help.
This question is partly motivated by the recent news announcement (Nature, NYTimes) that ten billion entangled pairs have been created.

4) Are these spin pairs also in a superposition?

5) Does this count as a cat state?

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Suggestion to the question(v3): Also ask about the difference with mixed and pure states while we are at it. –  Qmechanic Oct 16 '12 at 19:05
    
Fine with me. Or ask as a separate question. –  Jim Graber Oct 16 '12 at 19:40

3 Answers 3

up vote 15 down vote accepted

This goes on order from the more general to the more specific:

  1. If A and B are two states, then you can form any linear combination to get superposed state, let's call it C. Of course, any state C can be expressed as superposition of two other states A and B, so calling a state "superposed" is not all that meaningful, unless there is something distinguishing the A,B states from all their superpositions.

  2. You can only discuss entanglement if you first distinguish two (or more) sub-systems of the full system, for example if the distinguished sub-systems do not interact with each other, or if they are spatially separated. In this case states of system (1) can be entangled with states of system (2), if the total state cannot be written as a product $A=A_1 \times A_2$ where $A_1$ and $A_2$ are states of the separate subsystems. A more formal way to quantify that is the "entanglement entropy", which tells you roughly how much information you are missing if you have access to one sub-system only.

  3. The cat state is a particular example of an entangled state. Specifically in systems where two (or more) sub-systems consist of exactly one (qu)bit of information, which is labeled as 0 or 1, this is the state |00...0> +|11...1>. I am also told in comments there are generalizations beyond this system of n qubits, and more physical ways to characterize the state.

So, if you decide that your product states as above are distinguished over any linear combination thereof, there is a meaningful comparison between what you call "superposed" state (which I have to say is not a standard terminology) and the entangled states. An entangled state is one that cannot be expressed as product state, so it is necessarily a superposition of such states, say $A_1 \times A_2$ and $B_1 \times B_2$. The converse is not true: a superposition of product states may well be a product state itself.

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I won't pretend I understand the theory of entangled states, but this seems like a good explanation. –  Noldorin Jan 24 '11 at 0:04
    
Nice, Moshe, +1. –  Luboš Motl Jan 24 '11 at 7:27
    
I agree with 1 and 2, but the definition of a cat state is wrong. A cat state is the superposition of two classically different states. The tricky part of the definition is what constitutes "classicaly different" states. But there is no direct link with entanglement. –  Frédéric Grosshans Jan 24 '11 at 10:25
    
Thanks, I’ve edited the answer to hopefully a more accurate one. –  user566 Jan 24 '11 at 14:02
    
I looked at several Intro to QM textbooks, including Griffiths, Liboff and Merzbacher. I was amazed how little was there on this topic. In most cases, these terms were not even listed in the index. In common parlance(layman language), I believe superposition is usually used for contradictory states, such as spin up and spin down or live and dead or here and there. I found the distinction between one system and two distinguishable subsystems to be particularly helpful. Thanks to all. Jim Graber –  Jim Graber Jan 24 '11 at 17:11

Any discussion of entanglement implies you are thinking of (at least) two separate systems. Superpositions on the other hand, can also apply to situations where you are only looking at a single system. Keeping this in mind:

"Can a state be entangled without also being a superposition? (Please give an example)"

No. An entangled state is by definition a state which cannot be written as a separable product in any basis. Whatever basis you choose, the state will be a superposition of products of the separate systems.

"Can a state be a superposition without being entangled? (Again, an example please)"

Of course. A trivial example would be a single isolated system for which you can usually find some basis in which the state is in a superposition. And single isolated systems cannot be entangled with anything else because.. well.. they are single isolated systems - and like I said before, entanglement applies to a situation where you have divided your world into separate systems that can interact with each other.

"And what about a cat state?"

A Schrodinger Cat state is usually used to refer to a macroscopic system which is in a superposition of two or more states (in the eigenstate basis of some physical observable like energy or position). Real cats in geiger-counter-poison-rigged boxes probably won't be in such superpositions due to decoherence but there are some surprisingly large objects like buckyballs and macromolecules which have been shown to exist in superpositions.

"Are these spin pairs also in a superposition?"

Yes, in all bases, because they are entangled. Allegedly :)

"Does this count as a cat state?"

I'm not sure because I haven't read the whole paper and also because the definition of a cat state depends on what you consider macroscopic.

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I've often heard the term "cat state" used in contexts like quantum computing to refer to entangled states of microscopic systems. –  David Z Jan 24 '11 at 6:40
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+1 Almost everything I intended to write. Just want to add that the 10 billion pairs do not count as a cat state, because each of the pair is independent of the others. A 10 billion qubits cat state would be closer o a GHZ state :$|0\rangle^{\otimes n}+|1\rangle^{\otimes n}$. –  Frédéric Grosshans Jan 24 '11 at 10:30

The difference between a superposition and entanglement is the following. We consider a two slit experiment where a photon wave function interacts with a screen. The wave vector is of the form $$ |\psi\rangle~=~e^{ikx}|1\rangle~+~e^{ik’x}|2\rangle $$ as a superposition of states for the slits labeled $1$ and $2$. The normalization is assumed. The state vector is normalized as $$ \langle\psi|\psi\rangle~=~1~=~\langle 1|1\rangle~+~\langle 2|2\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle $$ The overlaps $\langle 1|2\rangle$ and $\langle 2|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate.

We now consider the classic situation where one tries to measure which slit the photon traverses. We have a device with detects the photon at one of the slit openings. We consider another superposed quantum state. This is a spin space that is $$ |\phi\rangle~=~\frac{1}{\sqrt {2}}(|+\rangle~+~|-\rangle). $$ This photon quantum state becomes entangled with this spin state. So we have $$ |\psi,\phi\rangle~=~e^{ikx}|1\rangle|+\rangle~+~e^{ik’x}|2\rangle|-\rangle $$ which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state. Now consider the norm of this state vector $$ \langle|\psi,\phi|\psi,\phi\rangle~=~\langle 1|1\rangle\langle +|+\rangle~+~\langle 2|2\rangle\langle-|-\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle\langle +|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle\langle-|+\rangle. $$ The spin states $|+\rangle$ and $|-\rangle$ are orthogonal and thus $\langle +|-\rangle $ and $\langle-|+\rangle$ are zero. This means the overlap or interference terms are removed. In effect the superposition has been replaced by an entanglement.

This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state. So one does not need to invoke an outright collapse to illustrate how a superposition is lost.

The cat state is a form of the GHZ state where you have $$ |\psi\rangle~=~C(|0\rangle^{\otimes n}~+~|1\rangle^{\otimes n}) $$ which are n-partite entanglements. These types of states can violate Bell inequalities with one state, which illustrates quantum mechanics is not a purely statistical theory. The statistics are derived from QM.

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