Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I recently came across two nice papers on the foundations of quantum mechancis, Aaronson 2004 and Hardy 2001. Aaronson makes the statement, which was new to me, that nonlinearity in QM leads to superluminal signaling (as well as the solvability of hard problems in computer science by a nonlinear quantum computer). Can anyone offer an argument with crayons for why this should be so?

It seems strange to me that a principle so fundamental and important can be violated simply by having some nonlinearity. When it comes to mechanical waves, we're used to thinking of a linear wave equation as an approximation that is always violated at some level. Does even the teensiest bit of nonlinearity in QM bring causality to its knees, or can the damage be limited in some sense?

Does all of this have any implications for quantum gravity -- e.g., does it help to explain why it's hard to make a theory of quantum gravity, since it's not obvious that quantum gravity can be unitary and linear?

S. Aaronson, "Is Quantum Mechanics An Island In Theoryspace?," 2004, arXiv:quant-ph/0401062.

L. Hardy, "Quantum theory from five reasonable axioms," 2001, arXiv:quant-ph/0101012.

share|improve this question
add comment

2 Answers

The no-cloning theorem prevents superluminal communication via quantum entanglement. Consider the EPR thought experiment, and suppose quantum states could be cloned. Alice could send bits to Bob in the following way:

If Alice wishes to transmit a $0$, she measures the spin of her electron in the $z$ direction (she could have chosen another direction if she had wished to transmit a $1$), collapsing Bob's state to either $|z+\rangle_B$ or $|z-\rangle_B$. Bob creates many copies of his electron's state, and measures the spin of each copy in the $z$ direction. Bob will know that Alice has transmitted a $0$ if all his measurements will produce the same result; otherwise, his measurements will be split evenly between $+1/2$ and $-1/2$. This would allow Alice and Bob to communicate across space-like separations (if the two components of the entangled state are space-like separated), potentially violating causality.

The linearity of quantum mechanics, however, prevents us from cloning arbitrary, unknown states (photoQopiers don't exist). Let's see it. Suppose the state of a quantum system A , which we wish to copy. The state can be written as

$$|\psi\rangle_A = a |0\rangle_A + b |1\rangle_A$$

The coefficients ''a'' and ''b'' are unknown to us. In order to make a copy, we take a system B with an identical Hilbert space and initial state $|e\rangle_B$ (which must be independent of $|\psi\rangle_A$, of which we have no prior knowledge). The composite system is then described by

$$|\psi\rangle_A |e\rangle_B$$

There are only two ways to manipulate the composite system. We could perform an observation, which irreversibly would collapse the system into some eigenstate of the observable, corrupting the information contained in A. This is obviously not what we want. Alternatively, we could control the Hamiltonian of the system, and thus the time evolution operator $U$, which is a linear operator. Then $U$ acts as a photoQopier provided

$$U |\psi\rangle_A |e\rangle_B= |\psi\rangle_A |\psi\rangle_B = (a |0\rangle_A + b |1\rangle_A)(a |0\rangle_B + b |1\rangle_B)= \\ a^2 |0\rangle_A |0\rangle_B + a b |0\rangle_A |1\rangle_B + b a |1\rangle_A |0\rangle_B + b^2 |1\rangle_A |1\rangle_B$$

for all $\psi$. This must then be true for the basis states as well, so

$$U |0\rangle_A |e\rangle_B = |0\rangle_A |0\rangle_B$$ $$U |1\rangle_A |e\rangle_B = |1\rangle_A |1\rangle_B$$

Then the linearity of $U$ implies

$$U |\psi\rangle_A |e\rangle_B= U (a |0\rangle_A + b |1\rangle_A)|e\rangle_B= a |0\rangle_A |0\rangle_B + b |1\rangle_A |1\rangle_B \\ \ne a^2 |0\rangle_A |0\rangle_B + a b |0\rangle_A |1\rangle_B + b a |1\rangle_A |0\rangle_B + b^2 |1\rangle_A |1\rangle_B$$

Thus, $U |\psi\rangle_A |e\rangle_B$ is generally not equal to $|\psi\rangle_A |\psi\rangle_B$ so that $U$ cannot act as a general photoQopier.

Edit. About the relation between no-cloning and Heisenberg uncertainty principle: In the 'quantiki; link below says:

The no-cloning theorem protects the uncertainty principle in quantum mechanics. If one could clone an ''unknown'' state, then one could make as many copies of it as one wished, and measure each dynamical variable with arbitrary precision, thereby bypassing the uncertainty principle. This is prevented by the non-cloning theorem.

This — at least superficially — seems to be a fallacy because, as Bruce Connor pointed out, regardless of if one has an infinite number of exact copies of an arbitrary, unknown state, non-compatible observables (non-commuting operators) cannot be measured simultaneously with arbitrary precision. So, if Alice measures the spin's $z$ component of her electron, and even if Bob could create many copies of his electron state (what is forbidden by the non-cloning theorem discussed above), Bob could not determine the value of the $x$ component with arbitrary precision — he would find a dispersion in the values.

Some other references claiming the opposite from a comment of mine: @BruceConnor I totally agree with you on this. I don't know why I copied that paragraph from the link. It seems to be an extended obvious fallacy. Or perhaps I'm wrong, because my knowledge about this is not so deep and it seems that this claim is even written in books ("no-cloning theorem basically implies uncertainty principle and viceversa", Quantum Computing Since Democritus, S. Aaronson) and in a PRL paper ("the no-cloning theorem yields a new formulation of the quantum uncertainty principle that applies to individual systems [19]", DOI: 10.1103/PhysRevLett.88.210601 . However in the cited reference [19], I don't see a clear statement about Heisenberg principle). – drake 18 mins ago

EDIT. Another comment:

Trimok: Quantum mechanics is linear, and respects no-signaling. It is a causal theory. Speaking about non-linear quantum mechanics is like speaking about non-linear "Lorentz" transformations.. –

Me: The issue is that linearity frequently involves some sort of approximation. Therefore, going beyond linearity or introducing non-linearities is a way of exploring new theories. To follow your examples: special relativity introduces a non-linearity in the law of composition of velocities with respect to Galilean physics. This allows the introduction of a new velocity scale, c, and its invariance under boost transformations. Likewise, the dependence of the power radiated by a blackbody on the temperature gets non-linearized in quantum mechanics with respect to classical physics. Of course, other laws in special relativity and quantum mechanics (such as the transformation of coordinates in special relativity) remain linear. Thus, where to introduce non-linearities in the known physical laws is a non-trivial task.

Almost copied from http://www.quantiki.org/wiki/The_no-cloning_theorem

share|improve this answer
    
So linearity implies no cloning implies no signalling (if we assume that signalling with cloning is the only way one could potentially signal). This means that signalling implies nonlinearity. If I understand the question though, the OP wants to know how/why nonlinearity implies signalling. –  SMeznaric Aug 9 '13 at 8:28
    
I do not agree. You cannot transmit information via an entangled state. Alice cannot control the outcome of its local measurement. –  Trimok Aug 9 '13 at 11:22
1  
This is a really good answer, but I disagree completely with the first paragraph (everything else is great). The uncertainty principle is a mathematical property of quantum states and non-commuting observables, it won't be violated by being able to measure the system many times. The reason is that even if you know a quantum state perfectly (the limit if infinite measurements) the uncertainties don't go to zero. –  Bruce Connor Aug 9 '13 at 12:10
    
@SMeznaric no cloning implies no signalling Not exactly. No cloning allows no signaling. And cloning would imply signaling. –  Bruce Connor Aug 9 '13 at 12:15
1  
@BruceConnor I totally agree with you on this. I don't know why I copied that paragraph from the link. It seems to be an extended obvious fallacy. Or perhaps I'm wrong, because my knowledge about this is not so deep and it seems that this claim is even written in books (Quantum Computing Since Democritus, S. Aaronson) and in a PRL paper ("the no-cloning theorem yields a new formulation of the quantum uncertainty principle that applies to individual systems [19]", DOI: 10.1103/PhysRevLett.88.210601 . However in the cited reference [19], I don't see a clear statement about Heisenberg principle). –  drake Aug 9 '13 at 18:47
show 10 more comments

Suppose you are given a nonlinear operator $N$. Then there exist at least two states $\rho$ and $\sigma$ such that $N(\alpha \rho + \beta \sigma) \neq \alpha N(\rho) + \beta N(\sigma)$. Here $\alpha + \beta = 1$, and $\alpha, \beta \geq 0$. So if you definitely have either a $\rho$ or a $\sigma$ with probabilities $\alpha$ and $\beta$ (say resulting from some kind of a measurement) and then apply $N$, you get $ \alpha N(\rho) + \beta N(\sigma)$. On the other hand if you cannot know which state you were given (resulting for example from having only a part of a pure state) and then apply $N$, you get $N(\alpha \rho + \beta \sigma)$.

Notice that this is an extremely undesirable property. There should be no statistical difference between being given a part of a purification of a mixed state, or the mixed state constructed by rolling a die and randomly dishing out states. However, nonlinear operators seem to have the property of being able to tell the two apart.

Here is how you can exploit this to signal superluminally. Let $|\psi\rangle$ be the purification of $\rho$ and $|\phi\rangle$ the purification of $\sigma$. In other words $Tr_A\left(|\psi\rangle\langle \psi|\right) = \rho$ and $Tr_A\left(|\phi\rangle\langle \phi|\right) = \sigma$. Since $\rho \neq \sigma$, we also have that $|\psi\rangle \neq |\phi\rangle$. Let's call A subsystem Alice and B Bob, as usual.

So suppose Alice creates a superposition of these two pure states and keeps the ability to keep track of them. In other words, she has $\sqrt{\alpha} |\psi\rangle \otimes |0\rangle + \sqrt{\beta} |\phi\rangle \otimes |1\rangle$. She sends Bob the part of $\psi$ and $\phi$ that reduces to $\rho$ and $\sigma$. Also suppose Alice and Bob have many such copies available prepared in the same way.

Now if Alice wants to send 0, she does nothing and Bob's part of the state is $\alpha \rho + \beta \sigma$. Bob now applies $N$ and has $N(\alpha \rho + \beta \sigma)$. Alternatively, if she wants to send a 1, she measures her ancilla. After that Bob has a definite $\rho$ with probability $\alpha$ or a definite $\sigma$ with probability $\beta$. Applying $N$ now results in a state that is statistically equivalent to $\alpha N(\rho) + \beta N(\sigma)$. Since these two states are different and Alice + Bob have many copies, Bob can now distinguish which of the two states he has, and has therefore received one bit of information superluminally. Notice that if $\rho$ and $\sigma$ are actually orthogonal pure states, then they only need a single copy to do that.

share|improve this answer
    
Good generalization of drake's argument +1 –  WetSavannaAnimal aka Rod Vance Aug 19 '13 at 3:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.