Most of the generic science news sources – see e.g. PC Magazine – report on a new result by experimenters at a Kavli-named institute in Delft, a historical academic town in Holland, that just appeared in Science:
Their evil device...
They have made some progress in the experimental work that could be useful for quantum computers in the future – potentially but not certainly foreseeable future. Two qubits – electron spins somewhere in two pieces of a diamond that are 10 feet away – are entangled, stored as nuclear spins, guaranteed to be sufficiently long-lived, and measured to be almost perfectly entangled.
I don't follow every experimental work of this kind and I won't pretend that I do. This prevents me from safely knowing how new their work is. I hope and want to believe it is sufficiently new, indeed. Obviously, quantum computers will require us to master many more operations than this one – potentially and probably more difficult ones.
However, the words chosen in the paper and in the popularization of the result are a mixed bag.
There have been too many recent TRF blog posts about quantum foundations so I will be brief. The popular articles usually correctly point out that experiments of this sort clearly show that Einstein was wrong etc.
On the other hand, they also tend to present "teleportation" as an actual transfer of information. Even the abstract of the Science paper starts with a comment about a "robust quantum information transfer". There is no actual transfer of information here in entanglement, however. Indeed, such long-distance influences would require unbelievably strong forces and their faster-than-light speed would contradict the laws and symmetries of the special theory of relativity.
What do the entangled bits look like? They are described by the state vector of the form\[
\ket\psi = \frac{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow} }{\sqrt{2}}
\] where the two characters inside the ket-vectors describe the state of the electrons in the left diamond and the right diamond, respectively. You see that if the left diamond spin is "up" (counter-clockwise if you look from the bird perspective), the right diamond spin is guaranteed to be "down", and vice versa.
Such a perfect anticorrelation could be achieved in classical physics, too. But the spins would have to objectively be either in the \((\uparrow\downarrow)\) classical state or the \((\downarrow\uparrow)\) state before the measurements of the spins. However, if this were so, one would easily see that if we perform measurements of the spin with respect to another pair of axes, like a horizontal axis, the outcomes \((\leftarrow\leftarrow)\), \((\leftarrow\rightarrow)\), \((\rightarrow\leftarrow)\), \((\rightarrow\rightarrow)\) would have the probability 25 percent each. It's because the two spins have to be independent.
\ket{\uparrow} = \frac{ \ket{\leftarrow} + \ket{\rightarrow} }{\sqrt{2}},\quad
\ket{\downarrow} = \frac{ \ket{\leftarrow} - \ket{\rightarrow} }{\sqrt{2}},
\] the state may be written as\[
\ket\psi = \frac{-\ket{\leftarrow\rightarrow}+\ket{\rightarrow\leftarrow} }{\sqrt{2}}
\] The relative phase, the absolute phase, and even the question whether it's a perfect correlation or perfect anticorrelation depends on some conventions for the basis vectors and the directions of the axes. I don't want to torture you with my particular choice here but I hope that my conventions have been internally consistent and if you rewrite the vertical spin basis in terms of the horizontal one (using the same absolute directions and the same formulae for both spins), you will get the result I wrote down.
But the more general and more important point is that instead of the 25%,25%,25%,25% distribution predicted by classical physics, one gets a perfect (anti)correlation i.e. 0%,50%,50%,0% again!
This (anti)correlation is guaranteed to be observed, so the reduced \(\ket{\psi_B}\) for the other spin may be immediately deduced from the measurement of the electron at \(A\). As long as you agree that relativity holds and material objects or genuine information can't propagate faster than light, the fact that \(\ket{\psi_B}\) instantaneously changed implies that \(\ket\psi\) encodes subjective information only. The change has only occurred in your head. You learned something about the spin \(A\), and because you knew that the spin \(B\) had a perfectly anticorrelated state, it follows that you may say what the spin at \(B\) is going to be.
Conceptually, this "collapse" is nothing else than the transition from the original complicated multivariate probability distributions such as \(p_{s_1,s_2}\) here or \(\rho(x_A,x_B)\) elsewhere (that exist even in classical statistical physics) to conditional probabilities such as \(p_{\leftarrow,s_2}\) here (or probability distributions of the kind \(\rho(17,x_B)\) elsewhere) that take the already known results of measurements ("conditions") into account. This transition to simplified probability distributions also occurs in your head only. It is not just an analogy. All the actual probability distributions that may be extracted from the state vector \(\ket\psi\) or \(\rho\) in quantum mechanics are doing exactly what the usual "adjustment of probabilities" is expected to do in Bayesian inference. Bayesian inference isn't dependent on classical physics in any way. It's a way of logical, probabilistic thinking and it's fully operational and unmodified in quantum mechanics, too.
This perfect (anti)correlation follows from the structure of the state \(\ket\psi\). And the reason why this state \(\ket\psi\) has the form it has isn't hiding in any event right before the measurements. Instead, the state \(\ket\psi\) has this form because that's how the two electron spins in the diamond were prepared. In the past, the spins came from the same small region of space – they were in a mutual contact. For example, \(\ket\psi\) is the only "singlet" i.e. two-spin state that has the total angular momentum \(|\vec J|^2=0\). So if you guarantee that the two spins are created from an object without any angular momentum, you automatically guarantee the perfect entanglement – the perfect (anti)correlation regardless of the type of the measurement that is made (if you make the same measurement on both spins).
So the reason behind all these correlations – correlations implied by quantum entanglement – is the same as the reason behind the anticorrelation of the two Bertlmann's socks. The color of the socks was decided by Herr Bertlmann in the morning, at the same piece of his brain, so that it was guaranteed that the two socks didn't have the same color (because he is a whackadoodle). The "only" new feature of the entanglement is that the observables in quantum mechanics don't commute with each other which may, perhaps ironically but demonstrably, increase (but also decrease) the degree of (anti)correlation between various pairs of quantities of the two electrons. Classical physics – a framework of physics that assumes the world to be perfectly described by some "objective reality" even in between any two measurements – wouldn't allow that but it's perfectly OK because classical physics is wrong.
Accuracy of the states
Now I want to mention something about the accuracy of the state \(\ket\psi\), the reliability of the correlation. Errors may arise
The Hilbert space of the two qubits is \(2\times 2\) i.e. four-dimensional. The entangled state \(\ket\psi\) I started with may be viewed as one basis vector of this four-dimensional space. It's likely that if you prepare the two spins in this state, there will be a small admixture of the other states, including those that are not anticorrelated. This admixture may also be created by some evolution between the production and the measurement – some deterioration of the state, so to say. The admixture will have coefficients of order \(\alpha,\beta\ll 1\):\[
\psi\to \ket{\psi'} = \psi + \alpha\ket{\uparrow\uparrow}+\beta\ket{\downarrow\downarrow}
\] You may think of \(\alpha,\beta\) as some small angles (in radians) with added phases in the four-dimensional complex Hilbert space. Imagine that \(\alpha,\beta\sim 10^{-10}\) because angles, at least if they're actual geometric angles, may be prepared this accurately.
The point I want to make is that if you measure the spins in the realistic initial state \(\ket{\psi'}\), you will sometimes – very rarely – see that the two spins are not anticorrelated because of the two new contributions where the spins are aligned rather than opposite.
But the key point is that \(\alpha,\beta\) are probability amplitudes so the actual probabilities that you will find both spins up or both spins down will be \[
P_{\uparrow\uparrow} = |\alpha|^2, \quad P_{\downarrow\downarrow}=|\beta|^2
\] and these probabilities are of order \(10^{-20}\), much smaller than the initial angles! Exactly the same comments apply to the imprecision in the final measurements of the spins. Imagine that you measure the spin of the electrons with respect to the axis \(z'\) that isn't exactly vertical but instead, is tilted at a small angle \(\alpha\) or \(\beta\) relatively to the vertical axis \(z\). It is equivalent to adding terms (ket vectors with the "wrong spin") of order \(\alpha\) or \(\beta\) to the state vector, too.
Again, the probability that you measure the "wrong" spin scales like the second power of the small angle!
So the probability of a mistake is really, really small.
I think that this point about quantum mechanics isn't being emphasized sufficiently often but I have started to do so. Quantum mechanics is often presented as being "more fuzzy" than classical physics, in all respects. But it ain't really true. Quantum mechanics is a different theory than classical physics. In some respects, it may be more fuzzy but in others, it is much more accurate.
If you take an object in classical physics and rotate it by a tiny angle \(\alpha\) around some axis, the errors in all generic observables will be of order \(\alpha\), right? The exactly vertical vector should have \(x,y\)-components equal to zero but the rotation makes them of order \(\alpha\). They enter linearly into many things that matter (although others may depend on \(x^2\) only if the linear terms happen to cancel) so you will find effects that were vanishing in the exactly vertical case but that will be non-vanishing, of order \(\alpha\), in the slightly tilted case.
However, if you consider a quantum system with a discrete spectrum, e.g. the spin, the rotation of a system by a small angle \(\alpha\) only changes some vanishing probability amplitudes from zero to \(O(\alpha)\) and it means that the actual probability of all results of a measurement or processes will be of order \(O(\alpha^2)\), much smaller than in classical physics! Quantum mechanics will entirely prohibit any errors whose probability would already scale as \(\alpha\).
So because of this squaring, quantum mechanics actually makes the behavior of physical systems with a discrete spectrum much more immune towards small transformations and rotations that may arise as errors. This is an example of the meme that quantum mechanics may be much less fuzzy than classical physics would ever allow.
The exact values of the discrete eigenvalues – e.g. the energy of an atom – are also something that classical physics could never achieve. Every fluctuation would always be allowed in classical physics. That's why nothing like atomic clocks that can measure time with the accuracy of \(10^{-16}\) would probably ever be built in a classical world!
You may also see that quantum mechanics is able to be "equally as" or "more determined than" classical physics on some quantum games. One may prepare initial states that guarantee some perfect correlations or anticorrelations that directly contradict with what you could derive from the classical logic. The wisdom of these games – infinitely many games – captures all the surprises in Bell's theorem but the contradiction becomes much cleaner and more strict.
(In general, people who are obsessed with Bell's theorem are lousy thinkers for whom group think and parroting of other lousy thinkers is much more likely than a penetrating individual thinking. It's just one particular setup where quantum mechanics shows predictions not reproducible by a classical model. But the truth is that virtually every small quantum system is predicted to behave differently by quantum mechanics than it is by classical physics. And the Bell's setup is neither among the cleverest ones nor among the sharpest or most impressive ones.)
Quantum mechanics is very different and despite its undeserved "fuzzy" image, its being different is actually making many things – including CHSH games, atomic clocks, and entangled pairs – more robust, sharper, and more immune against errors and drifting. Many things that are as sharp as they are in the world around us – perhaps including the reliable RNA/DNA code – would arguably never reach this level of accuracy in a world governed by a classical theory. I mean any classical theory – a theory based on the continuous evolution of objective variables as functions of time.
Thanks, quantum mechanics, for the life and everything that just works in our world.
Unconditional quantum teleportation between distant solid-state quantum bits (SciMag) by Pfaff, Hensen, Hanson, and 8 more co-authorsLet me emphasize that Hansen isn't among the authors. ;-)
Their evil device...
They have made some progress in the experimental work that could be useful for quantum computers in the future – potentially but not certainly foreseeable future. Two qubits – electron spins somewhere in two pieces of a diamond that are 10 feet away – are entangled, stored as nuclear spins, guaranteed to be sufficiently long-lived, and measured to be almost perfectly entangled.
I don't follow every experimental work of this kind and I won't pretend that I do. This prevents me from safely knowing how new their work is. I hope and want to believe it is sufficiently new, indeed. Obviously, quantum computers will require us to master many more operations than this one – potentially and probably more difficult ones.
However, the words chosen in the paper and in the popularization of the result are a mixed bag.
There have been too many recent TRF blog posts about quantum foundations so I will be brief. The popular articles usually correctly point out that experiments of this sort clearly show that Einstein was wrong etc.
On the other hand, they also tend to present "teleportation" as an actual transfer of information. Even the abstract of the Science paper starts with a comment about a "robust quantum information transfer". There is no actual transfer of information here in entanglement, however. Indeed, such long-distance influences would require unbelievably strong forces and their faster-than-light speed would contradict the laws and symmetries of the special theory of relativity.
What do the entangled bits look like? They are described by the state vector of the form\[
\ket\psi = \frac{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow} }{\sqrt{2}}
\] where the two characters inside the ket-vectors describe the state of the electrons in the left diamond and the right diamond, respectively. You see that if the left diamond spin is "up" (counter-clockwise if you look from the bird perspective), the right diamond spin is guaranteed to be "down", and vice versa.
Such a perfect anticorrelation could be achieved in classical physics, too. But the spins would have to objectively be either in the \((\uparrow\downarrow)\) classical state or the \((\downarrow\uparrow)\) state before the measurements of the spins. However, if this were so, one would easily see that if we perform measurements of the spin with respect to another pair of axes, like a horizontal axis, the outcomes \((\leftarrow\leftarrow)\), \((\leftarrow\rightarrow)\), \((\rightarrow\leftarrow)\), \((\rightarrow\rightarrow)\) would have the probability 25 percent each. It's because the two spins have to be independent.
You may have noticed that I didn't use the ket vector notation but rather parentheses in the previous paragraph. It is no mistake. On the contrary; it was absolutely deliberate because the states represented configurations in classical physics, not quantum states, and I used the logic of classical physics – its basic framework – to deduce what would be observed. Too many deluded people are using the word "quantum" and sometimes even quantum symbols but they still think classical. They still think that the physical system must be described by an objectively correct classical information at every moment, even before the measurements. That's too bad. If you aren't answering all questions in science – in principle – according to the quantum rules that compute probabilities from complex amplitudes obtained from bra and ket vectors and actions by linear operators and that urge you to avoid any "visualizations" of the "actual arrangement of the system" before the measurement, then you are not doing quantum mechanics, you are not thinking in agreement with modern physics, and you shouldn't use any words that quantum mechanics brought us! You should only use the vocabulary and symbols up to the 19th century because this is where you are hopelessly stuck and you only drown yourself in mud if you are using words you can't possibly understand.Quantum mechanics says something else. Even in this different basis used for the spins, the electrons are perfectly (anti)correlated because by linear combinations of the basis vectors (applied twice, using the distribution law etc.)\[
\ket{\uparrow} = \frac{ \ket{\leftarrow} + \ket{\rightarrow} }{\sqrt{2}},\quad
\ket{\downarrow} = \frac{ \ket{\leftarrow} - \ket{\rightarrow} }{\sqrt{2}},
\] the state may be written as\[
\ket\psi = \frac{-\ket{\leftarrow\rightarrow}+\ket{\rightarrow\leftarrow} }{\sqrt{2}}
\] The relative phase, the absolute phase, and even the question whether it's a perfect correlation or perfect anticorrelation depends on some conventions for the basis vectors and the directions of the axes. I don't want to torture you with my particular choice here but I hope that my conventions have been internally consistent and if you rewrite the vertical spin basis in terms of the horizontal one (using the same absolute directions and the same formulae for both spins), you will get the result I wrote down.
But the more general and more important point is that instead of the 25%,25%,25%,25% distribution predicted by classical physics, one gets a perfect (anti)correlation i.e. 0%,50%,50%,0% again!
This (anti)correlation is guaranteed to be observed, so the reduced \(\ket{\psi_B}\) for the other spin may be immediately deduced from the measurement of the electron at \(A\). As long as you agree that relativity holds and material objects or genuine information can't propagate faster than light, the fact that \(\ket{\psi_B}\) instantaneously changed implies that \(\ket\psi\) encodes subjective information only. The change has only occurred in your head. You learned something about the spin \(A\), and because you knew that the spin \(B\) had a perfectly anticorrelated state, it follows that you may say what the spin at \(B\) is going to be.
Conceptually, this "collapse" is nothing else than the transition from the original complicated multivariate probability distributions such as \(p_{s_1,s_2}\) here or \(\rho(x_A,x_B)\) elsewhere (that exist even in classical statistical physics) to conditional probabilities such as \(p_{\leftarrow,s_2}\) here (or probability distributions of the kind \(\rho(17,x_B)\) elsewhere) that take the already known results of measurements ("conditions") into account. This transition to simplified probability distributions also occurs in your head only. It is not just an analogy. All the actual probability distributions that may be extracted from the state vector \(\ket\psi\) or \(\rho\) in quantum mechanics are doing exactly what the usual "adjustment of probabilities" is expected to do in Bayesian inference. Bayesian inference isn't dependent on classical physics in any way. It's a way of logical, probabilistic thinking and it's fully operational and unmodified in quantum mechanics, too.
This perfect (anti)correlation follows from the structure of the state \(\ket\psi\). And the reason why this state \(\ket\psi\) has the form it has isn't hiding in any event right before the measurements. Instead, the state \(\ket\psi\) has this form because that's how the two electron spins in the diamond were prepared. In the past, the spins came from the same small region of space – they were in a mutual contact. For example, \(\ket\psi\) is the only "singlet" i.e. two-spin state that has the total angular momentum \(|\vec J|^2=0\). So if you guarantee that the two spins are created from an object without any angular momentum, you automatically guarantee the perfect entanglement – the perfect (anti)correlation regardless of the type of the measurement that is made (if you make the same measurement on both spins).
So the reason behind all these correlations – correlations implied by quantum entanglement – is the same as the reason behind the anticorrelation of the two Bertlmann's socks. The color of the socks was decided by Herr Bertlmann in the morning, at the same piece of his brain, so that it was guaranteed that the two socks didn't have the same color (because he is a whackadoodle). The "only" new feature of the entanglement is that the observables in quantum mechanics don't commute with each other which may, perhaps ironically but demonstrably, increase (but also decrease) the degree of (anti)correlation between various pairs of quantities of the two electrons. Classical physics – a framework of physics that assumes the world to be perfectly described by some "objective reality" even in between any two measurements – wouldn't allow that but it's perfectly OK because classical physics is wrong.
Accuracy of the states
Now I want to mention something about the accuracy of the state \(\ket\psi\), the reliability of the correlation. Errors may arise
- because the initial state isn't prepared exactly in the idealized state I described or because the state deteriorates between the production or the measurement or
- because the two individual electron spins aren't measured exactly
The Hilbert space of the two qubits is \(2\times 2\) i.e. four-dimensional. The entangled state \(\ket\psi\) I started with may be viewed as one basis vector of this four-dimensional space. It's likely that if you prepare the two spins in this state, there will be a small admixture of the other states, including those that are not anticorrelated. This admixture may also be created by some evolution between the production and the measurement – some deterioration of the state, so to say. The admixture will have coefficients of order \(\alpha,\beta\ll 1\):\[
\psi\to \ket{\psi'} = \psi + \alpha\ket{\uparrow\uparrow}+\beta\ket{\downarrow\downarrow}
\] You may think of \(\alpha,\beta\) as some small angles (in radians) with added phases in the four-dimensional complex Hilbert space. Imagine that \(\alpha,\beta\sim 10^{-10}\) because angles, at least if they're actual geometric angles, may be prepared this accurately.
The point I want to make is that if you measure the spins in the realistic initial state \(\ket{\psi'}\), you will sometimes – very rarely – see that the two spins are not anticorrelated because of the two new contributions where the spins are aligned rather than opposite.
But the key point is that \(\alpha,\beta\) are probability amplitudes so the actual probabilities that you will find both spins up or both spins down will be \[
P_{\uparrow\uparrow} = |\alpha|^2, \quad P_{\downarrow\downarrow}=|\beta|^2
\] and these probabilities are of order \(10^{-20}\), much smaller than the initial angles! Exactly the same comments apply to the imprecision in the final measurements of the spins. Imagine that you measure the spin of the electrons with respect to the axis \(z'\) that isn't exactly vertical but instead, is tilted at a small angle \(\alpha\) or \(\beta\) relatively to the vertical axis \(z\). It is equivalent to adding terms (ket vectors with the "wrong spin") of order \(\alpha\) or \(\beta\) to the state vector, too.
Again, the probability that you measure the "wrong" spin scales like the second power of the small angle!
So the probability of a mistake is really, really small.
I think that this point about quantum mechanics isn't being emphasized sufficiently often but I have started to do so. Quantum mechanics is often presented as being "more fuzzy" than classical physics, in all respects. But it ain't really true. Quantum mechanics is a different theory than classical physics. In some respects, it may be more fuzzy but in others, it is much more accurate.
If you take an object in classical physics and rotate it by a tiny angle \(\alpha\) around some axis, the errors in all generic observables will be of order \(\alpha\), right? The exactly vertical vector should have \(x,y\)-components equal to zero but the rotation makes them of order \(\alpha\). They enter linearly into many things that matter (although others may depend on \(x^2\) only if the linear terms happen to cancel) so you will find effects that were vanishing in the exactly vertical case but that will be non-vanishing, of order \(\alpha\), in the slightly tilted case.
However, if you consider a quantum system with a discrete spectrum, e.g. the spin, the rotation of a system by a small angle \(\alpha\) only changes some vanishing probability amplitudes from zero to \(O(\alpha)\) and it means that the actual probability of all results of a measurement or processes will be of order \(O(\alpha^2)\), much smaller than in classical physics! Quantum mechanics will entirely prohibit any errors whose probability would already scale as \(\alpha\).
So because of this squaring, quantum mechanics actually makes the behavior of physical systems with a discrete spectrum much more immune towards small transformations and rotations that may arise as errors. This is an example of the meme that quantum mechanics may be much less fuzzy than classical physics would ever allow.
The exact values of the discrete eigenvalues – e.g. the energy of an atom – are also something that classical physics could never achieve. Every fluctuation would always be allowed in classical physics. That's why nothing like atomic clocks that can measure time with the accuracy of \(10^{-16}\) would probably ever be built in a classical world!
You may also see that quantum mechanics is able to be "equally as" or "more determined than" classical physics on some quantum games. One may prepare initial states that guarantee some perfect correlations or anticorrelations that directly contradict with what you could derive from the classical logic. The wisdom of these games – infinitely many games – captures all the surprises in Bell's theorem but the contradiction becomes much cleaner and more strict.
(In general, people who are obsessed with Bell's theorem are lousy thinkers for whom group think and parroting of other lousy thinkers is much more likely than a penetrating individual thinking. It's just one particular setup where quantum mechanics shows predictions not reproducible by a classical model. But the truth is that virtually every small quantum system is predicted to behave differently by quantum mechanics than it is by classical physics. And the Bell's setup is neither among the cleverest ones nor among the sharpest or most impressive ones.)
Quantum mechanics is very different and despite its undeserved "fuzzy" image, its being different is actually making many things – including CHSH games, atomic clocks, and entangled pairs – more robust, sharper, and more immune against errors and drifting. Many things that are as sharp as they are in the world around us – perhaps including the reliable RNA/DNA code – would arguably never reach this level of accuracy in a world governed by a classical theory. I mean any classical theory – a theory based on the continuous evolution of objective variables as functions of time.
Thanks, quantum mechanics, for the life and everything that just works in our world.
The Dutch teleportation advance
![The Dutch teleportation advance]()
Reviewed by MCH
on
May 31, 2014
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