## Friday, July 14, 2017

### Classical simplex vs quantum ball

and another huge difference between classical and quantum physics

Bell's theorem is often hyped as a very important result about the foundations of quantum mechanics. There are two basic problems with this assertion: the theorem isn't a theorem about quantum mechanics at all; and it is not important. It is a theorem about local classical theories (those that were ruled out around 1925) and its claimed importance – the proven difference between classical and quantum theories – must be trivial to see for every intelligent schoolkid.

Here, I try to show you another elementary difference between classical and quantum physics that differs from the ideas in previous hundreds of blog posts.

For the sake of simplicity, consider one qubit, the simplest nontrivial quantum system, and let's describe it as the spin of an electron. And let's ask: What is the space of all possible mixed states? Those describe all possible states of the electron's spin that we may know before some observations. While pure states are of the form$\ket\psi = \alpha\ket{\uparrow}+\beta\ket{\downarrow},\quad \alpha,\beta\in\CC$ the mixed states are $$2\times 2$$ Hermitian matrices. All of them may be written as a combination of the unit matrix and three Pauli matrices:$\rho = \frac 12 \zav{ 1+\vec n \cdot \vec \sigma }$ The coefficient of the unit matrix had to be $$1/2$$ for the trace of $$\rho$$, the total probability, to be equal to one. Hermiticity guarantees that $$\vec n$$ is a real 3-dimensional vector. Also, the density matrix cannot have negative eigenvalues which tells us that$|\vec n| \leq 1.$ This inequality is saturated if and only if the qubit is in a pure state i.e. if $$\exists \ket\psi:\,\,\rho =\ket\psi\bra\psi$$. Excellent. So the density matrix is parameterized by a vector with $$|\vec n|\leq 1$$, i.e. by a three-dimensional ball!

If you want to know the probability that the spin will be measured "up" relatively to an axis $$\vec a$$ with $$|\vec a|=1$$, then you may define the projection operator$P_{\vec a} = \frac 12 \zav{ 1+\vec a \cdot \vec \sigma }$ which has an analogous form to the density matrix itself. And the probability is$P(\rho_{\vec n}=\uparrow \vec a) = {\rm Tr}(\rho_{\vec n}\cdot P_{\vec a}) = \frac 12\zav {1+\vec n\cdot \vec a}.$ a simple linear function of the inner product of the two 3-dimensional vectors.

Now, what would be the corresponding space classically? If there were a finite number of distinguishable classical states, their probabilities would be $$p_i\in\RR$$ and these parameters would play exactly the same role as the parameters $$\vec n$$ describing the density matrix above. The conditions obeyed by $$p_i$$ would be simply$\sum_i p_i = 1, \quad \forall i:\,\,0\leq p_i \leq 1.$ In the $$p_i$$-space, what are these conditions geometrically? Well, they define a simplex. Depending on the number of parameters (values of the index $$i$$), we get simplices of different dimensions.

Now, the question is whether you can distinguish these simplices (space of values of a probability distribution) from a ball (the corresponding space according to quantum mechanics).

Well, you should be able to distinguish simplices from balls, triangles from disks. You know, simplices are spiky and have linear faces, balls are smooth and all their boundaries are curved. In fact, increasing the dimension of the classical space (of the simplex, perhaps to infinite values) makes the difference more brutal. The simplices become "even spikier" and more distant from the ball. That's particularly true if you tried to linearly project a high-dimensional simplex to a lower-dimensional subspace, to mathematically convey the assumption that only some of the dimensions are measurable as the spin and other dimensions are hidden variables.

You could formulate and prove many particular theorems morally showing that a ball is different from a simplex, indeed, but I think it's obvious that these things have to exist. You may simply eyeball the difference between simplices and balls. Note that the ball's being "more inflated" is the reason why the correlations predicted by quantum mechanics may exceed those in classical physics. Bell's inequality is just an obscure, unreadable, would-be deep way to demonstrate something as simple as the fact that a ball differs from a simplex.

You must have realized that quantum mechanics is ball-like because the evolution operators are unitary "rotations" (where squares of probability amplitudes have constant sums, like in the Pythagorean theorem) while classical physics is simplex-like because it evolves probabilities by stochastic matrices (where sums of probabilities in the columns themselves have to be one).

Try to think about this simple ball-vs-simplex dichotomy and realize that the ball-like shape is easily proven experimentally. I don't need to consider any arbitrary contrived correlations of several spin measurements such as those in Bell's theorem. The difference between classical and quantum physics is absolutely obvious and experiments directly prove that the quantum mechanical answer is the right now.

Quantum mechanics doesn't allow you to distinguish things easily

The second difference I want to mention is a reinterpretation of the uncertainty principle closely related to the low heat capacities in our, quantum world that are impossible to be predicted by any realistic theories; and to the fact that non-orthogonal pure states in quantum mechanics aren't mutually exclusive.

Let us ask how many pure states (or states fully specified as points in the phase space, in classical physics) may be distinguished if you repeat e.g. 1 million measurements with a system that is prepared in an identical state. To have a particular system in mind, think about the electron's spin again (one qubit).

According to quantum mechanics, the pure states are given by the vector $$\vec n$$ I mentioned above with the extra condition of purity, $$|\vec n| = 1$$. So they belong to a sphere. How accurately can you measure the location on the sphere – the latitude and the longitude – if you are given the same initial state prepared 1 million times?

Well, to distinguish the axes $$\vec n_1$$ and $$\vec n_2$$ that only differ by a small angle $$\alpha$$ on the sphere, you need to realize that these states will behave identically with the probability given by $$\vec n_1\cdot \vec n_2$$ or so, or $$\cos\alpha$$. If you expand the cosine, you will obtain $$1-\alpha^2/2+\dots$$. So the probability is only $$\alpha^2/2$$ or so (which is much smaller than $$\alpha$$) that these two pure states produce some different result.

It means that according to quantum mechanics, you may only distinguish vectors $$\vec n_1,\vec n_2$$ whose relative angle is of order $$1/1,000$$, where the denominator is the square root of 1 million, the number of repetitions of the situation. The bad resolution and the effective indistinguishability of the states with different values of $$\vec n$$ is just another example of the uncertainty principle.

Needless to say, in classical physics where an objective state of the system exists prior to the measurement, it must be in principle possible to distinguish any two states, even after one measurement. If you want to prove that the world is fundamentally realist or classical, it's enough to realize a simple task. Show a method by which you may measure the coordinates describing the pure state with the accuracy much better than $$1/1,000$$ after 1 million measurements.

In classical or realist physics, it must be possible in principle. If there's something that prevents you from that, it either shows that the classical theory is completely wrong; or it is at least immensely fine-tuned so that it is hiding some degrees of freedom that should be observable in principle. In both cases, it's a serious enough problem for your classical theory.

If we describe the task roughly and qualitatively, it's simply "prove that the uncertainty principle isn't true". If you believe that any realist theory is right, it must be in principle possible to distinguish all the states and you should be able to do so experimentally. One may say that the historical efforts to get such extra information e.g. from the electron's spin is an experimental proof of the uncertainty principle, of the fact that you simply can't get any such accurate information about any hypothetical "objective properties" of the physical system.

You can only measure the properties probabilistically and you need many measurements ($$N\sim 1/ (\Delta\alpha)^2$$) to measure the parameters describing the pure or mixed states, just like quantum mechanics implies. If you're not a biased person who always tries to insist on a predetermined dogma regardless of the facts, you must agree that these observations basically mean that it is proven by direct simple experiments that the quantum mechanical framework is right while the classical or realist one is not.