Friday, November 20, 2015

Where are the complex numbers coming from in Quantum Mechanics?

So last time we produced the ingredients used to build quantum mechanics and today we will show how they combine just like Lego. And the main building pattern will turn out to be nothing but to complex number multiplication.

The mathematical structure will be that of a trigonometric coalgebra. 

So let us start by recalling the two products from last time: \(\alpha\) and \(\sigma\). \(\alpha\) respects the Leibniz identity (because it stems from infinitesimal time evolution) and apart from that we have no other information at this time about the two products. If we assume the real numbers to be the mathematical field corresponding to actual physical measurement values, the Lego operation of combining two physical systems into one is called a coproduct. Here is how we do it:

Let C be a R-space with {\(\alpha\), \(\sigma\)} as a basis. We define the coproduct \(\Delta : C\rightarrow C \otimes C\) as:

\(\Delta (\alpha ) = a_{11} \alpha \otimes \alpha + a_{12} \alpha \otimes \sigma + a_{21} \sigma \otimes \alpha + a_{22} \sigma \otimes \sigma\)
\(\Delta (\sigma) = b_{11} \alpha \otimes \alpha + b_{12} \alpha \otimes \sigma + b_{21} \sigma \otimes \alpha + b_{22} \sigma \otimes \sigma\)

There is also another operation called counit but that is only important for the mathematical point of view to complete what mathematicians call a coalgebra (S. Dascalescu, C. Nastasescu, and S. Raianu, Hopf Algebra: An Introduction, Chapman & Hall/CRC Pure and Applied Mathematics (Taylor & Francis, 2000)).

To get this abstraction more down to earth we need to see it in action and show what it means to construct the bi-partite products \(\alpha_{12}\) and \(\sigma_{12}\) from \(\sigma\) and \(\alpha\). If we have \(f_1, g_1\) elements belonging to physical system 1, and \(f_2, g_2\) elements belonging to physical system 2 we basically have the following:

\((f_1\otimes f_2) \alpha_{12} (g_1\otimes g_2) = a_{11} (f_1\alpha g_1) \otimes (f_2 \alpha g_2) + a_{12} (f_1 \alpha g_1) \otimes (f_2 \sigma g_2) \)                                                       \(+ a_{21}(f_1 \sigma g_1) \otimes (f_2 \alpha g_2) + a_{22} (f_1 \sigma g_1) \otimes (f_2 \sigma g_2)\)

\((f_1\otimes f_2) \sigma_{12} (g_1\otimes g_2) = b_{11} (f_1\alpha g_1) \otimes (f_2 \alpha g_2) + b_{12} (f_1 \alpha g_1) \otimes (f_2 \sigma g_2) \)                                                       \( + b_{21}(f_1 \sigma g_1) \otimes (f_2 \alpha g_2) + b_{22} (f_1 \sigma g_1) \otimes (f_2 \sigma g_2)\)

Basically we take all possible combinations of the original products, so what does this lead us? We have seem to make no progress whatsoever. However, there is hope to determine the 4+4  constants \(a\) and \(b\)!!! Why? Because \(\alpha\) respects the Leibniz identity. 

Because \(\alpha\) respects Leibniz identity (invariance of the laws of nature under infinitesimal time evolution) it is basically a derivation. And the derivation of the unit element (corresponding to "no physical system") is zero. So what if we take \(f_1 = g_1 = 1\)? [As a side note, because \(\alpha\) is distinct from \(\sigma\) (otherwise we find ourselves in the trivial case discussed last time), it can be normalized to respect: \(1\sigma f = f\sigma 1 = f\)]

So now let's plug in \(f_1 = g_1 = 1\) first in the \(\alpha_{12}\) equation above. On the left hand side we get:

\((1\otimes f_2) \alpha_{12} (1\otimes g_2) = f_2\alpha g_2\)

and in the right hand side we get:
\( a_{21}(f_2 \alpha g_2) + a_{22} (f_2 \sigma g_2)\)

which demands \(a_{21}= 1\) and \(a_{22} = 0\).

We can play the same game with \(f_2 = g_2 = 1 \) and get \(a_{12} = 1\) This cannot reveal anything about \(a_{11}\) at this time, but at least we have trimmed the possible combinations.

Same game on \(\sigma_{12}\) yields: \(b_{21} = 0\), \( b_{22} = 1\), and \(b_{12} = 0\). Again nothing on \(b_{11}\).

So now we have the reduced pattern:

\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha + a_{11} \alpha \otimes \alpha\)
\(\Delta (\sigma) = \sigma\otimes \sigma +  b_{11} \alpha \otimes \alpha\)

It turns out that \(b_{11}\) is a free parameter which can be normalized to +1, 0 , -1 resulting in 3 composition classes: hyperbolic (hyperbolic quantum mechanics), parabolic (classical mechanics), and elliptic (quantum mechanics). But we can eliminate \(a_{11}\) by applying Leibniz identity on the \(\alpha_{12}\).  This is tedious and I will skip it here, but take my word on it because in the bipartite identity \(\alpha_{12}\) will occur squared and by linearity of \(\alpha\) it must vanish.

If we now consider only the quantum mechanics case of \(b_{11} = -1\) we have:

\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha \)
\(\Delta (\sigma) = \sigma\otimes \sigma - \alpha \otimes \alpha\)

Does this remind you of something? How about:

Imaginary =  imaginary real  + real imaginary
Real          =  real real             - imaginary imaginary

This is how complex numbers start occurring naturally in quantum mechanics and this is why observables are Hermitean operators and generators are anti-Hermitean operators. The 1-to-1 map between observables and generators known as "dynamic correspondence" in literature is the 1-to-1 map between \(\alpha\) and \(\sigma\). It is this dynamic correspondence which is at the root of Noether theorem! Noether's theorem is baked in the quantum and classical formalism but you need to know where and how to look to uncover it.

So today we made good progress on the road to reconstruct quantum mechanics but there is quite a way to go. We do not know anything yet about the properties of the  \(\alpha\) and \(\sigma\) products, and without it we cannot hope to finds their concrete representation. But we'll get there. Please stay tuned.

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