## One product, two products?

### As I was posting this a large terror attack was unfolding in Paris. I stand in   solidarity with the victims of this barbarity. Je suis Parisienne.

Coming back to physics, I will now start a series of posts where I will rigorously derive quantum mechanics from physical principles. To prevent "abstraction indigestion", I will chop off this proof into easily manageable segments and today I want to discuss the minimum number of ingredients needed to cook the quantum soup.

The first physical principle needed is that of the invariance of the laws of nature under time evolution. This is a no-brainer, otherwise the search for physical laws would make no sense. But what can we get out of this? If there are algebraic operations which make sense in nature (and yes, there are plenty), those operations commute with time translation. Let us denote such operation with a generic symbol *, and let us use uppercase letters for the objects on which * operates. If we call T(A) the time translated version of the abstract object A, invariance under time evolution demands:

$$T(A*B) = T(A)*T(B)$$

So what? We did not get too far, unless... Unless we can do something more. Who says T has to be large? We can take as small time steps as we want. Infinitesimal steps. In other words we linearize $$T$$ into the identity and a small correction:

$$T= I + \epsilon D$$

So what do we get?

$$(I + \epsilon D)(A*B) = ((I +\epsilon D)A)*((I+\epsilon D)B)$$

which to first order yields the trivial $$A*B=A*B$$ but to first order in $$\epsilon$$ we get:

$$D(A*B) = D(A)*B + A*D(B)$$

Recognize this? It is the chain rule of differentiation. But we can do more. We can transform $$D$$ into a product which we will call $$\alpha$$:

$$A D B = A\alpha B$$

To put this abstraction in perspective, $$\alpha$$ will later become either the Poisson bracket or the commutator: those are concrete representation of the abstract product.

What we can do now is to rewrite the chain rule from above as the Leibniz identity:

$$A\alpha (B*C) = (A\alpha B)*C + B*(A\alpha C)$$

We know that one of the products * is $$\alpha$$, so the simplest case possible is that all there is in nature is this one and only product $$\alpha$$. What kind of world would we get is there is only one product possible in nature?

What we can do now is to apply the Leibniz identity on itself.

$$A\alpha (B\alpha C) = (A\alpha B)\alpha C + B\alpha(A\alpha C)$$

So we seem to be stuck, unless... Unless we can tell something about the abstract objects $$A, B, C, ...$$. Here is where category theory of k-algebras comes to the rescue. Now I can either go for the abstract math, or I can present the physical interpretation. Let's go the physical route and anticipate that what we ultimately want to consider are compositons of physical systems. Such compositions have a unit element: the "compose with nothing" element. Let's denote it's corresponding uppercase letter "I" and see what we obtain from Leibniz in this case:

$$I\alpha (I\alpha A) = (I\alpha I) \alpha A + I \alpha (I\alpha A)$$

from which we get:

$$(I \alpha I) \alpha A = 0$$

for any A!!! Which means $$I \alpha I = 0$$ or "nothing comes from nothing".

It is now time for the second physical principle: the invariance of the laws of nature under composition. What does this mean? It means the laws of nature do not change when we add additional degrees of freedom. For a silly "what if" game, how would the world would look like if the laws of nature would depend on the number of degrees of freedom? Imagine going to the airport and boarding a plane. When enough passengers board this plane increasing its internal degrees of freedom, the plane would suddenly become a bird which would fly to its destination. There it would revert back to a plane when the passengers would disembark. Sorry Harry Potter, no quantum soup for you. Come back one year.

So now consider a composite system  $$1\otimes 2$$. For this system we would have the bipartite product alpha: $$\alpha_{1 \otimes 2}$$

Now we need to construct this bipartite product just like Lego from the only ingredients we have available in this toy world: the ordinary products $$\alpha$$ for system 1 and 2:

$$(A_1\otimes A_2 ) \alpha_{1\otimes 2} (B_1 \otimes B_2) = a (A_1 \alpha_1 B1)\otimes (A_2 \alpha_2 B_2)$$

where $$a$$ is a generic proportionality constant. But what if we pick $$A_2=B_2 = I$$? In other words the second physical system is nothing.  On one hand we get:

$$(A_1\otimes I ) \alpha_{1\otimes 2} (B_1 \otimes I) = a (A_1 \alpha_1 B_1)\otimes (I \alpha_2 I)$$

which is zero because we showed earlier that: $$I\alpha I = 0$$. On the other hand:

$$(A_1\otimes I ) = A_1$$ and $$(B_1\otimes I ) = B_1$$ and we have:

$$(A_1\otimes I ) \alpha_{1\otimes 2} (B_1 \otimes I) = A\alpha B$$

So the product alpha by itself can only be trivial: $$A\alpha B = 0$$ for any A and B.

To get something non-trivial we need at least another product which we will call $$\sigma$$. For a rich quantum soup we need meat and potatoes. I mean the commutator and the Jordan product. We'll get there, please be patient. For now we only identified the soup ingredients:

• a product $$\alpha$$ which respects the Leibniz identity (due to invariance under time evolution)
• a second unspecified product $$\sigma$$
Can we consider more products? Of course, but it will turn out that we only need those ingredients. Invariance under composition will completely determine the properties of those two products. Please stay tuned for the next episode of cooking quantum mechanics.