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not 10, but 1 question

Today more than ever we ask ourselves: who bribed David Mills?

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Surprise

When he left this site a month ago, I thought it was an impromptu initiative that would end after a few weeks. But ...

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FM09010 (I)

So, let us start we the new lecture: Fourier methods and their applications to neuroscience .

Preliminary disclaimer: there are many many approaches to the topic. I will follow this one. In fact, Osgood's lecture is by far better than what I can hope to do in my life, but he has a slow pace which is not suitable for PhD students at the BCCN.

In the first lecture we explained the basic idea, which is simple: you have a complex periodic function (where complex stays for both "complicated" and "not real") and you want to represent it as weighted sum of simpler (but also not real) periodic functions. For definiteness, let us say that all functions have period 1.

We can choose to use trigonometric functions or complex exponentials . We will choose the complex exponentials for different reasons. To be precise, we want to express any periodic function of period 1 as the weighted (probably infinite) sum of the functions

  
 e_n(t):=e^{2\pi nit} 
  

where n ranges in the integers .
The difficult problem is: how to find the weights for the sum? It goes in the following way. You start from what you are looking for

  
 f(t)=\sum_{n \in \mathbb Z} c_n e^{2\pi nit} 
  

Here, the numbers c_n are the (unknown!) weights of the linear combination that you want to represent the function f. We now try to isolate a single coefficient, say the kth, to know whether is possible to get a formula for a single coefficient depending only on f and not on the other coefficients. So you get

  
 f(t)- \sum_{n\neq k} c_n e^{2\pi nit} = c_ke^{2\pi kit} 
  

You divide by the complex exponential obtaining

  
 f(t)e^{-2\pi kit}- \sum_{(n-k)\neq k} c_n e^{2\pi nit} = c_k 
  

For this you have to use all properties of the complex exponentials. Which is, by the way, only one. And in fact it is the same as for the real exponential: "sum is mapped into multiplication". This is one motivation for using complex exponentials instead of trigonometric functions.

You now have the problem that you have expressed one c_k in terms of the other ones, but you want to have c_k without its companions. Algebra gave us everything she could, so let us try with Analysis. He suggests to integrate between 0 and 1 (which is the period of all functions here) obtaining

  
 \int_0^1 f(t)e^{-2\pi kit} dt- \int_0^1\sum_{(n-k)\neq k} c_n e^{2\pi nit}dt = c_k 
  

The c_k, integrated, gives c_k! And now the magic. Compute the integral of the complex exponentials: it gives 0.
So, you are left with the famous expression

  
 \int_0^1 f(t)e^{-2\pi kit} dt- \int_0^1\sum_{(n-k)\neq k} c_n e^{2\pi nit} dt = c_k =: \\ hat {f} (k) 
 

Where We Have Introduced the symbol \\ hat {f} (k) . We will call this number by the suggestive name of the kth Fourier coefficient of f.

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My vote ...

... will Bersani.

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Today we are speaking part Unbeschränktheit Über die Quantenmechanik der der Operatoren Helmut Wielandt, appeared in the Mathematische Annalen in 1949.

The title, translated into Italian, means "Sull'illimitatezza operators of quantum mechanics." What are we talking about? Simplifying a bit, in quantum mechanics all the variables become operators. And if two sizes A, B are the Fourier transform of the other, then they must be such that

 [A, B] = i \\ hbar 

where we have denoted [A, B] = AB-BA the commutator of two operators.

This relationship is called the canonical commutation relation, more here. Wintner proved in 1947, using a technique developed by Rellich in 1946, that two operators satisfying the commutation relations must be unlimited. This demonstration was, however complicated, and in 1949 gave an elementary proof of Wielandt this fact. How does it work?

Wielandt on the observation that if

 [A, B] = 1 then 

 [A, B ^ {n +1}] = (N +1) B ^ n 

The proof of this fact is a simple algebraic manipulation and with an easy induction.

From this fact, and the reverse triangle inequality operatoriale applied the standard of A and B, yields an estimate of the standard

 B (n +1)