Momentum operator as the generator of translations.

[alexepascual]

Turin,
Your posts are not confusing. I just was unfamiliar with the notation you used. As a matter of fact, I think the only thing I believe I didn't understand was the square brackets. But your post made very clear where I was erring and how to fix it. The fact that you used time domain instead of space domain didn't confuse me at all either. Your explanation about the notation in your last post makes everything absolutely clear.
With respect to the kernel, I understand very well your definition. This definition coincides with the one given by Haelfix but I can't make a connection with the one given in my linear algebra book. I did a Google search on the "kernel of a transformation" and the definitions I saw coincided with the linear algebra book.
According to this definition, the kernel is that sub-set of the domain that is mapped by the transformation to zero. If we think of the transformation as represented by a matrix and the domain and range composed by vectors (functions), then this other definition would define the kernel as a set of vectors (functions) while your definition would consider it as an element of the transformation matrix. Maybe these are two completely different meanings of the same word, or maybe they are somehow connected or equivalent but I can't see the relationship.

 

[Eye_in_the_Sky]

Haelfix provides an explanation of my use of "kernel." Though, I am unclear how one can say this is not a linear algebraic application. I was always under the impression that that is exactly what the kernel is: the elements of the transformation matrix.

Haelfix isn't saying that T is not a linear transformation. Rather, there is a concept in linear algebra which is designated by the same term "kernel" (also "null space") but refers to something else. For that context, the definition is:

Defintintion: Let A be a linear transformation from a vector space V into a vector space W. Then, the "kernel" (or "null space") of A is the set
K_A = {v Є V | A(v) = 0}.

It turns out that K_A is a linear subspace of V, and A is invertible iff A is "onto" and K_A = {0}.

Thus, in your sense of "kernel", the "kernel" of T is e^-iωt, whereas in the other sense, the "kernel" ("null space") is the set consisting of [the equivalence class of functions corresponding to] the zero vector. (Note: You can ignore the preceding square-bracket remark if it troubles you (or, better yet, ask (if it troubles you)).)

 

[turin]

OK, I think I get it. The Haelfix version of "kernel" must be the set consisting of only the zero vector since there is an inverse Fourier transform? It doesn't "trouble" me, but I don't know what an "equivalence class" is. Care to enlighten us?

 

[Eye_in_the_Sky]

OK, I think I get it.

You got it.

----------

... I don't know what an "equivalence class" is.

When we choose to represent the "vectors" of our Hilbert space by (square-integrable) "functions", a potential difficulty may arise.

Suppose we have two functions, f and g, which are equal "almost everywhere" (i.e. the set of points for which f(x) ≠ g(x) has "measure zero"), then as far as all [of the "tame"] integrals are concerned, f(x) and
g(x), under the integral, will give the same result. Thus, for all practical purposes, f and g are considered to represent the same "vector" ... yet, at the same time, they may be distinct "functions" (i.e. we may not have f(x) = g(x) at every single point x).

In simple terms, the condition "f is almost everywhere equal to g" can be expressed as

[1] Integral { |f(x) - g(x)| dx } = 0 .

This condition puts our "functions" into "groups", or "classes", of "equivalent functions". These "groups" are the "equivalence classes" corresponding to the "equivalence relation" defined by [1]. In this way, a formal "vector" of the Hilbert space is represented by a particular "equivalence class", and when we wish to do a calculation, we can simply pick any "function" in the given "equivalence class" (... and it doesn't matter which one we pick).

[Note: In the above, a formal definition of "equivalence relation" has not been given. Nor has a formal demonstration been given that condition [1] satisfies such a definition. Neither has a formal definition of "measure" and "measure zero" been given, nor a formal demonstration that condition [1] is equivalent to "the set of points for which f(x) ≠ g(x) has measure zero".
... But these things are basically trivial to do, and yet, may be the cause for a "small headache" to some, as well as, being construed (by some) to be "a complete waste of time".]

 

[turin]

I don't want to deam the issue of equivalence class to be "a waste of time," but it seems to me that there can only be one of the functions out of the equivalence class that could survive the further restriction of being physically meaningful. Am I incorrect to interpret the distinctions as removable discontinuities?

 

[Eye_in_the_Sky]

Am I incorrect to interpret the distinctions as removable discontinuities?

It would be more accurate to describe the distinctions as occurring at "isolated points". Suppose that x = x_o is such a point of distinction. Then, if f(x_o-) = f(x_o+), yes, we are dealing with a "removable discontinuity". If, however, f(x_o-) ≠ f(x_o+), then, even though we have a rule for removing the "distinction", we cannot apply the term "removable discontinuity".

... it seems to me that there can only be one of the functions out of the equivalence class that could survive the further restriction of being physically meaningful.

Yes, I think that your statement is basically true. In a physical situation which demands that a function be continuous on some interval, all members of the corresponding "equivalence class" will differ on that interval only by "removable discontinuities". On the other hand, if a physical situation (probably involving an idealization of some kind) calls for some function to have a "step discontinuity", the physical context would probably tell us that the value of the function at the "step", say x = x_o, is quite irrelevant; we would probably just make f(x_o) = f(x_o-) or f(x_o+).

So ... it sounds to me like you may be suggesting that, in a general sort of way, the "physically distinct solutions" are, so to speak, in a one-to-one correspondence with the "equivalence classes". Yeah, this makes sense ... never thought of it, though.

 

[turin]

Perhaps this issue comes to its climactic import when decomposing a function that is physical into a basis of functions that may not themselves be physical (or the other way around).

 

[Eye_in_the_Sky]

Quite independently of whether or not such a climax - or one of similar import - can/cannot or will/will not occur, the real question (I would say) is: Why does the mathematician feel compelled to speak of "equivalence classes" instead of the "functions" themselves?

... Any idea?

 

[selfAdjoint]

Excuse me Eye, but it seemed to me that you yourself answered this question in post #54, earlier in this thread. You have ONE thing to represent, and several functions do it, and the differences between them you want to ignore, so you form equaivalence classes. You were so clear back there, I can't understand what your difficulty is here.

 

[Eye_in_the_Sky]

I apologize. I didn't realize that my mention of "equivalence classes" and initial egging-on was going to be the spearhead of such a lengthy tangent. I'm not having a difficulty here. I posed the question to Turin, because I thought, somehow, the main point was being lost. So, in order to return, once again, to the main point I posed the question:

Why does the mathematician feel compelled to speak of "equivalence classes" instead of the "functions" themselves?

I posed the question, not because I didn't have an answer, but rather, to give Turin something more to think about.

As for this question having already been answered in post #54, I see only two (distinct) statements there, each of which (only) appears provide an answer for why the mathematician feels "compelled":

(i) without equivalence classes "a potential difficulty may arise";

(ii) "for all practical purposes" a pair of almost-everywhere equal functions are considered to be the same.

But (i) only hints at an answer, while (ii) opts out of giving that answer.

That answer is: If we speak only of "functions", then a linear transformation, such as the Fourier transform, when construed as a mapping of "functions" to "functions" will be MANY-to-ONE, and therefore, have no well-defined inverse. By speaking of "equivalence classes" instead, this difficulty is removed.

This then explains why I felt compelled to [parenthetically] mention "equivalence classes" in the context of the "kernel" (i.e. "null space") of the Fourier transform back in post #52.

 

[turin]

Great discussion. I must embarassingly admit that I did not see this answer which now seems so obvious. This uncovers another concern of mine, though, which may or may not be related. I guess the best way to pose the issue is in terms of the null vector, but it is really a concern about the feasibility of referring to functions as vectors in the first place.

If I understand the null vector to be the function f(x) = 0, yet this is only one member of the equivalence class of functions f_i(x) such that:

integral {|f_i(x)|²} = 0,

then this does make me uncomfortable (especially when I try to throw the Dirac-delta into the mix). The main source of my concern, I suppose, is the permission for this null vector to have several (an arbitrarily large number of large) nonzero components (something that I don't imagine agrees with the notion of discrete vectors). If I just squint my eyes until the integral has done its job, then everything seems OK. But something just doesn't seem quite right with this in my gut.

OK, so we really mean that a particular equivalence class of functions is a vector. I think I just answered my own concern (after Eye put the words into my mouth, that is).

 

[selfAdjoint]

The main source of my concern, I suppose, is the permission for this null vector to have several (an arbitrarily large number of large) nonzero components (something that I don't imagine agrees with the notion of discrete vectors).

The null vector continues to have all-zero components. There may be in its equivalence class other vectors that do not share this characteristic, which is not essential to the equivalence relation. Consider that the man you know has a beard, but he is in a group labeled MEN with other men, and they don't all have beards. Likewise the number 2 is prime, but it is in an equivalence class, the even numbers, and none of the other numbers in that class are prime.

 

[turin]

It just seems strange to me that a vector with zero magnitude could be anything but a null vector.

 

[selfAdjoint]

A lightlike four-vector in relativity has its time component equal and negative to its spatial component, so they cancel out, even though neither of them is zero. This is the reason lightlike trajectories are called null trajectories.

 

[turin]

Now I feel like an idiot. I was completely thinking in terms of Euclidean 3-vectors when I made my comparison.

 

[Eye_in_the_Sky]

This uncovers another concern of mine, though, which may or may not be related.

It is related.

I guess the best way to pose the issue is in terms of the null vector, but it is really a concern about the feasibility of referring to functions as vectors in the first place.

Consider the set S = { f | f : R → C } (i.e. S is the set of (arbitrary) "functions" from "real numbers" to "complex numbers"). Let C be the "scalars". Then S is a vector space over C. (Review http://www.ncrg.aston.ac.uk/~jamescj/Personal/Downloads/AM20LM/AM20LM_Handout_A_2003.pdf.) This example shows quite simply and unambiguously that there should be no concern with regard to the feasibility of referring to "functions" as "vectors".

Next, consider our vector space of square-integrable functions, but without the modification induced by the equivalence relation. Let's call this space F (to emphasize that each "function" corresponds to a distinct "vector").

Now ... in F, what kind of sense be made out of a statement like [1] below?

[1] f(x) = Σ_n a_nφ_n(x) , φ_n a basis .

This statement has a serious problem. For suppose we have a candidate basis (say, for example, that the φ_n are the energy eigenfunctions for a simple harmonic oscillator). We can then set some of the a_n ≠ 0 in order to obtain some function f Є F. And now ... we take this function f and change its value at exactly one point. This gives us a new function, and if the φ_n are really a basis on F, then we must be able to get this new function by merely changing the values of the a_n without "touching" any of the φ_n's.
... Is that possible? ... How can we possibly cause the function f to change at one - and only one - point, merely by changing the a_n's? That is impossible. And from this, we see that a statement like [1] has no meaning in F ... because F has no such basis.

But once we modify F, by means of our equivalence relation, a statement like [1] can then make perfect sense. (... And this is yet another (related) reason why the mathematician is compelled to speak of "equivalence classes" instead of the "functions" themselves.)

Let us refer to this modified space as E (to emphasize that each distinct "equivalence class" corresponds to a distinct "vector").

--------------------

Next.

Here is how you expressed your concern in terms of the null vector:

The main source of my concern, I suppose, is the permission for this null vector to have several (an arbitrarily large number of large) nonzero components (something that I don't imagine agrees with the notion of discrete vectors).

You have now introduced another concept, that of "component". You have mentioned it in two distinct senses:

(c) relative to a continuous parameter "x" ;

(d) relative to a discrete index "n" .

In alluding to (d), you imply that a statement like [1] above makes sense. In that case, you definitely cannot be thinking of your vector space along the lines of F, but rather, more along the lines of E. ... Now, what about (c)? Were you thinking along the lines of F, or did you mean E?

Let's go one level deeper:

- "component" in the sense of (c) can live in F and can live in E ;

- "component" in the sense of (d) can live only in E, but not in F.

Let's go one more level deeper. Consider the following statement:

[2] "equal components" is a necessary and sufficient condition for "equal vectors"

(i.e. two "vectors" are the same iff their corresponding "components" are the same).

With regard to "components" in the sense of (c) (i.e. relative to a continuous parameter "x"), where does statement [2] hold? ... In F, or in E? Well, statement [2] is true only in F ... but not in E. And that is no surprise - for, in going from F to E, we decided to consider entire groups of "vectors" to be a single "vector". That is to say, in going from F to E, statement [2] has become

[2'] "equal components" is a sufficient, but not necessary, condition for "equal vectors".

... These remarks should be sufficient to clear up all levels of confusion to be found in the last quoted passage above. Specifically:

The null vector is granted permission to have (an arbitrarily large number of large) nonzero "x"-type "components" only in E, the space where those components don't matter ... and in F, where those components do matter, the concept of discrete "n"-type "components" has no meaning.

--------------------

I now leave you with a question:

What is a suitable redefinition of "component" in the sense of (c), whereby statement [2] does hold in E?

--------------------

 

[Eye_in_the_Sky]

It just seems strange to me that a vector with zero magnitude could be anything but a null vector.

That seems strange to me, too! Because, by definition, a vector with zero magnitude is nothing but a null vector.

Definition: u is a "null vector" iff ║u║ = 0 .

Turin ... you meant to say something different from what what you said meant. That is to say, what you meant to say was:

It just seems strange to me that a vector with zero magnitude could have anything but all of its "components" equal to zero.

So, selfAjoint came along gave you an example of both (what you meant to say as well as what what you said meant).

A lightlike four-vector in relativity has its time component equal and negative to its spatial component, so they cancel out, even though neither of them is zero. This is the reason lightlike trajectories are called null trajectories.

... But in order to accomplish that task, selfAdjoint was forced to depart from a Euclidian metric, to which you, Turin, responded ...

Now I feel like an idiot. I was completely thinking in terms of Euclidean 3-vectors when I made my comparison.

... and that just sends my head spinning, because the Hilbert-space metric is Euclidean ... implying that you were thinking in the right terms but just in the wrong "dimension" and "measure", and that what you really wanted answered wasn't.

 

[turin]

Simultaneously I have been clarified and confused by orders of magnitude. It is quite a strange mental sensation.

Eye,
In answer to the question with which you left me, I would suggest defining "compenents" as integrals over regions of the domain. That way the integration could eliminate the discrepancies between functions in an equivalence class as vectors (though the functions themselves would still remain distinct). Even if this is appropriate, though, I still have no clear picture in my mind of the specific details of such a definition.

 

[Eye_in_the_Sky]

Simultaneously I have been clarified and confused by orders of magnitude.

...

--------------------------

I would suggest defining "compenents" as integrals over regions of the domain. That way the integration could eliminate the discrepancies between functions in an equivalence class as vectors (though the functions themselves would still remain distinct).

Since an integral is defined over an interval, using that would eliminate more than just the "discrepancies". ... What we really need is an "infinitesimally small" interval - i.e. limits. So, I would suggest the following definition:

(c') the "x-component" of f is given by a joint specification of f(x-) and
f(x+) .

--------------------------

Now, regarding:

It just seems strange to me that a vector with zero magnitude could be anything but a null vector.

I think what you really wanted to say was:

It just seems strange to me that a vector with zero magnitude could have some nonzero "components".

According to the above definition (c'), we now have that every function in the equivalence class of the null vector does have all of its "x-components" equal to zero. ... But, of course, we didn't need (c') to realize that the "isolated discontinuities" in each of those functions were of no consequence. We already knew that they contribute nothing to an integral, implying, for each f in the "null class",

║f║ ≡ sqrt ( Integral { |f(x)|² dx } ) = 0 ;

i.e. those functions f really do have zero "length".

--------------------------

I have only one more (concluding) remark to make, and it relates to something you pointed out earlier:

... it seems to me that there can only be one of the functions out of the equivalence class that could survive the further restriction of being physically meaningful.

I return to this point, not because of the part about "physically meaningful", but because of the part about "restriction". Perhaps we could have "redefined" our space by means of specific "rules" telling us which functions are "in" and which functions are "out". A good start would have been to say, "Well, we only want piecewise continuous functions which have no removable discontinuities." Then we would have had to try to figure out what kind of "rule" to use at a step discontinuity ... which I won't even try to think about.

In short, equivalence classes are able fix the problem with very little effort.


(Note that equivalence classes have an import and utility over and above that (of our case here) of "removing" annoyances.)

--------------------------

 

[cathalcummins]

Doing space time symmetries through Group theory. I completely understand how do derive, say SO(2), from its generators.

And I understand how, as so(3) obeys the same commutation relations as Angular Momentum, that Angular Momentum is just the generator of rotations.

Now, I wish to use the same logic to show that the Generator of spatial translation is just linear momentum in spacetime. I tried the usual method:

Start with the group of all Spatial Translations, \mathrm{G} with x^i \in \mathbb{R} \{x^0=t,x^1=x,x^2=y,x^3=z\} and p^i \in \mathfrak{g} form the Lie Algabra.

A quick way would be to use the infinitesimal group element way:

g_i(x^i)=e^{ix^i p^i}=1+x^i \frac{ (i p^i)}{1!}+(x^i)^2 \frac{ (i p^i)^2}{2!}+ \cdots

and so if e^{ix^i p^i} acts on some f(X), and we wish to translate it by x^i to f(X+x^i) we would have to make p^i take the value:

ip^i = \partial_{x^i}

as per Taylor Series of f(X+x^i).

Is this sloppy/wrong?

Is there a nicer way to show this that follows the Rotational Transformations and Group Theory closer?

I'm looking for something analogous to the way infinitesimal generators are defined:

\tau^i = \frac{1}{i}\partial_{x^i}(F(x^i))\Big|_{\{x^i=0\}}

where F(x^i) become the Rotation Matrices,R_i(\theta), of SO(3) in the Angular Theory.