In this post, which is a continuation of Spin Chronicles Part 47, we'll get prepared for exploring the Tomita's "flow of time". It has very little, if anything at all, to do with "spin". But the two subjects both come under one umbrella - the umbrella of *-algebras, and it is somewhat instructive to discuss Tomita's flow on the simple example of the geometric Clifford algebra of space. Tomita's construction works for any von Neumann algebra with a cyclic and separating vector, whether finite or infinite-dimensional. The plan of this post is to discuss a general, though not the most general case. So we will deal with a possibly infinite-dimensional Hilbert space, and then see how things simplify (and become somewhat confusing at the same time) in the case of a finite-dimensional Clifford algebra A that was the subject of our study so far.
Let therefore H be a Hilbert space (complex, with a finite or countable orthonormal basis). For the spin 1/2 case we will take H = C2. Let A be the algebra A=B(H) of all bounded linear operators on H. This is our "algebra of observables" of some "quantum system". For spin 1/2 case A≃Mat(2,C). This is our von Neumann algebra. By "von Neumann" algebra we mean A=A''. So, it is a *-algebra of operators, equal to its double commutant.
It is here that we meet an example of the chicken and egg problem: what is more "primary", what should come first, the algebra, or the representation space on which the algebra acts? In geometrical algebra we usually start with the algebra, then we recover the representation space by examining minimal left ideals of the algebra. In quantum theory it is usually the other way around: we start with a Hilbert space and then decide which operators on this space represent "observables"? It would be more in tune with geometry if I would start with a possibly infinite-dimensional Clifford algebra. But that would take us too far from the problem at hand, and also would have to find a simple way how to present the concept, which I do not know at present. So let us stay as described above.
As I have mentioned above, we are not discussing the most general case. For instance it would be more general to take a direct sum of a finite or infinite number of Hilbert spaces ⨁iHi, and to take for A the direct sum ⨁i B(Hi) of corresponding B(Hi). This would let us to discuss "superselection sectors". We are not going to do it here, so let us stick with our simple case of just one H.
In previous posts we have discussed the GNS construction, which connects states of the algebra to its representations. We know that pure states lead to irreducible representations, mixed states to reducible ones. We know that a general state ω on B(H) is uniquely described by a "density" matrix:
ω(a) = Tr(ρa), a∈A,
where ρ is a non-negative operator on H, with Tr(ρ)=1.
Since we are including here the infinite-dimensional case, it is appropriate to recall that
Tr(ρa) = Σi (ei, ρa ei),
where ei is any orthonormal basis in H. The series is absolutely convergent, and its sum is independent of the choice of the orthonormal basis. Every density matrix has a purely discrete spectrum, so, in particular we can choose the orthonormal basis ei consisting of eigenvectors of ρ:
ρei = pi ei,
where pi, not necessarily all different (finite multplicities may occur - Why only finite for non-zero eigenvalues?), satisfy:
pi ≥ 0, Σi pi =1.
In this case we get (Why?)
Tr(ρa) = Σi pi (ei, a ei).
Thus, in general, ω is a mixture, a weighted sum (with probabilities pi) of pure states defined by the eigenvectors ei.
In an infinite-dimensional Hilbert space the landscape is much more interesting than in the finite-dimensional case. The algebra B(H) contains then several important non-trivial two-sided *-ideals (cf. Wikipedia Trace class):
{ finite rank} ⊂ { trace class } ⊂ { Hilbert-Schmidt } ⊂ { compact }.
For a finite-dimensional H, they all coincide and are equal to the whole B(H), but in the infinite-dimensional case all inclusions above are proper. For our purposes we dot need to worry now about finite rank and compact operators, but we need to know something about the trace class *-ideal, denoted B1(H), and the Hilbert-Schmidt *-ideal, denoted B2(H):
B1(H) = {T∈B(H): Tr(|T|) < ∞},
where |T| = √(T*T).
B2(H) = {T∈B(H): Tr(T*T) < ∞}.
If T,S∈B2(H), then TS∈B1(H). The ideal of of Hilbert-Schmidt operators B2(H) is a Hilbert space when equipped with the scalar product <S,T> defined as
<S,T> = Tr(S*T).
This Hilbert space plays the crucial role in the following. In the finite-dimensional case it coincides with the whole of B(H), but in the infinite-dimensional case it is a proper ideal of B(H).
Remark 1. It should be also noticed that while B2(H) is a complete Hilbert space when equipped with the scalar product <S,T>, it is not a closed subspace when considered as a subspace of B(H) equipped with the operator norm. The closure of B2(H) with respect to the operator norm is the ideal of compact operators.
Remark 2. This remark concerns our notation here. We were denoting the elements of the Clifford algebra A with letters like a,b,u,v etc. Now, when considering Hilbert spaces, we are using letters S,T, etc. for operators in B(H) - as it is done, for instance, in Wikipedia. This may be somewhat confusing, but once we know about it, we can live with it. Each symbol should always be considered within a context, since it is the context that gives meaning to the symbol.
Remark 3. I am assuming that the Reader knows the rudiments of quantum mechanics in a Hilbert space. Probably the best reference for that is Brian C. Hall, "Quantum Theory for Mathematicians, Springer 2013,
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B. C. Hall, Quantum Theory for Mathematicians |
in particular Ch. 19.2 Trace-Class and Hilbert-Schmidt operators, and Ch. 19.3 Density Matrices: The General Notion of the State of a Quantum System. But if you have any questions - don't hesitate to ask, and I will do my best to answer them.
The standard GNS construction
Now we need to return to the GNS construction. The door to this subject has been closed in Part 46, but now, dealing with "emergence of time", we need to reopen it, even if only a little bit to take a peek at it again, for the particular case when the state ω is a faithful state. For a general state ω we consider the left ideal Iω = {a∈A: ω(a*a)=0} and construct the quotient A/Iω. But for a faithful state ω, we have Iω={0}, and no quotient is needed in this case. Thus we define the inner-product (or "pre-Hilbert") space Hω = A, and equip it with the scalar product
<a,b>ω = ω(a*b).
For A=B(H) this becomes
<a,b>ω = Tr(ρa*b).
In the infinite-dimensional case Hω is not yet complete, so we define the Hilbert space Hω as the completion of Hω. So, we have
A = B(H) = Hω ⊂ Hω.
The scalar product <a,b>ω extends to Hω by continuity.
Notice that, for a faithful state ω, the space Hω, in fact, does not depend on ω, only Hω does. The same concerns the representation πω, which is defined on as Hω as
πω(a)b = ab,
and extended by continuity to Hω.
Similarly for the cyclic vector Ωω defined as
Ωω = 1∈A.
With these definitions we have
ω(a) = <Ωω, a Ωω>ω. (-1)
The advantage of the standard GNS construction is that the vector Ωω representing the state ω is independent of the state. The disadvantage is that the scalar product of the representation Hilbert space is ω-dependent. There is, however, an alternative construction in which the Hilbert space and the scalar product are fixed, ω-independent, but, instead, the cyclic vector representing ω is depends on ω. We will use this alternative construction in our discussion of the Tomita flow.
Alternative GNS construction for faithful states on B(H)
In the alternative construction of the representation unitarily equivalent to the standard construction we use one Hilbert space for all faithful states - the spaces B2(H) with the scalar product:
<a,b> = Tr(a*b). (0)
The *-representation is the same as before
π(a)b = ab, (1)
but the cyclic vector, which we denote Ωρ, depends on the state:
Ωρ = ρ½. (2)
Here we notice two things. In (1) we have a∈B(H) and b∈B2(H). Since B2(H) is a two-sided ideal of B(H), the product ab is again in B2(H), therefore the representation is well defined. In (2) we have ρ½, ρ being a positive trace class operator. Thus ρ½, is a Hilbert-Schmidt operator (Why?). Therefore Ωρ is well defined vector of B2(H), with (Why?)
<Ωρ , Ωρ> = 1.
Moreover, we have (Why?)
ω(a) = <Ωρ , π(a)Ωρ>.
Answer to the "Why?" above:
Using (2) we have
<Ωρ , π(a)Ωρ> = <ρ½, π(a)ρ½>.
Using (1) this becomes
<Ωρ , π(a)Ωρ> = <ρ½, aρ½>.
Using (0) and (ρ½)* = ρ½ we obtain
<Ωρ , π(a)Ωρ> = Tr(ρ½aρ½).
Using trace property Tr(ST)=Tr(TS) we obtain
<Ωρ , π(a)Ωρ> = Tr(ρa).
Using (-1) we finally get.
<Ωρ , π(a)Ωρ> = ω(a).
Therefore we can apply Theorem 4.5.3 in Part 42.
The vector Ωρ is cyclic for the representation π of B(H) on B2(H). (Why?)
Hint: We want to show that the subspace {Xρ½: X∈B(H)} is dense in B2(H). The subspace in a Hilbert space is dense if and only if its orthogonal complement is {0}. Thus we want to show that, with Y∈B2(H), <Y,Xρ½> = 0 for all X∈B(H)} implies Y=0. That means that Tr (Y*Xρ½) = 0 for all X∈B(H) implies Y=0.
Remark 4. According to Theorem 4.5.3 in Part 42 the representations πω on Hω and π on B2(H) are unitarily equivalent. But how can it be? In infinite dimension B2(H) is strictly smaller than B(H), and B(H) is a subspace of its completion Hω. How can a subset of some set be isomorphic to a superset of this set? Yet in infinitely dimensions this can easily happen. It certainly contradicts our intuitions based on finite-dimensional examples. We need to learn how to live with it, and how to make use of it.
Left and right representation
B2(H) is a two-sided ideal. But in (1) we have used only the fact that it is a left ideal. Our representation π defined in (1) is just the left regular representation, which previously we have denoted by L. So, instead of π we can as well use the notation L. But B2(H) is also a right ideal, therefore we can define another representation of B(H) on B2(H), let us denote it by π':
π'(a)b = ba.
Strictly speaking π' is the representation of the opposite algebra, since π'(ab)=π'(b)π'(a). We have then two subalgebras of B(B2(H)): π(A) and π'(A). Since left actions commute with right actions, the operators of π(A) commute with the operators of π'(A). In fact we have
π'(A) = π(A)', π(A) = π'(A)'.
Vector Ωρ is a cyclic and separating vector for both von Neumann algebras π(A) and π'(A) (Why?). So, we have now a perfect symmetry. This symmetry will be used in the construction of the Tomita's flow that we are ready to discuss. This will be the subject of the following post. If you have any questions - please, ask.
Exercise 1. Provide the explicit formula implementing the unitary isomorphism U: Hω→B2(H) mentioned in Remark 4. Show that U maps B(H) onto o proper subspace of B2(H).
P.S. 10-03-25 11:15 Reading "Self-Psychoanalysis: A System of PsychoSurgery", Institute of Advanced Thinking 1965. There, p.6:
"... the time is rapidly approaching when the government, airline companies, insurance agencies, and other independent organizations will employ "psychological research teams" to trace down the psychological backgrounds of each passenger aboard a plane that suddenly crashes to determine why and how each person, individually and collectively, "attracted" or "caused" the so called "accident.". The Law of Accident means that seeming "accidents" occur in one's life due to a confused Unconscious Mind and faulty thinking."
P.S. 12-03-25 10:21 Yesterday I attended an interesting and very lively seminar of the "Institute of Nature of Time". It was about plasmoids - mostly electromagnetic phenomena (like ball lightnings, big and small, usually connected to different theories of ether.
Ведущий: Колтовой Н.А.
1. Колтовой Николай Алексеевич, (koltovoi@mail.ru). Обзор моделей микроплазмоидов. (Текст доклада)
2. Сапогин Владимир Георгиевич, (sapogin@mail.ru). Полевой конфайнмент плазмоидов с однородной температурой. (Текст доклада)
3. Миркин Владислав Иосифович, (mirkinvlad@mail.ru). Плазмоиды (мю-плазмоиды) в униполярном эфире. (Текст доклада)
4. Шишкин Александр Львович, (avkbeta@mail.ru). Микроплазмоиды. (Текст доклада)
There is a lot of fascinating literature on related subjects that can be found on the personal web site of N.A. Koltovoi:
P.S. 12-03-25 13:30
Perhaps it is good to see an example in infinitely-dimensional case. So let en (n=1,2,....) be an orthonormal basis in H. Let ρ be defined by
ρ en = c(1/n2) en,
c = 6/π2.
Then ρ is positive of trace class since Σ(1/n2) = π2/6. ρ defines a faithful state since 0 is not an eigenvalue.
Now
ρ½ en = (6½/π) (1/n) en.
ρ½ is not of trace class since the series Σ(1/n) is divergent. But it is a Hilbert-Schmidt operator.
P.S. 12-03-25 14:47 Today I have finally received the rare book that I have ordered some time ago:
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Anna Sverdlik "How our Emotions..." |
An open door is much better than a closed one. What we find behind it is especially wonderful. Light and infinity.
ReplyDeleteI was recently told that my idea that "adding a dimension is like crossing infinity" is almost verbatim a statement by Peter Ouspensky in his "Tertium Organum". It is a pleasant surprise for me since i've almost lost hope to find understanding with such ideas.
Another related thesis is looking at our 3d world as a boundary (intersection) of worlds of higher dimensions. Generally, more complicated organization of matter can be reached via intersecting, interacting between, or going to the boundaries of more simply organized domains.
Tertium Organum, as far as I remember is mostly a theoretical vision. Fourth Way on the other hand, is practical. This is how I have found Gurdjieff. It is really a pity that Ouspensky ended up drinking too much!
Delete"and equip it with the scalar product ω = ω(a*a)"
ReplyDeleteshouldn't it be ω = ω(a*b) ?
Thanks. Fixed.
DeleteIt looks like i have too many questions to this post. So that it is even hard to formulate them properly...
ReplyDeleteHow can vector Ωω representing the state ω be independent of the state?
Why do we need the alternative GNS construction for considering Tomita's flow? Probably, a hint can be found in the analogy with the two alternative pictures of Heisenberg and Schroedinger, where time dependence is carried either by a state or by an operator?
Good question. We do not really "need". But it is good to know that we have the two constructions avail;able. It is much like knowing that we can choose the measure for calculating the integral, for instance
Delete∫ f(x) dμ(x) = ∫ f(x) (dμ(x) /dx) dx
In the first integral we integrate function f with respect to the measure μ. The function is the same for each measure μ. In the second integral the measure is the always the same = dx, but the dependence on the measure is in the function that we integrate - it is f(x) (dμ(x) /dx).
That is the analogy that is somewhat closer than between the Schrodinger and Heisenberg picture. In the literature you can find both ways of discussing Tomita's flow. Connes and Rovelli in their paper make use of Hilbert-Schmidt operators. I will make a comment on this subject in the next post.
Ark, thank you for the explanation. So, the difference between the two alternative GNS-constructions is rather formal...
DeleteI thought it is crucially important that we should apply the alternative construction; something like that in the standard version, we won't be able to get the Tomita flow at all.
Concerning the 'Why'-questions, for now i can only guess that
if ρ is a positive trace class operator, i.e.
Tr(|ρ|) < ∞, where |ρ| = √(ρ*ρ),
then Tr(√(ρ*ρ)) = Tr(√(ρ*) √ρ) < ∞
and we will get the definition of Hilbert-Schmidt operator if can show that √(ρ*)=√(ρ)*, which seems to be right for a positively determined ρ.
Good! In fact a positive operatory is, in particular, Hermitian ρ=ρ*, and its square root, by its definition, is also Hermitian.
DeleteThis comment has been removed by the author.
ReplyDeleteSo, since the norm in B2(H) is finite 1 is a finite number), we have indeed a well-defined vector in B2(H). And this the answer to this particular "Why?".
DeleteI deleted my comment above because the triangle brackets eat all the stuff between them. Now i will use { } instead of "<" and ">"
DeleteRepeating the post deleted above:
Delete"Therefore Ωρ is well defined vector of B2(H), with (Why?)
{Ωρ , Ωρ} = 1".
Does this 'why?' relate to the property of vector to be well-defined or to the equality {Ωρ , Ωρ} = 1?
"Therefore Ωρ is well defined vector of B2(H), with (Why?)
{Ωρ , Ωρ} = 1".
Does this 'why?' relate to the property of vector to be well-defined or to the equality {Ωρ , Ωρ} = 1?
Since i don't know what does it mean that vector is well-defined, i will check the equality:
For density matrix ρ, Tr(ρ)=1, therefore
1=Tr(ρ)=Tr(ρ½ ρ½)=(Tr(Ωρ Ωρ)={Ωρ* Ωρ}
by definition of scalar product in space B2(H): {a,b} = Tr(a*b)
Because ρ is Hermitian and ρ½ is Hermitian, we can write
{Ωρ Ωρ}=1 as required.
Ok, now i've learned that vector is well-defined if it has a finite norm, thank you!
DeleteWell, we are in special situation here. Vectors of B2(H) are bounded operators. We may have a bounded operator, with a finite operator norm, but infinite Hilbert-Schmidt norm. So it is not enough to have just a bounded operator to have a well defined element of B2(H).
DeleteBut it is true that some of my "Why"'s are far from being clear. You are pointing out this lack of my clarity pretty well!
Delete'an example in infinitely-dimensional case'
ReplyDeleteYes, it's a puzzling example. The matter is that the mysterious 1/n function is involved. There would be nothing extraordinary for higher degrees like 1/n^4, 1/n^6, etc., right?
I always wondered why the integral of a monomial function f(x) = x^n is again monomial of one degree higher for any n from Z, with only one exception of n=-1. The same sort of puzzle.
"Moreover, we have ω(a) = {Ωρ , π(a)Ωρ} (Why?)"
ReplyDeleteThis is probably by the theorem 4.5.2. from Part 42 https://ark-jadczyk.blogspot.com/2025/01/spin-chronicles-part-42-gns-promenade.html
We have a state ω(a) on *-algebra and, then, by theorem 4.5.2, there is a cyclic representation π(a) and a cyclic vector Ωρ, such that
ω(a) = {Ωρ , π(a)Ωρ}
But i am not sure this is sufficient to conclude that Ωρ is cyclic particularly for the representation π of B(H) on B2(H),
which is the next "Why?"
I added the answer to this particular "Why?" within the text after the question. It should help to make the reasoning clear. I realized that it was not clear before.
Delete"The vector Ωρ is cyclic for the representation π of B(H) on B2(H)"
ReplyDeleteArk, i'm afraid i have a problem with showing this fact.
I remember that "A vector is cyclic for an algebra, if acting with the algebra elements on this vector you get the whole space (in infinite-dimension case, a dense subspace)".
But i have no idea how to approach the proof in our case. Couldn't you give a hint please? Not the whole proof at once ... )
OK. Added a hint after the "Why?"
Delete"Hint: We want to show that the subspace {X∈B2(H)} is dense in B2(H)".
DeleteBoth times "B2(H)"? Probably, "...the subspace {X∈B2(H)} is dense in B(H)"? Or vice versa :)
You are right. Fixed.
DeleteHamilton's biggest mistake is that he assumed ij=k
ReplyDeletebut he should have assumed ji=k.
Why is human civilization so unlucky?
(Another example is that current flows in the opposite direction to electrons.)
(Yet another example is that the column and row numbers of a matrix start at 1 and not 0.)
"Why is human civilization so unlucky?"
DeleteIt is not THAT bad. In fact the human civilization is quite lucky to have Bjab who tirelessly notices many of these glitches.
"current flows in the opposite direction to electrons."
DeleteBut this may be only on the "our side of the Universe". On the other side, perhaps, the two directions agree?
Maybe ij=k might have something to do with that prefered orientation Cl(V) as space is showing?
DeleteRegarding the current thing, at the time guys from microelectronics claimed that while there is a charge flow, at the same time there is a flow of holes in the opposite direction if remember right. Maybe those dealing with currents would have us rather take the holes as the real thing?
And as it goes about counting 0 or not, luckily there is only one year 0 in a particular chronology calendar.
By the way, the whole ijk thing seems rather unfortunate as boxes in contrast to balls usually exhibit much larger friction and consequent heat emission while sliding then the balls do when rolling. And rolling the box more often than not gives us pain in the back and wobbly not so fluid movement. FWIW.
DeleteWhile at boxes and balls, a question about geometry.
DeleteSuppose we have chosen a representation where two vectors R1 and R2 define x-y plane and their cross product gives z direction in the sense that formally only z is kind of fixed, i.e. R1 and R2 are perpendicular to each other but can freely rotate in x-y plane at the same angular speed which would preserve their orthogonality.
Is there a way to rewrite combination (R1e1 + iR2e2) in sort of a polar notation that would reflect the arrangement just described?
Would it help if we additionally assume that their norm squared (R1^2 + R2^2) is related as in proportional to the angular frequency with which they can rotate in the x-y plane?
"rather take the holes as the real thing?"
DeleteI often think of it! The status of existence of a hole, i.e., a missing object, for example, a center of a vortex, is no less than that of any other, 'usual' object.
I would disagree on that one, as there seems to be an important qualitative difference on ontological level between the two.
DeleteThe existence of an object comes from its actual being, while the existence of a hole is defined in relation to its surroundings as a lack or a gap in the actual being of the medium where we perceive the hole is located.
FWIW.
P.S. In the sense of quality every hole is the same, while actual objects obviously can differ from each other in many possible ways.
DeleteThe hole relative to something can be unique like the hole for matter being antimatter which even turned out to be an actual object. Things like holes and quasiparticles can be useful to think about.
DeleteThank you, John! You anticipated my answer to Saša.
DeleteThe apophatic approach is a very old and deep tendency of human thinking, it is enough to mention Buddhism as the most famous doctrine of this kind. I see several arguments from physics side:
- Every experimenter waiting for a flash on the oscilloscope screen will say that if he saw a flash, this does not necessarily confirm that what he expected happened. The absence of a flash confirms much more reliably that there was no effect.
- Physics is essentially apophatic, we can measure a quantity in an interval, i.e. say what it should NOT be. With progress, the intervals become narrower, but in any case these are intervals between the excluded areas.
- In order to substantiate the absence of something, you need to have no less information about it than in order to substantiate its presence.
Vortices as duals to point objects is also a related topic. Once i tried to collect some of the remarkable properties of vortices; can send the list if someone wants to look at.
Thank you for elaboration, I see your point now
Delete"The vector Ωρ is cyclic for the representation π of B(H) on B2(H). (Why?)
ReplyDeleteHint: We want to show that the subspace {Xρ½: X∈B(H)} is dense in B2(H). The subspace in a Hilbert space is dense if and only if its orthogonal complement is {0}. Thus we want to show that, with Y∈B2(H), = 0 for all X∈B(H)} implies Y=0. That means that Tr (Y*Xρ½) = 0 for all X∈B(H) implies Y=0".
In Part 39 we have seen that
{a: f(a*a)=0} = {a:f(a*b)=0 for all b}
We used this equality in Part 40 to check that the only vector orthogonal to all vectors is the zero vector
(the property which is necessary in a Hilbert space):
"Suppose ([a],[b])f=0 for all b. That means f(a*b)=0 for all b. That means a is in If. That means [a]=[0]".
Here, the situation is much the same, isn't it?
If Tr(Y*Xρ½)=0 for all X∈B(H), then Tr(Y*Y)=0 and Y is in the ideal of B(H). We have already known that since Y∈B2(H), which is an ideal of B(H). So, the only thing that we should show is that if Tr(Y*Y)=0, then Y=0 ?
Indeed, this will do.
Delete"If Tr(Y*Y)=0, then Y=0" is more or less evident but i cannot prove it rigorously. It depends on the properties of norm in B2(H). For any Y∈B2(H), Y*Y = |Y|^2, the norm of 'Y' squared. Norm is finite and positive. By definition, trace is a sum of (e_i, Y*Y e_i), which are positive numbers. The sum of positive numbers can be zero only when all numbers are zero. But does it mean that Y is zero?
DeleteY is a bounded operator - therefore continuous.
Delete(e_i, Y*Y e_i)=0 implies Y e_i=0 for all i.
Finite combinations of e_i span a dense set.
Thus Y vanishes on a dense set.
But it is a continuous operator ....
If Y vanishes at a dense set, then for any b one can find a sequence of points y_n
Deletesuch that y_n→b
and Y(y_n)=0 for all n. And continuity implies that Y(b)=0.
Very good,. Anna!
DeleteThank you, Ark, for your patience and pedagogical talent.
DeleteBut do we really need such technical details to understand the essence of Tomita-Takesaki theory?
This is rather a personal taste. You did a wonderful job in your presentation on "Thermal Time as an example of the relational conception". Originally I have also planned to just give an overview and a finite-dimensional example. But with time I realized that it would not give a true "taste" of the whole story. On the other hand the really "true taste" would require an overwhelming number of techniques. So a middle line was required. Drawing such a middle line is always personal. You can shift it one way or another, it is a personal and collective adventure. And you are actively participating and shaping this adventure. Without your participation I would probably do it in an easy way!
DeleteIt is like with skating. One thing is to watch ice skating. But it adds a whole new dimension if you watch professional skaters on a screen while you yourself have some skating experience.
Delete"It's like ice skating"
DeleteYes, or like playing the piano. A completely different level of understanding. That's why I asked my friends theorists to let me do the tiniest part of the calculations under their guidance. But they all take their work very seriously...
Then I started doing student exercises, for example, from Lounesto's book, from Woit's lectures, etc. Of course, all this cannot be compared with the opportunity of live learning that you give us, it's really incredible and invaluable.
This comment has been removed by the author.
ReplyDeleteUsed wrong syntax for bolding, here's corrected version.
Delete"While at boxes and balls, a question about geometry."
Well, maybe some background in the sense of context might be in order.
The talk recently about the representations was a bit too abstract for my taste, so returned to the basics so to say and checked things starting from a general element of our algebra in the form of a biquaternion. Also that QM idealization in the form of ideals, but only left ones, and picking (1+e3) and (e1-ie2) as the basis vectors, apart from not feeling natural, seemed like sort of over-defined "problem" in one domain, as we don't need to specify all three (e1,e2,e3) because they form an orthonormal basis, while being under-defined in the other domain, as not knowing anything about imaginary part of the complex scalar.
So starting from q=r0exp(iφ0)+q^Re+iq^Im, calculating the whole product qq*, not just the scalar part of it,
qq* = r0^2 + (q^Re)^2 + (q^Im)^2 + 2r0cos(φ0)q^Re + 2r0sin(φ0)q^Im + 2(q^Re × q^Im),
and comparing it to the same if using general polar form of a biquaternion,
q=r0exp(iφ0)+r1exp(iφ1)e1+r2exp(iφ2)e2+r3exp(iφ3)e3,
noticed that the general minimal biquaternion form that would provide us with the complete information about our space, assuming we know the algebra behind it, would be,
q = r0exp(iφ0) + r1e1 + ir2e2,
or if written to explicitly show real and imaginary parts,
q = (r0cos(φ0)e0 + r1e1) + i(r0sin(φ0)e0 + r2e2).
This biquaternion form would then define sort of natural orthonormal basis, corresponding to the choice of φ1=0, φ2=pi/2 and r3=0, while φ3 in this choice remains kind of a free parameter.
In other words, as described in comment question above, our coordinate or reference system could freely rotate about the sort of a fixed e3 axis as the normal of the plane spanned by given e1 and e2. The rotation speed could, but does not necessarily need to be, related to frequency of an EM wave in the sense of angular frequency with which vectors of electric and magnetic fields are rotating. In principle, we would have basically infinitely many possibilities for reference systems even for a given (e1×e2)=e3 direction.
On top of that, there's that φ3 free parameter, which when think about it, might stand for spin, or different spin values, as it represents the angle in complex plane of e3 direction, which pretty much sounds like what's said about the spin in the sense of helicity being projection of the spin on the direction of momentum which would be our e3.
Any comments on this, and maybe any ideas how to get to sort of Maxwell equations from this general natural minimal form of biquaternion?
Differentiating the product qq* above by φ0 and assuming r0 being constant, harmonic oscillator diff. eq. obtained gives solution in the form,
Deleter0(q_r +- iq_i)exp(+-iφ0) + constant terms, which hints that by clever choice of spatial derivative as a sort of nabla, we might recover wave equation for EM field.
So, maybe there's no real need to rewrite vector part in polar notation, but more to see how to form nabla operator in appropriate manner. Once we have nabla, Maxwell eq. probably won't be far away.
Any suggestions and comments?
When using spatial derivatives as templates for possible nabla components, standard multiplication rules apply, right?