## Product by a complex numbergreenspun.com : LUSENET : quaternions : One Thread |

Hello Doug,I found the need to multiply a quaternion by a complex number; this does not seem to be covered in your text. Here is my proposal:

Call h to the complex imaginary unit and i, j, k to the quaternion "imaginary" units; the following rules apply:

h x h = i x i = j x j = k x k = -1

i x j = -j x i = k j x k = -k x j = i k x i = -i x k = j

h x i = h x j = h x k = -1

Do you agree with these rules? Is there another way of multiplying quaternions by complex numbers?

Thanks Jose B. Almeida University of Minho Portugal

-- Jose B. Almeida (bda@fisica.uminho.pt), September 05, 2002

Hello Jose:Real numbers are quaternions of the form (t, 0, 0, 0). There are three purely imaginary numbers, (0, x, 0, 0), (0, 0, y, 0), and (0, 0, 0, z), but these are not closed under multiplication like the reals. The flavors of complex numbers are (t, x, 0, 0), (t, 0, y, 0), and (t, 0, 0, z) [makes me wonder what I should call (0, x, y, 0), again a number that is not closed under multiplication but is under addition]. The next class of numbers beyond quaternions is the octonions, but they lack the associative property, so (a (b c)) does not equal ((a b) c). When I hear of a physical problem that requires a non-associative algebra, I'll become a fan of octonions, but not till then.

Dirac matrices are essential to quantum mechanics. People have often pointed out that a factor of i is all that separates the two. That makes it sound like there is a trivial difference. In my opinion, the difference is not minor. Dirac matrices do not form a division algebra. The product of two non-zero Dirac matrices can evaluate to zero. That property scares me (I am frightened by strange things). The reason is statistical. If two processes interact in some way that involves multiplying Dirac matrices together, how can one properly account for those where the product is zero? This is going to create an accounting issue only. The zero ads nothing to whatever is going on, but one cannot track these zeroes. Physics has some subtle problems with zeroes and infinities, and this could be a source: use a division algebra, no matter how painful. At least my work focuses on that problem.

If you want to think about how to manage spin, I do recommend at least giving my web page on spin a look (URL at the end). This was just a hugely fun mini project. I read Sakurai's "Modern Quantum Mechanics", chapter 3, and expressed the infinitesimal rotations with quaternions using rotation matrices in commutators. If you take the conjugate of the rotation matrices in the commutators, nothing changes. If you take what I call the first conjugate, which is (i q i)*, and put that in a commutator, one gets (i q i)* (i q' i)* = i q* i i q'* i = - i q* q' i. That minus sign in front because the two internal i's form a product give this rotation the twist needed to represent spin 1/2. At least that is a quick sketch of the idea.

So those are my biased thoughts :-) [note - for whatever reason, I was not notified of this thread, so that explains my delay. Wilke's answer was on target.]

doug

http://world.std.com/~sweetser/quaternions/quantum/spin/spin.html

-- Doug Sweetser (sweetser@theworld.com), November 06, 2002.

Hello José, each of the 3 imaginary units i, j, or k can be used for representing complex numbers. You have only to choose one of them. There is no need to get another imaginary unit. If you apply the quaternion multiplication rules to quaternions, having two imaginary parts set to zero, you will "automatically" get the multiplication rules for complex numbers. Try for example (s + 0*i + y*j + 0*k) * (t + 0*i + v*j + 0*k). In this example j is choosen as imaginary unit, hence the 1st and 3rd imaginary parts are set to zero. Complex numbers are a subset of quaternions! Greetings Gerd

-- Gerd Wilke (gerd-wilke@gmx.de), September 18, 2002.

Hello Gerd. Thanks for your answer; you are right, of course. I had already realized that my question was ill posed and what I was looking for was something like a 5-complex number with one real part and 4 complex parts. I may have found the answer already by turning to spinors and Pauli matrices but it is still too soon to write about it. I shall have to clarify my ideas beforehand.Ciao

Jose

-- Jose B. Almeida (bda@fisica.uminho.pt), September 18, 2002.

Easier way to remember the 'rules': ijk=-1 If you multiply both side on the left by i, you get -jk=-i => i=jk Same sort of thing to get the other pairs.

-- Tom Willis (thomas.willis@durham.ac.uk), November 06, 2002.

Doug,Thank you for your answer and also thanks again to gerd.

Things have become clearer in my mind and I don't have a problem anymore, because I chose to work with Dirac matrices; in any case, think about this:

If we take Dirac equation and rename \beta to \alpha^0, the following relation is valid:

\alpha^\mu \alpha^\nu + \alpha^\nu \alpha^\mu = 2 \delta^{\mu \nu}

So (i \alpha^\mu) represents 4 imaginary entities and not 3 as with quaternions. This is what I meant by 5-complex numbers.

Jose

-- Jose B. Almeida (bda@fisica.uminho.pt), November 07, 2002.

Hello Jose, Gerd and Doug,Gerd wrote: "Each of the 3 imaginary units i,j or k can be used to represent complex numbers. You have only to choose one of them. There is no need to get another imaginary unit." Jose agreed with Gerd.

But this is not correct.

The complex imaginary unit (which Jose calls h) can commute in a product with any quaternionic or complex quantity. But none of the three quaternion imaginary units (i,j,k) can replicate this commutativity.

Also, Jose's offered a set of postulates for h,i,j,k. The third says:

hxi = hxj = hxk = -1

This is not correct. This can be shown by pre-multiplying by h across the entire equation:

hxhxi = hxhxj = hxhxk = hx(-1)

(hxh)xi = (hxh)xj = (hxh)xk = (hxh)x(-1) By associativity.

(-1)xi = (-1)xj = (-1)xk = (-1)x(-1) Because -1 = h^2.

-i = -j = -k = +1 Which is in contradiction to Hamilton's postulates.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Hello Matthew:The h changes the algebra from quaternions to complex-valued quaternions (sometimes called biquaternions), which is not a division algebra. LOTS of work has been done with these. Personally, I stick with quaterions the division algebra.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Hi, Doug,The vector parts of biquaternions can be extremely useful. They form the basis of the phasor field formulation of Maxwell's Equations for sinusoidal steady-state undulatory fields at a single frequency.

This formulation is the approach taken in the textbook Applied Electromagnetism by Liang Chi Shen and Jin Au Kong. This text has gone through several editions and has numerous enthusiasts. However, it is infamous for its detractors too. Judging from the book reviews on Amazon, it is a wretched failure for teaching E&M. But I love it!

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Hello Matthew:There are BIG FANS of biquaternions out there. I am not one of them :-) The old theorem of Frobenius is what makes quaternions special. For me, there is no _rational_ way to pick out a Clifford algebra. Quaternions and biquaternions are just two Clifford algebras out of many. What makes quaternions special is that they are a finite dimensional division algebra that is associative (up to an isomorphism). It is my opinion that the subtle problems that plague quantum field theory may be directly related to ignoring the importance of division.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Hello, Doug,You mention subtle problems in quantum mechanics. Are you referring to apparently inescapable divergent series? I vaguely recall this was a problem when I studied QM (way back in the Old Stone Age.) I would be intersted in knowing if and how this issue is resolved.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 08, 2004.

Hello Matthew:The short, solid answer is that the problems in quantum field theory remain. Doing perturbation calculations requires regularization and renormalization. No, I cannot do wither type of calculation currently, one must be a youngster in a Ph. D. program to be able to work through the easy stuff, the higher order crunches all done by computers.

I do have a very specific dream about the origin of this long standing pea under the tall stack of mattresses. It comes down to the difference between Pauli matrices and the quaternions. The Pauli matrices are just a bad copy of the quaternions, twisted by a factor of i. They do not form a division algebra. This will necessarily cause artifacts in the process of doing exhaustive statistics. There will be two _non-zero_ states whose product _is_ zero. There is no accountant who could stay sane in such a situation. Multiply two perfectly fine numbers together, and do not include the result because it is the additive identity (fancy words for zero). One would have to go back and figure out how not to include that particular product.

This will never happen with quaternions. The only way to get a zero product is to have a zero as a factor. There are three solid technical reasons quaternions have not been adopted: quaternion analysis doesn't work, the Lorentz group cannot be represented by quaternions, nor can the Maxwell equations be written. OK, all three of those are incorrect _now_, but these changes are only very recent. The first and there are not in peer-review publications.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 08, 2004.

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