Matrix associated to a vector (how?)

[member 11137]

As I unfortunately could not get an answer to my technical question concerning the construction of a matrix associated to a vector within the Theory of Spinors in another thread (Theory development: recent questions) I try again here... Could someone give me more details about this association or some links to learn more about this subject? Thanks in advance.

 

[matt grime]

you won't get an answer since as stated your question has none. "associate" in what sense?

 

[CarlB]

The Pauli spin matrices form a vector. That is, there are three of these 2x2 matrices, and they correspond to the x, y, and z components of a vector. This is generalized in the Dirac theory to x, y, z and t, but requires 4x4 matrices.

The Pauli spin matrices are:

$$\sigma_x = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$

$$\sigma_y = \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)$$

$$\sigma_z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$

Now, given a vector, perhaps $$u = (u_x,u_y,u_z)$$, the matrix associated with it is given by:

$$\sigma_u = u_x\sigma_x + u_y\sigma_y + u_z\sigma_z$$

$$= \left(\begin{array}{cc} u_z & u_x-iu_y \\ u_x+iu_y & -u_z \end{array}\right)$$

The spinors would be the 1x2 vectors that are eigenvectors for the set of all possible matrices $$\sigma_u$$. They correspond to the complex 2x1 vectors (other than the zero vector). For any given $$\sigma_u$$, the eigenvectors of that matrix are the spinors that correspond to spin of $$\pm \hbar/2$$ in the direction $$(u_x,u_y,u_z)$$.

I don't have Cartan's work available to me at the moment. Is this about what you were looking for?

Carl

 

[member 11137]

I don't have Cartan's work available to me at the moment. Is this about what you were looking for?

Carl

I have this book and I read it. Thanks for the explanation concerning the Pauli's matrices. In deed a good example in relation with the Theory of Spinors.

you won't get an answer since as stated your question has none. "associate" in what sense? you won't get an answer since as stated your question has none. "associate" in what sense?

I try to ask better: What is the hidden sense or motivation followed by E Cartan in this manner to associate a matrix and a vector in the Theory of Spinors [E. Cartan; Doover books page 81 relation (5)]? I explain: the way to give a precise position to the different coordinates of the vector satisfying a given fundamental form "seems" to be entirely the result of an arbitrary choice. Why? What does it represent (in relationship with the geometry or with a special property for this matrix: orthogonality...)? Is there any geometrical signification in this way of doing?

I hope I could make my question more explicit.

 

[member 11137]

I have had the time to visit the recommanded links; just want to say thank you; I feel me like the smallest thing in the world ... Before having been seeing all these links I would never have believe that so many people were working around mathemathics... As said, I fell no more alone but... so so small. Best regards and thanks.

 

[CarlB]

I explain: the way to give a precise position to the different coordinates of the vector satisfying a given fundamental form "seems" to be entirely the result of an arbitrary choice. Why? What does it represent (in relationship with the geometry or with a special property for this matrix: orthogonality...)? Is there any geometrical signification in this way of doing?

Dear Blackforest;

There is another, deeper, way of looking at the Pauli spin matrices. It's called "Clifford algebra" or "geometric algebra", and perhaps that is what you are looking for.

The Clifford algebra for the Pauli spin matrices is called "complexified CL(2,0)" by the mathematicians.

The algebra amounts to supposing that we wish to create an "algebra" that defines the geometry of 3-dimensional space. In mathematics, an algebra is a collection of symbols that satisfy certain relationships. In the case of the Pauli spin matrices or complexified CL(2,0), the symbols and their relationship consists of the following:

$$\hat{1} = 1$$ = Scalar number. This can be multiplied by any complex constant, and the result is still a member of the algebra.

$$\sigma_x$$ = Vector for the x direction. This is not a 2x2 matrix, as in the Pauli spin matrices, but instead is to be thought of as a vector of unit length in the x direction. Any multiple of this vector by a complex number, for example $$(3-i)\sigma_x$$ is still a member of the algebra.

The same applies to y and z giving $$\sigma_y$$ and $$\sigma_z$$ as members of the algebra, along with all their multiples.

In addition to what's been described, the algebra also contains all possible sums of complex multiples of the above four elements. So a typical element of the algebra might be:

$$(2-i)\hat{1} + 7i\sigma_x + 22\sigma_y - \sigma_z$$

So far we've defined addition for the algebra and multiplication by complex numbers. An algebra also needs a rule for multiplication. In this case, the rule will give the products of the vectors as follows:

$$\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = \hat{1}^2 = \hat{1}$$

$$\hat{1} \sigma_n = \sigma_n$$ for n = x,y,z


$$\sigma_x \sigma_y = -\sigma_y \sigma_x = i \sigma_z$$

$$\sigma_y \sigma_z = -\sigma_z \sigma_y = i \sigma_x$$

$$\sigma_z \sigma_x = -\sigma_x \sigma_z = i \sigma_y$$

If you try these rules out with the Pauli spin matrices you will discover that the Pauli spin matrices also satisfy these rules. That means that the Pauli spin matrices are a "representation" of the algebra.

But you do not have to pick a representation in order to make calculations with the algebra.

In this way, the association of a vector with a matrix becomes instead an association of a vector with a member of the algebra. For example, if $$u_x, u_y, u_z$$ are complex numbers, then you can associate

$$(u_x,u_y,u_z) <=> u_x\sigma_x + u_y\sigma_y + u+z\sigma_z$$

In a lot of ways, it's easier to make calculations with Clifford algebra than with the Pauli spin matrices. There is a relationship between the Clifford algebra as defined above, and the spinors or vectors used with the Pauli spin matrices.

The relationship is subtle and beautiful, but is probably beyond the scope of this thread: It involves special elements of the Clifford algebra that are called "primitive idempotents", which means that when you square them, they don't change, that is, that $$\iota^2 = \iota$$, and they're particularly "primitive".

Among the Clifford algebra CL(2,0), an example of a primitive idempotent is:

$$(\hat{1} + \sigma_z)/2$$

Among the Pauli spin matrices, an example of a primitive idempotent (that corresponds to the above Clifford algebra element) is:

$$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)$$.

If you take this matrix above, and multiply an arbitrary 2x2 matrix by it, what you get is a matrix that is nonzero only in the left column. That left column is a spinor. The same thing happens with the Clifford algebraic idempotent, but it is harder to see. These idempotents are used as "projection operators" and they reduce the full algebra (either of 2x2 matrices or the CL(2,0)) down to a subalgebra that is half the size.

The arbitrary choice in the spinor amounts to the arbitrary choice in the selection of the idempotent used to create the spinor. In the above examples, the idempotent chosen was the one that corresponds to spin+1/2 in the z-direction.

The z-direction was arbitrarily chosen. Perhaps this is the arbitrary choice you are thinking of. The choice of the z-coordinate causes the spinor that corresponds to spin aligned in the z-direction to be particularly simple.

This all comes about because the idempotent is a way of selecting a "subalgebra" of the Clifford algebra. It is the choice of this idempotent (in this case, the one aligned in the +z direction) that is arbitrary.

One can use the idempotents themselves as replacements for the spinors, but this is not generally done in physics. I suppose the reason that it is not is because you end up with "unphysical degrees of freedom". That is, a complex 2x2 matrix or a complexified CL(2,0) Clifford algebra (where idempotents live) has 8 degrees of freedom. A spinor only has 4. Thus by using spinors instead of idempotents, one eliminates 4 unphysical degrees of freedom.

Having done this, the spinor still has two unphysical degrees of freedom. These correspond to multiplication by an arbitrary complex number. One of these can be removed by requiring that the spinor be "normalized". The other is left as an arbitrary complex phase.

Carl

 

[member 11137]

There is another, deeper, way of looking at the Pauli spin matrices. It's called "Clifford algebra" or "geometric algebra", and perhaps that is what you are looking for.

The Clifford algebra for the Pauli spin matrices is called "complexified CL(2,0)" by the mathematicians.

Dear Carl B
Before everything: thank you for the sympathetic effort you have made to understand my question. The Clifford algebra owns effectively a central position in several "modern" approaches; e.g.: as you explain it here; and as I could read it somewhere else, also in the general relativity (There exists a possibility to write the Einstein's equation within the Clifford algebra).

The question I was asking is arising from my own approach of physics which is absolutely not standart as every "avant-garde" research certainly is. It does not mean that I critic what is still known. I just try to find a new way through the accepted knowledges. The problem is that it is really difficult to "translate" my intuition in the modern language. That is why I developped what I called the "extended product" of two vectors; at least to get a link with the tensor calculus.

... Perhaps this is the arbitrary choice you are thinking of. The choice of the z-coordinate causes the spinor that corresponds to spin aligned in the z-direction to be particularly simple.

The arbitrary choice I am thinking of concerns 4 by 4 matrices. The need I have to know how Cartan has built his matrices comes from the fact that developments of the extended product yield the necessity to associate a matrix, a vector and a scalar to each situation... in accordance with the local geometry or with the manner to calculate at a given place at a given moment. In a 3D space, Euclidien situations are reducing the extended product to the usual cross product. Thus I postulate that the local geometry owns an influence on our way to make calculations and I try to test and to develop this idea along my theory (which is interesting nobody). Amazingly I get (in certain circumstances) a formulation of the Maxwell EM field tensor depending on the local geometry, on how the extended product is defined and on the EM field itself (of course). This formulation is indeed a 4 by 4 matrix. This formulation is suggesting an analyse with the Theory of Spinors. This is the reason of my question. To make this interpretation I must be in a situation that allows to associate a position vector to the metric tensor. I have a way (etgb31c2.pdf) but it induces a totally new analyse of the physical situations...

This all comes about because the idempotent is a way of selecting a "subalgebra" of the Clifford algebra. It is the choice of this idempotent (in this case, the one aligned in the +z direction) that is arbitrary.

One can use the idempotents themselves as replacements for the spinors, but this is not generally done in physics. I suppose the reason that it is not is because you end up with "unphysical degrees of freedom". That is, a complex 2x2 matrix or a complexified CL(2,0) Clifford algebra (where idempotents live) has 8 degrees of freedom. A spinor only has 4. Thus by using spinors instead of idempotents, one eliminates 4 unphysical degrees of freedom.

Having done this, the spinor still has two unphysical degrees of freedom. These correspond to multiplication by an arbitrary complex number. One of these can be removed by requiring that the spinor be "normalized". The other is left as an arbitrary complex phase.

Carl

The dilemna between mathematical (degrees of) freedoms and physical reality:... I must think longer about this; but you are right: its a very important point. I shall think a lot of days about your answer and come back to the discussion. I wanted to show that I have appreciated your help. Best regards