How Is the Matrix V Related to Dirac Spinors and Tensor Products?

In summary, the conversation discusses the derivation for Dirac spinors and the factorization of an arbitrary vector into two tensors. The vector is represented by a matrix with components that can be simplified using matrices ##R^-## and ##R^+##. The final step in the derivation involves factorizing the matrix into two tensors with symbolic variables ##\Psi_1, \Psi_2, \Psi_3, \Psi_4##. However, it is mentioned that this factorization is only possible for null vectors, and not for arbitrary vectors.
  • #1
James1238765
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TL;DR Summary
What are the derivation steps for the tensor product leading to Dirac spinors?
Could anyone help with some of the later parts of the derivation for Dirac spinors, please?

I understand that an arbitrary vector ##\vec v##

$$ \begin{bmatrix}
x \\
y \\
z
\end{bmatrix} $$

can be defined as an equivalent matrix V with the components

$$ \begin{bmatrix}
z & x - iy \\
x + iy & -z
\end{bmatrix} $$

Now let ##R^-## be defined as

$$ \begin{bmatrix}
e^{-i\frac{\theta}{2}} & 0 \\
0 & e^{i\frac{\theta}{2}}
\end{bmatrix} $$

and let ##R^+## be defined as

$$ \begin{bmatrix}
e^{i\frac{\theta}{2}} & 0 \\
0 & e^{-i\frac{\theta}{2}}
\end{bmatrix} $$

Next, consider the matrix product ##R^-VR^+##:

$$
\begin{bmatrix}
e^{-i\frac{\theta}{2}} & 0 \\
0 & e^{i\frac{\theta}{2}}
\end{bmatrix}

\begin{bmatrix}
z & x - iy \\
x + iy & -z
\end{bmatrix}

\begin{bmatrix}
e^{i\frac{\theta}{2}} & 0 \\
0 & e^{-i\frac{\theta}{2}}
\end{bmatrix}
$$

which simplifies component-wise to

$$ \begin{bmatrix}
z & x\cos\theta - y\sin\theta - i(x\sin\theta + y\cos\theta) \\
x\cos\theta - y\sin\theta + i(x\sin\theta + y\cos\theta) & -z
\end{bmatrix} $$

Substituting ##\theta## ("rotating") by 180 ##^\circ## , we obtain as per normal

$$ \begin{bmatrix}
z & -(x-iy) \\
-(x+iy) & -z
\end{bmatrix} $$

and substituting ##\theta## by 360 ##^\circ##, we obtain as per normal the original state

$$ \begin{bmatrix}
z & x - iy \\
x + iy & -z
\end{bmatrix} $$

Now, the matrix V is said to be decomposable into 2 tensor products.

$$ \begin{bmatrix}
z & x - iy \\
x + iy & -z
\end{bmatrix} =

\begin{bmatrix}
\Psi_1 \\
\Psi_2
\end{bmatrix} \otimes

\begin{bmatrix}
-\Psi_2 & \Psi_1
\end{bmatrix} $$

I am lost at this step: why are the ##\Psi_1, \Psi_2, \Psi_3, \Psi_4## left symbolic? I have tried to put concrete variables into ##\Psi_1, \Psi_2, \Psi_3, \Psi_4## but could not find what the correct arrangements should be.

Could anyone write down the explicit variables what ##\Psi_1, \Psi_2, \Psi_3, \Psi_4## should be set to, please?
 
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  • #2
Your matrix pertains to 2-spinors, and not Dirac 4-spinors. If you actually want to learn about these 2-spinors, which have applications outside of QM, they have applications in general relativity, you can look at chapter 41 of GRAVITATION Misner et al or you could look at chapter 2 of Advanced General relativity Stewart, J.

Appendix A of Advanced General relativity Stewart, J explains how the Dirac 4-spinor formalism can be formulated in terms of these 2-spinors.

p.s. you don't have to know about general relativity to read about these 2-spinors.
 
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  • #3
James1238765 said:
TL;DR Summary: What are the derivation steps for the tensor product leading to Dirac spinors?

Could anyone help with some of the later parts of the derivation for Dirac spinors, please?
[...]
Where are you getting this from? Please give a link or (precise) reference.
 
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  • #4
The precise steps follow exactly [this]. It's a derivation for SU(2) from group theory without any physics. Also [this] follows a similar path, but also left the factorization ##\Psi_1, \Psi_2, \Psi_3, \Psi_4## as vague symbolic variables (at 9m10s).

I am unable to search exhaustively for any factorization solution of even the simplest matrix V with x=1, y=1, z=1 for any 4 complex numbers ##\Psi_1, \Psi_2, \Psi_3, \Psi_4##, with the 8 components having values in the range [0, 3]. So it would seem to be difficult to find the factorization formula for any arbitrary matrix V.

Are the solutions ##\Psi_1, \Psi_2, \Psi_3, \Psi_4## ordinary complex numbers, or some purpose-built mathematical object made to satisfy this factorization?
 
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  • #5
$$ V ~=~ \begin{bmatrix}
z & x-iy \\
x+iy & -z
\end{bmatrix}$$
These videos seem to be talking in the context of null 3-vectors, i.e., where ##\|v\| = 0##. If the vector is written in matrix form, its norm##^2## is obtained via the determinant of the matrix, i.e., $$\|v\|^2 ~=~ x^2 + y^2 + z^2 ~=~ -\det V ~.$$
Now take the outer product that was suggested, i.e.,
$$ \begin{bmatrix}
-\Psi_1 \Psi_2 & \Psi_1^2 \\
-\Psi_2^2 & \Psi_2 \Psi_1
\end{bmatrix}
$$ The determinant of this matrix is $$-\Psi_1^2 \Psi_2^2 ~-~ (-\Psi_2^2 \Psi_1^2) ~=~ 0 ~,$$ i.e., always zero.

So, provided the original ##v## is a null vector, the factorization into ##\Psi##'s is trivial: ##\Psi_1 = \sqrt{x-iy}##, ##\Psi_2 = \sqrt{x+iy}##.

HTH.
 
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  • #6
@strangerep thank you! Based on the answer I managed to get a working factorization using slightly different prerequisites. For example,

Let $$ \vec v =

\begin{bmatrix}
\frac{1}{\sqrt 2} \\
\frac{1}{\sqrt 2} \\
1
\end{bmatrix} $$

Then $$ V =

\begin{bmatrix}
1 & \frac{1}{\sqrt 2} - i \frac{1}{\sqrt 2} \\
\frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2} & 1
\end{bmatrix} $$

And $$
\begin{bmatrix}
\sqrt{\frac {1}{\sqrt 2} - i \frac{1}{\sqrt 2}} \\
\sqrt{\frac {1}{\sqrt 2} + i \frac{1}{\sqrt 2}}
\end{bmatrix} \otimes

\begin{bmatrix}
\sqrt{\frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2}} & \sqrt{\frac{1}{\sqrt 2} - i \frac{1}{\sqrt 2}}
\end{bmatrix} =

\begin{bmatrix}
1 & \frac{1}{\sqrt 2} - i \frac{1}{\sqrt 2} \\
\frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2} & 1
\end{bmatrix} $$

The 2 prerequisites on ##\vec v ## are:

1. ##z = 1##

2. ##x^2 + y^2 = 1##

Thus, a very restricted set of ##\vec v ## satisfying the above conditions can be said to be factorizable into two tensors. Seems a little forced though to call these very restricted set "spinors", and disqualify all the rest of the [x,y,z] vectors?
 
  • #7
James1238765 said:
[Seems a little forced though to call these very restricted set "spinors", and disqualify all the rest of the [x,y,z] vectors?
It begins to make more sense, (and be more useful), after you move onto 1+3D spacetime instead just 3D space.

As with many such youtube videos, if you're serious about learning then it's advisable to study from a reputable textbook instead (and do the exercises therein).
 
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  • #8
@strangerep I'm afraid i am too old to spend months/years properly learning quantum theory as it should ideally be learned. But I try to learn what i can here and there.
 
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  • #9
James1238765 said:
@strangerep I'm afraid i am too old to spend months/years properly learning quantum theory as it should ideally be learned.
Nonsense. I, too, am no longer young, so I can reliably say that you just need to find a textbook at the right level, and work through it gradually.
 
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  • #10
@strangerep i have heard from many reputable sources that QFT is about the hardest theory to really grasp, how even professional mathematicians might struggle for multiple years to understand the technicalities [true story] , so i guess that factors as well into why I chose to approach it haphazardly.
 
  • #11
James1238765 said:
@strangerep i have heard from many reputable sources that QFT is about the hardest theory to really grasp, how even professional mathematicians might struggle for multiple years to understand the technicalities [true story] , so i guess that factors as well into why I chose to approach it haphazardly.
Coming from someone who had to leave graduate school and has spent the last 20+ years learning on my own, I have to say that your approach does not make much sense to me. It is okay(ish) if you just want to satisfy some curiosity, but you can't really learn it like that. Despite all of my years of "personal training" and efforts to do the job right I have missed any number of small details and they can quickly catch up to you.

QFT is a complicated subject and requires not only a lot of Physics knowledge but some rather advanced Mathematics. The only way you are going to understand this is by making the effort to learn it carefully, working (numerous) examples, and from the start. It is going to take time. Poking around with it is just going to get you seriously confused.

And if you think QFT is bad, you should try Fluid Dynamics! :)

-Dan
 
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  • #12
James1238765 said:
@strangerep i have heard from many reputable sources that QFT is about the hardest theory to really grasp,
QFT is indeed difficult, but can be (incompletely/imperfectly) understood on several levels. Of course, one should master ordinary QM and its associated math first (see the 3rd line in my signature block below).

James1238765 said:
how even professional mathematicians might struggle for multiple years to understand the technicalities [true story] ,
This is also true, but should not prevent one from attaining a basic understanding, e.g., by working through an "Intro to QFT" textbook (assuming one already has the necessary prerequisites).

James1238765 said:
so i guess that factors as well into why I chose to approach it haphazardly.
This does not make sense to me.

If you're not already proficient in ordinary QM, try the 1st chapter of Ballentine which presents some of the math that can otherwise be puzzling in QM. That should give you a feeling of whether it's feasible for you to continue into subsequent chapters.
 
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  • #13
Of course, if you try to learn QFT from the mathematicians' point of view, it's very difficult, because they have the goal to find a rigorous formulation of it or to prove that such a rigorous formulation is impossible (given certain criteria for what such a rigorous formulation might be), i.e., from this point of view QFT is an open research project.

From HEP phenomenologists' point of view it's not that difficult anymore. You just use what was developed over the past 80 years to describe the phenomena in terms of an "effective theory", i.e., the formulation in terms of renormalized perturbative QFT. To learn this, you should have some idea on non-relativistic quantum mechanics in terms of Dirac's formulation. For this I recommend to have a look at

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics, Addison Wesley (1993).

Then you can start studying relativsitic QFT from the HEP phenomenologists' point of view. For an introduction, see

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

For a more up-to-date presentation of the Standard Model, see

M. D. Schwartz, Quantum field theory and the Standard
Model, Cambridge University Press, Cambridge, New York
(2014).

For all the details from this pragmatic point of view:

S. Weinberg, The Quantum Theory of Fields, vol. 1,
Cambridge University Press (1995).

S. Weinberg, The Quantum Theory of Fields, vol. 2,
Cambridge University Press (1996).

and

A. Duncan, The conceptual framework of quantum field
theory, Oxford University Press, Oxford (2012).
 
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1. What are Dirac spinors?

Dirac spinors are mathematical objects used to describe the spin of particles in quantum mechanics. They are solutions to the Dirac equation, which is a relativistic wave equation that describes the behavior of fermions, such as electrons and quarks.

2. How are Dirac spinors derived?

Dirac spinors are derived from the Dirac equation, which is a combination of the Schrödinger equation and the relativistic energy-momentum relation. This equation is solved using techniques from differential equations and linear algebra to obtain the spinor solutions.

3. What are the properties of Dirac spinors?

Dirac spinors have four components, corresponding to the four possible spin states of a particle. They also have a complex conjugate, which is used in the calculation of probabilities in quantum mechanics. Additionally, they satisfy the Lorentz transformation, meaning they are invariant under rotations and boosts in special relativity.

4. How are Dirac spinors used in physics?

Dirac spinors are used to describe the spin of particles in quantum mechanics, particularly in the study of fermions. They are also used in the Standard Model of particle physics to describe the behavior of fundamental particles, such as electrons and quarks.

5. Are there any applications of Dirac spinors?

Dirac spinors have many applications in physics, including in quantum field theory, particle physics, and condensed matter physics. They are also used in technologies such as transistors and superconductors, which rely on the behavior of electrons described by Dirac spinors.

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