QFTD Chapter 1: Particle Physics and Special Relativity

14 Aug 2010

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Sat, 08/14/2010 - 22:13

doug

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Note: I will be updating this page for quite some time! Much thinking must be done...

The uncertainty principle for position/momentum and energy/time as written in eq. 1.1 and 1.2 - common as they are - hide 2 accounting issues. First is that one should not think of the variant in x separately from the variation in t, nor the separation in energy separately from that of momentum. Someone in a rocket ship will just say your change in energy is his change in momentum. If one went ahead and put these in the proper 4-vector, there would not be a way to form a product.
$\Delta R ===(\Delta t, \Delta \vec{R}/c)$
$\Delta EP ===(\Delta E, c \Delta \vec{P})$
$\Delta R \Delta EP = (\Delta E \Delta t - \Delta \vec{R}} \cdot \Delta \vec{P}, \Delta E \Delta \vec{R}/c + c \Delta \vec{P} \Delta t + \Delta \vec{R} \times \Delta \vec{P})$

The job of the hbar is to make this quantity dimensionless, i.e. in the normalization of plane waves. The exponentials that appear so often in field theory do not make sense unless the quantity in question is dimensionless. It is better accounting practice to keep the terms dimensionless throughout.

Quaternions contain an i, j, and k, so we do not need to add these to our expressions. This will be quite odd, since the factor of i is repeated almost as much as hbar. Prunning in good.

Commutator relations are presented as facts (which they are), but can be easily justified. Take the position operator and momentum operator. These are operators, symbolized by putting a hat on x or p. Operators must act on a wave fucntion to have any meaning, but this acounting detail is often skipped. We will not do that, ever. Just trying to keep the rules consistent. Another problem with such notation is that it hids the relationship between the two operators. In the position representation, $\hat{X}$ will just be x, but momentum will be the operator h times the change in momentum, $\hat{P} = \frac{\partial m c \psi}{\partial x}$ . Notice that we need to toss in a factor of c to get the units right. The commutator acting on a wave function brings back the opposite of the wave function.

$(\hat{x} \hat{P} \psi -\hat{P} \hat{x}\psi)/\hbar = (x \frac{\partial m c }{\partial x} - \frac{\partial x m c}{\partial x}) \psi /\hbar = -c m \psi/\hbar$

If one uses a different combination, say the position y operator with momentum in the x direction, that can and should always disappear:

$(\hat{y} \hat{P} \psi -\hat{P} \hat{y}\psi)/\hbar =0$

This is the product rule in action, nothing more or less. When one says two variables are "conjugates", it means there is some kind of link via calculus between the two: position and momentum in the x direction are linked weither one is using a position or momentum representation, but not position in y and momentum in x.

The most famous equation in all of physics as far as the popular culture is concerned, is not valid (true to all). Energy transforms like the first component of a 4-vector, while mass invariant under a Lorentz transformation and the speed of light is a constant. The equation is true for one and only one observer, where the relativisitic velocity is zero. And as is our practice, we will not write an equation with the energy without also including the momentum. Here is an expression written with energy, momemtum and mass that is always true:

$( E,\overset{\rightharpoonup }{P}c) = m c^2(\text{Cosh}\left[\text{ArcTanh}\left[\frac{v}{c}\right]\right],\text{Sinh}\left[\text{ArcTanh}\left[\frac{v}{c}\right]\right]\hat{V} )$

I have chosen to write $\gamma$ and $\gamma \vec{\beta}$ as hyperbolic functions because it is more apparent than when this quantity is squared, then this vector identity:

$\text{Cosh}[\alpha ]^2-\text{Sinh}[\alpha ]^2 = 1$

is the reason for the link between the Lorentz covariant energy and momentum with the invariant mass.

Do a few simple number games as a sanity check for the link between hyperbolic functions and gamma beta notation:

$\text{Cosh}[\text{ArcTanh}[.5]] = 1.1547$

$\frac{1}{\sqrt{1-.5^2}} = 1.1547$

$\text{Sinh}[\text{ArcTanh}[.5]] = 0.57735$

$\frac{.5}{\sqrt{1-.5^2}} = 0.57735$

There is an advantage to the gammas and betas when you think about this inequality:

$\gamma > | \gamma \vec{\beta} |$

This means that energy is always greater than the magnitude of the momentum.

What is more important than the definition of the 4-momentum is the square of the 4-momentum:

$(m c^2 ( \text{Cosh}\left[\text{ArcTanh}\left[\frac{v}{c}\right]\right],\text{Sinh}\left[\text{ArcTanh}\left[\frac{v}{c}\right]\right]\right.)^2 = (c^4 m^2,\frac{2 c^5 m^2 v}{c^2-v^2})$

$\{E,P c\}^2==\left(E^2-P^2c^2,2c E P\right)=m^2c^4\left(1, 2\gamma ^2\beta \right)$

This equation is both true in any reference frame and complete. It is a perfectly fine operation to multiply energy times momentum, yet no graduate level books I have paged through use this information. It is this equation that will be the basis for reformulating quaternion quantum field theory.

With all terms in the quaternion expression filled in, it is easy to spot a non-relativistic expression: the generator of the equation will have a constant somewhere. The Schrödinger equation is written like so:

$\frac{-\hbar ^2}{2 m}\frac{\partial ^2\psi }{\partial x^2}+\text{V$\psi $}=\text{i$\hbar $}\frac{\partial \psi }{\partial t}$

According to the rules of this method, the changes in time and changes in space should be kept together. Since we are working with quaternions, the factor of i should be dropped. Here is the rewrite:

$\left(\hbar \frac{\partial }{\partial t}+\frac{\hbar ^2}{2 m}\nabla ^2\right)\psi =\text{V$\psi $}$

Consider what is needed to generate the differential operator. Note this is a scalar operator, nothing in the vector is included. There is only one way to make the Lapacian:

$(0, \vec{\nabla} ) (0, \vec{\nabla} )^* = (\nabla^2,-\vec{\nabla} \times \vec{\nabla} )$

This has the Lapacian, and a curl of a del. One recurring theme is that physics as it is accounted for today ignores what goes on in the vector part of expressions. Another way to say it is that vector parts find their utility via the scalar. Since the operator is a scalar, the time derivative must end up there too. While it is easy enough to add a time derivative to one of the terms, the other gets the constant 1 which has different units from the 1/length provided by the spacial derivative. To make those units work out requires an hbar/mc because the units for Planck's constant are mass length^2/time. The entire expression has units of energy, hbar over time:

$\hbar ( (\frac{\partial}{\partial t}, \vec{\nabla}) (1, \hbar \vec{\nabla} / 2 m c) + (\frac{\partial}{\partial t}, \vec{\nabla}) (1, \hbar \vec{\nabla} / 2 m c)* \psi /2 = (V, \vec{0}) \psi$

$(\hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2 m c} \nabla^2, \vec{0} ) \psi = (V, \vec{0}) \psi$

Simple substition in the square of energy - momentum leads to the form of the Klein - Gordon equation :

$E \to \frac{1}{c} \frac{\partial }{\partial t },P \to \vec{\nabla} / c$
into
$\left(E^2-P^2c^2,2c E P\right)=m^2c^4\left(1, 2\gamma ^2 \vec{\beta} \right)$

$(\frac{1}{c^2} \frac{\partial}{\partial t^2} - \nabla^2, \frac{2}{c} \frac{\partial}{\partial t} \vec{\nabla}) \psi = m^2c^4\left(1, 2\gamma ^2 \vec{\beta} \right)$

The units are not right however. The operators have units of 1/length^2, so can be made dimensionless with hbar, m and c:

$\frac{\hbar^2}{m^2 c^2}(\frac{1}{c^2} \frac{\partial}{\partial t^2} - \nabla^2, \frac{2}{c} \frac{\partial}{\partial t} \vec{\nabla}) \psi = \left(1, 2\gamma ^2 \vec{\beta} \right)$

As we have come to expect, the phase is not used at this time, but it is well-formed.

The Dirac equation is sometimes described by the square root of the Klein-Gordon equation because squaring the Dirac equation results in the Klein-Gordon equation. This feat is accomplished by using the 16 4x4 gamma matrices (more information can be found here: http://mathworld.wolfram.com/DiracMatrices.html). Work by others (J. Löpez-Bonilla, L. Rosales-Roldàn) has show that these 16 gamma matrices can be written as quaternion triple products with a very simple form: pre and post-multipy a quaternion by one of the 4 basis vectors. Write the Dirac equation with the time and space operators on the same side, and drop the factor of i:

$\hbar(\frac{1}{c} \frac{\partial}{\partial t} - \vec{\alpha} \cdot \vec{\nabla}, \vec{0}) \psi = m c^2(1, \vec{0}) \psi$

If we hope to treat the time derivative as we do the space derivative, then it only makes sense to have a gamma matrix act on the time derivative, not the invariant mass. The goal now looks something more like this:

$\frac{\hbar}{m c^2} (\frac{\beta}{c} \frac{\partial}{\partial t} - \vec{\alpha} \cdot \vec{\nabla}, \vec{0}) \psi = (1, \vec{0}) \psi$

If a gamma matrix is implemented as a quaternon triple product, how can we write this as quaternion expression? The Del belongs in the vector, so forget about the alpha and beta for a moment:

$\frac{\hbar}{m c^2} (\frac{1}{c} \frac{\partial}{\partial t}, \vec{\nabla}) \psi = (1, \vec{0}) \psi$

This is something we can hit on both sides with a basis vector. The test is to see if the square of the triple product gives exactly the same scalar as generated by the Klein-Gordon equation. It should not be a surprise if hitting both sides of the 4-derivative with the identity element, one gets exactly the Klein-Gordon equation, even the phase:

$((1, \vec{0})\frac{\hbar}{m c^2} (\frac{1}{c} \frac{\partial}{\partial t}, \vec{\nabla})(1, \vec{0})^2 \psi = \frac{\hbar^2}{m^2 c^2}(\frac{1}{c^2} \frac{\partial}{\partial t^2} - \nabla^2, \frac{2}{c} \frac{\partial}{\partial t} \vec{\nabla}) \psi = \left(1, 2\gamma^2 \vec{\beta} \right)$

There are three other triples that generate the same scalar, but change the 3-vector:

Sun, 08/15/2010 - 01:10

Lowell

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Joined: 04/25/2009

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Re: QFTD Chapter 1: Particle Physics and Special Relativity

Come to think of it you could have used e.g. as opposed to i.e. as
the plane wave normalization in QFT is just an example.

<em>Quaternions contain an i, j, and k, so we do not need to add these to our expressions. This will be quite odd, since the factor of i is repeated almost as much as hbar. Prunning in good.

</em>

Ya get 3 complex numbers for the price of one!

<em>Commutator relations are presented as facts (which they are), but can be easily justified. Take the position operator and momentum operator. These are operators, symbolized by putting a hat on x or p. Operators must act on a wave fucntion to have any meaning, but this acounting detail is often skipped.
</em>

Differentiating operations on the wavefuction (as they exhibit the discrete characteristic or quantization of classical quantities, namely energy, momentum, location). Just as derivatives have no meaning per say independent of functions.

Shouldn't the momentum operator be P = -d/dx(mc psi)? Where's the minus?

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Lowell

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Lowell