Yang Mills Theory - Part - 3

We have our Lagrangian $$L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2\tilde\psi\psi$$
After the gauge transformation defined as, $$ \psi' = S\psi \\ where\,\,S = e^{-i\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}$$ $$ L' = ic\hbar\tilde\psi'\gamma^\mu\partial(S\psi) \\= ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\partial_\mu\psi + ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(\partial_\mu{S})\psi $$gives $$ L' = \left[ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi = L \right]+ {extra\,term} $$[assumed $\gamma^\mu$ commutes with S]

To compensate this extra term we use the previous mathematical tool, i.e. we define the covariant derivative as, $$ D_\mu = \partial_\mu + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A}_\mu$$
where the new $ \vec{A_\mu}$ term should cancel our previous extra terms on transformation. [The new term "$\vec{A_\mu}$" now has three components as to cancel each component in $\vec{\lambda(x)}$]

The purpose of this new covariant derivative is,

If we define our Lagrangian using this derivative as $$L = ic\hbar\tilde\psi\gamma^\mu{D_\mu}\psi$$ then we should have the same Lagrangian after the transformation $$L' = ic\hbar\tilde\psi'\gamma^\mu{D'}_\mu(S\psi) = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu{D_\mu\psi}$$ where we implied the condition, $$D'_\mu(S\psi) = S(D_\mu\psi)$$Remember, unlike the usual derivatives, Covariant derivatives also transform under the gauge transformation.

From our condition, the transformation rule for $\vec{A_\mu}$ is derived as, $$ D_\mu' \psi'=\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}\right)(S\psi) = S \left[\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A_\mu}\right)\psi\right] $$ Using the previous relations, $$\left(\partial_\mu{S}\right)\psi + S(\partial_\mu\psi) + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}(S\psi) = S(\partial_\mu\psi) + \frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\psi\right)$$ Cancelling the terms and taking out psi $$ \left[(\partial_\mu{S}) +\frac{iq}{c\hbar}\left(\vec{\sigma}\cdot\vec{A'_\mu}\right)S\right]\psi=\left[\frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)\right]\psi$$ Multiplying with the $ S^{-1}$ operator on the right hand side we get the transformation relation, $$ \vec{\sigma}\cdot\vec{A'_\mu} = S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)S^{-1}+ \frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}$$
Now, we can check the invariance of the Lagrangian, $$ L_{new} = ic\hbar\tilde\psi\gamma^\mu(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu + \frac{iq}{\hbar{c}}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi$$ After the transformation, $$ L'_{new} = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(D'_\mu(S\psi))\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[\partial_\mu(S\psi)+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A'_\mu})(S\psi)\right] \\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S\partial_\mu\psi+(\partial_\mu{S})\psi + \frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})S^{-1}S\psi+\frac{iq}{c\hbar}\frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}(S\psi)\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S(\partial_\mu\psi)+(\partial_\mu{S})\psi+\frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})\psi - (\partial_\mu{S})\psi\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\left[\partial_\mu\psi + \frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\psi\right]$$Gives finally the same, $$ L'_{new} = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi = L_{new} $$ Thus, we can check the invariance of our new Lagrangian.

Now, we expand our definition of S by the assumption $\lambda$ is very small so that all higher order terms can be neglected. After this approximation, substitution of the new "S" in our transformation equation for $ A'_\mu$ yields, $$ \vec{\sigma}\cdot\vec{A'_\mu} \approx \vec{\sigma}\cdot\vec{A_\mu} + \vec{\sigma}\cdot\partial_\mu\lambda + \frac{iq}{c\hbar}\left[(\vec{\sigma}\cdot\vec{A_\mu}),(\vec{\sigma}\cdot\vec{\lambda})\right]$$ Using the rule, $$(\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \vec{a}\cdot\vec{b} + i\sigma\cdot(\vec{a}\times\vec{b}) $$ we get, $$\vec{\sigma}\cdot \vec{A'_\mu} =\vec{\sigma}\cdot\vec{ A_\mu} +\vec{\sigma}\cdot\partial_\mu\vec{\lambda} + \frac{iq}{c\hbar}\left(2i\vec{\sigma}\cdot (\vec{A_\mu}\times\vec{\lambda})\right) $$gives$$ \vec{A_\mu'} = \vec{A_\mu} + \partial_\mu{\vec{\lambda}}+\frac{2q}{c\hbar}(\vec{\lambda}\times\vec{A_\mu})$$ Thus, we got our new Lagrangian and the new transformation rules for corresponding terms.
 
But, from the same argument for the vector potential in previous problem invokes its field term in our Lagrangian. Once again, we take the Proca Lagrangian term without mass to explain our vector potential term.
 
You can ask, why we can't take the Proca Lagrangian with mass and redefine our $A'_\mu$ such that the invariance holds for $$ A^\nu{A_\nu}$$ Yes. But we have already just finished defining the transformation rules for $A'_\mu$ and it doesn't have the invariance. Maybe, if you can find $ A'_\mu$ such that, it satisfies both the above transformation and the invariance of $A^\nu{A_\nu}$ it would be wonderful. To make a comment, I should try and workout completely and find the reason for the impossibility (if there is any).

Not even $ A^\nu{A_\nu}$ but also the $ F^{\mu\nu}F_{\mu\nu}$ term is not invariant when we redefine our $ F_{\mu\nu}$ for three vector potentials as, $$ L_{extra} = \frac{-1}{16\pi}F^{\mu\nu}_1F^1_{\mu\nu}-\frac{1}{16\pi}F^{\mu\nu}_2F^2_{\mu\nu}-\frac{-1}{16\pi}F^{\mu\nu}_3F^3_{\mu\nu} = \frac{-1}{16\pi}\vec{F}^{\mu\nu}\vec{F}_{\mu\nu}$$ We will separately deal with its transformation rules and how it can be restated with additional terms in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths

Yang Mills Theory - Part - 2

We previously constructed our Lagrangian in a simplified form using the matrix notation for the combined spinor field.

$$ L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2 \tilde\psi\psi$$ But, now we cannot define transformation as previous, since the new $\psi$ here is a column matrix. The transformation should be redefined in the matrix representation. So, we introduce an unitary operator U acting on our wave function to represent gauge transformation. $$ \psi' = U\psi \\ \tilde\psi' = \tilde\psi U^\dagger \\ where U^\dagger{U} =\mathbb{I}$$

For this transformation, the Lagrangian is invariant, if the $\gamma^\mu$ matrices commute with the Unitary matrix.

To apply the same case for Local gauge invariance, where we used to get an extra term, we need a new representation and Unitary matrix in general can be represented using Hermitian matrices as, $$ U = e^{iH}$$ where a general Hermitian matrix is defined using four parameters using Pauli matrices, $$ H = a_0\mathbb{I} + a_1\sigma_1 + a_2\sigma_2 +a_3\sigma_3 = a_0\mathbb{I}+a\cdot\sigma$$ where $ a\cdot\sigma = a_1\sigma_1+a_2\sigma_2+a_3\sigma_3$ in shorthand notation.

Thus, we get our Unitary matrix, $$ U = e^{ia_0\mathbb{I}} e^{ia\cdot\sigma} $$ where $ e^{ia\cdot\sigma}$ has determinant "1" and corresponds to SU(2).

With the same strategy as used in previous, we define $$ \lambda(x) = - \frac{\hbar{c}}{q} \vec{a} $$
actually it is not a vector in the usual sense. But, the notation is preferred here to differentiate it from others.

So, we have, $$ \psi' = e^{-\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}\psi$$ "$e^{ia_0\mathbb{I}}$" being a simple phase factor doesn't affect the final result even though it is a function of the coordinates.

For eg: We redefine $$ U = e^{ia_0\mathbb{I}}e^{ia\cdot\sigma} = e^{ia_0}S$$ where "S" is also unitary matrix and Identity is implied in the exponential of $a_0$. Under the unitary transformation.
$$ L' = i\hbar{c}\tilde\psi'\gamma^\mu\partial_\mu\psi'$$
where $mc^2\tilde\psi'\psi$ term doesn't affect the invariance in both local and global gauge tranformation. So, we have, $$ L' = ic\hbar\tilde\psi{U^\dagger}\gamma^\mu\partial_\mu(U\psi) $$ $$L' = ic\hbar\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\partial_\mu(e^{ia_0}S\psi)$$ Expanding the differential (call $ic\hbar = k)$ , $$L' =  k \tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\left( (\partial_\mu{e}^ia_0)S\psi + e^{ia_0}\partial_\mu(S\psi)\right)$$ $$L'= k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{i}e^{ia_0}\partial_\mu(a_0)(S\psi) + k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{e^{ia_0}}\partial_\mu(S\psi)$$ Defining $S\psi$ as our new $\psi'$ we have $ \psi' = S\psi$ we have, $$ L = ic\hbar\tilde\psi'\gamma^\mu{i}(\partial_\mu(a_0))\psi' + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu(\psi') $$
with the same definition of $\lambda$, $$L' = -q\tilde\psi'\gamma^\mu\psi'\partial_\mu\lambda(x) + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi'$$ where the first term is exactly what we got as additional term in the previous lagrangian for which we have already defined transformation rules
[It has its own lagrangian and own vector potential - $A_0$].
So, we just need to focus on the second term.

Thus, we have our new transformed Lagrangian as ,$$ L' = ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi' - mc^2\tilde\psi\psi $$
We will focus entirely on this new Lagrangian in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths

Yang Mills Theory - Part - 1

The usual Lagrangian used in our Field theory are,

The Klein-Gordon Lagrangian for scalar field - spin zero particles $$ L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}\left(\frac{mc}{\hbar}\right)^2\phi^2$$ that gives the Klein Gordon equation, $$ \partial_\mu\partial^\mu\phi + \left(\frac{mc}{\hbar}\right)^2\phi = 0 $$

The Dirac Lagrangian for spinor fields -  spin half particles $$L = i\hbar{c}\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2\tilde\psi\psi$$gives the dirac equations for $\psi$ and $\tilde\psi$ $$i\gamma^\mu\partial_\mu\psi- \frac{mc}{\hbar}\psi = 0$$$$ i\partial_\mu\tilde\psi\gamma^\mu+\frac{mc}{\hbar}\tilde\psi = 0 $$ And finally,

The Proca lagrangian for vector fields - spin one particles $$ L = \frac{-1}{16\pi}F_{\mu\nu}F^{\mu\nu} + \frac{1}{8\pi} \left(\frac{mc}{\hbar}\right)^2 A_\nu{A}^\nu $$ gives the vector field equation $$\partial_\mu{F}^{\mu\nu}+\left(\frac{mc}{\hbar}\right)^2A^\nu = 0 $$

These equations are not the derived ones, rather we take them as an abstract. You may ask how it can suddenly arise from nowhere?

It is more like the reverse engineering. We know the resultant Dirac equation and other field equations for the corresponding particles which is used to construct the appropriate Lagrangian for those fields. Once the Lagrangian is constructed, it is simple to solve the problem in a much formal way.

As far as concerned here, Yang Mills theory deals with the Local gauge invariance.

Let us take the lagrangian for spinor fields $$ L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2\tilde\psi\psi $$   
For the transformation, $$ \psi' = e^{i\theta}\psi $$ where $\theta$ some constant, then we can notice that new Lagrangian, $$ L' = ic\hbar\tilde\psi{e}^{-i\theta}\gamma^\mu\partial_\mu\psi{e}^{i\theta} - mc^2 \tilde\psi{e}^{i\theta}e^{-i\theta}\psi $$ which is invariant since, $e^{i\theta}$ being the constant commutes with all other variables. 

But, it is not the same case if we considered functional dependence for $\theta$ by considering local gauge invariance, $$ \theta \,\,\rightarrow\,\,\theta(x)$$ then we have, $$ \partial_\mu\left(e^{i\theta(x)}\psi\right) = e^{i\theta(x)}\partial_\mu\psi + \left[\left(\partial_\mu{e}^{i\theta}\right)\psi = ie^{i\theta(x)}\left(\partial_\mu\theta(x)\right)\psi\right]$$ which gives an extra term$$L ' = L + i^2c\hbar\tilde\psi{e}^{-i\theta(x)}\gamma^\mu\partial_\mu\theta(x){e}^{i\theta(x)}\psi = L - c\hbar\partial_\mu\theta(x)\tilde\psi\gamma^\mu\psi$$ which is rewritten as $$ L' = L + q\tilde\psi\gamma^\mu\psi\left(\partial_\mu\lambda(x)\right)$$ where $$\lambda(x) = -\frac{c\hbar}{q}\theta(x) $$
 
We need to remember that, $\theta$ and its derivatives are considered to be a parameter which is not a matrix, and so the order doesn't matter whether you put it in front or back, wherelse the wavefunction and $\gamma$ matrices don't commute - that makes the order important. 

Now, a new Lagrangian can be constructed to be invariant under a local gauge transformation $$ \psi' = e^{-\frac{iq}{c\hbar}\lambda(x)}\psi$$ as, $$ L_{new} = L_{old} - q\tilde\psi\gamma^\mu\psi\partial_\mu\lambda(x)$$
But, this $\lambda(x)$ is not a constant i.e. not the same everywhere as the previous. It takes different values at different positions. Only after we know the transformation function, we can decide this extra term. Before the transformation, its values are not determined. Thus, we need a function that transforms as,$$ X' = X +\partial_\mu\lambda $$ under the gauge transformation. 

So, far we just used the mathematical arguments involving the transformation rules. Now, we look for this new function X for which we already have a similar tool in our vector fields. $$ A_\mu' = A_\mu + \partial_\mu\lambda(x)$$ 
That is it! 

The New lagrangian is written as, $$ L_{new} = L_{old} - q\tilde\psi\gamma^\mu\psi{A}_\mu $$ 
But, it hasn't completed yet. The last term in our Lagrangian affects the field equations and so introduces its own free Lagrangian i.e.the Proca lagrangian without the mass term which would affect the invariance. 

From which, we arrive at our final lagrangian, $$ L = ic\hbar\tilde\psi\gamma^\mu\partial_mu\psi -  mc^2\tilde\psi\psi - q\tilde\psi\gamma^\mu\psi{A_\mu} - \frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu}$$
 
Things can be written in a compact way with the introduction of covariant derivatives defined as, $$ D_\mu = \partial_\mu + i\frac{q}{c\hbar}A_\mu$$ It is just a notation, not anything more. 



Yang Mills theory applies this same local gauge invariance to spinor fields that involves - SU(2) group.
Let us start with two spinor fields $\psi_1,\psi_2$ and the lagrangian for the combined system is written as, (without interaction term), $$L = ic\hbar\tilde\psi_1\gamma^\mu\partial_\mu\psi_1-m_1c^2\tilde\psi_1\psi_1 +ic\hbar\tilde\psi_2\gamma^\mu\partial_\mu\psi_2 - m_2c^2\tilde\psi_2\psi_2 $$ The combined Lagrangian in Matrix form, $$ L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - c^2\tilde\psi{M}\psi$$ where M is the diagonal matrix involving mass $m_1, m_2$ If we assume it to be equal, we can get rid of it by simple constant "m".
The matrix notation is $$ \psi = \left(\begin{matrix}\psi_1\\\psi_2\end{matrix}\right)$$
We will see how to implement the local gauge invariance in this SU(2) group in the next post.

Reference book: Introduction to Elementary Particle Physics - Griffiths

Relativistic Field theory - Euler Lagrange equation

To understand some of the advanced theories in monopoles and etc., we will take a look on the relativistic field theory.

Classically, the motion of a particle is understood by solving for the position of the particle as a function of time. Where else the fields are defined over a region in space as $\phi(x_i)\,\,\,i=0,1,2,3$ where $x_0$ is the time component taken together with the space coordinates.These fields give all the information we would like to know about the system. 

In classical method, the Lagrangian is defined as a function of coordinates and its time derivatives. Since, we take here time as one of the components, the new Lagrangian for our fields is a function of both $\phi$ and its derivatives about each of the components. 
We can obtain our new Euler Lagrange equation as,
$$ \delta \int\,L(\partial_\mu\phi,\phi) d\tau = 0 $$ where the L here is called the Lagrangian density and $d\tau = dx_0 \,dx_1 \,dx_2 \,dx_3$
On expansion, $$ \int\,\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)+\frac{\partial{L}}{\partial\phi}\delta(\phi)\right) d\tau = 0$$ using $\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$ and product rule we get, $$ \int \left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi) }\delta\phi\right) - \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi + \frac{\partial{L}}{\partial\phi}\delta{\phi}\right) d\tau = 0 $$
The first term on integrating and applying the condition $\delta\phi$ becomes zero at the end points becomes zero. Taking out the negative sign, $$\int\left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial{L}}{\partial\phi}\right)\delta\phi\,d\tau = 0$$
gives us the final Euler-Lagrange equation for relativistic field theory, $$ \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial{L}}{\partial\phi} = 0 $$
It needn't to be only one function $\phi$ but more functions of $\phi_j$.
Let us consider in specific, the Klein Gordon equation which describes the motion of particles with spin zero. The equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}\\ \vec{p} = -i\hbar\nabla$$ acting on a scalar field $\phi(x,t)$ to give, $$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$ where the d'alembertian operator is defined here as $$ \Box =  \frac{1}{c^2}\partial_t^2 - \nabla^2$$

We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ g_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have, $$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.

This can be derived from the Lagrangian, $$ L = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}\left(\frac{mc}{\hbar}\right)^2\phi^2 $$