Time dependent perturbation theory

Time dependent perturbation theory addresses the transition between one eigenstate to another eigenstate when there is a small perturbation with respect to time such that it doesn't change the eigenfunctions.

The Hamiltonian for these problems of perturbation is, $$ H = H_0 + \lambda {H'(t)}$$ where now the perturbing Hamiltonian is time dependent. Solution for $H_0$ is, $$\psi_n(x,t) = \phi_n(x) e^{i\frac{E_n}{\hbar}t} = \phi_n e^{-i\omega_nt}$$ where $$H_0\phi_n = E^0_n\phi_n$$ and $ \omega_n = E_n/\hbar$ and the general state vector is given by, $$ \Psi = \sum_nc_n\psi_n(x,t)$$
In our problem we start with a particular eigenstate and assume that, after switching on the perturbation the state evolves to a general state always expressed as the linear combination of initial eigenstates where the time dependence is accompanied with the coefficient as $$\Psi (x,t) = \sum_n c_n(t) \phi_n(x) e^{i\omega{t}}$$  Because of this assumption, the particle now can be found in any other eigenstate and needn't to be in the same eigenstate forever. As usual, the modulus square of the coefficient gives the probability of finding the particle in any one of eigenstates, except now it changes with perturbation time. Substituting our wave function in general Schrodinger's equation (to find the time dependence of c(t)), $$i\hbar\frac{\partial\Psi}{\partial{t}} = H\Psi$$ gives (taking the inner product with $\psi_k$)$$ i\hbar\dot{c_k} = \lambda\sum_n \langle\phi_k\vert{H'}\vert\phi_n\rangle e^{i(\omega_k-\omega_n)t} f(t) c_n$$ where f(t) is the time-dependent part of the perturbing Hamiltonian H' (in variable separable form).  In the limit $\lambda\rightarrow{0}$ the coefficients are all constant in time and so we seek a solution of the form, $$ c_k(t) = c_k^0 + \lambda{c_k^1(t)} + \lambda^2c_k^2(t) + \cdots$$ with corresponding order of correction in the upper index.  With this expansion we get, $$ i\hbar \dot{c_k}^0 = 0 $$ which means $c_k^0 = constant$ whatever the value of $c_n^0$ is at the beginning of the perturbation remains a constant throughout the perturbation. In specific, if we denote the starting time $t_0$ then $c_n^0(t_0) = \delta_{nl}$ where we assumed the state is in a specific eigenstate at the start.

Similarly, the first order equation gives, $$ i\hbar\dot{c_k}^1 = \sum_n H'_{kn} f(t) c_n^0 e^{i(\omega_k-\omega_n)t}$$ substituting for $c_n^0$ we get, $$ i\hbar\dot{c_k}^1 = H'_{kl}f(t)e^{i(\omega_k-\omega_l)t}$$ finally gives the expression for first order correction as,$$c_k^1(t) = \frac{H'_{kl}}{i\hbar} \int_{t_0}^t f(t) e^{i\omega_{kl}t'} \,dt'$$ where $\omega_{kl} = \omega_k-\omega_l$ From this we can transition probability for each eigenstate by taking modulus square of the above coefficient.

Time Independent Perturbation Theory - Degenerate

Similar to the non-degenerate case, we take our Hamiltonian $$H = H_0 + H'$$ except now, the initial unperturbed Hamiltonian has degenerate states. Assuming it is q- fold degenerate, if the symmetry giving rise to the degeneracy is disturbed, it will produce distinct 'q' non degenerate states. Our motive is to calculate the energy levels of these newly produced distinct energy levels. Let us consider two different initial eigenstates $E_1^0$ and $E_{q+1}^0$ [it is equal to the zeroth order term in the expansion] where $E_1^0$ is q-fold degenerate with states $$E_1^0 = E_2^0 = .... = E_q^0$$. As we have seen in non-degenerate perturbation theory, the coefficients $$c_{jn} = \frac{H'_{jn}}{E_n^0 - E_j^0}\,\,\,j\neq{n}\,\,and\,\, j,n\leq{q}$$ will become infinite for two different energy levels which has the same energy value. This situation can be avoided only if the numerator term is zero for all $j,n \leq {q}$ (Not the diagonal terms because $j\neq{n}$). It is equivalent to diagonalizing the sub-matrix $H'_{jn}\,\,\,\,for j,n\leq{q}$. This can be achieved by transforming to the new set of eigenfunctions from the initial set of eigenfunctions.
Thus, our problem for solving the energy values of the degenerate case reduced to diagonalizing the sub-matrix of H'. The diagonal elements are the change in energy levels of the initial degenerate states due to the perturbation. 
Representing the 'q' new eigenfunctions that diagonalize the $H'_{jn}$ as $\tilde{\psi_n}$ we write, $$ \tilde{\psi_n} = \sum_{m=1}^q a_{nm} \psi_m^0$$ Since, it will diagonalize, $$ \langle\tilde{\psi_n}\vert\tilde{\psi_j}\rangle = H'_{nj}\delta_{nj} $$ So, our new set of basis functions are, $$ Basis \,= \left\{\tilde{\psi_1},\tilde{\psi_2},\cdots\tilde{\psi_q},\psi_{q+1}^0,\cdots\right\}$$The diagonal element of the sub-matrix H' is the first order correction to the energy i.e. $$E'_n = \langle\tilde{\psi_n}\vert\tilde{\psi_n}\rangle = H'_{nn} \,\,n\leq{q}$$
Corresponding eigenfunctions are determined from $$H'\tilde{\psi_n} = E'_n\tilde{\psi_n}$$ substituting for new eigenfunctions in terms of initial wavefunctions, $$H' \sum_{m=1}^q a_{nm}\psi_m^0 = E'_n \sum_{m=1}^q a_{nm} \psi_m^0$$ Taking inner product with $\psi_j^0$ we have, $$\sum_{m=1}^q \left(H'_{jm} - E'_n\delta_{jm}\right)a_nm = 0\,\,\,n,j\leq{q} $$ It could be written simply in matrix form. For non-trivial solution for {a_nm} the determinant of coefficient matrix should vanish gives the secular equation in this case as, $$determinant\,\,\vert{H'_{jm}}-E'_n\mathbb{I}\vert = 0 $$ Using these the new eigenfunctions and eigenvalues are calculated.

Time Independent Perturbation Theory - Non-degenerate

When a given Hamiltonian of a problem is not so much different from the Hamiltonian of an absolutely solvable problem, we use the perturbation theory to find the new eigenfunction from the old  eigenfunctions with the assumption that the perturbing Hamiltonian is small compared to the original one. Let us jump into the mathematics by writing the total Hamiltonian as, $$ H = H_0 +\lambda{H' } \,\,\,\,...(1)$$ where H' is the perturbing Hamiltonian and $\lambda$ is used to denote how small the perturbation is with respect to the original. The eigenfunctions are expressed as, $$ H_0 \psi_n^0 = E_n \psi_n^0 \,\,\,\,...(2)$$ and $$H\psi_n = E_n \psi_n\,\,\,\,...(3)$$ When $\lambda\rightarrow{0}$ the perturbed eigenfunctions should reduce to the original eigenfunctions. So, we expand it in series as, $$\psi_n = \psi_n^0 + \lambda\psi_n^1+\lambda^2\psi_n^0+\cdots \,\,\,...(4)\\ E_n = E_n^0 +\lambda{En^1} + \lambda^2E_n^2+\cdots \,\,\,...(5)$$ where the upper index is used to denote the corresponding order correction on the specific eigenfunction and eigenvalue. Substituting this in (3) and to make this true for arbitrary $\lambda$ values we equate the corresponding terms of lambda in the equation to get, $$ H_0\psi_n^0 = E_n\psi_n^0\,\,\,...(6)\\ H_0\psi_n^1 +H'\psi_n^0 = E_n^1\psi_n^0 + E_n^0\psi_n^1\,\,\,...(7)\\ H_0\psi_n^2 + H'\psi_n^1 = E_n^1\psi_n^1 + E_n^2\psi_n^0+E_n^0\psi_n^2\,\,\,...(8) $$

Rearranging them gives, $$ H_0\psi_n^0 = E_n\psi_n^0\\ \left(H_0 - E_n^0\right)\psi_n^1 = \left(E_n^1-H'\right)\psi_n^0 \\ \left(H_0-E_n^0\right)\psi_n^2 = \left(E_n^1-H'\right)\psi_n^1 + E_n^2\psi_n^0 $$ To avoid $\psi_n^1+ \alpha\psi_n^0$ being a solution with first order energy correction, we choose the correction terms orthogonal to the initial eigenfunctions. $$\langle\psi_n^k\vert\psi_n^0\rangle = 0 \,\,...k>0$$ Assuming all correction terms are in the same Hilbert space of the initial eigenfunctions, the correction terms are expressed as a linear combination. Substituing the linear combination $ \psi_n^k = \sum_j c_{jn} \psi_j^0 $ in the first equation and taking the inner-production with $\psi_l^0$ we get, $$ (E_l^0 - E_n^0) + H'_{ln} = E_n^1\delta_{ln}$$ The first order correction in Energy eigenvalue and eigenfunction is obtained to be, $$ E_n^1 = \langle\psi_n^0\vert{H'}\psi_n^0\rangle \\ \psi_n^1 = \sum_{j\neq{n}} \frac{H'_{jn}}{E_n^0 - E_j^0}\psi_j^0 $$

Similarly solving for second order perturbation,by assuming $\psi_n^2 = \sum_j d_{nj}\psi_j^0$we get, $$ E_n^2 = \sum_{j\neq{n}}\frac{|H'_{nj}|^2}{E_n^0-E_j^0} $$ and $$\psi_n^2 = \sum_{j\neq{n}} \left[\sum_{l\neq{n}}\frac{H'_{jk}H'_{kn}}{\left(E_n^0-E_j^0\right)\left(E_n^0-E_l^0\right)}-\frac{H'_{nn}H'_{jn}}{\left(E_n^0-E_j^0\right)^2}\right]\psi_j^0$$ where $d_{nn} = 0$

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, $$ L = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}\left(\frac{mc}{\hbar}\right)^2\phi^2 $$ 

Monopoles - 9 - Dirac Monopoles in Quantum Mechanics - Part - 5

Nearly for a month, I searched for the solution of the differential equation we got in the previous post given by, $$ \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL}{dz}\right) + \left[\lambda - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right] L(z) = 0 $$
Finally I found the solution in a German paper by I.Tamm (which is eventually the same paper mentioned by Dirac).

It took a long time to get the paper and it took even more time to decode it into English (usual google translator buries all the meaning in it), so I took the hard way by translating each and everything as word by word.  Anyways, in the mean time I did learn a lot. 

Our equation looks similar to the general Associated Legendre equation except for the constant term "m", which is replaced with a variable term. 

Away from that, the general theory for solving a second order differential equation with variable coefficients starts from the characteristic equation. 
We will derive for the general differential equation, $$ a(x)y''(x) + b(x)y'(x) + c(x)y(x) = 0 $$ or simply $$ y''(x) + p(x)y'(x) + q(x)y(x) = 0$$ where we divided by a(x) and denote it with new variables. One should note that, a(x) shouldn't have any kind of singularities unless the differential equation itself become meaningless. 

Thus, there are different kinds of singularities and it goes with mathematical literature. Here, we focus only on regular singularities where p(x) or q(x) becomes singular at much slower rate than $\frac{1}{x}$ and $\frac{1}{x^2}$  where we take the singularity at x=0. 

For these regular singularities, it is advised to use the modified power series method known as Frobenius Power series method which we used it for Hermite polynomials, etc. $$ y = x^r \sum_{n=0}^\infty {c_n}x^{n}$$ where $c_0 \neq 0 $
Now, we define, $$ s(x) = xp(x) = \sum_{n=0}^\infty {s_n}\,x^n$$ and $$ t(x) = x^2q(x) = \sum_{n=0}^\infty {t_n}\,x^n$$ 
So, our differential equation becomes, $$ y'' +  \frac{s(x)}{x}y' + \frac{t(x)}{x^2}y = 0 $$ On substitution for y, $$ \sum_{n=0}^\infty(n+r)(n+r-1)c_n\,x^{n+r-2} +  \sum_{n=0}^\infty \frac{s(x)}{x} (n+r) c_n \,x^{n+r-1} + \sum_{n=0}^\infty \frac{t(x)}{x^2}c_n\, x^{n+r} = 0$$
or $$ \sum_{n=0}^\infty\left[(n+r)(n+r-1)+ (n+r)\,s(x) + t(x)\right] c_n\,x^{n+r-2} = 0 $$ Dividing by $x^{r-2}$ we get the equation in powers of $ x^n$ setting x=0 we get, $$ \left[(r)(r-1) + (r) s(0)+t(0)\right]c_0 = 0$$ since $c_0 \neq 0$ we have our indicial equation as, $$ r(r-1) + r s(0) + t(0) = 0 $$ where $$ s(0) = \lim\limits_{x\to{0}}\,s(x) = \lim\limits_{x\to{0}}\,x\,p(x) $$ and $$ t(0) = \lim\limits_{x\to{0}}\,x^2\,q(x) $$
With the same correspondence, our equation has singularities at two points namely, z=0 and z=2 and ofcourse $z=\infty$. Since, our domain lies from 0 to 2 we need to look out for the indicial equation at z=0 and z=2 which is obtained to be,
from our characteristic equation, 
 $r(r-1) + r zp(z) +z^2q(z) = 0$$
where $$ zp(z) = \frac{2(1-z) z}{z(2-z)}\\ z^2 q(z) = z^2\frac{\lambda}{z(2-z)} - z^2 \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{\left(z(2-z)\right)^2}$$
 
at z=0, we get for $r=r_1$ $$ r_1(r_1-1)+r_1 - \left(\frac{m+n}{2}\right)^2 = 0 \,\,\,\,\rightarrow\,\,\,\, r_1 = \pm\frac{m+n}{2}$$
and for z=2, we can change the variable by t = 2-z, 
$$L''(t) - \frac{2(t-1)}{t(2-t)}L'(t) + \left[\frac{\lambda}{t(2-t)} - \frac{\left(m+\frac{n}{2}t\right)^2}{t^2(2-t)^2}\right]L(t) = 0$$  
where use has been made that, $$ \frac{dL}{dx} = - \frac{dL}{dt} \,\,\,\,and\,\,\,\,\frac{d^L}{dx^2}=\frac{d^2L}{dt^2}$$
and substitute for t=0 to get, $ r=r_2$ from, $$ r_2(r_2-1)+r_2-\frac{m^2}{4} = 0 \,\,\,\,\rightarrow\,\,\,\,r_2 = \pm\frac{m}{2}$$

From this, we try for a similar solution we used to derive in associated legendre polynomials by the substitution, $$ ^nP^m_z = 2^{-\frac{S+M}{2}} z^{\frac{S}{2}} (2-z)^{\frac{M}{2}} \,^nV^m(z)$$ where $$ S = |n+m|\,\, and\,\, M=|m|$$ We will derive the resulting equation elaborately on next post.
[You may wonder the difference between regular and essential singularities in simple words - it is just that for a regular singularity, if you consider a plot of a function, you will find finite, continuous values for every point on the curve except for some unique points. Where else in essential singularity, all the nearby points itself tend to infinite or undefined value]

Associated Legendre Polynomial - Part - 1

The General Legendre differential equation as we know, $$ (1-x^2)y''-2xy'+n(n+1)y=0$$ is solved using the frobenius power series method and the solutions are obtained to be the general Legendre polynomials, where suitable normalization constants and boundary conditions are used. 
With this result, we proceed further to completely exploit all the possibile solutions can be obtained from this equation. 
Before that, we will have to use the orthonormalizability property of our solution which is proved as, 
From the generating function, $$ \frac{1}{\sqrt{1-2xz+z^2}}=\sum_n P_n(x) z^n$$ or $$ \frac{1}{1-2xz+z^2} = \sum_n \sum_m z^{n+m} P_n(x) P_m(x) \\ \int\frac{1}{1-2xz+z^2} \,dx = \sum_{m,n} z^{n+m} \int{P_n(x)}P_m(x) \,dx$$ The limits can change maximum from -1 to +1 as in Legendre polynomials and summation is usually implied from 0 to infinity. 
Left hand side term is evaluated to give,$$ \frac{-1}{2z}\int_{(1+z)^2}^{(1-z)^2}\frac{1}{u} du = \frac{-1}{2z} \ln{\frac{(1-z)^2}{(1+z)^2}}\\where\,\, u=1-2xz+z^2 and \,\,du = -2z\,dx$$
and $$ \frac{-1}{2z}\ln{\left(\frac{1-z}{1+z}\right)^2} = \frac{-2}{2z} \ln{\frac{1-z}{1+z}} = \frac{1}{z}\ln{\frac{(1+z)}{(1-z)}}$$
$$\ln{\frac{(1+z)}{(1-z)}} = \ln(1+z) - \ln(1-z)$$
using taylor expansion, 
$$ f(x) = f(a) + f'(a)\, (x-a) + \frac{f''(a)}{2!}\,(x-a)^2+...$$
where the function should be infinitely differentiable at x=a.
So, $$ \ln(1+z) = \ln(1+a) + \frac{1}{1+a}(z-a) - \frac{1}{2!(1+a)^2} (z-a)^2 +...$$
where we can take a=0, $$ \ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} +... $$ Similarly, $$ \ln(1-z) = -z - \frac{z^2}{2}  - \frac{z^3}{3} -..$$ This gives, $$ \ln(1+z) - \ln(1-z) = 2 \left[z+\frac{z^3}{3} + \frac{z^5}{5}+...\right] = 2\sum_{n} \frac{z^{2n+1}}{2n+1}$$
Thus we have our integral equal to, $$ \sum_{m,n} z^{n+m} \int_{-1}^{1} P_n(x)\,P_m(x) \,dx = \frac{1}{z} \sum_{n} \frac{z^{2n+1}}{2n+1} = 2  \sum_n \frac{z^{2n}}{2n+1}$$
From this, the left side terms will equal the right side terms only when n=m, so we get the result,
$$ when\,\, n=m\rightarrow\,\, \sum_n z^{2n} \int_{-1}^{+1}P_n(x)\,P_m(x) \,dx = 2\sum_n \frac{z^{2n}}{2n+1}\\ \rightarrow\,\,\,\, \int_{-1}^{1} P_n(x)\,P_m(x)\, dx = \frac{2}{2n+1}$$ $$ when \,\, n\neq{m}\rightarrow\,\, \int_{-1}^{1}P_n(x)\,P_m(x) \,dx = 0 \\ or \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \delta_{nm} \frac{2}{2n+1}$$ which is our orthogonality relation. 
We can also prove the completeness relation - that any arbitrary function can be expanded in terms of the linear combination of these Legendre polynomials. 
$$ f(x) = \sum_n a_n P_n(x)\\ \int_{-1}^{1} f(x)\,P_m(x)\,dx = \sum_n a_n \frac{2}{2n+1}\delta{nm} = \sum_n a_n \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2}{2m+1}a_m $$ amd $$ a_m = \frac{2m+1}{2} \int_{-1}^{1} f(x)\,P_m(x)\,dx$$
 Now, we will get back to our differential equation where we substitute our solution and differentiate it "m" times w.r.t. x 
(m < n), $$ (1-x^2)P_n''-2xP_n'+ n(n+1)P_n=0$$
$$ \frac{d^m}{dx^m}\left((1-x^2)\frac{d^2P_n}{dx^2}\right) = (1-x^2) \frac{d^{m+2}P_n}{dx^{m+2}} - m.2x.\frac{d^{m+1}P_n}{dx^{m+1}} - \frac{m(m-1)}{2!}2 \frac{d^mP_n}{dx^m}$$ 
where we made use of the Leibniz formula, $$ \frac{d^m}{dx^m} A(x) \,B(x) = \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{d^kA}{dx^k} \frac{d^{n-k}B}{dx^{n-k}}$$
Similarly, $$ -2\frac{d^m}{dx^m}\left[x\frac{dP_n}{dx}\right] = -2x \frac{d^{m+1}P_n}{dx^{m+1}} - 2m\frac{d^mP_n}{dx^m}$$
Which finally gives, $$ \frac{d^m}{dx^m}\left[Legendre\, eqn.\right] \\= (1-x^2)\frac{d^2}{dx^2}\left(\frac{d^mP_n}{dx^m}\right) - 2(m+1)x \frac{d}{dx}\left(\frac{d^mP_n}{dx^m}\right) + \frac{d^mP_n}{dx^m}\left[n(n+1)-m(m+1)\right] = 0 $$
If we assume, $$ \frac{d^mP_n}{dx^m} = V = \frac{W}{(1-x^2)^{\frac{m}{2}}}$$ So, $$ \frac{dV}{dx} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{dW}{dx} +\frac{mxW}{1-x^2}\right]$$ and $$ \frac{d^2V}{dx^2} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{d^2W}{dx^2} + \frac{2mx}{1-x^2}\frac{dW}{dx}+W\left(\frac{m}{1-x^2}+ \frac{mx^2(m+2)}{(1-x^2)^2}\right)\right]$$ Substituting and doing the necessary manipulation (multiply the whole equation by $(1-x^2)^{\frac{m}{2}}$, We get our associated Legendre equation, $$ (1-x^2)\frac{d^2W}{dx^2} - 2x\frac{dW}{dx} + \left[n(n+1) - \frac{m^2}{1-x^2}\right]W = 0 $$

Eigen values of $L^2$ and $L_z$ operator

The Angular momentum operator is defined by, $$ \vec{L} = \vec{r}\times\vec{p}= -i\hbar\vec{r}\times\nabla $$ In cartesian coordinates we have the usual known coordinates, $$ {L_x} = y{p_z}-z{p_y}\\ {L_y} = zp_x-xp_z\\ L_z = xp_y-yp_x$$ where in quantum mechanics, these terms are replaced by corresponding operators so we get the commutation relations as, $$ [L_x,L_y] = i\hbar{L_z} $$ with other components in cyclic form we have, $$ [L_i,L_j] = i\hbar\epsilon_{ijk}L_k$$ using levi-civita symbol. The components of the angular momentum operators are real, Hermitian since the terms on each coordinate commute with one another. And so, the angular momentum operator itself is a Hermitian operator. 
We can also express it in spherical polar coordinates as given in 
http://scientistech.blogspot.it/2015/11/angular-momentum-operator-in-spherical.html 

From, $$[r_j,p_k] = \delta_{jk}i\hbar$$ since these are operators, Identity operator is intrinsically assumed here. 
Using these we find $L^2$ operator commutes with every other component. 
$$ [L^2, L_x] =  [L_x^2 , L_x] + [L_y^2,L_x] + [L_z^2,L_x] $$
where, $$ [L_x^2,L_x] = 0 $$ $$[L_y^2,L_x] = [L_y,L_x]L_y + L_y[L_y,L_x] = - i\hbar{L_yL_z+L_zL_y}$$
$$ [ L_z^2,L_x] = [L_z,L_x]L_z + L_z[L_z,L_x] = i\hbar{L_yL_z+L_zL_y}$$ Combining all three we get, $$ [L^2, L_x] = 0$$ and similarly for other components. And so, $$ [L^2, \vec{L}]=0$$ 
So, $L^2$ and $L_z$ can be simultaneously diagonalized. It just a convention we use $L_z$. All that matters is that we can simultaneously diagonalize the $L^2$ with one of its components but not with all the three as we know other two are non-commuting with each other. 
Now, we define two new operators as. $$ L_+ = L_x + iL_y \\ L_- = L_x - iL_y$$ where one operator is the adjoint of the other. 
Once you define an operator, you can check for its commutation relation with all other operators, such as, $$[L_+, L_-] = [L_x+iL_y, L_x-iL_y] = 2\hbar{L_z}$$ and $$ [L_z,L_+] = [L_z,L_x+iL_y] = i\hbar{L_y}+ \hbar{L_x} = \hbar{L_+}$$ and $$[L_z, L_-] = -\hbar{L_-}$$ and finally, $$L_+L_- = L_x^2 + L_y^2 +iL_yL_x-iL_xL_y = L_x^2+L_y^2+\hbar{L_z} \\ \rightarrow L^2 = L_+L_- - \hbar{L_z} + L_z^2 $$ using the commutation between the two ladder operators, $$L^2 = L_-L_++ \hbar{L_z} +L_z^2$$

Remember, we have so far not used anything physical in our arguments. It is just that we define an operator as $$ \vec{L} = \vec{r}\times\vec{p} $$ and use the commutation relation $$[r_j,p_k] = i\hbar\delta_{jk}$$ for manipulation. Other than this, it can be or mean anything. It needn't even have any physical correspondence at all. 

Now, we make use of the commutation relation between $L^2,\,and\,L_z$ and seek simultaneous eigenstates[1] for both of these operator (a physical statement from the Heisenberg's principle[2]). 
We start by defining the eigenstate as, $$ L^2|X\rangle = \lambda\hbar^2|X\rangle \\ L_z|X\rangle = m\hbar|X\rangle $$
where $\hbar $ and its square is chosen for dimensional reasons. 
Note: We are just talking about the simultaneous eigenstates and we never checked whether it is the complete set of solution or not?[1].
Now, we check for the action of $L_+$ operator on any eigenstate by making use of the commutation, $$ [L_z,L_+] = L_+\hbar \\ \rightarrow L_zL_+ = L_+L_z + \hbar{L_+}$$ and it gives, $$ L_zL_+|X\rangle = \left(L_+L_z +\hbar{L_+}\right)|X\rangle = \hbar (m+1) L_+|X\rangle$$
similarly, $$L_zL_-|X\rangle = \hbar(m-1)L_-|X\rangle$$ From the above, it is understood that if $L_z$ operator acting on the general $|X\rangle$ gives the eigen value $m\hbar$. But the same operator acting on the wavefunction produced by $L_+ \,\,and\,\,L_-$ operator gives the new eigenvalue which is respectively unit value higher and lower than the old eigenvalue. i.e.These operators works as step operator which makes the eigenfunction to jump from one state to another in fixed unit values. From,$$ L^2 = L_x^2 +L_y^2 +L_z^2$$  we see that the for a fixed "$\lambda$" value there is a maximum and minimum value for "m" since the eigenvalues of Hermitian operators are real and its square is positive quantity $$ \lambda\geq{m^2}$$
Thus, there is a maximum and minimum value for "m" such that, $$ L_+|X_{max}\rangle = 0 \\ L_-|X_{min}\rangle = 0 $$ 
and the corresponding eigenvalue of $L^2$ operator is given by, $$ L^2 = \left(L_-L_++\hbar{L_z} + L_z^2\right)|X_{max}\rangle = \hbar^2m_{max}(m_{max}+1) $$
Similarly for the lowest state , $$ L^2 |X_{min}\rangle= \left(L_+L_- - \hbar{L_z} + L_z^2\right)|X_{min}\rangle = \hbar^2 m_{min}(m_{min}-1)|X_{min}\rangle$$
But $L^2$ operator should have the same eigenvalue and so $$ m_{min}(m_{min}-1) = m_{max}(m_{max}+1) \\ \rightarrow\,\,\, m_{min} = - m_{max}=-l \,\,and\,\, m_{min} = m_{max}+1$$ where the second possibility is excluded from physical reality. So, we do have, "m" varies from -l to +l in integer steps which forces the values of "l" to be the integers and the values of "l" are 0, 1/2, 1, 3/2,2..etc.(either positive or negative).   
Thus we can exploit the properties of the eigenvalues of $L^2$ and $L_z$ operators. 

Check for :
1.Simultaneous eigenstates
2.Heisenberg's Uncertainty principle - general statement

Monopoles - 8 - Dirac Monopoles in Quantum Mechanics - Part - 4

Let us start with the motion of an electron in the field of a magnetic monopole where the usual spherical polar coordinates are described by taking the monopole at the origin. In addition we know that Every wave function describing this system should have a singularity line starting from the origin, should pass through any closed surface.  We use the equation from previous posts (Part-1,2,3)
We consider the same wave function of the type, $$ \psi = \psi_1 e^{i\beta}$$ with corresponding definition on $\beta, \vec{k},\,etc.$
But, now it was introduced two separate things as nodal line and singular line. I couldn't understand it completely, but I just want to proceed with the next step. 
The magnetic field of the monopole is given by, $$ \vec{B} = \frac{q_m}{r^2} \hat{r} $$ and $$ \nabla\times\vec{K} = \frac{e}{\hbar{c}} \vec{B}  $$ On substitution, $$ \nabla\times \vec{K} = \frac{e}{\hbar{c}} \frac{q_m}{r^2}\hat{r} \\~\\ = \frac{e}{\hbar{c}}\frac{n\hbar{c}}{2er^2}\hat{r}\\~\\ = \frac{n}{r^2}\hat{r}$$
Thus, we get the curl of K as radial with magnitude $\frac{n}{2r^2}$ 
So, the solution of K could be worked out by expanding the curl in spherical polar coordinates as, $$ \frac{1}{r^2sin\theta} \left[ \frac{\partial(rsin\theta{k_{\phi}})}{\partial{\theta}} - \frac{\partial(rk_{\theta})}{\partial{\phi}}\right] \hat{r} + \frac{1}{rsin\theta}\left[\frac{\partial(k_r)}{\partial{\phi}} - \frac{\partial(rsin\theta{k_{\phi}})}{\partial{r}}\right] \hat{\theta}+ \\~\\\frac{1}{r}\left[ \frac{\partial(r{k_{\theta}})}{\partial{r}} - \frac{\partial(k_r)}{\partial{\theta}}\right]\hat{\phi} = \frac{n}{2r^2} \hat{r} $$
Equating the components we get a solution as, $$ k_\theta = k_r =  k_0 = 0 \\~\\ k_\phi = \frac{n}{2r} tan\frac{\theta}{2}$$
Then the Schrodinger for non-relativistic electron is given by, $$ \frac{-\hbar^2}{2m} \nabla^2\psi = E\psi$$ 
Applying $$ \psi = \psi_1 e^{i\beta}$$  we get, $$ \nabla\cdot\nabla(\psi_1e^{i\beta}) = \nabla\cdot\left[e^{i\beta}\nabla(\psi_1) + \psi_1 \nabla(e^{i\beta})\right] \\~\\= e^{i\beta} \nabla^2(\psi_1) + \nabla\psi_1\cdot\nabla(e^{i\beta}) + \nabla\cdot\left[\psi_1 \nabla (e^{i\beta})\right]$$ But $$ \nabla(e^{i\beta}) = \frac{\partial(e^{i\beta})}{\partial\vec{r}} = i e^{i\beta} \frac{\partial\beta}{\partial\vec{r}} = ie^{i\beta}\vec{k}$$ (These are 3 vectors - four vectors are separately indicated) Applying this we get, $$ \nabla^2\psi = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \nabla\psi_1\cdot \vec{k} + \nabla(\psi_1ie^{i\beta})\cdot\vec{k} + \psi_1ie^{i\beta} \nabla \cdot\vec{k} \\~\\ = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + \psi_1 \nabla(ie^{i\beta})\cdot\vec{k} + \psi_1 ie^{i\beta} \nabla\cdot\vec{k} \\~\\ = e^{i\beta} \nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + ie^{i\beta}\psi_1 \nabla\cdot\vec{k} - e^{i\beta}\psi_1 \vec{k}\cdot\vec{k}$$
which finally gives, $$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2\psi_1 + i\vec{k}.\nabla\psi_1 + i \left(\nabla\psi_1\cdot\vec{k} + \psi_1\nabla\cdot\vec{k}\right) - k^2\psi_1\right] $$$$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1$$
Now, our initial schrodinger equation can be rewritten as,
$$ \frac{-\hbar^2}{2m}\left[\nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1 = E\psi_1$$ 
Substituting for the values of k, we will get, 
$$\vec{k^2} = k_\phi^2 = \frac{n^2}{4r^2}tan^2{\theta/2} $$ and 
$$ \vec{k}\cdot\nabla = (\nabla\cdot\vec{k}) = \frac{k_\phi}{rsin\theta}\frac{\partial}{\partial\phi} = \frac{n\,tan\frac{\theta}{2}}{2r^2sin{\theta/2}cos{\theta/2}}\frac{\partial}{\partial\phi} = \frac{n\,sec^2{\theta/2}}{4r^2}\frac{\partial}{\partial{\phi}}$$ 
On substitution, $$ \frac{-\hbar^2}{2m}\left[ \nabla^2 + \frac{2ni}{4r^2} sec^2{\theta/2}\frac{\partial}{\partial{\phi}} - \frac{n^2tan^2{\theta/2}}{4r^2}\right]\psi_1 = E \psi_1 $$
Applying for Laplace operator in polar coordinates and using the regular separation of variables method we finally get (it is just manipulation), for Radial part, $$ \left[ \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} - \frac{\lambda}{r^2}\right]R(r) = \frac{-2mE}{\hbar^2} R(r)$$
and angular part, $$ \left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2} +\frac{ni}{2} sec^2{\theta/2}\frac{\partial}{\partial\phi} - \frac{n^2}{4}tan^2{\theta/2}\right]Y(\theta,\phi) = -\lambda{Y(\theta,\phi)}$$

From here, 
we need to solve two differential equations for which I searched for the solution in various places. I couldn't find the complete solution but just the preview of the starting of the solution by I.Tamm. Only first two pages are free to see and the complete paper costs more money. So, I tried my own to convert the above Angular equation into the usual simple equation of Spherical Harmonics.  

We will start with the Angular part by assuming the solution of type, $$ Y(\theta,\phi) = L(\theta) e^{im\phi} $$ Upon substitution, $$e^{im\phi}\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}-\frac{mn}{2} sec^2{\theta/2} - \frac{n^2}{4}tan^2{\theta/2}\right]L(\theta) = -\lambda{e^{im\phi}}{L(\theta)} $$ With slight alterations, we can again rewrite the above as, $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = -\lambda{L(\theta)} $$ and $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right)+\lambda - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = 0 $$Now, we will try to convert this into usual equation by replacing with suitable new variable $$ z = 1+cos\theta$$,
So that, $$ \frac{dL}{d\theta} = \frac{dL}{dz} \frac{dz}{d\theta}$$ and the first term becomes $$ \frac{dz}{d\theta} = -sin\theta$$ and $$ \frac{1}{sin\theta}\frac{d}{d\theta}\left(sin\theta\left(-sin\theta\frac{dL}{dz}\right)\right) = \frac{1}{sin\theta}\left[\frac{d}{dz}\left(-sin^2\theta\frac{dL}{dz}\right)\right]\frac{dz}{d\theta}\\~\\ = \frac{d}{dz}\left(sin^2\theta\frac{dL}{dz}\right) = \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL}{dz}\right)$$ where we used the fact that, $$ cos\theta = z - 1\\~\\ sin^2\theta = 2z-z^2$$ and the second term becomes, $$ \left[\lambda - \frac{m^2}{sin^2\theta} -\frac{m}{1+cos\theta} - \frac{n^2}{4} \frac{(1-cos\theta)}{(1+cos\theta)}\right] = \left[\lambda - \frac{m^2}{2z-z^2}-\frac{m}{z}-\frac{n^2}{4}\frac{2-z}{z}\right]\\~\\ = \lambda - \left[\frac{m^2+mn(2-z)+\frac{n^2}{4}(2-z)^2}{z(2-z)}\right]\\~\\ = \lambda - \left[\frac{\left(m+\frac{n}{2}(2-z)\right)^2}{z(2-z)}\right]$$
Finally we get our differential equation as, $$ \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL(z)}{dz}\right) + \left[\lambda - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right]L(z) = 0$$
  

Monopoles - 7 - Dirac Monopoles in Quantum Mechanics - Part - 3

From our previous post we found that, the gauge invariance allows to reconsider our ordinary wave function with the phase factor as the wavefunction of an electron in the Electromagnetic potentials given by, $$ \vec{A} = \frac{\hbar{c}}{e}\vec{k}$$ and $$ V = \frac{-\hbar}{e}k_0$$ 
So, the Magnetic and Electric fields are given by, $$ \nabla\times\vec{A} = \vec{B} $$$$\rightarrow\,\,\,\,\,\,\nabla \times\vec{k} = \frac{e}{\hbar{c}} \vec{B}$$ and $$ E = -\nabla{V} - \frac{1}{c}\frac{\partial{A}}{\partial{t}} = \frac{\hbar}{e}\nabla{k_0} - \frac{\hbar}{e} \frac{\partial{\vec{k}}}{\partial{t}} $$ $$ \rightarrow\,\,\,\,\,\,\nabla{k_0} - \frac{\partial{\vec{k}}}{\partial{t}} = \frac{e}{\hbar}\vec{E}$$ 

Looking at these relations we see that, a new physical meaning is given to "k" in terms of the potentials of the Electromagnetic field. So, whatever the mathematical manipulation we do in "k" will give rise to a new physical meaning in terms of Electric and Magnetic fields. That makes our problem physically more significant as it is in Mathematics.

So, we need to go back and take a re look at our initial definitions of phase factor and its non-integrability and to check whether we could impose some more conditions, so that its applicability can be generalized.  

We have seen in the beginning that the change in phase (the change in $\beta$) of the wave function around a closed curve is same for all the wave functions. But even at the single point, there is some arbitrary freedom in choosing the phase of the wavefunction, so that we can add any integer multiples of phase $2\pi{n}$ to the wave function and we will still get the same result. 
$$ e^{i\beta} = e^{i (\beta + 2\pi)} = e^{i(\beta+4\pi)} = e^{i(\beta+2n\pi)}$$
So, it is always the phase and the change in phase is undetermined up to the addition of multiples of $2\pi$. 

From this we can conclude, the change in phase of any wave function around a closed loop should be equal the Electromagnetic flux penetrating through the area enclosed by the loop and with the addition of arbitrary multiples of $2\pi$.  

$$\oint_{(4d)}\,d\beta = 2n\pi + \oint_{(4d)}\vec{K_{(4d)}} \,dl_{(4d)} \\~\\= 2n\pi + \int (\nabla\times\vec{K}) \vec{da_{(6d)}}$$
Using Stokes' theorem for 4-dimensional vectors. Finally, we get the change in phase, $$ \Delta\beta = 2n\pi + \frac{e}{\hbar{c}}  \int \vec{B}\, \vec{da}$$
Only, magnetic flux comes into play, as we are considering the region where the monopole is enclosed. 
In a special case, if you take a very small closed curved where the functions are smooth, then the change in phase cannot be in terms of the integer multiples of $2\pi$. Only, the flux term will play the role, which by itself is unique and cannot vary as we take the small curve. This wouldn't be the case, if there is any singularity in the function that is in the potentials. 

It can be concluded that, for a very small closed curve with smooth functions (potential), the change in phase for different wave functions cannot arbitrarily vary in multiples of $ 2\pi$ but is determined by the flux penetrating through the surface. But the same wave function, with singularity will give the change as, $$ \Delta\beta = 2n\pi + \int \vec{B} \,\vec{da} $$ For a given monopole the second term cannot change for various wave functions. And so, the change in phase of various wave functions should depend on the various values of the first term "n" which itself depends on the singularities enclosed within the surface. 

Now, if make my closed curve very small approaching near to zero, then the change in phase should be reduced to give zero (from the simple logic that the wave function at a single point should have definite phase - and so definite probability). 

Remember, it was not considered an empty space, but the space with singularity passing through our point. 

From the above discussion we get, $$ \Delta\beta = 0 $$ So, $$2n\pi = \frac{-e}{\hbar{c}} \int \vec{B}\cdot\vec{da} $$
where $$\int\vec{B}\cdot\vec{da}$$ in three dimensions gives the net magnetic flux penetrating through the closed surface, which is equal to $ 4\pi{q_m}$. The positive and negative sign depends on the nature of singularity. 
On equating, we finally get the quantization condition for the charges as, $$ 4\pi{q_m} = \frac{2n\pi\hbar{c}}{e}$$ and $$ q_m = \frac{n\hbar{c}}{2e}$$

Thus, Dirac proved that even if one magnetic monopole exist in Nature, it would explain why all the electric charges in the Universe are Quantized.  

Let us consider next, the problem worked out by Dirac with a little more explanation on its solution.  

Statistical and Thermal Physics - Summary of the contents - Part - 2

Let us start with the more abstract mathematical formalism of the statistical mechanics by introducing some new concepts. Let us start with defining $P_i$ is the probability of finding the system in some specific state "i". It is not the combination of outcomes for a particular event but the probability of a single outcome i.e. "i" refers to one particular microstate with energy $E_i$ 
So, the probability for the system to be found in this state is, $$ P_i = C\, N(E_0 - E_i) = C\, N(E)$$ where $ E_i + E = const. = E_0$ You may ask why we don't treat this event as the combination of two independent events and write the probability, as we did previous in two systems S,S' with energy E, E' as P(E) = C N(E) N(E') 
because now we consider the first one as a single microstate. 
Now, we assume E act as a reservoir for $E_i$, and it gives, on Taylor expansion and with suitable approximation, $$ \ln{N(E_0 - E_i)} = \ln{N(E_0)} - \beta{E_i} \\ N(E_0-E_i)= N(E_0) e^{-\beta{E_i}}$$ or simply, $$ P_i = C e^{-\beta{E_i}} = \sum_E N(E) e^{-\beta{E}}$$ We know in addition, the sum of all probabilities should give unity, $$ \sum_i P_i = C \sum_i e^{-\beta{E_i}} = 1 $$ or $$ C = \frac{1}{\sum_i e^{-\beta{E_i}}} \\ \rightarrow\,\,\,\,\,\, P_i = \frac{e^{-\beta{E_i}}}{\sum_i e^{-\beta{E_i}}}$$ Note that, the probabiliy depends exponentially on the value of energy E. 
Defining a new term called Partition function and denoting with a new symbol, $$ X = \sum_i e^{-\beta{E_i}} $$ we try to express all other variable using this X,
The average of Energy is given by, $$ \langle{E}\rangle = \frac{\sum_i e^{-\beta{E_i}E_i}}{\sum_i e^{-\beta{E_i}}} $$ but using X, $$ \sum_i e^{-\beta{E_i}}E_i = -\frac{\partial{X}}{\partial{\beta}}$$ and $$ \langle{E}\rangle = -\frac{1}{X} \frac{\partial{\ln{X}}}{\partial\beta}$$ The generalized force is given by (similar procedure), $$ f = \frac{1}{\beta} \frac{\partial\ln{X}}{\partial{x}}$$ with the same notations used in the previous post. In particular, f = p when x = V. When we talk about pressure, we mean the average pressure. 
Since, X is function of both $\beta$ and x, we get $$d(\ln{X}) = \beta{dW} - E d\beta \\ d\left(k\ln{X} + k\beta{E}\right) = \frac{dQ}{T} \\\rightarrow\,\,\,\,\,\, S = k\left(\ln{X} + \beta{E}\right)$$ where S - entropy. 
The partition for an ideal gas can be calculated as, $$ X = \frac{V^N}{h^{3N}} \left[\int_{-\infty}^{\infty}e^{-\beta{\frac{p^2}{2m}}} \,\,d^3p_i\right]^{3N} $$ The gaussian integral gives, $$\int_{-\infty}^{\infty} e^{-\alpha{x^2}} \,dx = \sqrt{\frac{\pi}{\alpha}} $$
So, $$\int_{-\infty}^{\infty} e^{-\frac{\beta}{2m}p^2}\,dp = \sqrt{\frac{2m\pi}{\beta}}$$
And so, $$ X = V^N \sqrt{\frac{2\pi{m}}{\beta{h^2}}}^{3N} \\ \ln{X} = N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}\right] $$
And the mean energy is calculated to be $$ E = \frac{3}{2}kT$$
with this, the Entropy is calculated to be, $$ S = k (\ln{X}+\beta{E}) = k N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}+ \frac{3}{2}\right] \\~\\= kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T} + \frac{3}{2}\right]$$ But, it has some small correction due to Gibbs paradox, and it is corrected as, $$ X' = \frac{X}{N!} $$ So that, our equation will get modified to (using sterling's approximation), $$ \ln{X'} = \ln{X} - \ln{N!} = \ln{X} - N \ln{N} + N $$ $$ S = kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T}+\frac{3}{2} - \ln{N} + 1 \right] \\~\\= kN \left[\ln{\frac{V}{N}} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) + \frac{3}{2} \ln{T} + \frac{5}{2}\right]$$
which is the final result for Entropy. 
Now, we can calculate Maxwell's velocity distribution law using these ideas and by assuming the molecules to be non-interacting classical ideal particles.
The probability of finding the particle in the position range r and r+dr and momentum range p and p+dp is given by the canonical distribution as we used in the partition function, we can convert it into position and velocity and obtain equation, $$ M(r,v) d^3r \,d^3v = C e^{-\frac{\beta{mv^2}}{2}} d^3r \, d^3v $$ and integrating through all over the space and for all velocity range, we get the total number of molecules , $$ N = C V \left[\int_{-\infty}^{\infty} e^{-\frac{\beta{mv_i^2}}{2}}dv_i \right]^3 \\~\\= CV \left(\frac{2\pi}{\beta{m}}\right)^{\frac{3}{2}} = N $$ 
And so,we get the value of C , and the Maxwell's velocity function (as it does depend only on "v"), $$ M(v) \,d^3r\,d^3v = \frac{N}{V} \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} e^{-\frac{mv^2}{2kT}} \,d^3r\,d^3v $$ Once we get this, we can ask for the various ways of velocity distribution. 
We can calculate the old school stuff - the three kinds of velocities - but now with the correct mathematical formalism.
Average velocity: To calculate the average of anything, first we need the probability density function. With our equation, the density function is calculated by taking small volume in the velocity space, $$ m(v) dv = 4\pi{M(v)} v^2 dv $$ and the function should be normalized, and so, $$ \int_0^\infty m(v) dv = \frac{N}{V}$$
gives the mean velocity as, $$ \langle{v}\rangle = \frac{V}{N} \int_0^\infty m(v) v dv = \frac{4V\pi}{N} \int_0^\infty M(v) v^3 dv$$
Applying the value of M(v), we get, $$ \langle {v}\rangle = 4\pi \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{mv^2}{2kT}}v^3 \,dv  = \sqrt{\frac{8kT}{\pi{m}}}$$
Similarly, the mean square speed is $$ \langle{v^2}\rangle = \sqrt{\frac{3kT}{m}}$$
Finally the most probable speed is calculated by finding the maximum of the function m(v). $$ \frac{dm}{dv} = 0 $$
Apart from the constants, $$ m(v) = c v^2 e^{-\frac{mv^2}{2kT}} = kv^2 e^{(\,\,)} $$ and so the maximum condition gives, $$ 2v e^{(\,\, )} - \frac{m}{kT} v^3 e^{(\,\,)} = 0$$ $$ v_{(mp)} = most \,\,probable\,\,speed\,\,=\sqrt{\frac{2kT}{m}} $$
Specifically, if the particles are moving in one dimension (let us say "x"), then we get, $$\frac{dm_x}{dx} = 0 $$
In one dimension, the factor $4\pi{v^2}$ would not come in the equation of m(v), so it is just $$ m(v) = c e^{-mv^2}{2kT} $$ and maximum condition gives, $$ c \left(-\frac{m}{kT}\right) v = 0 \\\rightarrow \,\,\,\,\,\, v_{(mp)} = 0 $$ 
So, the most probable speed and also the mean speed in one dimension is calculated to be zero. Since, the distribution curves attain maximum at v = 0 and it is also symmetrical about v = 0.

Statistical and Thermal Physics - Summary of the contents - Part - 1

I just want to make a brief summary of the mathematics and concepts involved in Statistical Mechanics. 
Microstate is the state of the system where the phase space is split into unit cells and labelled by some indices.
Instead of analyzing the states of all single particles, we introduce the concept of Ensemble, where we take a large number of identical particles but at different states characterized by various corresponding parameters like spin, pressure, angular momentum, etc. And we ask for the probability of the particular value of that parameter - and so the probability arguments comes into the play.  

To make a probability argument, first we need to define the system and its behaviour as a postulate. For example, in an event of throwing the dice, we are intrinsically assuming the dice is a perfect one so that no outcome is preferred than any other outcome.
Similarly, here we assume that, when an isolated system is in equilibrium (the probability of the state of the particles are independent of time), the system has equal probability for any of its accessible states. 
The accessible states are determined by the initial conditions which can be imposed arbitrarily by the observer - the part where Experimental physicist like to play very much.   
Let us consider an ensemble of particles with energy ranges from E and E+$\delta{E}$ and N(E) denote the total number of possible states of the system in this range (similar to the number of possible combination of the outcomes for an event). Out of these states, some "N(E,s_j)" number of states correspond to some other event corresponding to a physical quantity with the value "$s_j$".   Then the probability of getting the value of the physical quantity "$s_j$", $$ P(s_j) = \frac{N(E,s_j)}{N(E)}$$
[Note: The reason I am using different notation for each time is to practice with the flexibility of our mind - So, don't stick with the notation. Everytime, it will be stated the meaning of the representation] 
Macroscopic system which consists of many microscopic states and characterized by external parameters like Pressure, Volume, etc.
Let us considering two systems S and S' in two different case where in the first place it is allowed only the exchange energy due thermal interactions, and in the second part the energy is transferred purely on mechanical interactions. When we talk about large number of particles, the mean Energy is implied.
First case gives, $$\langle{E}\rangle + \langle{E'}\rangle = constant \\~\\ Heat\,\,\,\,\rightarrow\,\,\,\,\,\,Q + Q' = 0 $$ 
Second case gives, Work done by S = - Work done by S' ,$$W + W' = 0$$
In the general case where both Energy and external parameters are changed, the change in mean energy is given by, 
Differential Change in Mean Energy (dU = d$\langle{E}\rangle$)= differential change in mean energy due to external parameters (mechanical work done on the system "W") + small amound of Heat given to the system "Q"
Which gives us, 
dU = dQ + W(on the system) 
or in reverse, using W (mechanical work on the system) = -W(mechanical by the system), we get $$ dQ = dU + W $$ which is known to be the first law of Thermodynamics. 
If we considered two systems S, S' with energy E, E', then the number of states for the energy E is N(E) and for E' is N(E'). Then the probability of  happening two at the same time (where both are independent), so $$P(A\cap{B}) = P(A).P(B)$$
So, the probability for the event with total energy E + E' = const. is, $$P(E) = C N(E) N(E')$$ where C - proportionality constant. 
To get the maximum probability w.r.t. Energy, we get, $$ \frac{\partial{P}}{\partial{E}} = 0 \\~\\ \frac{\partial\ln(N(E))}{\partial{E}} = \frac{\partial\ln(N(E'))}{\partial{E'}} $$
For the mathematical purpose, we take log, so that function will become more smooth. And we denote, $$\beta(E) =\beta(E') \\~\\ \beta(E) = \frac{\partial\ln(N(E))}{\partial{E}} $$ where we define this quantity as the inverse of temperature, so that $$ \beta = \frac{1}{kT} $$ and we define another new and useful quantity named Entropy as, $$ S = k \ln(N(E)) $$
And thus we can see that, when thermal equilibrium is attained between any two states, they should have the same $\beta$ value or the same temperature. Here comes the zeroth law of thermodynamics, states that if two system A, B are in thermal equilibrium with a third system C then all the three system will be in thermal equilibrium with each other i.e. If A with C and B with C implies A with B. All of them will have same temperature characterized by $\beta$. This temperature has a special name which is Absolute Temperature.  
On a special occasion, if we consider one system as a heat reservoir then we have for the reservoir (if we give some heat Q') then using taylor expansion, $$ f(x) = f(a) + f'(a) (x-a) + \frac{f''(a) (x-a)^2}{2!}+... $$ 
Then, if I replace f(x) = ln(N(E+Q)) where x = E+Q and a = E $$ \ln {N(E+Q)} = \ln(E) + \frac{\partial\ln{N(E)}}{\partial{E}} Q + \frac{1}{2}\frac{\partial^2\ln{N(E)}}{\partial{E^2}}Q^2 +...\\~\\ = \beta{Q} + \frac{1}{2}\frac{\partial\beta}{\partial{E}}Q^2 + ...$$
From the definition of reservoir, we can neglect higher order terms and we get, $$ \ln(N(E+Q)) - \ln(N(E)) = \beta{Q}= \frac{Q}{kT} \\~\\ \delta{S} = \frac{Q}{T} $$ If we take infinitesimal amount of heat energy $Q \,\,\rightarrow\,\, dQ$ then we get, $$ ds = \frac{dQ}{T}$$ which is the formal definition of Entropy. The third law is mostly based on the physical properties of this Entropy when the absolute temperature goes to zero. 
Let us analyze the ideal gas.. We know that for an ideal gas, $$ N \propto V^N \Phi(E)$$ or $$ \ln{N} = N\ln{V} + ln{\Phi(E)} + const. $$ and it is defined the generalized force, $$ f = \frac{1}{\beta}\frac{\partial\ln{N}}{\partial{s_i}}$$ where s- external parameter, and f-generalized force. When x=V we get the mean pressure as,  $$ \langle{p}\rangle = \frac{1}{\beta}\frac{\partial\ln{N}}{\partial{V}} $$ Using this, $$ \langle{p}\rangle = \frac{N}{\beta} \frac{\partial\ln{V}}{\partial{V}} $$ other terms don't depend on "V" and give zero.
We finally get, $$ \langle{p}\rangle = \frac{N}{\beta{V}} = \frac{NkT}{V} \\ \langle{p}\rangle V = nRT $$ where k-boltzmann constant.
N - number of particles
R = Avogadro number * boltzmann const. 
and n- number of moles = N/Avogadro number
and specifically $\beta$ depends only on energy $\Phi(E)$ and Temperature depends only on $\beta$ which gives us, for an ideal gas Energy is a function of Temperature. 
Now, we define a new quantity named specific heat capacity - that is the measure of the heat required to raise the temperature of the system to a unit value keeping a parameter at a constant value, $$ C_s = \left(\frac{dQ}{dT}\right)_s$$ 
Let us look at the relation between $ C_p = \left(\frac{dQ}{dT}\right)_p $ and $ C_v = \left(\frac{dQ}{dT}\right)_v$ where it is measured at constant pressure and volume. Using this in the first law we get, $$ dQ = dU + pdV $$ at constant volume $$dQ_v = dU = c_v dT $$ similarly from the equation of state of an ideal gas at constant pressure, $$ p\, dV = nR\, dT$$ and $$ dQ_p = c_p dT $$ Combining the equations, $$ c_p = c_v + nR$$ and if we change to molar specific heat capacity we get, $$ c_p - c_v = R $$  
We can also measure the microscopic calculation of specific heats from our assumption of monatomoic ideal gas, where the interaction between the particles is negligible. It implies the number of states can be written as, $$ N(E,V) = C V^N E^{\frac{3N}{2}}$$ 
where N is the number of particles and 
C - proportionality constant.
V - volume,
E - energy,
N - number of states. 
Then, $$ \ln{N(E,V)} = \ln{C} + N \ln{V} + \frac{3N}{2} \ln{E}$$ but we know, $$\beta = \frac{\partial\ln{N}}{\partial{E}} = \frac{3N}{2E}$$ and it gives us $$ E = \frac{3N}{2\beta} = \frac{3}{2}NkT = \frac{3}{2} n RT$$ where "n" is the number of moles.
In addition, the molar specific heat at constant volume is given by, $$ C_v = \frac{1}{n} \left(\frac{\partial{U}}{\partial{T}}\right)_v$$ where at constant volume $ dQ = dU$ so, it gives $$ C_v = \frac{3}{2}R$$  for an ideal gas. And so we can find $ C_p,\,\,and\,\,\, \gamma = \frac{C_p}{C_v}$
We can explore more with these. Now various situation may arise depending on which two variables are changed. Let us start with the first law, $$ dQ = dU + pdV$$ Using the relation of entropy, $$ dU = T dS - pdV $$ where the independent variable is S, V.
Now, if we have the independent variables as S, P then we can make use of the legendre transform as, $$ dU = T dS - d(pV)+ VdP \\ d\left(U + pV\right) = T dS + VdP $$ where we call H = U + pV = Enthalpy.
$$ dH = T dS + VdP$$ Next, if we take chose T and V as the independent variables then, $$ dU = TdS - pdV = d(ST) - pdV - SdT \\ d \left(U-TS\right) = - pdV - SdT \\ dF = -SdT - pdV$$ where F = U - TS is called the Helmholtz free energy. 
Finally if we take the independent variables T, P then $$ dU = d(TS) - SdT - d(pV) + VdP \\ d \left(U-TS+pV\right) = -SdT + VdP \\ dG = -SdT + VdP$$ where G = U -TS +pV = F + pV is called the Gibbs Free energy.