Fourier series and Fourier transform

The concept of Fourier series is based on the orthogonality of sine and cosine functions. So, it is a perfect thing to start from these properties.
A sine function with time period $2\pi$ radians is given by $sin(t)$ and the integral of the function over its time period is zero. In general, for an arbitrary sine function with Time period $T_0$, the function is given by $sin (\frac{2n\pi{t}}{T_0})$, where the time integral over this new time period is zero. Everything applies to "cosine" function as much as the same.

In functional form, $$sin\left(\frac{2n\pi}{T_0}(t+T_0)\right) = sin(\frac{2n\pi{t}}{T_0})$$ and $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2m\pi{t}}{T_0}) = 0  \\ for \,\,\,\,n,m > 0 $$
From our trigonometric identities, we can write $$ sinA cosB = \frac{sin(A+B)+ sin(A-B)}{2}$$ Using this in our case, we will get (n+m) and (n-m) where n and m are integers. This gives, $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$
Similarly, with the other identity $$ sinA sinB = cos (A-B) - cos(A+B) \\ cosA cosB = cos(A+B) + cos(A-B)$$ we can prove $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ except now, there arise a problem for the case n = m as we will get an extra non-zero term $$\int_{t_0}^{t_0+T_0} \,dt \frac{cos(\frac{2(n-m)\pi{t}}{T_0})}{2} = \int_{t_0}^{t_0+T_0} \frac{1}{2}\,dt = \frac{T_0}{2}$$
Finally we get the following identities,
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ and
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) sin(\frac{2m\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = \delta_{nm}\frac{T_0}{2} $$
This is the fundamental orthogonality relation we require for the sine and cosine function to serve as the basis vectors.
Now, the Fourier series is defined as, $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{2n\pi{t}}{T_0}) + \sum_{n=1}^\infty b_n sin(\frac{2n\pi{t}}{T_0})$$
The only unknown terms in this series are the Fourier coefficients $a_0, a_n, b_n$ which can be determined using our orthogonality relations as, $$ a_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) f(t) \,\,\,\, n = 0,1,2..\\ b_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) f(t) $$
Once we are able to write a function in terms of sine and cosine, we proceed to write it in terms of exponential functions using Euler's formula. And we can contract the series in a simple form as, $$ f(t) = \sum_{n=-\infty}^\infty c_n e^{i\frac{2n\pi{t}}{T_0}}$$
Now, the only unknown in this series is $c_n$ and it can also be determined using the orthogonality relation of exponentials $$ \int_{t_0}^{t_0+T_0}\,dt\,e^{i\frac{2n\pi{t}}{T_0}}e^{-i\frac{2m\pi{t}}{T_0}} = \delta_{nm}T_0 $$ which gives finally, $$ c_m = \frac{1}{T_0} \int_{t_0}^{t_0+T_0} f(t) e^{\frac{i2n\pi{t}}{T_0}} \,dt $$

Fourier Transform:

Substituting for $ \omega_n = \frac{2n\pi}{T_0}$ we get $$ f(t)  = \sum_{n= -\infty}^\infty c_n e^{i\omega_nt}$$ and $$ c_m = \frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} f(t) e^{-i\omega_nt}\,dt$$ (from the substitution $t_0 = -\frac{T_0}{2}$)

It is easily seen from the above equation that, it fails for $T_0 = \infty$ For a function with infinite period, the coefficient cannot be calculated from above equation.

Since, many of the real function used in Physics has this structure (Time period infinity), it is necessary to look our for an alternative to deal with these functions. This is where the integral comes about to play the significant role and known as fourier transform.

For very large time period, the change in $\omega$ - the step size becomes negligible i.e. $$\omega_{n+1} -\omega_n = \frac{2\pi}{T_0} \\ \rightarrow \,\,\,\, \delta\omega = d\omega\\ \frac{1}{T_0} = \frac{d\omega}{2\pi}$$ and there is no need for the index 'n'.

Combining everything together, $T_0\rightarrow\infty$ and substituting for $c_n$ with some dummy variable, $$ f(t) = \sum_{n= -\infty}^\infty\frac{d\omega}{2\pi} \left[\int_{-\infty}^{\infty}f(\tau) e^{-i\omega_n\tau}\,d\tau\right]e^{i\omega_n{t}}$$ As "n" changes $d\omega$ is very small and the sum can be replaced by Integral,

$$ f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega_nt}\,d\omega \left[\int_{-\infty}^\infty f{\tau} e^{-i\omega_n\tau}\,d\tau\right] $$ the bracketed term depends only on $\omega$. Thus, we have finally arrived at our fourier and inverse fourier transform. It doesn't matter which one you call the fourier transform and which one is the inverse one. By splitting $\frac{1}{2\pi}$ factor in two different ways, the final equations are, $$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,dt \,e^{-i\omega{t}}\,f(t) \,\,\,\,\\or\,\,\,\,\\ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \int_{-\infty}^{\infty}\,dt e^{-i\omega{t}}\,f(t)$$ or you can put the $2\pi$ factor in the other way. 

Volume of a Hypersphere in N - dimension

    To find the number of micro states in a statistical system for N number of particles in a fixed volume and energy within the range E and $E+\delta{E}$, we require the knowledge on how to find the volume for N dimensional hypersphere in momentum space.

In the simplest way, the problem is to find the volume of an N-dimensional hypersphere. If we try to find the general formula by mathematical induction, using the equation in Cartesian form, it will produce a complex equation.

    Instead there is a clever trick to simplify the procedure based on the concept of infinity. The equation of a sphere in Cartesian form is $$ x_1^2+x_2^2+x_3^2+...+x_N^2 = r^2$$ where 'r' is the radius of the sphere.

    In general, volume of an n-dimensional sphere should be proportional to $r^N$ and its surface is proportional to $r^{N-1}$. So, $$V \propto r^N\\ S \propto r^{N-1} $$ or $$ S = A r^{N-1}$$ where A is the proportionality constant that is equal to the surface area of a unit sphere. If we could find somehow this value of A in n-dimension, it is then a simple task to proceed to the volume. The special characteristic of the surface of a sphere is that, once we know the surface, the volume is just a simple line integral of the surface value along 'r'.

$$ dV = S dr  = A r^{N-1} dr $$

The generalization of the equation of sphere in Cartesian form is a straight task. To connect all these things, we use the property of the infinity given as,

    Integral of a function over the infinite volume is the same independent of whether the integral is done with Cartesian cubic volume element (cubic volume extending to infinity) or spherical volume element (spherical volume extending to infinity), the result is the same.  This is not a completely new one because we use this same fact to find the integral of $$\int_{-\infty}^{\infty}e^{-\alpha{x^2}}\,dx = \sqrt{\frac{\pi}{\alpha}}$$

    Using the same integral we have, $$ I = \int_{-\infty}^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} \,dx_1\,dx_2\,...\,dx_N = \pi^{N/2} $$ Now,the same function integrated over spherical volume element gives, $$ I = \int_0^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} A r^{N-1}dr = \int_0^{\infty} A e^{-{r^2}} r^{N-1} dr $$ Substituting $y = r^2 $ we have $$I  = \int_0^{\infty} A e^{-y} y^{\frac{N-1}{2}} \,\frac{1}{2y^{1/2}}\,dy = \int_0^{\infty} A \frac{1}{2} e^{-y} y^{\frac{N}{2}-1}dy = \frac{1}{2} A \Gamma(N/2)$$ then we have, by equating the value of I in two ways, $$ A = \frac{2\pi^{N/2}}{\Gamma(N/2)}$$ Now, the volume is evaluated to be $$V  =  \int _0^R \,S \,dr  = \int_0^R \,A r^{N-1}\,dr \\ V =  \int_0^R\,\frac{2\pi^{N/2}}{\Gamma(N/2)} r^{N-1} \, dr = \frac{\pi^{N/2}}{\Gamma(N/2)} \frac{R^{N}}{N/2} = \frac{\pi^{N/2} R^N}{\Gamma(\frac{N}{2}+1)}$$ where we used the fact  $$\frac{N}{2} \Gamma(\frac{N}{2}) = \Gamma(\frac{N}{2}+1)$$

Thus we can determine the volume and so the surface area of an n-dimensional hypersphere in a simple form. 

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, $$ (1-x^2)\frac{d^2W}{dx^2} - 2x\frac{dW}{dx} + \left[n(n+1) - \frac{m^2}{1-x^2}\right]W = 0 $$

Hermite Polynomials - Derivation

Let us look into the formal solution of Hermite polynomial equation. 
Hermite differential equation is given by, $$ \frac{d^2y}{dx^2} - 2x \,\frac{dy}{dx} + 2ny = 0 $$
Using Frobenius method, we assume an infinite series solution of ,
 $$ y(x) = \sum_{m=0}^{\infty} C_m x^{m+r} ........ where C_0 \neq 0 $$ Substituting this on our differential equation, our differential equation gets modified into $$ \sum_{m=0}^\infty \left[ (m+r)(m+r-1) C_m x^{m+r-2} + 2 [ n - (m+r)] C_m x^{m+r} \right] = 0 $$ This should be equal to zer0, which can be obtained only if and only if all the coefficients are zero. 

  Equating to zero, the coefficients of,
$$ (1) x^{r-2}, \ldots... r(r-1) = 0 \rightarrow r=0 \; or \; r=1 \; since \; C_0 \neq 0 $$ $$ (2) x^{r-1}, \ldots... (r+1)rC_1 = 0 .... \; if \;\; r=0, \; C_1\neq 0 ; and \; if \;\; r=1, \; then \; C_1=0 $$ $$ (3) x^{m+r}, \\~\\ for \; r=0 \dots......... \frac{C_{m+2}}{C_m} = \frac{-2(n-m)} {(m+1)(m+2)} \\~\\ for \; r=1 \dots...........\frac {C_{m+2}}{C_m} = \frac{-2[n-m-1]}{(m+2)(m+3)}$$

  Considering r=0, the general expression for even coefficients expressed as, 
$$ C_{2s} = \frac{(-1)^s 2^s n (n-2).....(n-2s+2)}{(2s!)} C_0 $$
and the odd coefficients are expressed as,
$$ C_{2s+1} = \frac{ (-1)^s 2^s (n-1) (n-3).....(n-2s+1)}{(2s+1)!} C_1$$
The general solution is therefore,

$$ y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right] + \\~\\ C_1x \left[ 1 + \sum_{s=1}^\infty \frac{(-1)^s 2^s (n-1)(n-3)....(n-2s+1)}{(2s+1)!} x^{2s+1} \right] $$   

For r=1, if we proceed like the same, we will get a solution exactly similar to the second part of the series in the general solution with other coefficient. Since the coefficients are arbitrary, the series itself already inherent in the general solution, so we don't need to put separate attention on that. 

   That's it. We arrived to our general solution. Depending on the    
nature of the problem, we can make the series to converge or stop by appropriately choosing the values of $ \; C_0 \;and \;C_1$ 

You can choose the constants in various ways, where all those solutions will obey our Hermite differential equation - It is true, because I have tried!

But all of the solutions satisfying our equation is not physically meaningful.  

   There is a conventional way of choosing the constants, such that the terms in the series have the properties known as orthogonality, completeness,etc. And those specific functions in our series is known as the Hermite polynomials. 

In the general solution, we make the convergence test by taking the ratio of consecutive elements either in the odd series or in the even series, i.e. ratio of the terms $x^{2s}, x^{2s+2}$,
eg.In Even series $$\left|\frac{t_{s+1}}{t_s}\right| = \left| \frac{ C_{2s+2}}{C_{2s}} \right| = \left| \frac{-2 (n-2s)}{ (2s+2) (2s+1)} x^2 \right| $$
As "s" becomes very large,
$$ \lim_{s\rightarrow\infty} \left| \frac{C_{2s+2}}{C_{2s}}\right| = \left|\frac{4s}{4s^2} x^2\right| = \frac{x^2}{s} $$
[this is the same convergence of $ e^{x^2} $ series]. The series is an infinite series for all values of x and it will be meaningful only if it is terminated to finite terms. It can be achieved by choosing suitable values for "n".
Choosing the other constant as zero, we can always work with either odd or even series. Once we take an odd or even series, we choose the value of "n" such that,

for even series,  the coefficients of $x^{2s+2}$ is zero. i.e. n = 2s
and $ C_1 = 0 $ the general series reduces to, 
     $$y(x) = C_0 \left[ 1 + \sum_{s=1}^\infty \frac{ (-1)^s 2^s n (n-2)....(n-2s+2)}{(2s!)} x^{2s}\right]$$ 
Substituting, n = 2s, $$ y(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{(-1)^s 2^s (2s)(2s-2)...(2s-2s+2)} {(2s!)} x^{2s} \right] $$ 
$\rightarrow$ $$ y_{even}(x) = C_0 \left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!} {2s!} x^{2s} \right]$$

for odd series, n = 2s+1 such that there is no terms involving power of x greater than 2s+1 and $C_0 = 0 $, then the series becomes,
$$ y_{odd}(x) = C_1x\left[ 1 + \sum_{s=1}^s \frac{ (-1)^s 2^{2s} s!}{(2s+1)!} x^{2s+1} \right]$$

Now, the conventional choice states that the constants $C_0\,,C_1$ chosen in a way such that the power of $x^n$ has the coefficient $2^n$ , then,
$$ for \,\,C_{2n} \rightarrow C_0 \frac{(-1)^n 2^{2n} n!}{2n!}x^{2n} = 2^{2n}  x^{2n} $$ which implies, $$ C_0 = \frac{2n! (-1)^n}{n!} $$
$$ for \,\,C_{2n+1} \rightarrow C_1 \frac{(-1)^n 2^{2n} n!}{(2n+1)!}x^{2n+1} = 2^{2n+1} x^{2n+1} $$$\rightarrow$ $$ C_1 = (-1)^n \frac{(2n+1)! \,(2)}{n!} $$ 
That is all we need to know!
Now, we will derive the first few Hermite polynomials as an example, 
n=0 $$ H_0(x) = C_0 \,\,where\,\, C_0 = \frac{2(0)!}{0!}(-1)^0 = 1 $$ Therefore , $$ H_0(x) = 1 $$
n=1 $$ H_1(x) = C_1x \,\,where\,\, C_1 = \frac{(-1)^0 1! 2}{0!} =2 $$ $\rightarrow$$$ H_1(x) = 2x $$
n=2 $$ H_2(x) = C_0 + C_0 \frac{(-1) 2^2 1!}{2!} x^2 \,\,where\,\, C_0 = \frac{2(1)! (-1)^1}{1!} = -2 $$$\rightarrow$ $$ H_2(x) = -2 + (-2) (-2x^2) = 4x^2 -2 $$
n=3, $$ H_3(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 $$ and $$ C_1 = \frac{(-1)^1 3! 2}{1!} = -12 $$ $\rightarrow$ $$ H_3(x) = -12 x + (-12) \frac{(-4)}{(6)}x^3 = 8x^3 - 12x $$ 
n=4, $$ H_4(x) = C_0 + \frac{(-1) 2^2 1!}{2!} x^2 C_0 + \frac{(-1)^2 2^4 2! }{4!} x^4 C_0 $$ and $$ C_0 = \frac{2(2)! (-1)^2}{2!} = 12 $$ $ \rightarrow$ $$ H_4(x) = 12 + 12 (-2) x^2 + 12 \frac{4}{3} x^4 = 16x^4 - 24x^2 +12 $$
n=5 $$ H_5(x) = C_1x + C_1 \frac{(-1) 2^2 1!}{3!} x^3 + C_1 \frac{(-1)^2 2^4 2!}{5!}x^5 $$ and $$ C_1 = \frac{(-1)^2 5! 2}{2!} = 120 $$ $\rightarrow $ $$ H_5(x) = 120 x + (-1) 80 x^3 + 32x^5 = 32x^5 - 80x^3 +120x $$ 
Thus we can find all the Hermite polynomials. 


You may ask "Why this choice of Hermite polynomials?" 
It is because of the special properties followed by these polynomials. For eg. These polynomials can be simply written in one line using a formula known as Rodrigues formula. And there is a specific generating function for these and orthogonality property for specific conditions and etc.  All these make Hermite polynomials more special in real life. 

Cartesian to Spherical coordinate system - Coordinate Transformation

In general orthogonal curvilinear coordinate system, general position in 3 dimension is given by, $$\vec{s} = \vec{s}(u_1, u_2, u_3)$$ and small displacement "ds" is can be written as, $$ \vec{ds} = \sum_i^3 \frac{\partial{\vec{s}}}{\partial{u_i}} du_i $$ where $ \frac{\partial{\vec{s}}}{\partial{u_i}} = h_i \hat{e_i} $ and $\hat{e_i} $ is the unit vector along the 'i'th direction. Rewriting it in simple form, we have, $$ \vec{ds} = \sum_i^3 h_i du_i \hat{e_i} $$
For cartesian coordinate system, $h_i = 1$
$\rightarrow$ 
$$ \vec{ds} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} $$
But there is no unique choice of coordinate system, we can also choose spherical coordinate system as, 
$$ \vec{ds} = dr \hat{e_r} + r \,d\theta\hat{e_\theta} + r \,sin\theta \,d\phi \hat{e_\phi} $$ with corresponding scaling factors. 

To go from one coordinate system to another, we use the relations, $$ x = x(r,\theta,\phi) \\~\\ y = y(r,\theta,\phi) \\~\\ z=z(r,\theta, \phi) $$ 
The only relation we know from our conventional assumption is that, $$ x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\, $$ 
Using the chain rule, $$ dx = \frac{\partial{x}}{\partial{r}}dr + \frac{\partial{x}}{\partial{\theta}} d\theta + \frac{\partial{x}}{\partial{\phi}}d\phi $$
$\rightarrow$ $$ dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi $$
Similarly, $$ dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi $$ and $$ dz = cos\theta \, dr - r\, sin\theta \, d\theta $$
Applying it in our Cartesian equation for "ds", we get, $$ ds =(\,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi )\hat{e_x} \\~\\+ (\, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi)\hat{e_y} \\~\\+ (cos\theta \, dr - r\, sin\theta \, d\theta) \hat{e_z} $$  

Thus we expressed the infinitesimal displacement in cartesian system using spherical measurements such $ r, \theta, \phi $. Now, from our definition of unit vector, $$ \hat{e_r} = \frac{{\frac{\partial{\vec{s}}}{\partial{r}}}}{|\frac{\partial{\vec{s}}}{\partial{r}}|} $$ and $$ \hat{e_\theta} = \frac{{\frac{\partial{\vec{s}}}{\partial{\theta}}}}{|\frac{\partial{\vec{s}}}{\partial{\theta}}|}$$ and $$\hat{e_\phi} = \frac{{\frac{\partial{\vec{s}}}{\partial{\phi}}}}{|\frac{\partial{\vec{s}}}{\partial{\phi}}|}$$ 
Using the above, equation, we can write the unit vectors along $r,\,\theta,\,\phi $ directions as follows, $$ \hat{e_r} = sin\theta\, cos\phi\, \hat{e_x} \,+ \,\sin\theta\,sin\phi\,\hat{e_y} \,+\, cos\theta\,\hat{e_z} $$ and $$ \hat{e_\theta} = cos\theta\, cos\phi \,\hat{e_x} \,+\, \cos\theta\, sin\phi \,\hat{e_y} - sin\theta\,\hat{e_z} $$ and $$ \hat{e_\phi} = \,-sin\phi \,\hat{e_x} \,+ \,cos\phi\,\hat{e_y} $$ From these three unit vectors we can solve for the other three unit vectors as following,
Solving the first two,
 $$ sin\theta\,\hat{e_r} + \, cos\theta\,\hat{e_\theta} = \,cos\phi \hat{e_x} + \,sin\phi \,\hat{e_y} $$

Combining with the third we can solve for x and y, $$ \hat{e_x} = sin\theta\,cos\phi\,\hat{e_r} + cos\theta\,cos\phi\,\hat{e_\theta} - \,sin\phi\,\hat{e_\phi} $$ and $$ \hat{e_y} = \,sin\theta\,sin\phi \hat{e_r} +\, cos\theta\, sin\phi\, \hat{e_\theta} + \,cos\phi\, \hat{e_\phi} $$ and finally solving for z, $$ \hat{e_z} = cos\theta\,\hat{e_r} - sin\theta\,\hat{e_\theta} $$

That is all we do need to derive for the coordinate transformation from cartesian to spherical. 

Canonical Transformations

In this section, we are primarily concerned about the problems where the Hamiltonian is independent of all the $ q_i $ coordinates i.e. cyclic coordinates. In those cases, the canonical conjugate momentum of the system is constant over time. Let us call those constants as, $$ p_i =  c_i $$ Then the Hamiltonian is rewritten as the function of these constants, $$ H = H ( c_1, c_2,... c_n ) $$   In this simple form, the corresponding equations of motion are linear and easy to solve. In general all of the coordinates in the problem is not cyclic in Nature. 

But it was known that, the number of cyclic coordinates in a given system depends on the choice of generalized coordinates. That is why there are various coordinates systems are defined like spherical, cylindrical, ellipsoidal and etc. So we try to find a coordinate system such that all of the coordinates are cyclic. And the procedure is popularly known as canonical transformation which is the transformation procedure from our initial coordinate system to the desired cyclic coordinate system. 

Before going to the transformation, we prefer to maintain the structure of Hamilton's equations i.e. we would like to get a new Hamiltonian with new set of Q and P but with old structure which follows the variational principle and its results. 

From variational principle we know that the Lagrangian is not unique and can differ from one another about the differential of a function that vanishes at the end points, given by,
$$ L' = L + \frac{dF}{dt} $$ Similarly, the variation principle gives us the freedom, $$ p_i\dot{q_i} - H = P_i \dot{Q_i} - H' + \frac{dF}{dt} \,\,\,...eq.(1)$$ with necessary scale transformations. 

Depending on the functional form of F, there are various ways of obtaining the transformation rules. Let us look the at the four basic functional forms. 
First let us assume that F is a function of old and new coordinates i.e.q and Q. $$ F _1= F(q,Q,t) \,\,\,...eq.(2)$$ 
Applying it in eq.(1) we get, 
$$ p_i\dot{q_i} - H = P_i \dot{Q_i} - H' + \frac{dF(q, Q, t)}{dt} $$
using the chain rule, 
$\rightarrow$ $$ p_i\dot{q_i} - H = P_i\dot{Q_i} - H' + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{Q_i}} \dot{Q_i} + \frac{\partial{F_1}}{\partial{t}} \,\,\,...eq.(3)$$
$$ p_i \dot{q_i} - H = \left( P_i + \frac{\partial{F}}{\partial{Q_i}} \right) \dot{Q_i} + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{t}} -H'$$
Equation the coefficients, (since all q,p,Q,P are independent variables) we get, $$ \frac{\partial{F}}{\partial{q_i}} = p_i \,\,\,...eq.(4)$$ and $$ P_i + \frac{\partial{F}}{\partial{Q_i}} = 0 \,\,\,...eq.(5)$$ and $$ H' = H + \frac{\partial{F}}{\partial{t}} \,\,\,...eq.(6)$$

From these relations we can solve for 2n new coordinates and momentum provided we know the generating function. Similarly, from the inverse method that is if we know the transformed equation we can seek the corresponding generating function. 

Sometimes, it needn't to be the generating function of the above type that completely solves the problem but rather be in a different form like $$ F_2 = F (q,P,t)\,\,\,...eq.(7) $$ But the function should depend only on q, Q,t and not on p or P because as a whole it behaves like Lagrangian and so the derivatives of p and P should not exist. 

Thus we make the corresponding Legendre transformation such that the dP element should be canceled out,
 $$ dF = f_1 dq_i + f_2 dP_i + f_3 dt \,\,\,...eq.(8)$$  
we know $$ d((f_2)_iP) = P (df_2)_i + f_2 dP_i \,\,\,...eq.(9)$$
Subtracting eq.(9) from eq.(8) we get, $$ dF - d((f_2)_iP) = f_1dq_i - P (df_2)_i + f_3 dt \,\,\,...eq.(10)$$  If we call $ F - (f_2)_iP $ as F' then, F' is the function of only f_2, q and t. There is a special condition that $$ \frac{\partial{F}}{\partial{P_i}} = f_2 $$ 
Let us test the condition for this new function F' by again substituting this in our initial definition eq.(1),

$$  p_i\dot{q_i} - H  = P_i\dot{Q_i} - H' + \frac{dF'}{dt} $$
$\rightarrow$ 
$$ p_i\dot{q_i} - H  = P_i \dot{Q_i} - H' + \frac{dF}{dt} - (f_2)_i \dot{P} - P_i\dot{(f_2)_i} \,\,\,...eq.(11)$$
If we make the further assumption that, $$ (f_2)_i = \frac{\partial{F}}{\partial{P_i}} = Q_i $$
$\rightarrow$
$$ p_i\dot{q_i} - H = -Q_i\dot{P_i} - H' + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{P_i}} \dot{P_i} + \frac{\partial{F}}{\partial{t}}\,\,\,...eq.(12)$$

Comparing the coefficients we get the necessary conditions, $$ \frac{\partial{F}}{\partial{P_i}} = Q_i \\~\\ \frac{\partial{F}}{\partial{q_i}} = p_i  \\~\\ H' = H + \frac{\partial{F}}{\partial{t}} $$

Similarly we can derive other generating functions and corresponding relations by suitable legendre transform.  

Let us assume the third generating function as the function of p, Q, t. 
dF = f dp + g dQ + h dt
and
d(wp) = w dp + p dw 
Subtracting with the assumption $\frac{\partial{F}} {\partial{p}} = f = -w = -q $, we get,  d(F + qp) = g dQ + p dq + h dt 

Substitution in eq.(1)  where $$ F_3 = F(p,Q,t) + p_iq_i $$ because subtraction or negative sign would result contradiction. 
Finally we will get, $$ \frac{\partial{F}}{\partial{p_i}} = q_i \\~\\ \frac{\partial{F}}{\partial{Q_i}} = -P_i \\~\\ H' = H + \frac{\partial{F}}{\partial{t}} $$

Finally, the fourth generating function is given by, 
$$ F_4 = F(p,P,t) + q_ip_i - Q_iP_i $$ 
and the resultant solutions are, $$ \frac{\partial{F}}{\partial{p_i}} = -q_i \\~\\ \frac{\partial{F}}{\partial{P_i}} = Q_i  \\~\\ H' = H + \frac{\partial{F}}{\partial{t}} $$ 

Linear Vector Space - Introduction

The concept of Vector space is not straight forward as it sounds. Unlike the usual ones, it doesn't have a perfect physical basis in reality starting with the question "Why". 

But the ideas are not completely abstract as you think,  it was created not from a specific topic in physics, but from the generalization of all the usual mathematical concepts. 

I will try to go with my own formal introduction, where everything could be started and understood from the beginning of Quantum Mechanics. 

When the idea of Wave function and operations of quantum mechanics are introduced, people really don't understand the insights. They just used all the arithmetical manipulations and concepts from the known classical mechanics and applied it into quantum mechanics in terms of operators. 

But they never know, why those operators behave in a classical form and why it explains the Nature so beautifully and so on with many philosophical questions. 

You may ask, then why people work with a mathematics for which, they themselves don't know the reason "why" it works.  

As scientists, they have other things to create and work with in real life instead of simply getting into the philosophical questions. 
After all, applications are more important than the complete reasoning.

A single Hydrogen atom in Earth gives a spectrum that is exactly as same as in the Jupiter. We can use this property to communicate, even if we don't know the answer for the question "why exactly it works the same?".

So they just said, "The mathematics works fine. What do we need extra other than that, to apply it in real life!!" . 

Eventually, they stopped about thinking "Why" and proceeded to define "How" things can be developed in this new mathematics with the help of introducing some new abstract concepts. 

This is the reason why, Quantum Mechanics looks as it has more postulates than any other field in Physics. 

From these numerous abstract postulates, they developed a whole lot of other concepts and succeeded with it in real life. And so, Quantum Mechanics was formulated. 

As we did go along, we found that these abstract definitions and operations plays crucial role not only in Quantum Mechanics but also in many other places. 
From this, it was believed that, may be there is some basic mathematics intrinsically hidden in Nature. 

To understand more, accordingly, they combined it together and found out some of the most common basic rules followed by all those abstract quantities and initiated the concept of Vector space. 

Linear Vector Space:   

[Note that, the following concepts are not the first and newly defined but they are just the compilation of basic concepts you can find it anywhere in physics.]


A vector is a mathematical notation or an entity used to denote a concept. As we used to express the whole of Nature itself using numbers, these concepts are also intrinsically related with one another with the help of numbers. 

The numbers can be either real, imaginary or complex, etc. These numbers form a field, i.e. just a new name to denote the set of numbers that is used to related these vectors. If they are scalar numbers, then it is called scalar field. 

Let us denote the set of vector elements by V and the set of field elements by F. They should obey some fundamental axioms to be defined as the Vector space. 

To be considered as a field, the set F should follow these axioms, 

Closure: For all two elements 'a' and 'b' , $ a\,,b\, \in F $ then $ a*b \in F $ where  *  denotes any binary operation. The most usual one is addition and multiplication.  

Associativity: For all three elements $a,\,,b\,,c \in F $ there exists an equality $$ a*(b*c) = (a*b)*c $$

Commutativity: For all two elements $ a,\,b \in F $ there exists an equality, $$ a*b = b*a $$ 

Existence of Identity: For all elements $ a \in F $ there exists an identity element "e" such that, $$ a * e = a $$

Existence of Inverse: For all element $ a \in F $ there exists an inverse element $ a^{-1} $ such that $$ a * a^{-1} = e $$ where "e" is the identity element. 

Distributivity: If two operations are considered i.e. addition and multiplication then distributivity is defined by the condition that for all three elements $ a,\,b\,,c \in F $ there is an equality $$ a(b+c) = ab + ac $$ 

The operations needn't be addition and subtraction but can be any binary operation. 

Once the above axioms are satisfied, the set is called a "field". Now, proceeding to the next set of vector elements "V" it has to satisfy the following axioms similar to the old one, but now we take two vector elements. 
Associativity, commutativity, identity and inverse are defined as the same as previous. 
The new properties are , 
compatibility with scalar multiplication with a field element. For $ a,b\in F $ and $ \vec{u} \in V $
$$ a(b\vec{u}) = (ab) \vec{u} $$
Distributivity with scalar multiplication with vector addition, For all $ a \in F $ and $ \vec{u}, \vec{v} \in V $ there exists, $$ a(\vec{u} +\vec{v}) = a\vec{u} + a\vec{v} $$ 

When I denote vector elements with vector notation, it doesn't mean it is the usual three dimensional vector. I am just using it for the notation consistency and nothing more!   

Idea of Hyperbolic trigonometric functions in complex analysis

Complex numbers complete the description of numbers as we know, by without losing or spoiling any of the known data in the physical world. It is just a developed notation to handle the extra numbers came along with the solutions of equations, and named to be the "so called" physically meaningless solutions. 

The notation used to represent the most general form of complex number is, $$ z = a + ib \,(or)\, x+iy $$ where "i" is the imaginary root i.e. the root of "-1". With this new notation, any known number in our Nature, can be written in this form. 

Once we expand the number system with this new notation, it eventually expands all the fundamental definitions used by those numbers in any field. It leads to subsequent changes in all of the functions that is defined over the real numbers. For example, we can analyze the effect in our usual well behaved functions such as trigonometric function, exponential functions, etc. 

We need to remember one thing that, the new real functions we are going to define in a complex domain is simply just the extension of the foreknown concepts and they are all just axiomatic definitions. So, it is not possible to ask for the proof of these definitions! 

For example, in the polar form, a complex number is denoted by $$ z = r\,e^{i\theta} $$ where $\theta = \theta_p \,+\, 2\pi\,k$ k= 1,2,3,..
$\theta_p$ is called the principle angle measured from the positive x-axis. 

Now, we extend this concept of trigonometric functions into complex functions by replacing the real x-values with new complex numbers. It is achieved with the help of the handful tool i.e. series expansions of all powers of x. 

Since complex part is in the form of addition, all these powers just adds extra terms into our real expansion series. So that, the essence of old functional forms are not affected in anyway due to this new definition transformation, except that it was just incorporated in a larger domain.  

Using Euler's formula, the general form of a complex number can be denoted as, $$ z\, = \, re^{i\theta} = r (cos{\theta} + i sin{\theta}) $$

From Euler's formula, we tend to write sin($\theta$) and cos($\theta$) in terms of exponential functions, where each exponential form is used to represent a complex number. 

Euler's formula also gives, $$ e^{-i\theta}\, = \,cos(\theta)\, -\, i\,sin(\theta)\,$$ Thus we get,
$$ cos(\theta) \,=\, \frac{e^{i\theta} + e^{-i\theta}}{2} $$ and $$ sin(\theta) \, =\, \frac{e^{i\theta} - e^{-i\theta}}{2i} \,$$ 
As the right hand side of this equation only deals with exponentials, if we replace the $\theta$ with z - complex number, the definition of sine function expands to wider regions which includes all the complex numbers. 
And so, a new name and definition is given, where the independent variable $\theta$ is replace by the complex number "z". And they are hyperbolic trigonometric functions. 
The complex number "iy" is used in the euler equation to give,    
$$ e^{\{i(iy)\}}\, = \,e^{-y} \,= \,cos(iy)\, + \,i \,sin(iy)\, $$ Similarly, $$ e^{\{-i(iy)\}}\, = \,e^y \,= \, cos(iy) \, - \, i\, sin(iy) \,$$ 

We get, $$ cos(iy) \,=\, \frac{e^y\, + \,e^{-y}\,}{2} = cosh(y) $$

We choose this cos(iy) as cosh(y) since it has the similar form of cosine in Euler formula.  

But sine is defined from, $$ sin(iy) = \,\frac{e^{-y}\,-\,e^{y}\,}{2i} = - (\frac{e^{y} - e^{-y}}{2i} = -i sinh(y)$$ so that the both structures of the equation will look similar. 

As we ourselves changed the basic definition, we cannot expect the same results of usual trigonometry in here. For example, the maximum value of sine and cos is equal to one when dealing real numbers, but with complex number it can have any value

That is all we need to know about definitions.. now we can proceed further to define all other identities from this basic concept and all other things can be sought out from those definitions. 

Rigid Body Motion - Euler Angles

A rigid body is defined from the idealized concept that, distance between any two mass points remains constant throughout the motion. 
     Since, it is satisfied by the most of the objects we use in real life (Not absolutely, but perfectly applicable), the kinematics of these rigid bodies plays significant role in many areas. 

The special property of this rigid body is that, we need only 6 independent coordinates to completely define the state of a rigid body in 3-Dimensional Space. No matter how many particle it contains, it can be always applied, entirely due to the constraints. 
Other than the distance constraints, it is also possible to add additional constraints in any rigid body motion, and so the number of independent coordinates will be more reduced. 

It is customary to use the first set of 3 independent coordinates as the "Space fixed coordinates" and the second set of 3 coordinates as the "Body fixed coordinates" but with the same origin. 
     So that, you can explain the state of the body (general position) with the Space fixed coordinates and its orientation relative to this Space fixed coordinates using the Body fixed coordinates. 

Specifying one set of coordinates relative to another needs the basic rules of "Coordinate Transformation".   

Let us say the $(x_1,x_2,x_3)$ and $(x'_1,x'_2,x'_3)$ are the components of same vector in two sets of orthogonal coordinate system with same origin. If $ \{ \hat{e_1}, \hat{e_2}, \hat{e_3} \}$ and $\{ \hat{e'_1} , \hat{e'_2} , \hat{e'_3}\}$ are the corresponding unit vectors,

Direction cosines are defined by,

$$ cos(\theta)_{ij} = cos (\hat{e'_i}\cdot\hat{e_j}) = \hat{e'_i}\cdot\hat{e_j} \,\,\,\ldots...eq.(1)$$

Using the direction cosines, new primed unit vectors in terms of old non primed unit vectors written by, 
$$ \hat{e'_i} = \sum_j (\hat{e'_i}\cdot\hat{e_j}) \hat{e_j} = \sum_j cos(\theta)_{ij} \hat{e_j} \,\,\ldots...eq.(2)$$

Let me write the general vector in cartesian coordinates as,
$$ \vec{r} = x_1\hat{e_1} + x_2 \hat{e_2} + x_3\hat{e_3} = x'_1\hat{e'_1} + x'_2\hat{e'_2} + x'_3\hat{e'_3} \,\,\ldots...eq.(3) $$ 

Then each of the new coordinates in general can be written using eq.(2),

$$ x'_i = \vec{r}\cdot\hat{e'_i} = \sum_j x_j\hat{e_j} \cdot \hat{e'_i} = \sum_j x_j cos(\theta)_{ij} \,\,\ldots...eq.(4)$$

The domain of this definition can be expanded for any general vector. 

Note: In three dimension, all indices run from 1 to 3

Accordingly in 3 dimension, we need 9 direction cosines to make the transformation from one to another namely,
$$ cos(\theta)_{ij} = \hat{e_j}\cdot\hat{e'_i} $$ where "i and j" each runs from 1 to 3 and so it gives 9 components. 

But only three of them are needed to specify the orientation. What about the others?
It so happens, there are quite few extra relations exists due to the orthogonality property of the coordinates. 

The orthogonality relations are given by, 
$$ \hat{e_i}\cdot\hat{e_j} = \delta_{ij} \\~\\ \hat{e'_i}\cdot\hat{e'_j} = \delta_{ij} \,\,\, eq.(5) $$
Expanding the unit vector in one system in terms of the unit vectors of other system using eq.(2) and making use of direction cosines, it can be rewritten as, 

$$ \sum_{j=1}^3 cos(\theta)_{i'j} cos(\theta)_{ij} = \delta_{ii'} \,\,\, eq.(6)$$  
where $\delta $ is the Kronecker-delta symbol.    

To make life simpler, we use the Einstein summation convention, and we denote the direction cosines $$ cos(\theta)_{ij} = a_{ij} $$

Now, eq.(4) becomes,
$$ x'_i = a_{ij} x_j \,\,\,\,\,\,\ldots...eq.(7) $$
where summation is assumed.
The magnitude of the vector $\vec{r} $ in both cases are the same. Using that property with the Pythagoras relation for orthogonal coordinates, we can write,

$$ x'_{i} x'_i = a_{ij} x_j a_{ik} x_j x_k $$
therefore,  $$  a_{ij} a_{ik} = \delta_{jk} \,\,\,\,\,\ldots...eq.(8)$$
where both j,k runs from 1 to 3.

This is the exact same condition obtained from orthogonality relation but in a new representation. 

Since both i and j runs from 1 to 3 in discrete sense, we can write the set of direction cosines in a general matrix form where i used for row and j for column. Let me call the matrix as A,
$$ A = \left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] $$

The general transformation from one system to other can be thought of as an Matrix operation on one system which results the coordinates of other system.

Using matrices, 
$$ \left[\begin{matrix} x'_1 \\ x'_2 \\ x'_3 \end{matrix}\right] = \left[\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right] \,\,\,\ldots...eq.(9)$$ 

Euler Angles: 

Lagrangian formalism is created based on the concept of degrees of freedom and constraints. Lagrangian itself is defined in terms of Independent coordinates. 

Finding the independent coordinates for a system in motion is essential for solving the Lagrangian. 

From orthogonality relations, we can be sure that, we don't need all the 9 direction cosines as independent coordinates. All we need is some three independent functions of these direction cosines. 

It will therefore necessary to define a set of three new independent functions to describe the orientation of rigid body in space using the Lagrangian. 

[But there is also another condition for our Matrices, which says the determinant of the matrix should be +1. Otherwise it would  be an inversion of the coordinates.]  

There are really a number of different independent functions obtained in a number of different ways to describe the Lagrangian. But Euler angles is the customary one, where we will make out transform from one to another by making three successive rotations in a specific way. 

The sequence in each step can be thought of a matrix operating on the coordinate system at that particular instant. 

Let me start the initial transformation from the $\{ \hat{e_1},\hat{e_2},\hat{e_3}\}$ coordinate system, by making angle $\phi$ counter clock wise about $\hat{e_3}$ axis. And let me denote the resultant coordinate system by, $\{\hat{e_1}^1, \hat{e_2}^1,\hat{e_3}^1\}$  

In the second stage, we make the transformation of $\{\hat{e_1}^1, \hat{e_2}^1,\hat{e_3}^1\}$ by rotating it about $\hat{e_1}^1 $ axis in the counter clock wise direction by an angle $\theta$ 
Consequently we arrive at a newer system denoted by $ \{\hat{e_1}^2, \hat{e_2}^2, \hat{e_3}^2\}$. 

Now, we make the final transformation by rotating $\{\hat{e_1}^2,\hat{e_2}^2,\hat{e_3}^2\}$ by an angle $\psi$ with respect to $\hat{e_3}^2$ axis.    

Thus, we finally arrived our desired transformation that is $$\{\hat{e_1}^3,\hat{e_2}^3,\hat{e_3}^3\} = \{\hat{e'_1},\hat{e'_2},\hat{e'_3}\} $$

In Matrix Notation, each transformation can be written as follows,

First transformation, $$ E_1 = A_1 E \,\,\,\,\ldots...eq.(10)$$

where "E" is the set of coordinate elements and "A" is the transformation Matrix.

Second transformation, $$ E_2 = A_2 E_1 \,\,\,\,\ldots...eq.(11)$$

Final transformation, $$ E_3 = A_3 E_2 = E' \,\,\,ldots...eq.(12)$$

Hence therefore, the combined transformation is denoted by the product of respective matrices,
$$ E' = A_3 A_2 A_1 E = R E \,\,\,\,\ldots...eq.(13) $$
where $ A_3 A_2 A_1 = R $

Writing each transformation in terms of its matrix element values i.e. direction cosine values, 
Since $A_1$ represents the counter clock wise rotation of "E" by an angle $\phi$ about $\hat{e_3}$ axis, it can be written in the matrix form as,

$$ A_1 = \left[ \begin{matrix} cos\phi &sin\phi &0\\ -sin\phi & cos\phi &0 \\ 0 & 0 & 1 \end{matrix}\right] \,\,\,\ldots...eq.(14) $$

Similarly, $A_2$ is the rotation of $E_1$ by an angle $\theta$ with respect to $\hat{e_1}^1$ axis, 

$$ A_2 = \left[ \begin{matrix} 1&0&0\\ 0& cos\theta &sin\theta\\ 0 & -sin\theta & cos\theta\end{matrix} \right] \,\,\,\ldots...eq.(15) $$

Finally, $A_3 $ is the rotation of $ E_2$ by angle $\psi$ with respect to $\hat{e_3}^2 $ axis,

$$ A_3 = \left[ \begin{matrix} cos\psi &\sin\psi & 0\\ -sin\psi & cos\psi & 0 \\ 0 &0&1 \end{matrix}\right] \,\,\,\ldots...eq.(16)$$

Combining together, three transformations can be written using a single matrix R as follows, 
$$ R = A_1A_2A_3 $$ 
subsituting for $ A_1, A_2, A_3 $ gives,
$$ R= \left[ \begin{matrix} cos{\psi} cos{\phi} - cos{\theta} sin{\phi} sin{\psi} & cos{\psi} sin{\phi}+ cos{\theta} cos{\phi} sin{\psi} & sin{\psi}sin{\theta}\\ -sin{\psi}cos{\phi} - cos{\theta}sin{\phi}cos{\psi} & -sin{\psi}sin{\phi}+cos{\theta}cos{\phi}cos{\psi} & cos{\psi}sin{\theta}\\ sin{\theta}sin{\phi} & -sin{\theta}cos{\phi} & cos{\theta} \end{matrix}\right] \,\, \ldots...eq.(17)$$

The inverse transformation from body fixed to space fixed coordinates is just given by the inverse of R i.e. $R^{-1}$ which is the transpose of R.

You can ask, why I choose this specific order of rotations. It needn't be. You can choose many other ways. But the only condition is, two consecutive rotations shouldn't be about the same axis. 

Quantum Mechanics - Postulates (Part -1) - Wave function, Hermitian Operators

I am not going to give the Postulates as it is in the books or anything, but I just want to postulate and speak its' mathematical importance, in a way I understood. 

From the Classical Physics of Lagrangian and Hamiltonian, we know that any system [it can be single particle or multi particle or anything] can be associated with a function so called Lagrangian or Hamiltonian, such that all the information about the system can be extracted from this function using the corresponding Equations of motion. 
Mathematically, we assume that the Lagrangian or Hamiltonian function contains all the necessary information we need to describe the system completely. 

In the same way, here we assume that "Every Quantum Mechanical System is completely described by an arbitrary State Vector or a Wave function $ \vert{\psi(t)}\rangle $ , read as "ket - psi" is an element of complex linear vector space called Hilbert Space. The State vector contains all the information about the system and it changes only with time.  

The state vector is an abstract concept and you can never measure this state vector or imagine it in a physical manner. 

From the concept of Vector space, we assume that it is always possible to define a set of vectors which are linearly independent and forms the basis for the Vector Space.  

Note: You needn't to panic on hearing the term Vector Space. Your Euclidean space follows the rules of Vector space. Whenever you get in trouble understanding vector space, you can always make a comparison with your 3 dimensional Euclidean space. 

The set of basis vectors needn't to be unique, but it is always possible to represent any vector in the Vector space as a linear combination of these basis vectors. 


As a consequence, you can imagine this arbitrary state vector as the linear combination of all the basis vectors.
We don't know what are these basis vectors, since there is many possible ways of choosing a set of basis vectors from different possible sets. Let us consider this as a general linear combination.  

It is represented as, $$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle = A_1(t) \vert{\phi_1}\rangle + A_2(t) \vert{\phi_2}\rangle + \ldots.....eq.(1)$$

All the basis vectors are ket vectors, after all left side should be equal to right side. And we can always make the time dependence of ket vector to come into the coefficients.  

We already said that, these are in abstract Hilbert space, so we cannot measure anything about them. 
To measure anything, we need to make the projection of this abstract quantities in the known space where we could describe the wave function completely. 

To measure the projection, we make the dot product of desired known parameter with this abstract Wave function. 
So that, the wave function and all its basis vectors are now described using our desired known parameter. 

For example, if the desired known parameter is position, then all the Wave function and its basis vectors will be projected into position space (where position is the parameter). And so, the new projected wave function is called "Position Space Wave function".

$$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle $$

Dotted with x to give the projection in Position space, 

$$ \langle{x}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{x}\vert{\phi_b}\rangle   \,\, \ldots...eq.(2)$$

Now, the new projection of Wave function in Position space, i.e. Position Space wave function is,
$$ \psi(x,t) = \sum_b A_b(t) \phi_b(x) $$

Where $\langle{x}\vert{\psi(t)}\rangle = \psi(x,t)$ and $ \langle{x}\vert{\phi}\rangle = \phi(x)$ 

If we choose momentum as the desired known parameter, then we can dot momentum with the general wave function. It will result into, $$ \langle{p}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{p}\vert{\phi_b}\rangle \,\,\ldots...eq.(3)$$ 
And the new wave function is called Momentum Space Wave function, $$ \psi(p,t) = \sum_b A_b(t) \phi_b(p) $$

That is all we can do with the first Postulate. 

The second Postulate is stated as, "Each dynamical variable that relates to the motion of the particle can be associated with a linear operator". 

An operator is called to be linear if it satisfies the condition, $$ \hat{Q}(c_1\psi_1 + c_2\psi_2) = c_1 \hat{Q}\psi_1 + c_2 \hat{Q}\psi_2 \,\,\,....\ldots.eq.(4)$$

With Each operator, it can be associated a linear eigen value equation such that $$ \hat{Q} \psi_i = \lambda_i \psi_i \,\,\,\,\ldots..eq.(5)$$
where $\psi_i $ is called the eigen state and 
$\lambda_i$ is called the eigen value. 

A linear operator is also an abstract concept, which is represented using a matrix. A linear operator is determined by how it acts on the basis vectors because any vector can be expanded as the linear combination of these basis vectors. 

If we know how an operator acts on the basis, then it gives us everything we need to know about the operator on that Vector Space. 

Let me represent the basis vectors as $$\vert{e_1}\rangle, \vert{e_2}\rangle, \ldots...$$

Therefore, $$ \hat{Q}\vert{\psi}\rangle = \hat{Q} \vert{e_1}\rangle + \hat{Q} \vert{e_2}\rangle + ... \,\,\ldots...eq.(6)$$


If we represent the linear operators with the matrix, knowing the matrix elements is knowing the operator itself. 

Let me take a basis vector $\vert{e_i}\rangle$ in the Hilbert Space. To understand how a linear operator works on this basis vector, we operate it on this basis and it will result some new vector. 

For example, you can consider the rotation of the coordinates as an operation that acts on the basis vectors. 

Due to linear property, $\hat{Q}\vert{e_i}\rangle$ - the new vector itself can be written again as a linear combination of the basis vectors, represented as $$ Q\vert{e_i}\rangle = \sum_k Q_{kj}\vert{e_k}\rangle $$

You can compare it with the coordinate transformation rules.   

Now, the third postulate says that, "Any observable in Quantum Mechanics is a linear Hermitian operator on the Hilbert space, where the eigenvalues are the only possible results of a precise measurement of that observable. 
Definition of a Hermitian operator:
$$ \int \psi_i^* (\hat{Q} \psi_i)\, dx = \int (\hat{Q}\psi_i)^* \psi_i \,dx \,\,\,\ldots...eq.(7)$$

Eq.(7) which gives a special property on expanding with eigenvalue eq.(5) as follows,
$$ \int \psi^* (\lambda_i \psi_i)\,dx =  \int (\lambda_i \psi_i)^* \psi_i\,dx $$
which gives, $$ \lambda_i \int \psi_i^*\psi_i \,dx = \lambda_i^* \int \psi_i^* \psi_i \,dx $$

$$(\lambda_i - \lambda_i^*) \int \psi_i^*\psi_i \,dx = 0 $$

But $\int\psi_i^*\psi_i \,dx = \int {|\psi_i|}^2 \,dx > 0 $ and it is equal to zero only when $|\psi_i| = 0 $ where wave function itself vanishes and that is not a desirable solution. 

So, the only solution is $$ \lambda_i - \lambda_i^* = 0 $$ or $$ \lambda_i = \lambda_i^* $$ It is only possible when $"\lambda_i"$ is a real number. This is a characteristic result of any Hermitian operator, which states that "The eigenvalues of an Hermitian Operator is always a real number". 
This is the reason why, eigenvalues of an operator is the only possible results on a precise measurement, because measurement should give a real number. 

There are much more things to talk about an operator and important relations like Completeness, Orthogonality, Hermiticity of an operator and etc. It should be dealt separately. 

Curvilinear Coordinate System and General expression for Gradient, Curl, Divergence and Laplacian

Curvilinear Coordinate system, which in fact is the most general coordinate system used to describe the motion of any particle. It includes all our usual systems like Cartesian, Spherical and Cylindrical coordinate systems. 

     With one to one correspondence, it is always possible to define a set of transformation rules like, $$ x_1 = x_1(u_1, u_2, u_3) \\~\\ x_2 = x_2(u_1, u_2, u_3) \\~\\ x_3 = x_3(u_1, u_2, u_3)$$

to write each of the cartesian coordinates in terms of the general coordinates. We can also define inverse transformation rules like,

$$ u_1 = u_1(x_1, x_2, x_3) \\~\\ u_2 = u_2(x_1, x_2, x_3) \\~\\ u_3 = u_3(x_1, x_2, x_3) $$

to go from one system to another. These transformations are unique, since they have one to one correspondence. 

The surfaces $ u_1 = const., u_2 = const., u_3 = const.$$ are called coordinate surfaces and the curve formed from the intersection of pair of two surfaces is called coordinate curves. The point where the tangent lines drawn to these coordinate curves intersect is chosen to be the origin of the coordinate system. 

For the sake of simplicity, we often used to deal with the coordinate systems where the coordinate surfaces intersect at right angles. They are called "Orthogonal coordinate system". 

Now, we can formulate the general rules for describing a point and its motion and to describe various vector operations in this new curvilinear coordinate system.  

But the formulation is going to be more general to apply in any system at any point. It should apply to Cartesian, Cylindrical, Spherical, Paraboloidal, Ellipsoidal and etc. 

Note: There are nearly more than 10 types of orthogonal coordinate systems we are using in mathematics. 

To start, first we will consider the example of describing small differential element in 3D Euclidean Space. $$\vec{dr} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} ..... \ldots eq.(1)$$ 

Using the chain rule, we can write the same differential element as,
$$ \vec{dr} = \frac{\partial\vec{r}}{\partial{x}} dx + \frac{\partial\vec{r}}{\partial{y}} dy + \frac{\partial\vec{r}}{\partial{z}} dz \ldots... eq.(2)$$  

Comparing eq.(1) and (2) we get that $$\frac{\partial\vec{r}}{\partial{x}} = \hat{e_x} \, , \,\frac{\partial\vec{r}}{\partial{y}} = \hat{e_y} \, , \, \frac{\partial\vec{r}}{\partial{z}} = \hat{e_z} $$   In a similar way, if the differential element is written in terms of curvilinear coordinates as $$ \vec{r} = \vec{r} (u_1, u_2, u_3)$$ Tangent vector to $u_1$ curve at some point P is, $ \frac{\partial\vec{r}}{\partial{u_1}}$ 

Therefore, the unit tangent vector in this direction given by,  $$ \frac{\partial\vec{r}/\partial{u_1}}{|\partial\vec{r}/\partial{u_1}|} = \hat{e_1} $$

We call $$ |\partial\vec{r}/\partial{u_1}| = h_1 = scaling factor $$

Since these coordinates needn't necessarily have the dimension of distance, these parameters are used to make them all to same dimension - after all we cannot add mangoes and apples together to single count. 

To complete, we write, $$\frac{\partial\vec{r}}{\partial{u_1}} = h_1 \hat{e_1} \ldots... eq.(3) \\ \frac{\partial\vec{r}}{\partial{u_2}} = h_2 \hat{e_2} \ldots... eq.(4) \\ \frac{\partial\vec{r}}{\partial{u_3}} = h_3 \hat{e_3} \ldots... eq.(5) $$


These basis vectors are tangent vectors to the curves. Similar to that, we can always form another basis whose unit vectors are normal to the coordinate surfaces, where the normal vectors are represented in terms of gradient operator $ \nabla{u_1} , \nabla{u_2}, \nabla{u_3}$ 

After normalizing, we get a new basis unit vectors represented as, $$ \hat{E_1} = \frac{\nabla{u_1}}{|\nabla{u_1}|} , \hat{E_2} = \frac{\nabla{u_2}}{|\nabla{u_2}|}, \hat{E_3} = \frac{\nabla{u_3}}{|\nabla{u_3}|} $$  It can be shown separately that, these two set of basis vectors constitute reciprocal system of vectors under coordinate transformation. It leads to the concept of Co-variant and Contra-variant vectors.

Thus, any vector can be expressed as either in terms of first set of basis vectors or in terms of second set of basis vectors. 

The square of the magnitude of the differential element in terms of first set of unit basis vectors, $$ ds^2 = \vec{dr}\cdot\vec{dr} = {h_1}^2 {du_1}^2 + {h_2}^2 {du_2}^2 + {h_3}^2 {du_3}^2 $$ Since we got the basic things we need to work, now we can start defining the general relation for operations like Gradient, Divergence, Curl and Laplacian. 

Gradient:
Gradient from the definition, $$ df = \nabla{f} \cdot \vec{dr} $$

Using the chain rule, $$ df = \frac{\partial{f}}{\partial{u_1}} du_1 + \frac{\partial{f}}{\partial{u_2}} du_2 + \frac{\partial{f}}{\partial{u_3}} du_3 \dots... eq.(6) $$  and we can also write the differential element $ \vec{dr} $ using eq. (3), (4), (5) as, $$ \vec{dr} = h_1 du_1 \hat{e_1} + h_2 du_2 \hat{e_2} + h_3 du_3 \hat{e_3} $$

Still we don't know what is the form for gradient operator, but we do know, from the definition of gradient that it would give "df" when dotted with $ \vec{dr}$ . So, 

$$ \nabla{f} \cdot \vec{dr} = \nabla_1{f} h_1 du_1 + \nabla_2{f} h_2 du_2 + \nabla_{f} h_3 du_3 \ldots... eq.(7) $$

where $ \nabla_1{f} , \nabla_2{f} , \nabla_3{f} $ are the components of Gradient operator when it is written in terms of the general basis vectors $ \hat{e_1}, \hat{e_2}, \hat{e_3} $. 
Again comparing eqs.(6) and (7), the components of the gradient operator found out to be, $$ \nabla_1{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}},\nabla_2{f} = \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}}, \nabla_3{f} = \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} $$  

Hence the general form of Gradient operator in any curvilinear orthogonal coordinate system is given by,
$$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} \,\, \ldots...eq.(8)$$  

Divergence:

Let us analyze the first term we will get, when we apply the divergence operator on any vector function $\vec{A} $,
$$ (\nabla\cdot\vec{A})_1 = \nabla \cdot (A_1\hat{e_1})  \ldots.....(9)$$

We don't know, what we will obtain when we apply the divergence operator on $ \hat{e_1} $. But, if we could write $\hat{e_1}$ in terms of some gradient operations, then there is a real possibility of obtaining the expression for Divergence with our prior knowledge of Gradient.     

According to write the unit vectors in terms of gradient relations, 

We apply the gradient operator for the functions $ u_1, u_2, u_3 $ in eq.(8) from which, we will get $$ \nabla{u_1} = \frac{\hat{e_1}}{h_1}\,, \,\nabla{u_2} = \frac{\hat{e_2}}{h_2}\, ,\, \nabla{u_3} = \frac{\hat{e_3}}{h_3} $$ 

The resultant unit vectors using gradient relations are,
$$ \hat{e_1} = h_1 \nabla{u_1} \, ,\, \hat{e_2} = h_2 \nabla{u_2} \, ,\, \hat{e_3} = h_3 \nabla{u_3} $$  

But we need to relate it with $ \hat{e_1}$, So we apply the volume relation $$ \hat{e_1} = \hat{e_2} \times \hat{e_3} = h_2 h_3 \nabla{u_2} \times \nabla{u_3} $$
Applying this in eq.(9),
$$ \nabla \cdot (A_1\hat{e_1}) = \nabla\cdot[A_1h_2h_3 \nabla{u_2} \times \nabla{u_3}] \, \, \, \ldots...eq.(10)$$
Using the vector relation, $$ \nabla\cdot {f\vec{A}} = \nabla{f}\cdot\vec{A} + f \nabla\cdot\vec{A} $$
where f- scalar function, $\vec{A} = vector function $.

Eq.(10) becomes, $$ \nabla\cdot(A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) + A_1h_2h_3 \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) \, \, \, \ldots...eq.(11)$$

But using the vector identity, $$ \nabla\cdot(\vec{A}\times\vec{B}) = \vec{B}\cdot(\nabla\times \vec{A}) - \vec{A}\cdot (\nabla\times\vec{B})$$
$$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = \nabla{u_3}\cdot(\nabla\times\nabla{u_2}) - \nabla{u_2}\cdot(\nabla\times\nabla{u_3})$$

But, Curl of gradient is always zero for any scalar function, which implies $$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = 0 $$  

Eq.(11) gives, $$\nabla \cdot (A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) \dots..eq.(12) $$  

Again writing $ \nabla{u_2}\times\nabla{u_3} $ in terms of basis vectors that is $$ \nabla{u_2}\times\nabla{u_3} = \frac{\hat{e_2}\times\hat{e_3}}{h_2h_3} = \frac{\hat{e_1}}{h_2h_3} $$  

Eq.(12) results into $$\nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3} \cdot \nabla(A_1h_2h_3) $$
Using our prior knowledge of Gradient, it can be expanded as,
 $$ \nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3}\cdot\left[ \frac{\hat{e_1}}{h_1} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} + \frac{\hat{e_2}}{h_2} \frac{\partial(A_1h_2h_3)}{\partial{u_2}} + \frac{\hat{e_3}}{h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_3}}\right] $$  While, we are dealing with orthogonal basis, dot product between any two different basis gives zero and dot product of same vector gives unity. 

Making using of the orthonormality, we finally arrive at the result,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{1}{h_1h_2h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} $$

Similar procedure gives the expression for other coordinates. 

The final expression for the divergence operator in general curvilinear coordinates is,

$$\nabla\cdot\vec{A} = \frac{1}{h_1h_2h_3}\left[ \frac{\partial{A_1h_2h_3}}{\partial{u_1}} + \frac{\partial{A_1h_2h_3}}{\partial{u_2}} + \frac{\partial{A_1h_2h_3}}{\partial{u_3}}\right] \ldots...eq.(13)$$  

Curl:

As the same, first we will take single component, write it in terms of gradient and apply the curl,

$$ \nabla \times (A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) \, \, \ldots...eq.(14)$$
Since there is already a curl operator, we don't need to use volume relation but we could just simply write $ \hat{e_1} $ in terms of its own gradient relation i.e. $ \hat{e_1} = h_1 \nabla{u_1} $

Using the vector identity, $$ \nabla \times (f\vec{A}) = f \nabla\times\vec{A} + \nabla{f}\cdot\vec{A} $$

Eq.(14) gives, $$ \nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} + A_1h_1\nabla \times \nabla{u_1} $$
but curl of gradient is zero.

So, Eq.(14) becomes, $$\nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} $$  

With the help of eq.(8) we can rewrite the above into,
$$ \nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \nabla{u_1} $$

Again using the relation, $ \nabla{u_1} = \frac{\hat{e_1}}{h_1} $

$$\nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \frac{\hat{e_1}}{h_1} $$


Using the cross product rule for the positive volume element,  we finally get, 

$$\nabla \times (A_1\hat{e_1}) = \frac{1}{h_1h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{-e_3} + \frac{1}{h_1h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_2} \, \ldots... eq.(15)$$ 

Similar procedure gives for other components the following result,
$$\nabla \times (A_2\hat{e_2}) = \frac{1}{h_1h_2} \frac{\partial{A_2h_2}}{\partial{u_1}} \hat{e_3} + \frac{1}{h_2h_3} \frac{\partial{A_2h_2}}{\partial{u_3}} \hat{-e_1} \, \ldots... eq.(16)$$

$$\nabla \times (A_3\hat{e_3}) = \frac{1}{h_1h_3} \frac{\partial{A_3h_3}}{\partial{u_1}} \hat{-e_2} + \frac{1}{h_2h_3} \frac{\partial{A_3h_3}}{\partial{u_2}} \hat{e_1} \, \ldots... eq.(17)$$

Combining the components from eq.(15),(16),(17) we get the General expression for the Curl operator in Curvilinear coordinates as,

$$ \nabla\times\vec{A} = \frac{1}{h_2h_3} \left[\frac{\partial(A_3h_3)}{\partial{u_2}} - \frac{\partial(A_2h_2)}{\partial{u_3}}\right] \hat{e_1} + \\~\\ \frac{1}{h_1h_3} \left[\frac{\partial(A_1h_1)}{\partial{u_3}} - \frac{\partial(A_3h_3)}{\partial{u_1}}\right] \hat{e_2} + \\~\\ \frac{1}{h_1h_2} \left[\frac{\partial(A_2h_2)}{\partial{u_1}} - \frac{\partial(A_1h_1)}{\partial{u_2}}\right] \hat{e_3} \, \, \, \ldots...eq.(18)$$ 

Or simply we can write this in Matrix form as,

$$ \nabla \times \vec{A} = \frac{1}{h_1h_2h_3}\begin{vmatrix} h_1\,\hat{e_1} & h_2 \,\hat{e_2} & h_3 \, \hat{e_3} \\ \frac{\partial}{\partial{u_1}} & \frac{\partial}{\partial{u_2}} & \frac{\partial}{\partial{u_3}} \\ h_1\vec{A_1} & h_2 \vec{A_2} & h_3 \vec{A_3} \end{vmatrix} \,\,\, \dots...eq.(19) $$

Laplacian:

Unlike others, we don't need to find anything extra for Laplacian since it is just the combination of gradient and divergence. 

Let us take a scalar function "f" and write its gradient from eq.(8), $$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} $$
Now, applying the Divergence operation for the resultant outcome, we get the General expression for Laplacian in curvilinear coordinates, 

$$ \nabla^2f = \frac{1}{h_1h_2h_3} \left[ \frac{\partial\left(\frac{h_2h_3}{h_1}\frac{\partial{f}}{\partial{u_1}}\right)}{\partial{u_1}} + \frac{\partial\left(\frac{h_3h_1}{h_2}\frac{\partial{f}}{\partial{u_2}}\right)}{\partial{u_2}} + \frac{\partial\left(\frac{h_1h_2}{h_3}\frac{\partial{f}}{\partial{u_3}}\right)}{\partial{u_3}} \right] \ldots...eq.(20)$$  




That is all we need to derive. We need to remember that, these derivations are done for general orthogonal curvilinear coordinate system. 
For the most general curvilinear coordinate system (i.e. which are not orthogonal), we will need Tensors and its analysis.