Angular Momentum Operator in spherical cooridnate system - derivation

Classical definition of angular momentum is given by, $$\vec{L} =\vec{r} \times \vec{p} \,\,\,...eq.(1)$$
Writing the components in terms of Cartesian coordinates we get,
$$L_x = yp_z - zp_y \\~\\ L_y =  zp_x -  xp_z \\~\\ L_z = xp_y - yp_x$$

In quantum mechanics, the momentum operator in position basis for x component is given by $ - \hbar \frac{\partial}{\partial{x}} $ using this, we rewrite the above equations as,   $$L_x = -i\hbar ( y\frac{\partial}{\partial{z}} - z\frac{\partial}{\partial{y}}) \\~\\ L_y = -i\hbar ( z\frac{\partial}{\partial{x}} - x\frac{\partial}{\partial{z}}) \\~\\ L_z = -i\hbar ( x\frac{\partial}{\partial{y}} - y\frac{\partial}{\partial{x}})$$

But, spherical symmetric potentials are often useful in quantum mechanics where we might need to express the equations in spherical coordinate system for simplicity. So, we do need to make the coordinate transformation rules to these equations to write this in terms of spherical polar coordinate system.

From our previous post of Cartesian to Spherical Coordinate transformation, we know that $$dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi$$
$$dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi$$  
$$dz = cos\theta \, dr - r\, sin\theta \, d\theta$$
where $$x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\,$$
Solving for $ dr, \,d\theta\,, d\phi $ as follows,
first we solve for $$d\phi = -\frac{1}{r}\frac{sin\phi}{sin\theta} dx + \frac{1}{r}\frac{cos\phi}{sin\theta} dy$$ Now we solve for $$dr = sin\theta\,cos\phi\, dx + sin\theta\, sin\phi\,dy + cos\theta\, dz$$ and finally we get, $$d\theta = \frac{1}{r} cos\theta\, cos\phi\,dx + \frac{1}{r} cos\theta\,sin\phi \, dy - \frac{1}{r}sin\theta\, dz$$
Using the above three equations with chain rule,
$$\frac{\partial}{\partial{x}} = \frac{\partial}{\partial{r}}\frac{\partial{r}}{\partial{x}} +\frac{\partial}{\partial{\theta}}\frac{\partial{\theta}}{\partial{x}} +\frac{\partial}{\partial{\phi}}\frac{\partial{\phi}}{\partial{x}}$$  
or $$\frac{\partial}{\partial{x}} = sin\theta\,cos\phi\, \frac{\partial}{\partial{r}} + \frac{1}{r} cos\theta \,cos\phi\, \frac{\partial}{\partial{\theta}} - \frac{1}{r}\frac{sin\phi}{\sin\theta} \frac{\partial}{\partial{\phi}}$$  Similary for others, $$\frac{\partial}{\partial{y}} = sin\theta\,sin\phi\, \frac{\partial}{\partial{r}} + \frac{1}{r} cos\theta \,sin\phi\, \frac{\partial}{\partial{\theta}} + \frac{1}{r}\frac{cos\phi}{\sin\theta} \frac{\partial}{\partial{\phi}}$$ $$\frac{\partial}{\partial{z}} = cos\theta\,\frac{\partial}{\partial{r}} - \frac{1}{r} sin\theta \, \frac{\partial}{\partial{\theta}}$$
Applying our equations in our definition of angular momentum, the angular momentum operator in terms of spherical coordinate system can be written as,
$$L_x = i\hbar \left( sin\phi \,\frac{\partial}{\partial{\theta}} + cot\theta\,cos\phi\, \frac{\partial}{\partial{\phi}}\right) \\~\\ L_y = i\hbar \left( -cos\phi\, \frac{\partial}{\partial{\theta}} + cot\theta\, sin\phi\,  \frac{\partial}{\partial{\phi}} \right) \\~\\ L_z = -i\hbar \frac{\partial}{\partial{\phi}}$$ and $$\vec{L}^2 = L_x^2 + L_y^2 +L_z^2 = -\hbar^2 \left[ \frac{1}{sin\theta} \frac{\partial}{\partial{\theta}}\left(sin\theta\frac{\partial}{\partial{\theta}}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial{\phi^2}}\right]$$  

We will be often using this expression in quantum mechanics when we deal hydrogen atom, spherically symmetrical potential problem, etc. 

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, $$\frac{\partial{F}}{\partial{p_i}} = -q_i \\~\\ \frac{\partial{F}}{\partial{P_i}} = Q_i  \\~\\ H' = H + \frac{\partial{F}}{\partial{t}}$$