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.