21. MHD: Some Important Equations of Electrodynamics (1)

In most of our cases, we are concerned with non-magnetic and conducting materials. We assume that the material properties, such as conductivity, σ, are spatially uniform and incompressible. In this chapter, we are going to discuss about the electric force and the Lorentz force.

Generally, a particle moving with velocity u and carrying electric charge q, is subjected to electromagnetic forces and governed by the following equation:

$\mathbf{f}=q{{\mathbf{E}}_{s}}+q{{\mathbf{E}}_{i}}+q\mathbf{u}\times \mathbf{B}$                                              (21.1)

Here, E_s is the electrostatic field and the first term is the electrostatic force or Coulomb force which arises from the mutual attraction or repulsion of electric charges. The second term is the force which arises in the presence of time-varying magnetic field, since E_i is the electric field induced by the varying magnetic field. Due to the motion of the charge in a magnetic field, Lorentz force arises, and this is given by the third term $q\mathbf{u}\times \mathbf{B}$.

Coulomb’s law says that electrostatic field E_s is irrotational, and Gauss’s law fixes the divergence of E_s. These laws are respectively given by:

$\nabla \cdot {{\mathbf{E}}_{s}}={}^{{{\rho }_{e}}}/{}_{{{\varepsilon }_{0}}}$                                                     (21.2)

$\nabla \times {{\mathbf{E}}_{s}}=0$                                                        (21.3)

ρ_e is the total charge density (free plus bound charges) and ε₀ is permittivity of free space. Introducing electrostatic potential V, defined by ${{\mathbf{E}}_{s}}=-\nabla V$, equation (21.2) becomes:

${{\nabla }^{2}}V=-{}^{{{\rho }_{e}}}/{}_{{{\varepsilon }_{0}}}$                                                   (21.4)

On the other hand, induced electric field E_i has zero divergence, and finite rotational component. This is governed by Faraday’s law:

$\nabla \cdot {{\mathbf{E}}_{i}}=0$                                                      (21.5)

$\nabla \times {{\mathbf{E}}_{i}}=-\frac{\partial \mathbf{B}}{\partial t}$                                                      (21.6)

Conveniently, we define the total electric field as: E = E_s + E_i and so we have:

Gauss’s law: $\nabla \cdot \mathbf{E}={}^{{{\rho }_{e}}}/{}_{{{\varepsilon }_{0}}}$                                            (21.7)

Faraday’s law: $\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$                                              (21.8)

Electrostatic and Lorentz force: $\mathbf{f}=q(\mathbf{E}+\mathbf{u}\times \mathbf{B})$                           (21.9)

Equations (21.7) and (21.8), i.e. Gauss’s and Faraday’s law, determine uniquely the electric field since the divergence and curl of the field are known and the boundary conditions are specified. Equation (21.9) defines the electric field E and magnetic field B, where E is the force per unit charge on a test charge at rest in the observer’s frame. If the charge is moving, the additional force $q\mathbf{u}\times \mathbf{B}$ appears which is used to define B. However, Newtonian relativity which is all that is required for MHD tells us that the electric force relative to a moving frame f_r is equal to the electric force due to electric field at rest f, i.e. f_r = f.

QM Solved Problems #1

1.      A particle moves along the x-axis in a potential field:

V(x) = 0, |x| < a

V(x) = ∞, |x| > a

(I.e. a box with closed, isolated walls at x = ± a).

a. Calculate the eigen-values of energy and define the corresponding normalized eigen-functions.

b. At t = 0, the particle is in interval |x| < c, (c < a). Inside the interval, all positions of the particle have the same chances. Assume that the corresponding wave function is real and positive. Please find the chance at energy measurement of finding the particle at:

i.   the ground state;

ii.  the first excited state;         

c. Assume that at the energy measurement, the ground state was found and then, we suddenly remove the walls of the box to x = ± b (b > a). What is then the probability of finding the particle at:

            i.   the ground state?

            ii.  the first excited state?

           

            Solution:

                       

 

16. Formalism of Quantum Mechanics: Linear Transformations

A Linear transformation ($\hat{T}$) is an operation in a vector space that takes each vector and ‘transforms’ it into another ‘form’ of vector, i.e. $|\alpha >\to |\alpha '>=\hat{T}|\alpha >$. One of the important properties of this operation is that, this operation is linear for any vector: $|\alpha >,|\beta >$, and any scalars a, b:

$\hat{T}(a|\alpha >+b|\beta >)=a(\hat{T}|\alpha >)+b(\hat{T}|\beta >)$                                  (16.1)

For instance, what a linear transformation $\hat{T}$ does to a set of basis vectors $|{{e}_{1}}>,|{{e}_{2}}>,...,|{{e}_{n}}>$ is as follows:

$\hat{T}|{{e}_{1}}>={{T}_{11}}|{{e}_{1}}>+{{T}_{21}}|{{e}_{2}}>+...+{{T}_{n1}}|{{e}_{n}}>$

$\hat{T}|{{e}_{2}}>={{T}_{12}}|{{e}_{1}}>+{{T}_{22}}|{{e}_{2}}>+...+{{T}_{n2}}|{{e}_{n}}>$

…

$\hat{T}|{{e}_{n}}>={{T}_{1n}}|{{e}_{1}}>+{{T}_{2n}}|{{e}_{2}}>+...+{{T}_{nn}}|{{e}_{n}}>$

Or,

$\hat{T}|{{e}_{j}}>=\sum\limits_{i=1}^{n}{{{T}_{ij}}|{{e}_{ij}}>}$ For $j=1,2,...,n$                                  (16.2)

For an arbitrary vector:

$|\alpha >={{a}_{1}}|{{e}_{1}}>+{{a}_{2}}|{{e}_{2}}>+...+{{a}_{n}}|{{e}_{n}}>=\sum\limits_{j=1}^{n}{{{a}_{j}}|{{e}_{j}}>}$                        (16.3)

The linear transformation is:

$\hat{T}|\alpha >=\sum\limits_{j=1}^{{}}{{{a}_{j}}(\hat{T}|{{e}_{j}}>)=\sum\limits_{j=1}^{n}{\sum\limits_{i=1}^{n}{{{a}_{j}}{{T}_{ij}}|{{e}_{i}}>=\sum\limits_{i=1}^{n}{\left( \sum\limits_{j=1}^{n}{{{T}_{ij}}{{a}_{j}}} \right)|{{e}_{i}}>}}}}$                   
(16.4)

Evidently, the linear transformation $\hat{T}$ transforms a vector with components ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}$ into a vector with components $a_{i}^{'}$ :

$a_{i}^{'}=\sum\limits_{j=1}^{n}{{{T}_{ij}}{{a}_{j}}}$                                                       (16.5)

In the same fashion as n components are uniquely characterize vector $|\alpha >$, there are n² elements T_ij those are uniquely characterize linear transformation $\hat{T}$ with respect to the same given basis vector:

$\hat{T}\leftrightarrow ({{T}_{11}},{{T}_{12}},...,{{T}_{nn}})$                                                 (16.6)

When the basis vector is orthonormal, each element T_ij becomes:

${{T}_{ij}}=<{{e}_{i}}|\hat{T}|{{e}_{j}}>$                                                     (16.7)

In the form of a matrix the linear transformation can be expressed as follows:

$T = \begin{pmatrix}
T_{11} & T_{12} & ... & T_{1n} \\
T_{21} & T_{22} & ... & T_{2n} \\
... \\
T_{n1} & T_{n2} & ... & T_{nn}
\end{pmatrix}$

(16.8)

The sum of two linear transformations is defined in the same way as the usual rule for adding two matrices, as follows:

$\left( \hat{S}+\hat{T} \right)|\alpha >=\hat{S}|\alpha >+\hat{T}|\alpha >$                                          (16.9)

i.e. by adding their corresponding matrix elements:

U = S + T ↔ U_ij = S_ij + T_ij                                           (16.10)

And the product of two linear transformations $\left( \hat{S}\hat{T} \right)$ is the net effect of performing the multiplication in succession: first $\hat{T}$ and then$\hat{S}$:

$|\alpha >\to |\alpha '>=\hat{T}|\alpha >\to |\alpha ''>=\hat{S}|\alpha '>=\hat{S}(\hat{T}|\alpha >)=\hat{S}\hat{T}|\alpha >$                  (16.11)

Hence, the matrix U, that represents the combined linear transformation $\hat{U}=\hat{S}\hat{T}$ can be given:

${{{a}''}_{i}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}{{{{a}'}}_{j}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}\left( \sum\limits_{k=1}^{n}{{{T}_{jk}}{{a}_{k}}} \right)=\sum\limits_{k=1}^{n}{\left( \sum\limits_{j=1}^{n}{{{S}_{ij}}{{T}_{jk}}} \right){{a}_{k}}=\sum\limits_{k=1}^{n}{{{U}_{ik}}{{a}_{k}}}}}}$                  
(16.12)

Then,

U = ST $\Leftrightarrow {{U}_{ik}}=\sum\limits_{j=1}^{n}{{{S}_{ij}}{{T}_{jk}}}$                                           (16.13)

This is the standard rule for matrix multiplication, i.e. the corresponding ik^th element of the product is found by multiplying the i^th row of S and the k^th column of T then add.

The n-tuple components of $|\alpha >$ can be written as $n\times 1$ column matrix:

a $ = \begin{pmatrix}
a_{1} \\
a_{2} \\
... \\
a_{n}
\end{pmatrix}$

(16.14)

and the transformation rule in matrix form can be written:

a’ = Ta                                                               (16.15)

Another useful terminology in the theory of matrix is the transpose of a matrix: T^T is the same set of elements, but with rows and columns interchanged:

T^T $ = \begin{pmatrix}
T_{11} & T_{21} & ... & T_{n1} \\
T_{12} & T_{22} & ... & T_{n2} \\
... \\
T_{1n} & T_{2n} & ... & T_{nn}
\end{pmatrix}$

 (16.16)

And the transpose of column matrix (16.14) is a row matrix a^T :

a^T = $\left( {{a}_{1}}{{a}_{2}}...{{a}_{n}} \right)$                                                        (16.17)

A square matrix is symmetric (T^T = T) when the matrix is equal to its transpose and anti-symmetric (T^T = -T) when its transpose is reversed the sign.

The complex conjugate T* of a matrix is constructed by taking the complex conjugate of each element of the matrix:

\begin{pmatrix}
T^{*}_{11} & T^{*}_{12} & ... & T^{*}_{1n} \\
T^{*}_{21} & T^{*}_{22} & ... & T^{*}_{2n} \\
... \\
T^{*}_{n1} & T^{*}_{n2} & ... & T^{*}_{nn}
\end{pmatrix}

                                                      (16.18)

And also for column matrix:

a* $ = \begin{pmatrix}
a^{*}_{1} \\
a^{*}_{2} \\
... \\
a^{*}_{n}
\end{pmatrix}$

(16.19)

A matrix is real when the conjugate is equal to the matrix itself: T* = T and imaginary when: T* = -T.

The Hermitian conjugate of a matrix is the transposed conjugate of a matrix, which is indicated by a dagger: T^†

$\begin{pmatrix}       T^{*}_{11} & T^{*}_{21} & ... & T^{*}_{n1} \\
       T^{*}_{12} & T^{*}_{22} & ... & T^{*}_{n2} \\
       ... \\
       T^{*}_{1n} & T^{*}_{2n} & ... & T^{*}_{nn}
       \end{pmatrix}$

(16.20)

And the row matrix:

a^† = $\left( {{a}^{*}_{1}}{{a}^{*}_{2}}...{{a}^{*}_{n}} \right)$                                          (16.21)

A square matrix is Hermitian if it is equal to its Hermitian conjugate: T^† = T and skew Hermitian if: T^† = -T. With this notation, the inner product can be expressed in matrix form as follows:

$<\alpha |\beta >=$a^†b                                                     (16.22)

The commutator is defined as the difference between two ordering of matrix multiplication, since matrix multiplication is not commutative (ST ≠ TS):

[S,T] ≡ ST – TS                                                  (16.23)

The transpose of a product is the product of transposes in reverse order:

(ST)^T = T^TS^T                                                      (16.24)

And for the Hermitian conjugates:

(ST)^† = T^†S^†                                                      (16.25)

An inverse of a matrix T^-1 is defined:

T^-1T = TT^-1 = 1                                                      (16.26)

Here, 1 is a unit matrix, i.e. a matrix consists of ones on the main diagonal and zeroes everywhere else. A unit matrix represents every vector into itself. In other words:

1_ij = δ_ij                                                           (16.27)

The inverse of a matrix exists if and only if the determinant is non-zero:

T^-1 = (1/det T) C^C                                                  (16.28)

C^C is the cofactor of the matrix, the cofactor of the element T_ij can be calculated by multiplying the determinant of the sub-matrix obtained from T by erasing the i^th-row and j^th-column. A matrix without an inverse is said to be singular. The inverse of a product is the product of the inverses in reverse order:

(ST)^-1 = T^-1S^-1                                                    (16.29)

A matrix is said to be unitary if its inverse is equal to its Hermitian conjugate:

U^† = U^-1                                                         (16.30)

If the basis is orthonormal, the columns and the rows of unitary matrix constitute also an orthonormal set.

The vector components depend on the choice of basis as do the elements in the matrix representing a linear transformation. When we switch to a different basis, the components will also change. The old basis vectors are linear combinations of the new ones:

$|{{e}_{j}}>=\sum\limits_{i=1}^{n}{{{S}_{ij}}|{{f}_{i}}>}$  for $j=1,2,3,...,n$                               (16.31)

And the components with respect to new basis:

$a_{i}^{f}=\sum\limits_{j=1}^{n}{{{S}_{ij}}a_{j}^{e}}$                                                    (16.32)

Here, the superscripts indicate the basis vectors. This equation can be also expressed in matrix form:

a^f = Sa^e                                                          (16.33)

The matrix representing a given linear transformation is also modified by the change of basis. In the old basis we had:

a^e’ = T^ea^e                                                          (16.34)

Since a^e = S^-1a^f, by multiplying both sides with S^-1, we obtain:

a^f’ = Sa^e’ = S(T^ea^e) = ST^eS^-1a^f                                         (16.35)

Hence,

T^f = ST^eS^-1                                                     (16.36)

Generally, two matrices T₁ and T₂ are similar, if : T₂ = ST₁S^-1, given that the matrix S is non-singular. If the first basis is orthonormal, the second will also be orthonormal if and only if the matrix S is unitary.

While the elements of a matrix change in the new basis, the determinant and trace of the matrix are unchanged. For the determinant of a product is the product of the determinants:

det(T^f) = det(ST^eS^-1) = det(S)det(T^e)det(S^-1) = det(T^e)                           (16.37)

And the trace is the sum of the diagonal elements:

Tr(T) = $\sum\limits_{i=1}^{m}{{{T}_{ii}}}$                                                 (16.38)

With the property:

Tr(T₁T₂) = Tr(T₂T₁)                                             (16.39)

Hence,

Tr(T^f) = Tr(ST^eS^-1) = Tr(T^e)                                        (16.40)