Appendix B

Some Supplementary Topics

B.1  FOURIER TRANSFORM

B.1.1  Fourier Series

Let us consider a periodic single valued function f(x) in the interval –π ≤ x ≤ π [such that f(x + 2π) = f(x)]. The Fourier series corresponding to f(x) is defined as

This series converges in the said interval (–π, π), if f(x) and f′(x) [= (∂f /∂x)] are continuous. Multiplying Eq. (B.1) by cos(mx) and integrating over –π to π, we get

Again multiplying (B.1) by sin (mx) and integrating over −π to π, we get

The series (B.1) may also be written in an alternative form of exponentials as cos (nx) and sin (nx) may be written in terms of e^±inx,

Here, factor has been extracted from the coefficient C_n, as we shall see later, only for symmetry appearance. Multiplying Eq. (B.4) by and integrating over x, we get

Now for n = m, we have

and for n≠m,

Therefore, we may write

where

 

is the Kronecker delta function.

If, instead, a function f(x) is defined in the interval (–L, L) with periodicity f(x + 2L) = f(x), it can be expanded in a Fourier series with the change of variable x ⇒ (πx/L). Then Eq. (B.4) gives

with coefficients C_n given as

B.1.2  Fourier Transform

Quite often, we come across functions f(x) which are not periodic and are defined for all real values of x, –∞ < x < ∞. Such functions may also be expressed in terms of exponentials by simply taking the limit L → ∞ in Eqs (B.10) and (B.11). For L → ∞, the successive terms in Eq. (B.10) differ by a very small amount and therefore the sum over n may be replaced by integration and we get

Let us introduce a new variable k

Then integral (B.12) may be written as

where new function g(k) is defined as

Putting C_n from Eq. (B.11), we get for g(k)

or

Equations (B.14) and (B.16) are called the Fourier transform pair; g(k) is called the Fourier transform of f(x) and function f(x) is called Fourier (inverse) transform of g(k).

The Fourier transforms (B.14) and (B.16) may be generalized to three dimensional cases, as

and

Equations (B.17) and (B.18) may be written in vector notations as

and

The Fourier transforms may also be written for a time-dependent function f(t), as

and

B.2  DIRAC DELTA FUNCTION

Let us put expression of g(k) from Eq. (B.16) in Eq. (B.14). We get

or

where δ(x – x′) is defined as

The function δ(x – x′) was introduced by Dirac and is known as Dirac delta function. Equation (B.23) may be satisfied for a particular choice of δ(x –x′);

 

However, this is not the general form of delta function δ(x – x′). Let us analyze the general form starting with its definition (B.24). The integral (B.24) as such does not converge. However, we may replace it by the limiting integral (taking x′ = 0)

The R.H.S. of Eq. (B.27) with finite l is

Now

and

The integral of the function δ_l(x) over the whole range is unity (independent of value of l). For l → ∞ value of δ_l(x) is very large at x = 0, its integral is unity and this function (in the said limit) is negligible everywhere except at x = 0.

The representation of delta function by Eq. (B.26) is not unique. Various useful representations of δ(x) may be obtained as:

Some properties of δ-functions:

B.3  BLOCH THEOREM

If we consider an electron in a one-dimensional potential box of width L, under periodic boundary condition ψ(x + L) = ψ(x), as we saw in Section 5.5, the electron wave functions which are the solutions of one-dimensional Schrodinger equation.

are given as

 

Here the values of the wave vectors, as dictated by the periodic boundary condition, are given as

Next let us consider an electron in a one-dimensional box of width L and having periodic potential V(x) with periodicity a, that is

 

This case really corresponds to the case of an electron put in a one-dimensional periodic lattice with lattice constant a and with its total length L = Na; N being the number of lattice sites or number of unit cells. To find out wave functions of electron in this case, we shall have to solve Schrodinger equation

with periodic potential

 

and with periodic boundary conditions

 

Now we know that if the potential V(x) were constant, the wave function of the electron would have been,

 

In presence of the periodic potential, we may expect the wave function to be of similar type [Eq. (B.33)] but simply modulated by the periodic field. Now with the use of periodic boundary conditions, the end effects or the surface effects are not there, so we should expect the modulating function to have the period of the lattice. In other words, we expect the wave function to have the form like

 

where the modulating function has periodicity of the lattice, that is

 

It was Block who in 1928 proved that the wave functions of an electron in a periodic field are really those given by Eq. (B.39) along with Eq. (B.40). This is known as Bloch theorem in solid state physics, though mathematically the same was known much earlier in the form of Floquet theorem. We shall now discuss an elementary proof of Bloch theorem.

Equation (B.36) may be written as

 

where

Let us consider a translation operator whose role is to change x to x + a, for example

Then, we have

or

So, operator commutes with Ĥ, and thus both the operators have simultaneous eigenfunctions. Therefore, we may write

where λ is a constant.

Now

 

and, therefore,

 

Hence

 

 

or

 

So the wave function must satisfy the condition

 

Now we know

Therefore, the functions

 

 

satisfy Eq. (B.47), only if

 

B.4  THE VARIATIONAL METHOD

Let us consider a system, which may have one or more particles, described by the time-independent Hamiltonian Ĥ. Let us denote by E_n and ψ_n the set of bound state energies and corresponding normalized eigenfunctions. Then

and

In particular, the ground state energy is given by

Let us consider an arbitrary single valued and square integrable function ϕ of position coordinates of the system, which is normalized, so that

Then the functional E[ϕ] defined as

is greater than or equal to the ground state energy E₀ of the system. This is the variational principle, which is, in fact, the basis of the variational method. The variational method is quite helpful in case it is not possible to solve Schrodinger equation (exactly) to get exact results of ground state wave function and corresponding energy.

Firstly, let us prove the variational principle: we can expand the function ϕ in terms of the complete, orthonormal set of eigenfunctions {ψ_n} as

Due to normalization of ψ_n and ϕ, we have

Putting the expansion (B.53) in (B.52), we get

Now, subtracting ground state energy E₀ from both sides,

In above equation, the quantities (E_n – E₀) and |a_n|² are positive or zero, so we have

 

So, the function E[ϕ] gives an upper bound for the ground state energy. In fact, same results hold good even if ϕ is not normalized to unity. The property [Eq. (B.57)] forms the bases of Rayleigh–Ritz variational method for the approximate calculation of E₀. The variational method consists of finding the quantity E[ϕ] by using trial function ϕ which depends on a certain number of variational parameters. The function E[ϕ] thus becomes a function of these variational parameters. E[ϕ] is then minimized to obtain the best approximation of E₀; the minimized value, in fact, depends upon the form chosen for ϕ. A good choice of the trial function is one which may contain as many features of the exact wave function ψ₀ as possible. Symmetry arguments may provide some features for trial function ϕ. Wherever it is difficult to assess ϕ with required features, several different forms of ϕ may be tried; the one which gives the lowest value of E [ϕ] may be taken to be the best one.

Example 1 Ground State of Harmonic Oscillator

Even though Schrodinger equation of a one-dimensional harmonic oscillator may be solved exactly for eigenvalues E_n and eigenfunctions ψ_n(x), let us try to use variational method to estimate the ground state energy. We have the Hamiltonian operator Ĥ of a harmonic oscillator

Here, the potential function is symmetric about the origin, so the ground state wave function is expected to be symmetric about the origin. In addition, the wave function should not have any nodes except at x = ±∞. With these features, the simplest function may be like

 

where α is a positive constant and works as a variational parameter.

Now, we have

and we may get

(using results of Appendix C.6 for definite integrals). Therefore, we get

Minimizing E(α) gives

which gives

Putting this α in Eq. (B. 62), we get minimum value of energy

We are able to find the ground state energy and the wave function exactly same as given by the (exact) solution of the Schrodinger equation. This could be possible fortunately with the right choice of the trial function ϕ. This type of correct guess happens only rarely.

Example 2 The Ground State of Helium Atom

Helium atom has two electrons which interact with each other. It is not possible to solve the two-electron Schrodinger equation analytically because of the presence of the interaction term in the potential energy function. Therefore, it would be interesting to use variational method to find out ground state energy and wave function of Helium atom. The Hamiltonian of the helium atom may be written as

If we ignore the Coulomb interaction (e²4π∊₀|r₁ – r₂|) between electrons, in Eq. (B.66), the remaining Hamiltonian may be written as the sum of two hydrogenic Hamiltonians. In this situation, the ground state wave function of each electron may be written like e^–2(r/a₀); therefore, the combined wave function of two electrons, which is the product of individual ground state wave functions, may be written as e^{–2(r₁ + r₂)/a₀}.

When interaction between the electrons is taken into account, let us start with a trial wave function

 

where α is a variational parameter, which is assumed to be real and positive. Value of α is obtained by minimizing E(α) given by

Let us evaluate the denominator

We note that the trial wave function ϕ(r₁, r₂) does not depend on angles θ and ϕ, therefore operator ∇² in Ĥ should be

So, we have

Thus, we can easily get

Now

As shown in Appendix C.7, we have

So, we get for E(α)

or

gives

Putting this α in Eq. (B. 69), the minimum value of E is

The experimental value of the ground state energy, that is the minimum energy required to remove both the electrons from helium atom is 79.01 eV. So the calculated value is differing only by about 2 per cent.

EXERCISES

Exercise B. 1

Find the Fourier transform of the function f(x) given by

Exercise B.2

Find the Fourier transform of the Gaussian distribution function

 

Exercise B.3

Show that the three-dimensional Fourier (exponential) transform of a radially symmetric function f(r) may be written as a Fourier (sine) transform

Exercise B.4

If f(r) = (1/4πr), show that its Fourier transform is

SOLUTIONS

Solution B.1

Using Eq. (B. 16), we have

We may note here [see Fig. B. 1] that function f(x) has width ∆x = 2a and its Fourier transform has width ∆k ≈ (π/a); ∆x ∆k ≈ π

or 

 

Figure B.1 The rectangle function f (x) given by Eq. (B.79) and its Fourier transform g(k) given by Eq. (B.81).

Solution B.2

Using Eq. (B.16)

where we have used the result

and the fact that second integral is zero as it is the integral of an odd function from –∞ to ∞.

So, the Fourier transform of a Gaussian function is a Gaussian [see Fig. B.2], We note

and

Therefore, ∆x ∆k ≈ 1, a general characteristic of Fourier transform pair.

Figure B.2 The Gaussian function f (x) given by Eq. (B.80) and its Fourier transform g(k) given by Eq. (B.82).

Solution B.3

Solution B.4

Using the result of above exercise