Solution of the Schrödinger equation for the Kronig-Penney potential

We assume,$$E<V_0$$

For  $$0\leq x\leq a$$

$$ψ=A sin⁡αx+B cos⁡αx$$ 

  $$\frac{d\psi}{dx}=A\alpha\;\sin\alpha x+B\alpha\;\cos\alpha x$$

where, $$\alpha=\frac{\surd(2mE)}ℏ$$

For $$-b≤x≤0$$

$$\frac{-\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+(E-v_0)\psi=0$$

Solution is periodic, barrier of limited thickness- solution doesn’t continue decay to zero. There is tunneling between wells.

$$\psi=C\;\sin h(\gamma x)+D\;\cos h(\gamma x)$$

$$\frac{d\psi}{dx}=C\gamma\;\sin h(\gamma x)+D\gamma\;\cos h(\gamma x)$$

$$\gamma=\frac{\surd(2m(V_0-E))}{\hbar}$$

ψ must be contuinues at x=0, and B=D

$$Also,\;\;\frac{d\psi}{dx}\;\;must\;be\;continuous\;at\;x=0,\;so\;A\backslash alpha=\backslash gamma\;C$$

Since the Kronig-Penney potential exhibits translational symmetry, the energy eigenfunctions of the Schrödinger equation will simultaneously be eigenfunctions of the translation operator.

we proceed by seeking the eigenfunctions of the translation operator. The translation operator T shifts the solutions by one period,

$$T\psi(x)\;=\;\psi(x\;+\;a)$$

Notice that any function of the form

$$\psi_k(x)=e^{ikx}U_k(x)$$

Is an eigen is an eigen function of translational operator with eigenvalue $$e^{ikx}$$

$$\psi_k(x)=e^{ik(x+a)}U_k(x+a)=e^{ikx}U_k(x)$$

In solid state physics, functions like this are said to have Bloch form. The convention for describing electron waves in a periodic medium is to express them in terms of the eigenfunctions of the translation operator.

These eigenfunctions have a well defined frequency and form a complete set that can be used to describe any wave.

An eigenfunction is specified by the k that appears in the expression for the eigenvalue. It turns out that the physical interpretation of k is nearly the same as the wave number of harmonic waves (usually also called k).

The eigenfunctions of the translation operator can be readily constructed from any two linearly independent solutions of the one-dimensional Schrödinger equation. A convenient choice is,

$$\frac{d\psi(x)}{dx}=e^{ik(x+a)}\frac{d\psi(x+a)}{dx}$$

There for,

$$Asin\alpha x+Bcos\alpha x=e^{ik(x+a)}(Asinh\gamma x+Bcos\alpha a)$$

Now take X= a and a=-b

Then,

$$A\;\sin\alpha a+B\;\cos\alpha a=e^{ik(a+b)}(A\;\sinh(-\gamma b)+B\;\cosh(-\gamma b))$$

$$A\lbrack\;\sin(\alpha a)+e^{ik(a+b)}\;\sinh(\gamma b)\rbrack+B\lbrack \;\cos(\alpha a)-e^{ik(a+b)}\;\cosh(\gamma b)\rbrack=0—–(1)$$

By considering (a+b)=c and taking the derivative,

$$A\alpha\lbrack\;\cos\;(\alpha c)+e^{ik(a+b)}\;\cos h\;(\gamma b)\rbrack+B\lbrack\;-\sin\;(\alpha c)-\;e^{ik(a+b)}\;\sin h\;(\gamma b)\rbrack=0—–(2)$$

Equation 1 & 2 have a non trivial solition (i.e, a spolution other than A=B=0 only)

If

$$\frac{\gamma^2-\alpha^2}{2\gamma\alpha}\;\sin (\alpha c)\;\sinh (\gamma b)+\cos (\alpha c) \cosh (\gamma b)= \cos (k(b+c)—–(3)$$

At this point it is convenient to consider the special case in which 0b and infinity

While the product  remains constant.

Let

$$P=(\frac{(mV_0bc)}{H^2})$$

Then eqn 3 became

$$\frac P{\alpha c}\;\;\sin(\alpha c)+\cos(\alpha c)=\cos(kc)\;\;——(4)$$

This can be solved graphically

The solution for $$P=\frac{3\mathrm\pi}2$$

(corresponding to the higher barrier) shown below, it plotted by using the 4th equation RHS and LHS