### Solution to schrodinger equation with dirac comb potential

• 1. SOLUTION OF SCHRÖDINGER EQUATION FOR DIRAC COMB POTENTIAL Zulfidin Khodzhaev AUTH Physics Department ABSTRACT I have studied periodic potential formed by a sequence of Dirac comb functions and solved Schrödinger equation for this potential. At the end energy band of the potential is determined. The potential is given as: (x) (x )V = m ħ2 ∑ P n=1 δ − na Where a negative represents an attractive potential while a positive represents repulsiveδ δ one, potential is infinite for and positions of the Dirac delta functions are positive.x < 0 an INTRODUCTION Floquet’s theory This theory is used to solve general second order differential equation: where is coefficients of the equation, and it is complex-valued, piecewise(x) (s , , )as = 0 1 2 continuous and periodic, all with period a, where a is a non-zero real constant. Hence, if is a solution to differential equation, then is also a solution to the above(x)ψ (x )ψ + a equation. Theorem 1: There exist a non-zero constant ρ and a non-trivial solution of differential(x)ψ equation such that the below equation holds: Theorem 2: There are linearly independent solutions and of differential equation(x)ψ1 (x)ψ2 such that where and are constants, not always distinct, and andm1 m2 (x)p1 (x)p2 are periodic with period a; or where m is a constant and and are periodic with period a.(x)p1 (x)p2 The above results and their proofs are known as the Floquet theory after G. Floquet​(5)​ .
• 2. Schrödinger equation The Schrödinger equation plays the role of Newton's laws and conservation of energy in classical mechanics etc., it predicts the future behavior of a dynamic system. It is a wave equation in terms of the wavefunction which predicts analytically and precisely the probability of events or outcome. The detailed outcome is not strictly determined, but given a large number of events, the Schrödinger equation will predict the distribution of results ​(1)​ . The wave function has all the information about a system or a particle. For one particle that only moves in one direction in space, the Schrödinger equation looks like: where is the square root of -1i is the reduced Planck's constantℏ is timet is positionx is the wave function(x)Ψ is potential energy(x)V And the left hand side is equivalent to the Hamiltonian energy operator acting on (2)​ .(x)Ψ But, this differential equation is general, we need equation independent of time and that depends only on given potential. Assuming that the wave function, , is separable. In other assuming the function of two(x, )Ψ t variables can be written as the product of two different functions of a single variable: then, using standard mathematical techniques of Partial differential equations, it can be shown that the wave equation can be rewritten as two distinct differential equations: where the first equation is solely dependent on time , and the second equation depends(t)T only on position , and where is just a number. The first equation can be solved(x)ψ E immediately to give:
• 3. where is Euler’s number.e Solutions of the second equation depend on the potential energy function, , and so cannot(x)V be solved until this function is given. It can be shown using quantum mechanics that the number is actually the energy of the system:E (x)E = p2 2m + V so these separable wave functions describe systems of constant energy​(2)​ . Dirac comb function It is a periodic equally spaced dirac delta functions. In other words, Dirac comb is an infinite series of Dirac delta functions spaced at intervals of T​(4)​ : Dirac comb function is given by: (x) (x )V = m ħ2 ∑ P n=1 δ − na For​ , ​repulsive potential shown as:+ δ For​ ,​ ​attractive potential shown as:− δ
• 4. The Kronig-Penney model It is a simplified model for an electron in a one-dimensional periodic potential. The possible states that the electron can occupy are determined by the Schrödinger equation. In the case of the Kronig-Penney model, the potential V(x) is a periodic square wave​(3) and the potential function is approximated by a rectangular potential: RESULTS AND DISCUSSION The Time Independent Schrödinger equation after some manipulations, it takes the form: (a)(E (x))ψ(x)dx2 d ψ(x)2 + ℏ2 2m − V = 0 In between delta functions ,1 )( < x < a (x)v = 0 We introduce a new constant k Ek2 = ℏ2 2m For , repulsive potential+ δ The general solution of (a),which is one period of periodic function, where and V(x+a)=V(x); x=x+na, n=1, 2, 3 is:(x) + (x a)V = m ħ2 ∑ P n=1 δ − n for(x) e eu = A ikx + B −ikx 1 < x < a Then, according to Floquet theory (x ) u(x)u + a = λ ⇒ (x) [Ae e ]u = eiKx ik(x−a) + B −ik(x−a) for aa < x < 2 Boundary condition at ⟹ 1) :x = a (a) (a)ux<a = ux>a =e eA ika + B −ika [Ae e ]eiKa ik(a−a) + B −ik(a−a) = (b)e eA ika + B −ika [A ]eiKa + B
• 5. 2) Derivative of are continuous everywhere except where V is infinite:(x) and u (x)ux<a x>a | | − ψ(a)dx dψ a+ε − dx dψ a−ε = ℏ2 2mα In other words, by integrating the Schrödinger equation: from to and take the limit .x = a − ε x = a + ε ε → 0 Each term of the integration will be: Adding them up and dropping common factor we will get the second boundary condition: So, derivatives of 2) derivative equation results in: ik)Ae − k)Be( ikx + ( i −ikx = [(ik)Ae − k)Be ]eiKx ik(x−a) + ( i −ik(x−a) After inserting , the equation becomes:x = a ik)Ae − k)Be( ika + ( i −ika = [(ik)Ae − k)Be ]eiKa ik(a−a) + ( i −ik(a−a) ik)Ae − k)Be( ika + ( i −ika = [(ik)A − k)B]eiKx + ( i Combining with all the conditions we get: (c)ik)e [A ]( iKx − B = ik)[Ae e ] Ω(Ae e )( ika − B −ika + 2 ika + B −ika Fundamental solution for periodic function propagating is:0 < x < P , where is a wavenumber(x) e eu = A ikx + B −ikx k
• 6. This (b)(c) homogeneous system can be solved for A and B if determinant is zero: By expanding determinant of the matrix and using trigonometric identities: it is concluded that:os(x) and sin(x) ,c = 2 e +eix −ix 2 e −eix −ix os(Ka) os(ka) sin(ka)c = c + k Ω And bands of the energy eigenvalues are: | |os(ka) sin(ka)c + k Ω ≤ 1 Then we define , then the expression becomes:an(θ)k Ω = t = sin(θ) cos(θ) | os(ka) sin(ka)| cos(ka) sin(ka) [cos(ka)cos(θ) in(θ)sin(ka)]c + k Ω ⇒ + sin(θ) cos(θ) ⇒ 1 cos(θ) + s From trigonometric identity: the inequality becomes:os(a ) os(a)cos(b) in(a)sin(b)c − b = c + s | |cos(ka )1 cos(θ) − θ ≤ 1 cos(ka ) os(θ)⇒ − θ ≤ c To find , we will use our definition ,os(θ)c an(θ)k Ω = t = opposite adjacent cos(θ)⇒ = adjacent hypotenuse = k √k ( +1)2 k2 Ω2 and , alsok √k ( +1)2 k2 Ω2 ⇒ k k √( +1) k2 Ω2 ⇒ 1 √( +1) k2 Ω2 an(θ)k Ω = t ⇒ rctan( )θ = a k Ω Then, the inequality becomes: cos(ka ) os(θ)− θ ≤ c ⇒ cos[ka rctan( )]|| − a k Ω ≤ 1 √(1+ ) k2 Ω2 ro | |os(ka) sin(ka)c + k Ω ≤ 1 ⇒ cos[ka rctan( )]|| − a k Ω ≤ 1 √(1+ ) k2 Ω2 but we will make constantk Ω k a Ω a
• 7. Now, we need to graph band energy of the result: We will take , and ,aΩ = 4 ak = x , , ...x = 0 1 2 Then, first equation will be: and second equation is:y1 os[x rctan( )]= c − a x 4 y2 = 1 √1+( )x 4 2 The graph of band structure is: Figure 1: Graph of band structure The condition holds in lines below two expression intersection.cos[ka rctan( )]|| − a k 1 ≤ 1 √(1+ )1 k2 Furthermore, band limits that are determined in Figure 1, are shown as energy scales will be: Band limits are determined by | |os(ka) sin(ka)c + k Ω ≤ 1
• 8. Figure 2: Band structure of Dirac comb potential The bands become broader with increasing energy, coming closer to continuum. Also, colored region of Figure 2 are allowed bands. For , attractive potential− δ The general solution of (a), which is one period of periodic function, where and V(x+a)=V(x); x=x+na, n=1, 2, 3 is:(x) − (x a)V = m ħ2 ∑ P n=1 δ − n for(x) e eu = A kx + B −kx 1 < x < a Then, according to Floquet theory (x ) u(x)u + a = λ ⇒ for(x) [Ae e ]u = eiKx k(x−a) + B −k(x−a) aa < x < 2 Boundary condition at ⟹ 1) :x = a (a) (a)ux<a = ux>a =e eA ka + B −ka [Ae e ]eiKa k(a−a) + B −k(a−a) = (d)e eA ka + B −ka [A ]eiKa + B
• 9. 2) Derivative of are continuous everywhere except where V is infinite:(x) and u (x)ux<a x>a | | ψ(a)dx dψ a+ε − dx dψ a−ε = ℏ2 2mα In other words, by integrating the Schrödinger equation: − (x))u(x) u(x)( ħ 2 2m d2 dx2 − V = E Integrating from to and taking the limit , we will get the secondx = a − ε x = a + ε ε → 0 boundary condition: (a ) (a ) Ωu(a)u′ + 0 − u′ − 0 + 2 = 0 So, derivatives of 2) boundary condition results in: k)Ae − )Be( kx + ( k −kx = [(k)Ae − )Be ]eiKx k(x−a) + ( k −k(x−a) After inserting , the equation becomes:x = a k)Ae − )Be( ka + ( k −ka = [(k)Ae − )Be ]eiKa k(a−a) + ( k −k(a−a) k)Ae − )Be( ka + ( k −ka = [(k)A − )B]eiKx + ( k Combining all the conditions we get: (e)k)e [A ]( iKx − B = k)[Ae e ] Ω(Ae e )( ka − B −ka − 2 ka + B −ka Fundamental solution for periodic function propagating is:0 < x < P , where is a wavenumber(x) e eu = A kx + B −kx k This (d)(e) homogeneous system can be solved for A and B if determinant is zero: By expanding determinant, we find that​: ke [e ] Ωe [e ] ke Ω Ω− 2 (iKa) (ka) + e(−ka) + 2 (iKa) (ka) − e(−ka) + 2 (2iKa) + k − 2 + k + 2 = 0 By using Hyperbolic identities: the equation becomes,cosh(x) , 2sinh(x) ,2 = ex + e−x = ex − e−x 2cosh(ka) sinh(ka)− k 2Ω = eiKa + e−iKa Then, by using trigonometric identity: , the equations becomes:cos(x)2 = eix + e−ix os(Ka) osh(ka) sinh(ka)c = c − k Ω To find band structure of this equation, we will manipulate with the equation: rccos[cos(Ka)]a = rccos[cosh(ka) sinh(ka)]a − k Ω aK = rccos[cosh(ka) sinh(ka)]a − k Ω
• 10. We will graph versusa rccos[cosh(ka) sinh(ka)]K = a − k Ω ak So, we will make as constantk Ω k a Ω a Now, to graph this equation, we need to define that and , ,aK = y ak = x , , ...x = 0 1 2 The equation becomes: rccos[cosh(x) sinh(x)]y = a − x Ωa Since depending on the value of , the solution changes. We will graphos(Ka) ,− 1 ≤ c ≤ 1 aΩ assuming that .a .8Ω = 0 The graph of band structure is: Figure 3: Graph of band structure
• 11. CONCLUSION In this report periodic potential formed by a sequence of Dirac comb functions are solved by Schrödinger equation. The potential is given as: (x) (x )V = m ħ2 ∑ P n=1 δ − na Where is arbitrary real number (where a positive represents an repulsive potential while aδ δ negative represents attractive one), potential is infinite for and positions of the Diracδ x < 0 delta functions are positive.an First the Schrödinger equation was manipulated to evaluate for time independence form. After, the constant was introduced and general solution for schrödinger equation was introduced for two periods of potential function. Also, boundary conditions for potential was evaluated, and boundary value was inserted into the final two boundary conditions. Resulted homogeneous system from boundary conditions determinant was equalled to zero, and the expression that was obtained from the determinant gave us information about band energy. Some trigonometric identities was applied to the inequality that was obtained from the expression. Using the inequality, band structure was graphed and band limits obtained. At last, band limits were graphed in energy scale and it gave visual perspective on allowed and prohibited bands of the energy. REFERENCE 1. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html 2. https://simple.wikipedia.org/wiki/Schr%C3%B6dinger_equation 3. R. de L. Kronig and W. G. Penney, Proc. Roy. Soc. (London) A 130 (1931) 499; D. A McQuarrie, The Chemical Educator, vol. 1 (1996) S1430-4171. 4. https://en.wikipedia.org/wiki/Dirac_comb 5. http://frontslobode.org/vedad/pdf/diss.pdf