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AN INVESTIGATION ON THE SOLUTION OF SIMPLE HARMONIC OCILLATOR problem USING SCHRODINGER METHOD A PROJECT REPORT Submitted in the partial fulfillment of the requirement for the degree of BACHELOR OF SCIENCE In PHYSICS DEPARTMENT OF PHYSICS CHAPTER-1 Schrodinger Equation and Wave function Time dependent Schrödingerequation: Thenon-dissipation of the wave packet of the material particle has been explained by assuming the necessity of the guiding wave obeying Schrödinger wave equation which we shall derive here. Consider a system of stationary wave to be associated with the particle. Let (r , t ) be the wave displacement for the de Broglie wave at any location r i x j y kz at time t. Then the differential equation of the wave motion in three dimension in accordance with wax well’s wave equation can be written as 1 2 2 2 2 1 2 I.e. 2 2 or 2 2 2 2 2 u t x y z u t 2 Where u is thewave velocity. The solution of equation (1) gives as a periodic displacement in terms of time i.e. ( r , t ) = 0 (r )e it . (2) Where 0 is the amplitude at the point considered. Itis a function of position r i.e. of coordinate (x y z) and not of time t. The equation (2) May be expressed as (r , t ) 0 (r )e it (3) Differentiating equation (3) twice with respect to t we get 2 2 0 (r )e it 2 t 2 (r , t ) Substituting this in (1) we get 2 2 2 2 2 x 2 y 2 z 2 u (4) 2 v But I.e. Also 2 u 2 u (5) 2 2 2 2 2 2 2 x y z (6) Using (5) and, (6) equation (5) becomes 2 4 2 2 0 (7) So far we have not introduced wave mechanical concept and so the treatment is general. For introducing the concept of wave mechanics we must put from de Broglie equation. h mv (8) Substituting this equation (7), we get 2 4 2 m2 v 2 0 h2 (9) If E and V are the total and potential energies of the particle respectively then its kinetic energy 1 2 mv is given by 2 1 2 mv E V 2 This gives m2u 2 2m( E V ) Substituting this equation (9), we get 2 4 2 m ( E V ) 0 h2 (10) The above equation is called Schrödinger time independent wave equation .The quantity is usually referred as wave function. Let us now substitute, inequation (10) h 2 (11) Then the Schrodinger time independent wave equation, in usually used from, and may be written as 2 2m 2 ( E V ) 0 (12) 2) Schrödinger equation for a free particle For the free particle V=0; therefore if we put V=0 in equation (12) it will become the Schrödinger equation for a free particle i.e. 2 2mE 2 0 3) Time dependent Schrodinger equation Time dependent Schrodinger equation Differentiating equation (3) with respect to t we get i 0 (r )e it t i (2 v) 0 (r )e it 2 iv (r ) iE i i Which gives 2 iE h (13) t E i (14) Substituting value of E from above equation in (12), we get 2 2m i V 0 2 t Or 2 I.e. 2 2m 2m i V 2 t 2 V i t (15) This equation contains the time and hence is called time independent Schrodinger equation Equation (15) may written as ( The operator ( 2m 2m 2 V ) i t (16) 2 V ) is called Hamilton and is represented by H operated on. Gives E which may be seen from (14).This equation (16) t may be written as While operator i H E (17) The above forms of Schrodinger equation describe the motion of a non-relativistic material particle. PHYSICAL INTERPRETATION OF WAVEFUNCTION It would be now interesting to interpret wave-function appearing in Schrodinger equation, physically in terms of the observable properties of the system. In the beginning it was considered that the wave-function is merely an auxiliary mathematical quantity employed to facilitate computations relative to experimental results. This is true to a certain extent in direction of the detector; but it does not seem reasonable to introduce an isolated mathematical function without inquiring into its physical significance. Schrodinger himself attempted the physical interpretation of in terms of charged density. As the beams of electrons exhibit diffraction phenomenon like X-ray, one might use the optical analogy to arrive at the physical significance of . In any electromagnetic wave system if A is the amplitude of the wave, them the energy density, i.e., energy per unit volume is equal to A2 , so that the number of photons per unit volume, i.e. Photon density is equal to A2 / hv where hv is the energy of a photon. This means that the photon density is proportional to A2 . As hv is constant. By analogy, if is the amplitude of matter wave at any point in space, then the number of material particles per unit volume i.e., the particle density must be proportional to 2 Thus the square of absolute value 2 of ,i.e., , is a measure of the particle density. If q is the electric charge on a particle, the charge density will be equal to the product of the charge q and the particle density. Thus the 2 quantity. Is measure of charge density as Schrodinger considered. Usually * is written instead of , where * .is the complex conjugate of 2 This interpretation was found to lead to very satisfactory results when wave mechanics was applied to the directional distribution of photo-electrons, intensity distribution in Compton scattering, the stable states of Bohr atom, the emission of spectral lines etc., but the difficulty arises when we wish to follow the flight of a single electron or any other material particle. The wave packet associated with material particle dissipates in a course of time, so that it cannot represent for long the material particle which maintains its identity. Moreover in such a case it is natural to ask. “Where is the particle in relation to the wave packet?”. “To remove the above discrepancy another physical interpretation of the wave function, generally accepted at present, was suggested by Max - Bron in 1926 and then developed by 2 Bohr,Dirac, Heisenberg and others. According to this view. * = represents the probability density of the particle in the state Then the probability of finding the particle in a volume element dx dy dz about any point r at time t is expressed as P(r ) (r , t ) 2 The function is sometimes called probability amplitude for the position of the particle. The postulate suggested by Born shows that the quantum mechanical laws and the results of their measurements can be interpreted on the basis of probability considerations NORMALISED WAVE FUNCTIONS. We have seen that * represents the probability density and so * or. 2 .represents the probability of finding the particle in the volume element In physical problems we come across situations where the particle is bound by forces to a limited region. The examples of this kind are: - the electron in an atom, the particle in a box with impenetrable walls. In such cases the total probability of finding the particle in the entire space is of course, unity, i.e. P(r ) (r , t ) 1 2 Where the integration extends over all space. (r , t ) * ( r , t ) 1 v A wave function which satisfies the above equation is said to be normalized to unity (or briefly, normalised). When the particle is bound to the limited regions the probability of finding the particle at infinite distance is zero, i.e., [psi-psi star] at x=plus or minus infinity or zero. Any solution of the wave equation may be normalised by multiplying or dividing by a constant and it can be readily seen that the result is also a solution of the wave function. ORTHOGONALITY OF WAVE FUNCTIONS An orthogonal wave function is of paramount importance in quantum mechanics. When two wave functions are orthogonal, it means that they represent two mutually exclusive physical states. We state this in quantum mechanics by taking the inner product, or the integral of the two wave quality means that the two "vector spaces" which are being described by the wave functions are not linear combination of each other. The problem of liner harmonic oscillator is of importance since many systems of interest can be 1 approximated to it. It potential energy V kx 2 , k is the force constant, is a continuous function 2 of the coordinate x and therefore is completely different from systems we considered so far where the potential is constant over a region. WAVE EQUATION In SHM restoring force is proportional to displacement F kx According to Newton's second law F md 2 x dt 2 md 2 x kx dt 2 md 2 x kx 0 dt 2 d2x k x0 dt 2 m This equation represents a periodic motion of angular frequency k m , or frequency 1 k 2 2 m The potential energy of oscillator is x V kxdx 0 1 2 kx 2 With the force constant expression in equation (1.40), the potential V is given by V 1 1 4 Vo2 x 2 m 2 x 2 2 2 (4.72) The time-independent Schrodinger equation of the linear harmonic oscillator is then d 2 2m 1 2 E m 2 x 2 0 2 2 dx Or d 2 m 2 2 2 dx mk E 1 kx 2 0 2 k 2 1 d 2 2 mk dx 2 m mk 2 E x 0 2 2 k Or 2 Let mk 2 2E m 2 k Then 1 d 2 2 x 2 0 2 2 dx Let y x d d dy d dx dy dx dy 2 d 2 d d 2 dx dx dx d d dy dy dx dx d d dy dy d 2 2 dy 2 d 2 y 2 0 2 dy I.e.equation (4.73) reduces to 2 y 2 0 x 2 (4.75) ASYMPTOTIC SOLUTION We shall investigate first the solution of equation (4.75), when y when y is very large, y 2 y 2 And equation (4.75) becomes 2 y 2 0 2 x Its asymptotic solutions are ( x) e y 2 /2 2 2 d 2 y 2 e y /2 e y /2 2 dy y 2 1 e y 2 /2 As y is very large y 2 1 y 2 2 d 2 y 2 e y /2 y 2 2 dy y 2 /2 y Out of the two asymptotic solutions, e .is not acceptable as it diverges when the exact solution of equation (4.75) may be written as e y /2 and e y 2 2 /2 e y /2 Is not acceptable since it increases with x 2 While e y /2 Satisfies the condition 2 The exact solution =(complimentary function +particular integral) H ( y) e y e y 2 /2 2 /2 H ( y) Since substitution of equation (4.77) in equation (4.76) gives 2 2 y 2 1 e y /2 y 2 2 x e y /2 H ( y ) Where H(y) is a function of y and the product e y y /2 H ( y ) tends to zero as SERIES SOLUTION Substitution of equation (4.78) in equation (4.75) gives 2 d 2 y 2 /2 e H y 2 e y /2 H 0 2 dy 2 2 d y 2 /2 dH e He y /2 y y 2 e y /2 H 0 dy dy e y e 2 /2 y 2 /2 2 2 2 dH d 2 H dH y 2 /2 e y e y /2 He y /2 y y y 2 e y /2 H 0 2 dy dy dy 2 2 2 2 2 d2H dH dH ye y /2 e y /2 y 2 e y /2 H e y /2 H y 2 e y /2 H 0 2 dy dy dy d2H dH dH y y y2 H H y2 H 0 2 dy dy dy d2H dH 2y H 2 y2 H 0 2 dy dy d2H dH 2y H H 0 2 dy dy d2H dH 2y 1 H 0 2 dy dy Hermite Equation H " 2 yH ' 2kH 0 Solution H ( y ) an y n n 0 H '( y ) nan y n 1 n 1 H "( y ) n(n 1)an y n 2 n2 We shall look for H(y), a series solution of the type H ( y ) an y n n 0 Equation (4.80) when substituted in equation (4.79), we have k 1 k 2 a k 0 k 2 2k 1 ak y 2 0 n(n 1)an y n2 2 y nan y n1 ( 1) an y n 0 n2 n 1 n 0 n 0 n 0 ie (n 2)(n 1)an 2 y 2 2 y an y n 0 ie (n 2)(n 1)an 2 2nan ( 1)an y n 0 n 0 ie (n 2)(n 1)an 2 2nan ( 1)an 0 an 2 2n 1 a n 1 n 2 n For the validity of this equation, coefficient of each power of y must vanish separately. k Coefficient of y when equated to zero gives the recurrence relation ak 2 2k 1 a k 1 k 2 k a This formula allows the calculation of all even coefficients in terms of 0 and the odd a a 0 coefficients in terms of. 1 Equation (4.80) will have only odd coefficients if 0 and odd a 0 coefficients if 1 Thus we have two independent solutions for equation and a linear combination of the two will be the most general solution. The solutions are: H e ( y ) a0 a2 y 2 a4 y 4 ... And H 0 ( y ) y a1 a3 y 2 a5 y 4 ... ENERGY EIGEN-VALUES equation (4.82),we get when k ak 2 2k 1 a k 1 k 2 k e y 1 y2 2 y4 y4 ... 2! 3! ak 2 2k 2 2 ak k k ak 2 2 ak k 2 Consider the Taylor series expansion of ( y ) exp( y 2 ) 1 y 2 1 k / 2 !y y4 y6 ... 2! 3! k 0,2,4 (In general for even solutions) The ratio of the coefficients of the successive terms in equation (4.86) is k / 2 ! ak 2 1 2 ak k / 2 1 ! k / 2 1 k 2 Where k is large. Therefore, for large values of k, exp( y / 2) H ( y ) tends to behave like exp( y 2 / 2) , if the series is even, and y exp( y 2 / 2) if the series is odd; which is not acceptable. This unrealistic solution can be avoided if the series in equation (4.80) terminates after a finite number of terms. In such a situation ( y ) 2 Will tend to zero as y because of the factor exp( y / 2) .The series can be terminated by selecting in such a way that 2k 1 vanishes for k=n. Thus one of the series becomes a polynomial and the other can be eliminated by setting the first coefficient to zero. Substitution of the value of gives consider k th term 2k 1 ak 2n 1 2E 0 The series can be terminated by selecting in such a way that 2k+1vanishes for k=n i.e. 2n 1 2E 0 i.e. 1 En n , n 0,1, 2,... 2 The energy value of an oscillator based on quantum theory is En n , n 0,1, 2,... From the two expressions, it is evident that the quantum mechanical energy value is higher than 1 the quantum theory value by 2 , which is the energy possessed by the lowest state n = 0. 1 This energy of 2 is called the zero-point energy which is a manifestation of uncertainty principle. E = 0 means that both position and momentum are well defined oscillator without 1 violating uncertainty principle is 2 Figure 4.11(a) illustrates the energy values as given by equation (4.88). ENERGY EIGENFUNCTIONS When 2n 1 , equation (4.79) reduces to d 2 H n ( y) dH n ( y ) 2y 2nH n ( y ) 0 2 dx dy The solutions of equation (4.89) are Hermite polynomials........... of degree n. The energy Eigen functions can now be expressed as y 2 2 m 2 x , y 2 n ( y ) N n H n ( y ) exp y2 m x2 The first-four Hermite polynomials are y 2 H 0 ( y) 1 A Hermite Polynomials of degree n is defined by d n e y Hn( y ) (1) .e . dy n n 2 y2 H 0 ( y) 1 H1 ( y ) 2 y H 2 ( y) d y2 ( e H 2 y ) dx 2 2 2 e y 2 y.e y . 2 y e y 1 e y 2 e y 4 y 2 .e y 2 2 4 y2 2 2 d 3 y2 H 3 ( y ) (1) e . 3 (e ) dy y2 3. 1 e e y .2 2 y2 d 2 y2 e 2 y dy 2 2 d y2 ye 2 y e y 1 dy 2 2 2 2 2e y e y 2 y 2 y 2 e y 2 y e y 4 y 2e y e y 2 y 2 y 2 e y 2 y e y 4 y 3 4 y 8 y 8 y 2 2 2 2 8 y 3 12 y The normalization condition leads to 1/2 Nn 2 m H 2 n ( y ) exp( y 2 )dy 1 Or 1/2 Nn 2 1/2 n 2 (n !) 1 m Or 1/2 1/2 1 N n n m 2 n ! The normalized Eigen functions and the ground state Eigen function are given by 1/2 1/2 1 n ( y ) n m 2 n ! y2 H n ( y ) exp 2 m 1/4 0 ( x) m x 2 exp 2 The wave functions................. and the probability density ................. of the lowest four states are also illustrated in figure 4.11. It may also be noted from figure that oscillator Eigen functions are even for even values of n 2 and odd for odd values of this is understandable as exp( y / 2) is always even and. H n ( y ) is even or odd according as n is even or odd. It may also be noted that the quantum oscillator can be found outside the parabolic potential barrier since does not vanish at the classical turning points. In other words, the particles can penetrate the barrier to some extent. This barrier penetration is an important feature of quantum mechanics. Another interesting point to be noted in connection with the results is the nature of probability distribution of classical and quantum oscillators. Classically, the probability of finding the oscillator at a given point is inversely proportional to its velocity at that point. The total energy E 1 2 1 2 mv kx 2 2 Or 2E kx m Therefore the classical probability Pc m 2 E kx 2 GRAPHICAL REPRESENTATION OF RESULT. REFERENCE: 1. 2. 3. 4. Aruldhas - Quantum Mechanics NouredineZettili- Quantum Mechanics-Concepts and Applications Townsend- A Textbook of Quantum Mechanics Arthur Beiser - Concept of Modern Physics, Sixth Edition.