Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
particle in a box Potential problem Wave functions Significance of wave function Normalisation The time-independent Schrodinger Equation Solutions of the T.I.S.E Particle in a 1-Dimensional Box Time Dependent Schrödinger Equation 2 d 2 V ( x) E 2m dx 2 Region I Region II Region III KE PE TE wavefunction is dependent on time & position function: V(x)=∞ V(x)=0 V(x)=∞ 1 ( x, t ) f (t ) ( x) Time Independent Schrödinger Equation L 0 V(x)=0 for L>x>0 V(x)=∞ for x≥L, x≤0 Classical Physics: The particle can exist anywhere in the box and follow a path in accordance to Newton’s Laws. Quantum Physics: The particle is expressed by a wave function and there are certain areas more likely to contain the particle within the box. x 2 d 2 ( x) V ( x) E 2 2m dx Region I and III: Applying boundary conditions: 2 d 2 ( x) * E 2 2m dx Region II: 2 d 2 ( x) E 2m dx 2 0 2 Finding the Wave Function d 2 ( x) 2m 2 E dx 2 2 d 2 ( x) E 2m dx 2 This is similar to the general differential equation: d 2 ( x) k 2 2 dx A sin kx B cos kx So we can start applying boundary conditions: x=0 ψ=0 0 Asin 0k B cos 0k x=L ψ=0 0 Asin kL A 0 Calculating Energy Levels: 2mE k 2 2 k 2 2 E 2m h 2 n 2 2 h 2 E 2 L 2m4 2 k 2h2 E 2m4 2 n2h2 E 8mL2 II A sin nx L But what is ‘A’? Normalizing wave function: L 2 ( A sin kx ) dx 1 0 L x sin 2kx A 1 4k 0 2 2 0 0 B *1 B 0 kL n where n= Our new wave function: * n sin 2 L 2L L A 1 n 2 4 L Since n= 2 L A 1 2 * A 2 L Our normalized wave function is: II 2 nx sin L L Particle in a 1-Dimensional Box II 2 nx sin L L Applying the Born Interpretation II 2 2 nx sin L L n=4 n=3 E x/L 2 n=4 E n=3 n=2 n=2 n=1 n=1 x/L de Broglie Hypothesis In 1924, de Broglie suggested that if waves of wavelength λ were associated with particles of momentum p=h/λ, then it should also work the other way round……. A particle of mass m, moving with velocity v has momentum p given by: p mv h Kinetic Energy of particle 2 2 2 2 p h k KE 2 2m 2m 2m If the de Broglie hypothesis is correct, then a stream of classical particles should show evidence of wave-like characteristics…………………………………………… Standing de Broglie waves Eg electron in a “box” (infinite potential well) V= V=0 Electron “rattles” to and fro V= V= V= V=0 Standing wave formed wavelengths of confined states In general, k =nπ/L, n= number of antinodes in standing wave 2L 3 ;k 3 L 2 L;k L 2L ; k L energies of confined states k n E 2 2m 2mL 2 2 2 En n 2 E1 E1 2 2mL 2 2 2 2 Energies of confined states En n E1 2 E1 2 2mL 2 2 particle in a box: wave functions From Lecture 4, standing wave on a string has form: y ( x, t ) ( A sin kx) sin( t ) Our particle in a box wave functions represent STATIONARY (time independent) states, so we write: ( x) A sin kx A is a constant, to be determined…………… interpretation of the wave function The wave function of a particle is related to the probability density for finding the particle in a given region of space: Probability of finding particle between x and x + dx: ( x ) dx 2 Probability of finding particle somewhere = 1, so we have the NORMALISATION CONDITION for the wave function: ( x) 2 dx 1 interpretation of the wave function Interpretation of the wave function Normalisation condition allows unknown constants in the wave function to be determined. For our particle in a box we have WF: nx ( x) A sin kx A sin L Since, in this case the particle is confined by INFINITE potential barriers, we know particle must be located between x=0 and x=L →Normalisation condition reduces to : L ( x) 0 2 dx 1 normalisation of wave functions ( x) 0 nx A sin dx 1 L 0 L L 2 dx 1 2 2 nx ( x) sin L L 2 Solving the SE :an infinite potential well V ( x) 0 0xL So, for 0<x<L, the time independent SE reduces to: 2 d 2 ( x) E ( x) 2 2m dx d 2 ( x) 2mE ( x) 0 2 2 dx General Solution: 1/ 2 2mE ( x) A sin 2 1/ 2 2mE x B cos 2 x 1/ 2 2mE ( x) A sin 2 1/ 2 2mE x B cos 2 x Boundary condition: ψ(x) = 0 when x=0:→B=0 1/ 2 2mE ( x) A sin 2 x Boundary condition: ψ(x) = 0 when x=L: 1/ 2 2mE (0) A sin 2 n E 2 2mL 2 L0 2 2 nx ( x) A sin L In agreement with the “fitting waves in boxes” treatment earlier………………..