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Properties of the wavefunction: Consider a one dimensional Schrödinger Equation, for 1 particle in 1 dimension: H ( x) ( x) E ( x) n n n where n is the quantum number (1 dimension, 1 quantum number) Normalization: n ( x) n ( x) dx 1 * Orthogonality: m ( x) n ( x) dx 0 * mn (Using complex conjugates just incase the wavefunction is imaginary.) This is a very important property and is much used to derive very powerful statements concerning spectroscopy and chemical bonding without having to perform difficult (impossible) calculations. Orthogonality always applies. For 2 coordinates (1 particle in 2 dimensions or 2 particles in 1 dimension) or larger systems, where degeneracy is possible all wavefunctions are orthogonal. E.g. Particle in the box: ( x) A sin( nx / L) n ( x) ( x) dx A sin( mx / L) sin( nx / L) dx m n L L * 0 2 m n 0 A / 2 cos(m n x / L) cos(m n x / L)dx L 2 0 A / 2 ( L / )sin m n x / L / m n sin m n x / L / m n 2 0 sin( p ) 0 p integer L L 0 0 Rotation in 2 dimentions: () (1 / 2 ) exp(im ) 1/ 2 m 2 2 ( ) ( ) d (1 / 2 ) exp(im ) exp(in ) d * m 0 n 0 2 (1 / 2 ) exp(i (n m) ) d 0 (1 / 2 )exp(i (n m) ) /(n m) 2 0 (1 / 2 )(cos(n m)2 i sin( n m)2 ) (cos(n m)0 i sin( n m)0)/(n m) (1 / 2 )(1 i 0) (1 i 0) /(n m) 0 (n m) integer Rearranging the Schrödinger Equation: H ( x) ( x) E ( x) n n n ( x) H ( x) ( x) ( x) E ( x) E ( x) ( x) ( x) H ( x) ( x) dx E ( x) ( x) dx E ( x) H ( x) ( x) dx * * n n n * n n * n n n * n n n n n * n n n for a normalized wavefunction or: E ( x) H ( x) ( x) dx / ( x) ( x) dx * n n * n n n for an un-normalized wavefunction. Expectation values For any physical property q there is an operator Q such that: Qψ=qψ or < q > = ∫ ψ* Q ψ dτ < q > is called the ‘Expectation Value’ of the property. It may be an exact value, for example for energy, or a ‘most probable value’, for example for position. The operator Q is simply the physical quantity itself except that in the case of momentum the substitution: p i is made. For example: What is the average position of a particle in a box in any energy level? Q=x ( x) (2 / L) sin( nx / L) 1/ 2 n x ( x) x ( x) dx (2 / L) x sin (nx / L) dx L L * 2 n 0 n 0 x 1 cos(2nx / L) dx 2 (1 / L)x / 2 sin 2nx / L L / 4nx L (2 / L) 0 2 L L 0 0 L/20 L/2 This is obviously correct for any state of the particle in the box if you look at the diagrams of the wavefunctions. (It is more interesting to consider the average position of the particle in the left hand side of the box, between 0 and L/2. This is L/4 for n = even and is a function of n for n = odd. However it is necessary to take into consideration the fact that only part of the ‘space’ is being considered, and use: < q > = ∫ ψ* Q ψ dτ / ∫ ψ* ψ dτ as the ∫ ψ* ψ dτ integral is not being integrated over all space.) What is the radius of the 1s orbital of the hydrogen atom? ( r , , ) (1 / a ) e 3 1, 0 , 0 1/ 2 r / a 0 0 r ( r , , ) r * 1, 0 , 0 1, 0 , 0 (r , , ) d d r sin dr d d 2 r (1 / a ) r e 3 0 3 2 r / a 0 0 2 0 0 dr sin d d (1 / a )(3!/(2 / a ) )(2)(2 ) 3a / 2 3 0 4 0 0 a 0 = 5.291 772 108(18)×10-11 m , the radius of the hydrogen atom is ~ 1.5 x 10-10 m