Wave Functions and the Particle in a Box PDF

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Dr David Cooke...

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Wave Functions and the Particle in a Box Hayward Chapters 1 and 2

A 1D Wave Function • Consider a beam of particles moving along the x-axis at constant velocity • Wave – Particle duality tells us the particles will behave like a wave but how do we describe the wave mathematically? • Some combination of cosine and sine waves seems the logical choice? – e.g. (x) = Acos(kx) + Bsin(kx)

A 1D Wave Function • As you study more quantum mechanics you will see that the maths is easier and more versatile if we use a complex wave of the form – (x) = Aeikx = A(cos kx + i sin kx)

• Two sinusoidal waves 90˚ out of phase one in the real plane and one in the imaginary plane • A is the amplitude of the wave and k = 2 /  • If we were working in 3D k would be a wave vector

Relating Waves to Particles • As p = h /  it follows p = h k / 2 So the particles described by the wave have a momentum of kℏ – ℏ (h bar) = h / 2

• As KE = T = ½ mv2 and p = mv It follows the particles have a KE of T = (ℏk)2 / 2m – The larger the KE the smaller the wavelength becomes

Relating Waves to Particles • If k is positive then a particle moving in the positive direction is described by the wave function (x) = Aeikx and has momentum ℏk • If k is positive then a particle moving in the negative direction is described by the wave function (x) = Ae-ikx and has momentum -ℏk • The complex conjugate of the wave function represents a particle moving in the opposite direction

Probability and the Wave Function • As the wave Function contains a real and imaginary part it is perhaps more useful to think about the quantity  * which is a real function and is related to the probability of finding the particle at a particular point. • But P(X = x) = 0 for a continuous function Correctly P(x > X > x+x) =  * • This point is often only considered implicitly

Probability and the Wave Function • For a wave moving with momentum ℏk (x) = Aeikx • For a wave moving with momentum -ℏk *(x) = Ae-ikx • The probability of finding the particle in some small interval of the wave is P = Aeikx . Ae-ikx = A2 • We are equally likely to find the particle anywhere along the wave. Or the location of the particle is unknown

Well Behaved Waves • The mathematical form of a wave function must satisfy three criteria. 1. There can only be one value of (x) for every value of x. 2. It must vary smoothly with x so d(x) /dx is defined at all values. 3. The integral over all space of (x) *(x) must be 1.

Well Behaved Waves • Why? 1. Otherwise there would be more than one probability of the particle being somewhere. 2. There must be a way of defining the wave-function at all points in space 3. The particle must be somewhere.

Example • Which of the following are well behaved?

Particle in a 1D Box • Consider a particle trapped in a 1D potential well with infinitely high sides: • The probability must be zero at the walls • The particle must be equally likely to be moving in either direction • To fulfil the second point the wave function needs to be of the form: (x) = Aeikx  Ae-ikx = 2A cos kx or 2iA sin kx

Which Wave Function?

Clearly the sine function fits the boundary conditions

N = 2i A and n =1,2,3...etc

Properties of the Particle in a Box • Previously we showed: p =ℏk and T = (ℏk)2 / 2m • So substituting k = (n / L) gives: p =(nℏ / L) T = (nℏ / L)2 / 2m or T = n2h2 / 8mL2 where n=1,2,3.... • The momentum and energy of the particle are quantized

• As n  0 (otherwise the particle would disappear). The smallest amount of energy it can have is h2 / 8mL2. • Allowed Solutions are therefore: 1

0

(x)

2

3

4

L

0

2(x)

L

Probability Distribution • Recall P(0 < X < L) = 1 This can be achieved by choosing a suitable value of N • With some maths we can show that: N = (2 / L)1/2 So

n=1,2,3,etc

Probability Distribution • In classical physics we would expect it to be equally likely to find the particle anywhere in the box. • When n=1 we now predict the particle will spend most time in the centre of the box • Higher values of n give results impossible in a classical world – For n=2 how can the particle get from one side of the box to the other without ever being in the middle?

A Real Example  – electrons in butadiene • Butadiene CH2=CH-CH=CH2 • One of the 2p orbitals in each carbon (usually defined as 2pz) are able to overlap to form delocalised  orbitals that extend over the whole molecule • If we ignore any effect from the molecule not been linear we can consider the properties using the particle in a 1D box model

• If the box was exactly 4 carbons long then there would be a node on each of the carbons and therefore never any electron density • This is solved by extending the box by half a bond length in each direction so it is exactly 4 C-C bond lengths long

Predicted structure • Butadiene has 4  electrons so the first two levels will be occupied. n = 2 is the HOMO and n = 3 is the LUMO

C

C

C

C

• When n=2 we have CH2 = CH – CH = CH2 Stable arrangement with  bonds on 1-2 and 3-4. • When n = 3 we have CH2 – CH = CH – CH2  bond on 2-3 and unsatisfied valence on end carbons

Energy an Example Given d(C-C)=154 pm and d(C=C)=135 pm • Calculate the energy that must be adsorbed for an electron to move from n=2 to n=3. • Calculate the corresponding wavelength? • What type of radiation does this correspond to? • Repeat the calculation using: a) hexa,1,3,5,triene and b) -Carotene (a molecule consisting of 22 conjugated carbon atoms.)...