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# Lecture 8

By Martha Gibson,2014-09-04 04:35
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Lecture 8Lect

Lecture 8: Particle in a Box

Properties of

1) is single-valued (at any point in space, it has only one value)

2) and ’ are continuous: waves in nature are always continuous and always smooth

3) is normalized: ?：*(x)(x)=1 or ? P(x)dx=1

The probability of finding the particle somewhere in space is 1

Particle in a Box

Motivation

Conjugated polyene: -electrons are constrained to move in the 1D conjugated

polyene (Chem 346 lab)

3D Quantum dots: Dots ~100 Å in diameter. Change color based on size; used

as fluorescent tags in biochemistry (Blue Book page B-24)

Handout: Particle in a 1-D Box

Consider a particle confined to a 1-D Box:

~

~ x < 0 U(x) = 0 0 x a U(x) ~ x > 0

0 x 0 a

The particle is in an infinite potential well: the potential is infinite except for where

the particle is allowed to be

Classical Mechanics: solve Newton’s Laws for energy E and position x of the particle

Lecture 8 1/6 Chemistry 344

12EKEPEmvU 02

Quantum Mechanics: Solve Schrodinger Equation for energy E and wavefunction (x)

of the particle

ˆH(x)E(x)0 22d(x)U(x)(x)E(x)22mdxU(x)=0 if 0 x a 22d(Particle is in the box) ()()0xEx22mdx

2d2mE (x)(x)022dx;2This is a linear homogeneous

differential equation with

(xconstant coefficients Try(x)e

2(x2(x(0ee

(?i

ixixThus(x)cece12Euler formula ccosxisinxccosxisinx；？；？；？；？；？；？12

Cos/sin are even/odd cccosxcicisinx；？；？；？；？1212Regroup constants

；？；？(x)AcosxBsinx

12；？2mEwhere

The probability of the particle being outside the box is zero, i.e., (x)=0 if it is outside the box. Thus by continuity of (x), the boundary conditions are:

(x=0)=0 and (x=a)=0

must be zero at the ends of the box because is zero outside the box and must be continuous

Apply boundary conditions:

(x=0)=0 ; Acos(0)+Bsin(0)=0 ; A?1+ B?0=0 ; A=0

Ignore trivial solution B=0 (x=a)=0 ; Bsin(a)=0 ; a=n where n=1,2,3…

Lecture 8 2/6 Chemistry 344

nx) Thus, (x)Bsin?na?

Quantization arises naturally from boundary conditions!

Find B from normalization of (x), i.e., the probability of the particle being inside the box

equals one.

？~*()()1xxdx?~

anx)22sin1Bdx??Inside back cover a?0 McQuarrie problem 3.24 a)21B?2?

2Ba

Thus,

2nx)；？xsinn1,2,3 ?naa?

and

222mEEDefinition of n22m

22Boundary condition n)? 2ma?

22hnEn1,2,3n28ma

Lecture 8 3/6 Chemistry 344

n4n2n3n1

22216h24h9h1h En22228ma 8ma8ma8ma

(x) n

P=*(x) (x) nnn

Notes on energy levels

2 ? n, i.e., spacing increases with n 1) En

2) E > 0. Lowest energy state has “zero point energy”. n=1

Quantum particles cannot remain stationary, as this would violate the

Heisenberg uncertainty principle because since the particle is not moving,

the position and momentum are both known with infinite precision.

23) E ? 1/a. Energy level spacing depends on size (and shape) of the box. n

The smaller the box, the more observable quantum mechanics becomes.

For a bigger box, the energy levels are more closely spaced and quantum

mechanics is not as important, in accordance with the Bohr

correspondence principle.

Lecture 8 4/6 Chemistry 344

E5

E4

E3 E

E2

E1

0

Notes on wavefunctions

(x) has n-1 nodes (more nodes ? higher energy) 1) n

2 2) P(x)=| (x) | has a maximum at x=a/2 and minima at x=0 and x=a. The 11

most probable region for finding the particle is at x=a/2 and the least probable

regions are at x=0 and x=a.

area=1 (x) is Classically, PCl

independent of x (x) PCl

x x=a x=0

3) Consider P(x) for large n. Nodes are so close together that they are n

indistinguishable, approaching the classical result:

Bohr Correspondence

Principle: in the limit of large P(x) ;~n, QM ; CM

x x=a x=0 Lecture 8 5/6 Chemistry 344

4) To localize a particle, add wavefunctions together to create a wave packet,

then square:

m

(x)N(x)P(x)*(x)(x)(nn1n

m=1 m=10 m=50

P(x)

Energy of (x) is not “well-defined” because each (x) has a different energy.

Heisenberg uncertainty principle: as position becomes better defined, uncertainty

in momentum (KE) increases

Quantum Brain-Teaser: Consider a particle in a box described by . How does the 22particle get from one side of the box to the other through the node where P==0?

Lecture 8 6/6 Chemistry 344

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