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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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