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
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
Quantum Mechanics: Solve Schrodinger Equation for energy E and wavefunction ：(x)
of the particle
：：ˆH(x)！E(x)0 22：：：，d，(x)？U(x)(x)！E(x)22mdxU(x)=0 if 0 ； x ； a 22，d(Particle is in the box) ：：()()0，x，Ex！22mdx
2d2mE：： (x)？(x)！022，dx;；？2；！This is a linear homogeneous
differential equation with
(xconstant coefficients ：Try(x)！e
；ix，i；x：！？Thus(x)cece12Euler formula ；；；；！？？，？，ccosxisinxccosxisinx；？；？；？；？；？；？12
；；Cos/sin are even/odd ！？？，cccosxcicisinx；？；？；？；？1212Regroup constants
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
n，x)： 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
a，nx)：22sin1！Bdx?，?Inside back cover a，?0 McQuarrie problem 3.24 a)：21！B?，2，?
22；，2mE；E！！Definition of ； n22m，
22Boundary condition ，，n)：！?， 2ma，?
Lecture 8 3/6 Chemistry 344
22216h24h9h1h En22228ma 8ma8ma8ma
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
Lecture 8 4/6 Chemistry 344
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:
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,
m=1 m=10 m=50
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