Physics 249 Lecture 12, Oct 1st 2012
Reading: Chapter 6
HW4 posted on the course homepage, due Friday
1) The infinite square well potential.
It is useful to consider infinite potentials. While infinite potentials will introduce
discontinuities in the properties of the wave function, they give easy to understand
boundary conditions simplifying solutions. The probability in the infinite region of the
potential is zero and only a limited number of functions can satisfy this boundary
condition. The infinite potential can be representative of physical situations involving
potentials large enough that a particle cannot escape the potential well. However, if we
want to understand the behavior at the boundary we will have to analyze the problem
more accurately.
Aside
Similarly we may be interested in time dependent systems but treat them in time
independent ways. When an electron transitions between initial and final states in those
two states the system are time independent, but transition itself if time dependent. If we
want to understand the transition we will have to employ methods to analyze that
transition more accurately.
Consider:
V(x) = 0, 0<x<L and infinity elsewhere.
The potential is time independent so we can use the time independent version of the
Schrodinger equation.
ℏ2 𝜕 2 𝜓(𝑥)
𝐸𝜓(𝑥) = −
+ V(x)𝜓(𝑥)
2𝑚 𝜕𝑥 2
In this case the second derivative of the wave function in space plus a constant is equal to
infinity at the boundaries. Thus the wave function itself can and should be continuous,
while the first derivative, slope, of the wave function is discontinuous. For a continuous
wave function representing the probability the boundary condition specified by this
potential requires require the wave function to be zero at the boundaries. There is a
discontinuity at the boundary so we can transition from some function to zero at that
point.
Inside the box
ℏ2 𝜕 2 𝜓(𝑥)
𝐸𝜓(𝑥) = −
2𝑚 𝜕𝑥 2
We need wave function that is zero at 0 and L and gives a negative constant when
differentiated twice. In genera the solution is a sin or cos or imaginary exponentials (note
the imaginary exponential solution is just a specific superposition of sin and cos or vice
versa). However, only the sin can be set to zero at x=0.
𝜓𝑛 (𝑥) = 𝐴𝑛 𝑠𝑖𝑛(𝑘𝑛 𝑥)
where
𝜋
𝑘𝑛 = 𝑛 , 𝑛 = 1,2,3, …
𝐿
𝐸=
𝑘𝑛2 ℏ2
𝜋 2 ℏ2
= 𝑛2
2𝑚
2𝑚𝐿2
We solved for A in the homework from Friday
∞
∞
𝐿
𝜋
= ∫ 𝐴2𝑛 sin2 (𝑛 𝑥) 𝑑𝑥 = 1
𝐿
−∞
𝑜
𝐿 2
2
𝐴𝑛
𝜋
𝐴𝑛
𝜋
𝐴2𝑛
=∫
(cos (2𝑛 𝑥) − 1) 𝑑𝑥 =
(𝐶𝑠𝑖𝑛 (2𝑛 𝑥) − 𝑥) |𝐿0 =
𝐿=1
𝐿
2
𝐿
2
𝑜 2
∫
|𝜓(𝑥)|2
𝑑𝑥 =
∫ 𝜓𝑛∗ (𝑥)𝜓𝑛 (𝑥)
−∞
A is independent of n.
2
𝐴=√
𝐿
2
𝜋
𝜓𝑛 (𝑥) = √ 𝑠𝑖𝑛 (𝑛 𝑥) , 𝑛 = 1,2,3, …
𝐿
𝐿
2) Finite potential well.
Consider the same problem with a finite potential well and energy low enough that the
object is confined to the well.
In general the solutions must be symmetric around the center of the well so it is useful to
consider a well centered at 0 with walls at +-L/2
The boundary is not infinite so the wave function value and slope should be continuous.
In fact for the wave function to be continuous and be a valid solution of the Schrodinger
equation we will find that the probability outside the box does not have to be zero.
Inside the square well the Schrodinger equation is:
ℏ2 𝜕 2 𝜓(𝑥)
𝐸𝜓(𝑥) = −
2𝑚 𝜕𝑥 2
The general solution can be expressed as sin and cos or imaginary exponentials. The
book used qualitative arguments about the curvature of the functions to determine the
correct functions. Using the superposition of exponentials this can also be understood.
𝜓𝑛 (𝑥) = 𝐴1𝑛 𝑒 𝑖𝑘𝑛𝑥 + 𝐴2𝑛 𝑒 −𝑖𝑘𝑛𝑥
Outside the box the Schrodinger equation is
𝐸𝜓(𝑥) = −
ℏ2 𝜕 2 𝜓(𝑥)
+ 𝑉0 𝜓
2𝑚 𝜕𝑥 2
with 𝐸 < 𝑉0
or
ℏ2 𝜕 2 𝜓(𝑥)
−(𝑉0 − 𝐸)𝜓(𝑥) = −
2𝑚 𝜕𝑥 2
for which the general solution are exponentials
𝜓𝑛 (𝑥) = 𝐵1𝑛 𝑒 𝛼𝑥 + 𝐵2𝑛 𝑒 −𝛼𝑥
𝛼=
1
√2𝑚(𝑉0 − 𝐸)
ℏ
Outside the box at infinity we will want the probability to go to zero. Therefore:
𝜓𝑛 (𝑥) = 𝐵1𝑛 𝑒 𝛼𝑥 , 𝑥 < −
𝜓𝑛 (𝑥) = 𝐵2𝑛 𝑒 −𝛼𝑥 >
𝐿
2
𝐿
2
Next enforce continuity of the wave function on either side of the box.
𝐿
𝐿
𝐿
𝐿
𝜓𝑛 (− ) = 𝐵1𝑛 𝑒 −𝛼2 = 𝐴1𝑛 𝑒 −𝑖𝑘𝑛2 + 𝐴2𝑛 𝑒 𝑖𝑘𝑛2
2
𝐿
𝐿
𝐿
𝐿
𝜓𝑛 ( ) = 𝐵2𝑛 𝑒 −𝛼2 = 𝐴1𝑛 𝑒 𝑖𝑘𝑛2 + 𝐴2𝑛 𝑒 −𝑖𝑘𝑛2
2
The sum of imaginary exponentials is equal to an exponential. Therefore, the sum must
be such that the imaginary exponentials add exactly to a sin or cos function to eliminate
the imaginary potions. Thus there are two independent sets of solutions. Proof of
independence will come below.
𝐴1𝑛 = 𝐴2𝑛 , cos, and 𝐴1𝑛 = −𝐴2𝑛 , sin
Lets absorb the constants A in the imaginary exponential decomposition in into constants
C
𝜓𝑛 (𝑥) = 𝐶1𝑛 sin⁡(𝑘𝑛 𝑥) or 𝜓𝑛 (𝑥) = 𝐶2𝑛 cos(𝑘𝑛 𝑥)
𝐿
𝐿
for the sin solution: sin (𝑘𝑛 2) = − sin (−𝑘𝑛 2)
Therefore 𝐵1𝑛 = −𝐵2𝑛
𝐿
𝐿
For the cos solution: cos (𝑘𝑛 2) = cos (−𝑘𝑛 2)
Therefore 𝐵1𝑛 = 𝐵2𝑛
Lets call these constants D
𝐿
𝐿
𝜓𝑛 (𝑥) = 𝐶1𝑛 sin⁡(𝑘𝑛 𝑥), 𝜓𝑛 (𝑥 < 2) = −𝐷1𝑛 𝑒 𝛼𝑥 ,⁡𝜓𝑛 (𝑥 > 2) = 𝐷1𝑛 𝑒 −𝛼𝑥
𝐿
𝐿
2
2
𝜓𝑛 (𝑥) = 𝐶2𝑛 cos⁡(𝑘𝑛 𝑥), 𝜓𝑛 (𝑥 < ) = 𝐷2𝑛 𝑒 𝛼𝑥 ,⁡𝜓𝑛 (𝑥 > ) = 𝐷2𝑛 𝑒 −𝛼𝑥
Consider the cos solution. Continuity of the wave function and first derivative gives:
𝛼𝐿
𝐿
𝑘𝑛 𝐿
𝜓𝑛 (± ) = 𝐶2𝑛 cos (
⁡) = 𝐷2𝑛 𝑒 − 2
2
2
𝛼𝐿
𝜕
𝐿
𝑘𝑛 𝐿
𝜓𝑛 (± ) = 𝑘𝑛 𝐶2𝑛 sin (
⁡) = 𝐷2𝑛 𝛼𝑒 − 2
𝜕𝑥
2
2
Dividing the bottom by the top equation
tan (
𝑘𝑛 𝐿
𝛼
⁡) =
2
𝑘𝑛
for the sin solution
−cot (
𝑘𝑛 𝐿
𝛼
⁡) =
2
𝑘𝑛
For the cos solution. The tangent is a periodic function that runs upward vertically as a
function of k, while alpha over k is positive and approximately declines as the inverse of
k. Note exactly because k is related to the energy, which is in alpha as well. There will
be periodic, quantized, solutions where the two cross.
Rewrite this slightly as:
𝑘𝑛 𝐿
𝑘𝑛 𝐿
𝐿1
𝑚𝐿2 𝑉0 2𝑚𝐿2 𝑘 2 ℏ2
𝑚𝐿2 𝑉0
𝑘𝑛 𝐿 2
√
tan (
⁡) =
−
=
−(
)
√2𝑚(𝑉0 − 𝐸) = √
2
2
2ℏ
2ℏ2
4ℏ2 2𝑚
2ℏ2
2
or
−
𝑘𝑛 𝐿
𝑘𝑛 𝐿
𝑚𝐿2 𝑉0
𝑘𝑛 𝐿 2
cot (
⁡) = √
−
(
)
2
2
2ℏ2
2
where 𝑣 =
𝑘𝑛 𝐿
2
Plotting the right and left hand sides of these equations. We can see that when k is zero
the right hand side of the equation has a maximum value and declines parabolic-ly
starting there. There is also a maximum x value corresponding to E=V. Above that
value the particle in the potential has too much energy and can exist outside the well.
As V gets larger there are more valid solutions in the box and the solutions will be closer
𝑘 𝐿
𝑛𝜋
to when the trigonometric functions are maximum at 𝑣 = 𝑛2 = 2 as above for the
infinite solution!
Looking graphically at some solutions. Depending on the value of the potential and L
there can be substantial probability for the particle to exist outside the box. More on this
when we study tunneling.