Physics 249 Lecture 14, Oct 5th 2012
Reading: Chapter 6
HW 4 due today. No HW for next week.
Exam scheduled Friday Oct 12th. 50 minutes, in class
Covering chapters 3-6 excluding reflection and transmission of waves.
Closed book, 3-5 index card for equations allowed. Integrals and derivatives more
complex than polynomials, trig functions or first order exponentials will be provided.
1) Harmonic oscillator
The quantum harmonic oscillator problem is of interest because diatomic molecules
vibrate somewhat like two masses on a spring with a potential energy that depends upon
the square of the displacement from equilibrium. To good approximation for lower
energy this solution is also valid for molecular bonds in larger, more complex molecules
ℏ2 𝜕 2 𝜓(𝑥)
𝐸𝜓(𝑥) = −
+ V(x)𝜓(𝑥)
2𝑚 𝜕𝑥 2
1
1
𝑉(𝑥) = 𝐾𝑥 2 = 𝑚𝜔2 𝑥 2
2
2
or
2𝑚 1
𝜕 2 𝜓(𝑥)
2 2
(
𝑚𝜔
𝑥
−
𝐸)
𝜓(𝑥)
=
ℏ2 2
𝜕𝑥 2
Where the change of constants from the force constant to the angular frequency is
initially motivated from classical physics.
The second derivative of the wave function gives a constant plus a function of x2. One
solution is easy to guess.
𝜓(𝑥) = 𝐴𝑒 −𝛼𝑥
2 /2
The constants need to be determined. The x2 is to get the correct power of x and the ½ to
make the factors simpler.
Plugging into the Schrodinger equation.
2𝑚 1
𝜕
( 𝑚𝜔2 𝑥 2 − 𝐸) 𝜓(𝑥) =
(−𝛼𝑥𝜓(𝑥))
2
ℏ 2
𝜕𝑥
2𝑚 1
( 𝑚𝜔2 𝑥 2 − 𝐸) 𝜓(𝑥) = (−𝛼 + 𝛼 2 𝑥 2 )𝜓(𝑥)
ℏ2 2
and
𝛼=√
𝐸=
2𝑚2 𝜔 2 𝑚𝜔
=
2ℏ2
ℏ
ℏ2
ℏ𝜔
𝛼=
2𝑚
2
Is this the lowest energy state?
Consider the uncertainty principle. A particle in a harmonic oscillator potential is
essentially bouncing back a forth by ∆𝑥 with momentum ±𝑝 = ∆𝑝. Using these ideas I
can write the total energy and minimize it.
(∆𝑝)2 1
𝐸=
+ 𝑚𝜔2 (∆𝑥)2
2𝑚
2
rewrite in terms of x using the uncertainty principle where these values should be
ℏ
minimum: ∆𝑥∆𝑝 = 2
𝐸=
ℏ2
1
+ 𝑚𝜔2 (∆𝑥)2
2
(∆𝑥) 8𝑚 2
and minimize by setting the derivative equal to zero
−
ℏ2
+ 𝑚𝜔2 ∆𝑥 = 0
(∆𝑥)3 4𝑚
∆𝑥 = √
𝐸=
ℏ
2𝑚𝜔
ℏ𝜔 ℏ𝜔 ℏ𝜔
+
=
4
4
2
The solution we just found!
The complete solution can be expressed as the set of Hermite polynomials.
𝛼 1/4 1
2
𝜓𝑛 (𝑥) = ( )
𝐻𝑛 (√𝛼𝑥)𝑒 −𝛼𝑥 /2 ; 𝐻𝑛 = 1,2√𝛼𝑥, 4𝛼𝑥 2 − 2, … ; 𝑛 = 0,1,2 …
𝑛
𝜋
√2 𝑛!
1
and 𝐸𝑛 = (𝑛 + 2) ℏ𝜔; 𝑛 = 0,1,2 … - a linear progression in energy levels!
The energy dependence and the functional forms of the waves are quite different than the
particle in a box case. Though graphing them reveals that there are similar wave like
behaviors.
The wave functions are symmetric for the even values of n and anti-symmetric for the
odd values or one. However, the wave function is not a directly observable object. The
probability density |𝜓(𝑥)|2 is completely symmetric around x=0. This was expected
since the potential was symmetric.
2) Operators
In deriving the Schrodinger equation in a systematic way we introduced the concepts of
operators. Operators are mathematic operations that when performed on the wave
function can extract a given value, assuming the wave function has a definite value of
that quantity. If the wave function has a definite value for a certain quantity then the
wave function is an Eigenstate of that quantity and applying the operator to the wave
function will extra that value times the original wave function.
Common operators:
𝑝 → −𝑖ℏ
𝜕
𝜕𝑥
a plane wave in free space is an eigenstate of p. Though a realistic particle in free space
that is represented by a wave packet is not.
𝐸 → 𝑖ℏ
𝜕
𝜕𝑡
and the related operator, the Hamiltonian
𝐻=
𝑝2
ℏ2 𝜕 2 𝜓
+𝑉 ⇒−
+𝑉
2𝑚
2𝑚 𝜕𝑥 2
Since the wave functions of the systems we are currently investigating are simple
solutions to the Schrodinger equation they will be Eigenstates of the Hamiltonian.
𝐻𝜓 = 𝐸𝜓
3) Expectation values
Even if a particle is not in a definite state of some quantity you can calculate an average
value or expectation value. Expectation values are the weighted averages - weighted by
the probability distribution.
For functions of position you just use x and for function of momentum the momentum
operator.
The expectation value for a function f is calculated as follows.
∞
⟨𝑓⟩ = ∫ 𝜓 ∗ (𝑥, 𝑡)𝑓𝜓(𝑥, 𝑡)𝑑𝑥
−∞
For any wave function it is possible to calculate expectation values. Some interesting
values are
⟨𝑥⟩: the average position
√⟨(𝑥 − ⟨𝑥⟩)2 ⟩ = √⟨𝑥 2 − 2𝑥⟨𝑥⟩ + ⟨𝑥⟩2 ⟩ = √⟨𝑥 2 ⟩ − ⟨𝑥⟩2 : the standard deviation from the
average position
∞
∞
𝜕
⟨𝑝⟩ = ∫−∞ 𝜓 ∗ (𝑥, 𝑡)𝑝𝜓(𝑥, 𝑡)𝑑𝑥 = ∫−∞ 𝜓 ∗ (𝑥, 𝑡) (−𝑖ℏ ) 𝜓(𝑥, 𝑡)𝑑𝑥 : the average
𝜕𝑥
momentum.
These values can provide characteristic information about quantum systems. For
instance, after solving for the atom if we find the average deviation from r=0 is large then
that means the electron is typically far from the center making it more available for
forming bonds. Also if the distribution is not spherically symmetric this will give define
an orientation for that atom along which it will bond. The chemical properties of all the
atoms are defined by these type of characteristics.