Physical Chemistry 20130508 week 6 Wednesday May 8, 2013 page 1
−ℏ2 d2 ψ(x)
+V(x)ψ(x)=Eψ(x)
2m dx 2
equation I
For n particles:
-ℏ2 ∂2 ψ ∂2 ψ ∂2 ψ
ℏ2 ∂2 ψ ∂2 ψ ∂2 ψ
+
+
( 2 + 2 + 2 ) -…(
) +Vψ=Eψ
2m ∂x1 ∂y1 ∂z1
2m ∂xn2 ∂yn2 ∂zn2
That’s assuming all particles have the same mass.
V depends on each particle’s position.
V=V(x1,y1,z1,x2,y2,z2,…,xn,yn,zn)
time independent
E is total energy for system = KE + PE = kinetic + potential energy for all particles in the system
States (having a definite energy) for which the probability distribution is independent of time are called
stationary states.
For stationary states:
d|Ψ|2
=0
dt
d|ψ|2
=0
dt
Stationary state = probability distribution doesn’t change with time (although particle is still moving)
Also, for stationary states, the energy is constant with time.
𝛹and 𝜓 must be:
1. single valued (has to pass vertical line test) so no loops
2. continuous
(can’t have any gaps in curve)
3. quadratically integrable
(can’t go to infinity) so y=x2 is ruled out
∫|𝜓|2dτ=1 If y=x2 then:
∞
∞
1
∞
∫ |ψ|2 dx = ∫ x 4 dx = x 5 | =∞--∞=∞ so f(x)=x 2 is not quadratically integrable
5
-∞
-∞
-∞
If a function meets all 3 criteria, we say it’s well behaved. The wave function which meets all of the
above conditions is said to be well-behaved.
How does total energy relate to 𝜓? 𝜓 is well behaved only for certain values of total energy (E).
𝜓=𝜓(x1,y1,z1,…,xn,yn,zn,E)
How about potential energy?
For isolated system (not influenced by external field like electric field or magnetic field) (molecule,
atom…), V is independent of time and the system can exist in stationary state.
Note: For an isolated system, there may be several stationary states, each represented by a
wavefunction, 𝜓.
Particle in a one dimensional box
- very crude, primitive model
Consider a graph with potential energy on the vertical axis and position on the horizontal axis. The box
starts at location 0 on the left and goes to location a on the right. This means the length of the box is a.
Left of the box is region I. Right of the box is region III. Inside the box is region II. At locations 0 and a,
the height of the box wall is infinite, so V is infinite. Outside the box (in regions I and III) V is infinite.
Inside the box, V is zero. There is only one particle in the box and the particle has mass m.
0
V{
∞
-ℏ2 d2 ψ
+Vψ=Eψ
2m dx 2
𝜓I=0
0≤x≤a
x<0 or x>a
one dimensional box so ψ=ψ(x)
𝜓III=0 since 𝜓 is well behaved
|𝜓I|2=0
𝜓III|2=0
For region II: 𝜓≠0 but V=0
-ℏ2 d2 ψ
=Eψ
2m dx 2
d2 ψ -2mE
= 2 ψ
dx 2
ℏ
since Vψ=0
0≤x≤a
equation I
That’s a second order differential equation to solve.
𝜓=Asin(rx)+Bcos(sx)
constants.
This is our trial function. A,B,r, and s are all constants. A and B are integration
dψ
=Arcos(rx)-Bssin(sx)
dx
d2 ψ
=-Ar 2 sin(rx) -Bs 2 cos(sx)
dx 2
equation II, substitute into equation I
-Ar 2 sin(rx) -Bs 2 cos(sx) =
-2mE
ψ
ℏ2
equation III
1⁄
2
(2mE)
Let s=r=
ℏ
-A
so s 2 =r 2 =
2mE
2mE
-2mE
sin(rx) -B 2 cos(sx) = 2 ψ
2
ℏ
ℏ
ℏ
1⁄
2
(2mE)
ψ=Asin (
ℏ
2mE
ℏ2
divide by
1⁄
2
(2mE)
x) +Bcos (
ℏ
x)
-2mE
ℏ2
0≤x≤
Now we find A and B by imposing boundary conditions. 𝜓=0 when x=a and when x=0 (boundaries) so:
0=Asin0+Bcos0
so B=0.
1⁄
2a
(2mE)
0=Asin (
ℏ
) +0
1⁄
2a
(2mE)
For that to be zero:
ℏ
1⁄
2
(2mE)
ℏ
ψ=Asin (±
nπx
nπx
) =±Asin (
)
a
a
nπx
ψ=Asin (
)
a
Next we find energy…
=±
nπ
a
since B=0
=±nπ
n=0,1,2,…
equation IV
±give the same solution we we can drop the-
n≠0 because then ψ≡0 so n=1,2,3,…