Study Guide
Physics TDSE and TISE
-
University:
Amherst College -
Course:
PHYS 348 | Quantum Mechanics I Academic year:
2008
-
Views:
145
Pages:
12
Author:
Luca Conley
Physics 48 – January 30, 2008
• Apply separation of variables to the
TDSE to obtain the TIDS.
• Consider general characteristics of the
solutions to the TIDS.
• Solve the infinite square well. Solving the TDSE
2
Ψ( x, t )
∂Ψ( x, t )
∂
h
−
+ U ( x)Ψ( x, t ) = ih
2
∂t
2m ∂x
We assume that the solution is separable. i.e. Ψ( x , t ) = ψ ( x ) f (t )
2
δf (t )
h 2 f (t )∂ 2ψ ( x )
−
+ U (x )ψ (x ) f (t ) = ihψ ( x)
2
δx
δt
2m
Divide by Ψ( x , t ) = ψ ( x ) f (t )
ih δf (t )
h 2 ∂ 2ψ (x )
+ U (x ) =
This yields −
2
f (t ) δt .
2mψ ( x ) δx
Left side only depends on x, right only depends on t. Separating the equations
h 2 ∂ 2ψ ( x )
ih δf (t )
−
+ U (x ) =
2
2mψ ( x ) δx
f (t ) δt
To be true for all times and positions we require that both sides
equal a constant. Call it E.
df (t ) E
= f (t )
So
dt
ih
(Ordinary differential equation!)
−iEt / h
f
(
t
)
=
e
Which has a solution
.
If we identify E = hω .
−iω t
f
(
t
)
=
e
then
. And for the time dependent part
h 2 ∂ 2ψ ( x )
ih δf (t )
−
+ U (x ) =
2
2mψ ( x ) δx
f (t ) δt
ψ must satisfy the TISE
h2 d 2ψ (x)
−
+U (x)ψ (x) = Eψ (x) .
2
2m dx General characteristics of TISE solutions
h 2 d 2ψ ( x )
−
+ U ( x )ψ ( x ) = Eψ ( x )
2
2m dx
∂2ψ
∂x
When E > U(x) (i.e. K > 0)
2
=
∂2ψ
−2 m
h
2
∂x 2
( E − U ( x ) )ψ
= − cnst . × ψ
In this case the wave function always curves towards the axis (oscillations).
When E < U(x) (i.e. K < 0)
∂2ψ
∂x 2
= cnst . × ψ
In this case the wave function always curves away from the axis.
To avoid diverging it must asymptote to the axis for large absolute value of x. Note: Curvature is proportional
to the product of the Kinetic
Energy and the amplitude of the
wave function. Liboff Figure 7.3 Solving the Time-independent Schrodinger
Equation
1)
Write the SE with the appropriate potential in all
regions of space.
2)
Solve the resulting differential equation in each
region of space.
3)
Apply boundary conditions (this usually determines
the allowed energies).
4)
For discontinuous potentials – require continuity of
the wave function, and usually its derivative.
5)
Evaluate all undetermined constants.
6)
Normalize – the probability of finding the particle
somewhere should be one.
7)
Discard any solutions with infinite probabilities
(they are non-physical).
8)
Calculate the expectation values of the quantities of
interest.
∞
f ( x) =
∫
−∞
f ( x ) Ψ ( x ) dx
2 Particle in a Box
With U(x) = 0 if 0 ≤ x ≤ L
and U ( x ) = ∞ if x < 0 or if x > L
Inside the box we have
2
2
ψ
∂
ψ( x)
∂
x
(
)
h
2m
2
−
=
ψ
=
−
E
ψ
x
≡
−
k
ψ( x)
E
x
(
)
(
)
or
2
2
2
2m ∂ x
∂x
h
2
The equation is that of a harmonic oscillator with solutions
ψ ( x ) = A sin( kx ) + B cos( kx ) .
We match these solutions within the well to the Boundary Condition that the
wavefunction must vanish a 0 and at L. These imply that B=0 and that the
constant k = nπ L , with n an integer. Thus k n = nπ / L = 2mEn / h with n = 1, 2, 3, ..
Applying the normalization condition on the solutions yields the eigenstates:
⎛ nπ x ⎞
2
ψn ( x) =
L sin⎜⎝ L ⎟⎠ where n = 1, 2, 3, ...
k 2 h 2 n 2π 2 h 2
=
And the corresponding energies are En =
2 .
2m
2mL
Any linear combination of these solutions is also a solution. Including the
time dependence we have the general solution:
Ψ( x , t ) =
∑
cnψ n ( x )e −iEnt / h
n
The cn are the coefficients that determine the admixture of the different states. ⎛ nπ x ⎞
2
(
)
⎟ where n = 1, 2, 3, ...
ψn x =
L sin⎜⎝
⎠
L
It can be easily shown that the ψn form a complete orthonormal set of
functions on the interval between x = 0 and L (Fourier’s theorem).
i.e.
∫ψ
*
m
( x ) ψ n ( x ) dx = δ mn
(Orthonormal) Any arbitrary function can be written as a linear combination of these eigenfunctions.
f ( x ) = ∑ cnψ n ( x )
(Complete)
n
One may use Fourier’s trick to determine the coefficients.
∫ψ
∞
( x) f ( x)dx = ∑ cn ∫ψ m ( x) ψ n ( x)dx
*
m
*
n =1
∞
= ∑ cnδ mn = cm
n =1
Once the coefficients are determined (usually from the initial
conditions) the general solution of the TDSE is known.
Ψ (x, t ) = ∑ cnψ n (x )e − iEnt / h
n
Physics TDSE and TISE
of 12
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document