Energy Eigenfunctions and Eigenvalues of the Finite Well: Solving the Schrödinger Equation, Slides of Microwave Engineering and Acoustics

An in-depth analysis of the energy eigenfunctions and eigenvalues of a finite quantum well. It covers the solution of the schrödinger equation for different potentials, the classification of states as bound or unbound, and the behavior of the wave function in different regions. The document also discusses the normalization of the wave function and the determinant condition for a solution.

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ENERGY EIGENFUNCTIONS &
EIGENVALUES OF THE FINITE WELL!
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Download Energy Eigenfunctions and Eigenvalues of the Finite Well: Solving the Schrödinger Equation and more Slides Microwave Engineering and Acoustics in PDF only on Docsity!

ENERGY EIGENFUNCTIONS &

EIGENVALUES OF THE FINITE WELL

Solve the energy eigenvalue equation for different potentials

and for examples where there are many solutions with different

energies.

V ( x ) =

V 0

x > a

0 x < a

!

"

$

-a ax

Energy

V 0

0

E

E

Region 1 Region 2 Region 3

" (^) ( x ) =

1

x < # a

2

a < x < a

3

x > a

unbound states

(continuum states)

bound states

" 1

( x ) =^ Ce

ik

1

x

  • C ' e

  • ik 1

x

Focus first on the case E < V 0

("bound states")

In regions 1 & 3, k is imaginary

! 3

( x ) =^ D^ ' e

" ik 3

x

  • De

  • ik 3

x

Imaginary k means

exponential growth or

exponential decay!

(classically forbidden region)

!

" 2

( x ) =^ Ae

ik

2

x

  • Be

  • ik 2

x

Real k means oscillatory

behavior

(classically allowed region)

In region 2, k is real

It would not be physically reasonable to

allow an infinite probability of finding a

particle in a classically forbidden region.

-a ax

Energy

0

V 0

E

(^1 2 )

2

( x ) =^ Ae

ik 2

x

  • Be

ik

2

x

3

( x ) =^ D ' e

ik

1

x

  • De

ik 1

x

1

( x ) =^ Ce

ik

1

x

  • C ' e

ik 1

x

" (^) ( x ) =

1

x < # a

2

a < x < a

3

x > a

ϕ is continuous everywhere

ϕ has a continuous derivative

ϕ goes to zero at ±∞

ϕ is normalized

0 0

k 1

= i

2 m

!

2

V 0

(!^ E )

k 2

2 mE

2

4 equations, 5 unknowns ( A , B , C , D, E ). ( E is buried in k 1

and k 2

)

Normalization gives fifth condition.

e

" ik 2

a

e

ik 2

a

" e

ik 1

a

e

ik 2

a

e

" ik 2

a

0 e

ik 1

a

ik 2

e

" ik 2

a

" ik 2

e

ik 2

a

" ik 1

e

ik 1

a

ik 2

e

ik 2

a

" ik 2

e

" ik 2

a

0 ik 1

e

ik 1

a

A

B

C

D

This set of equations has a solution when the determinant of the 4x

matrix is zero. Tedious! See Liboff for details. When the

determinant condition is set up, we get a condition on E! This

condition can be satisfied in 2 sets of ways. One set has A = B

(even solutions) and the other set has A = - B (odd solutions).

2 mEa

2

!

2

Here is one condition for the determinant to be zero

(Eqn 5.83 in McIntyre):

tan

2 mEa

2

!

2

=

2 m V 0

(!^ E ) a

2

!

2

2 mEa

2

!

2

one value of E

3 values of E

-0.

0

1

-0.1 -0.05 0 0.05 0.

-0.

0

1

-0.1 -0.05 0 0.05 0.

-0.

0

1

-0.1 -0.05 0 0.05 0.

tan

2 mE 1

a

2

!

2

=

2 m V 0

! E 1

( ) a

2

!

2

2 mE 1

a

2

!

2

tan

2 mE 3

a

2

!

2

=

2 m V 0

! E 3

( ) a

2

!

2

2 mE 3

a

2

!

2

! cot

2 mE 2

a

2

!

2

=

2 m V 0

! E 2

( ) a

2

!

2

2 mE 2

a

2

!

2

-0.

0

1

-0.1 -0.05 0 0.05 0.

-0.

0

1

-0.1 -0.05 0 0.05 0.

-0.

0

1

-0.1 -0.05 0 0.05 0.

This set corresponds to the green

curves on the previous graphs - the

value of V 0

that yields 3 solutions (

even and 1 odd).

Note the size of the decay length for

the state corresponding to each

energy. Wave function "leaks" into

forbidden region. We call this an

evanescent wave.

E n

=

n

2

!

2

!

2

2 m (^) ( 2 a )

2

-0.

0

1

-0.1 -0.05 0 0.05 0.

! n

( x ) =

2

2 a

cos

n " x

2 a

-0.

0

1

-0.1 -0.05 0 0.05 0.

! n

( x ) =

2

2 a

sin

n " x

2 a

n =1,3,5 (symmetric or even solutions)

n =2,4,6 (antisymmetric or odd solutions)

n

( x ) = (^0) n =1,2,3,4,5… for x<-a and x>a

x=-a x=a

x=-a x=a

E n

=

n

2

!

2

!

2

2 m (^) ( 2 a )

2

-0.

0

1

-0.1 -0.05 0 0.05 0.

-0.

0

1

-0.1 -0.05 0 0.05 0.

! n

( x ) =

2

2 a

sin

n " x

2 a

n =1,2,3,4,5… and 0 < x < 2a

(neither symmetric nor

antisymmetric solutions - about

x=0)

n

( x ) = (^0) n =1,2,3,4,5… for x < 0 and x > 2a

x=0 x=2a

x=0 x=2a

  • Hamiltonian - set up with piecewise potential
  • Solve energy eigenvalue equation
  • Matching boundary conditions - continuity of φ and φ '
  • Graphical solutions will suffice for now
  • Discrete energies for bound states
  • Limiting case is well-known infinite square well problem
  • Mathematical representations of the above

ENERGY EIGENFUNCTIONS &

EIGENVALUES OF THE FINITE WELL

REVIEW