Solving the Schrodinger Equation

Time-dependent Schrödinger Equation:

! (x,t) = "(x)#(t)

Time-independent Schrödinger equation:

This equation is not always easy to solve analytically, but can besolved numerically on a computer.However we can analytically solve some special cases….

!!2

2m

d2

dx2"(x) +U(x)"(x) = E"(x)

i!!

!t" (x,t) = #

!2

2m

!2

!x2" (x,t) +U(x)" (x,t)

!(t) = e" i# t , # =E

!

Stationary Solutions:

Solving the Schrodinger Equation

Consider a particle inside a box of length L with impenetrable walls:Outside the box the particle wave function is 0.Inside the box, U(x) = 0Inserting into the TISE,

!!2

2m

d2

dx2"(x) = E"(x)

Solutions inside the box:

!(x) = Acos(kx) + Bsin(kx), k =2mE

!

Time-independent Schrödinger Equation: Particle in a box(infinite square well potential):

!!2

2m

d2

dx2"(x) +U(x)"(x) = E"(x)


The wave function is continuous so the wave function inside the boxmust match the wave function outside the box at x=0 and x=L.Outside the box, the wavefunction is always 0.

Solutions inside the box at x=0 and L:!(0) = Acos(k0) + Bsin(k0) = A = 0

!(L) = Acos(kL) + Bsin(kL) = 0


A = 0, kL = n! , n = 1,2...

Acceptable solutions are thus

!n(x) = Bsin

n" xL

#$%

&'(, n = 1,2,3..

(For n=0, ψ(x)=0, particle not found in the box, hence not acceptablesolution)


Solutions :


Energy is quantized:

!n(x) = Bsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

Non-zero minimum energy E1 (zero-point energy)

En =1

2mv

2=p2

2m=!2k2

2m=!2n2!2

2mL2

This satisfies the uncertainty principle. If the minimum energy were0, then the momentum would be precisely 0, and then the location ofthe particle would be unknown - it would not be confined to the box.


Solutions :


Normalization:

!n(x) = Bsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

!n(x)

2dx

0

L

" = B2sin

2 n# xL

$%&

'()dx

0

L

" = B2 1

21* cos

2n# xL

$%&

'()

$%&

'()dx

0

L

"

=B2x

20

L

*L

2n# xsin

2n# xL

$%&

'()0

L

=B2L

2= 1

B =2

L


Solutions :


!n(x) =

2

Lsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

En=!2n2!2

2mL2, n = 1,2,3...


A particle is in a box of length L in the ground state (lowest energystate). Where is the particle most likely to be found? What is theprobability of finding the particle in the middle half of the box?


!1(x) =

2

Lsin

" xL

#$%

&'(, n = 1

The particle is most likely to be atx=L/2 where the probability densitycurve is a maximum.



!1(x) =

2

Lsin

" xL

#$%

&'(, n = 1

The middle half of the box is the region from x=l/4 to x=3L/4.Probability of finding the particle in the middle half of the box:

P(L / 4 ! x ! 3L / 4) = "1(x)

2dx

L /4

3L /4

#

2

Lsin

2 $ xL

%&'

()*dx

L /4

3L /4

# =x

L+1

$sin

$ xL

%&'

()*cos

$ xL

%&'

()*

L /4

3L /4

, P(x) =1

2+1

$= 0.818 = 81.8%




Solutions :

!n(x) =

2

Lsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

Orthonormality: Notice that

!n(x)" !

m(x)dx = #

mn

i.e. the above integral is zero except when m=n.For the case of m=n, the integral is 1 (normalization).This property holds for stationary solutions of other potentials as well.



Solutions :

!n(x) =

2

Lsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

Completeness: Any function f(x) can be written as a linear combination of ϕn(x)

f (x) = cn!n (x)n=1

"

#

This is nothing but the use of Fourier series or Dirichlet’s theorem



Solutions :

!n(x) =

2

Lsin

n" xL

#$%

&'(, n = 1,2,3.., 0 ) x ) L

!(x) = 0, x > L, x < 0

Completeness: Any function f(x) can be written as a linear combination of ϕn(x)

f (x) = cn!n (x)n=1

"

#

How to calculate the coefficients cn ? Well,

cn = cm!mn =

m=1

"

# cm $n

*(x)$m (x)dx% =

m=1

"

# $n

*(x) f (x)dx%


Pclassical

(x) =1

L

Correspondence principle: Recall that the correspondence principleas formulated by Bohr states that generally when the quantum numberof a system becomes large it should behave classically.

Average classical probability densitymatches the quantum value for largequantum numbers n

For a particle moving with constant speedv inside a box, the classical probabilitydensity is constant:


Solving the Schrodinger Equation

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