Friday, October 3, 2014

Introduction to Quantum Mechanical Operators (Pt. 2)

Continued from Part 1:
    2. Particle Confined in a 3-Dimensional Box or Square Well

One of the more interesting applications is for a particle confined to a 3-dimensional square ‘well’ or box, as shown:
No automatic alt text available.
 Using this we can set out the rectangular regions defined according to:

0 < x < a

0 < y < b

0 < z <  c

The potential energy inside the region is zero and outside is infinite so the Schrodinger equation becomes:

T op y  =  Eop  y

Where T op is the kinetic energy operator which is written in 3D as:

T op   =    p^ p^ / 2m  =  [p^ x 2  +  p^ y 2    + p^ z 2  ] / 2m

=  - ħ2/ 2m [ 2 / x2  +   2 / y2  + 2 / z2 ) = - ħ2/ 2m Ñ 2

Where Ñ is the Laplacian operator.

The Schrodinger equation then becomes:
2 y/ x2  +   2 y/ y2  + 2 y/ z2  + 2mE/  ħ2 y = 0

3. Solution of the 3D Square Well - Box Schrodinger Equation:

This is a partial differential equation easily solved by the separation of variables:

y  =   X(x) Y(y) Z(z)


X’ =   X/ x    Y’ =  Y/ y  and Z’ =  Z/ z   

This leads to the equation:

X”YZ  +  XY” Z   + XYZ” + 2mE/  ħ2  XYZ  = 0

The required normalization equation is then:

òao  òbo òc o  y2  dz dy dx = 1

Then, dividing the Schrodinger equation by XYZ:

X”/ X + Y”/ Y  + Z”/Z + 2mE /  ħ2   = 0

We let:

X”/ X =  - a2,  Y”/ Y  =  -b2,    Z”/Z =  -g2

With a, b   and g constants.

Then, we have:

x2   +  a2 x  = 0,   y2   +  b2 y  = 0,  z2   +  g2 z 

So the independent solutions will be:

x= Ö(2/a)   sin ( n xpx/a)] 

y= Ö(2/b)   sin ( n y px/b)] 

z= Ö(2/c)   sin ( n zpx/c)] 

where:  n x   = 1, 2, 3, 3 etc.


y  = (8/ abc)1/2 sin ( n xpx/a) sin ( n y px/b) sin ( n zpx/c)

And further, to obtain the quantized energy, E:

n2 x p2 /a2  -  n2 y p2 /b2  -  n2 z p2 /c2  +  2mE/  ħ2   = 0


E =   p2  ħ2 / 2m [n2x  /a2  + n2 y  /b2  + n2 z  /c2 )

If the box is a cube, so a = b = c:

E =   p2  ħ2 / 2m a2 (n2x  + n2 y    + n2 z )

(It should be noted that in this system, and for states higher in energy than the ground state, many levels-distinct states have the same energy. This property is called the degeneracy of states.)

4. The Hydrogen Atom

The Hamiltonian for the hydrogen atom Schrodinger equation can also be written in concise form as:

H op y  =  Eop  y

For which we only need to know the operators in order to expand it to its proper representation.  In this case:

H op    = p^r 2/ 2m +  ^2/ 2mr2 + V(r)

And: p^r    = - i ħ (1/r  2 / r2 )

The operators are radically different because we are dealing with an atom and treating it along the lines of radial symmetry – so the operator must be a function of r. (Spherical coordinate system)

We also note the wave function:  y = y (r, q, j)

Again, this is to preserve spherical symmetry.  The full Schrodinger equation for the hydrogen atom can then be written as:

[p^r 2/ 2m +  2/ 2mr2 + V(r)] y (r, q, f) =  Ey (r, q, j)

This equation can now be used, after further expansion of the assorted operators, to obtain the properties of the hydrogen atom, including its energy levels.

Problems (Continued from earlier set):

3)     Find the expectation value for the momentum of a particle with a wave function:
U(x) =  A exp [i(a - a2 ħ t/ 2m]

4)Using the operators,   p^r    and  ^2   

= - ħ2/ sin2 q [ sin q / q (sin q / q)  + 2 / j2 ]

Write out the full form of the Schrodinger equation for the hydrogen atom in spherical coordinates.  Thence, obtain the final form such that H op    =

- [E – V] y (r, q, j)

Hint: Replace m by the reduced mass,

m = mM/ m+M

5) In the form shown, the variables r, q, j may be separated by letting:

y (r, q, j)  =  R(r)  Q (q ) F( j)

And performing suitable manipulations. Do this in a similar way to the approach used to solve for the 3D box and then show the solutions which result are:

i) F( j) =  exp (i m j)

ii) Q (q ) = Q m =  sin m q F | m | (cos q)

iii) R nℓ    = exp(-Zr/nao) [Zr/ ao]  G   (Zr/ ao)

Explain how your solutions help to account for the Table entries below:

6) In the case of the first solution (i) we demand that the function be single-valued and continuous so that:
F( j + 2p)  = F( j)

So that: 

 exp [ i m   (j + 2p)] = exp (i m j)

where  m   is the magnetic quantum number. Dividing by
exp (i m j) we obtain:   

exp [ i m   (2p)] = 1

Indicate the condition on  m   for which this is satisfied.

7) The solutions Q m  are associated Legendre polynomials and require two quantum numbers, the magnetic and the azimuthal, ℓ. If ℓ is quantized and can only be a positive integer, show its acceptable values based on the values of m.

8) The solutions R nℓ    are Laguerre functions which also require two quantum numbers, ℓ and n to identify acceptable solutions. (Note: n is the principal quantum number)

The eigenfunctions y (r, q, j) are formed by taking the product of the 3 types of functions so that:
y nℓ m (r, q, j) = R nℓ   (r) Q m (q)Fm   ( j)

By inspection of the table (Prob. 4), identify the part or whole of the expression for the wave function y 210  that comes closest to the form for R nℓ   (r).

8) Complete the normalization condition for the H-atom:
ò ¥o  òpo ò2p o  y* nℓ m (r, q, j) y nℓ m (r, q, j)  dV = ? = 1

dV =   (               ) dr dq dj

9) Let the ground state wave function of the hydrogen atom be given by:  U =  exp (-ra)
Assume there is no relative motion of the nucleus so that q = 0 and  j = 0

a)     Show that the relevant Schrodinger equation would now be:  
- ħ2 / 2m [1/ r2  / r ( r2 ( / r)]U = [E – V(r))]U

b) If  V(r) =  - e2 / r   for the hydrogen atom, show that the quantized energy for this system is:
E = - ħ2 / 2m [a2  - 2a/ r] - e2 / r  

c) Find the specific energy in the ground state if:

r = ħ2 / m e2 

and:   a =    m e2 / ħ2

10) The Hamiltonian operator for the quantum harmonic oscillator is: 

H^ =  p^2/ 2m + mw2 x2/ 2.  

Also the momentum operator: p^ = i h  ( /x). Use these to write out the applicable Schrodinger equation then indicate the form of the expected solutions – with a diagram for the potential.

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