Wednesday, October 8, 2014

Selected Solutions for Quantum Mechanical Operators – Part 2


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]

Solution:

By definition:  [p]  =   ò ¥-¥   U(x) p^x U(x)*  dx

Where:   p^x    = -i ħ / x

Then:

[p]=

 ò ¥-¥     A exp [-i(a - a2 ħ t/ 2m] (-i ħ / x) A exp [i(a - a2 ħ t/ 2m] dx

[p]=  

-i ħ ò ¥-¥     A exp (-ia x) / x  A exp (ia x) dx

But:  / x  [A exp (ia x)]   =  ia A exp (ia x)

Therefore:

[p] =  (-i ħ ) (ia )  ò ¥-¥     AA* dx 

 =  ħ a ò ¥-¥     AA* dx 

But, ò ¥-¥     AA* dx   = 1 is just the normalization condition so:

[p]=  ħ a

(N.B. Square brackets [] used for expectation values because blogger code rejects the standard ones)

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
 

Solution:

 
Write out:

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

And  y  =  y (r, q, j)

Then:  H op   y =  E y  =

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

Where:

p^r 2 =   (-iħ 1/r ( (r)/ r ) 2 = - ħ2   (1/r 2  /r  (r 2 /r )

But  ^2  =

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

Then we may write out the Schrodinger equation in spherical coordinates:

- ħ2/ 2m ( /r  (r 2 /r ) -

ħ2/ 2mr2 sin2 q [ sin q / q (sin q / q) 

+ 2 / j2 ] y (r, q, j) +  V(r) y (r, q, j)
 

Replace m by the reduced mass, m = mM/ m+M, then:

- ħ2/ 2m [ /r  (r 2 /r ) - 1/ r2 sin2 q   sin q / q (sin q / q) 

+ 2 / j2 ] y (r, q, j) +  V(r) y (r, q, j) = Ey (r, q, j)

Which is the appropriate Schrodinger equation in spherical coordinates.

 
6) In the case of the first solution (5(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.

 
Solution:

 
exp [ i m   (j + 2p)] = exp (i m j) =  1

 
Then by de Moivre’s theorem:

cos 2p m   +  i sin 2p m   = 1


But:   sin 2p m   = 0  so:  cos 2p m   = 1

Which requirement is satisfied provided the absolute value of  m    has one of the following values:

ç m ç  =   0, 1, 2, 3, 4……

 
(10) The Hamiltonian operator for the quantum harmonic oscillator is:  H^ =  p^2/ 2m + mw2 x2/ 2.   Also: 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.

Solution:

We have for the Hamiltonian:

H^ =  p^2/ 2m + mw2 x2/ 2

And the kinetic energy operator: p^ = i h  ( /x).

It is useful to sketch the quantum potential at this point and also get the normalization condition. We have:

Harmonic oscillator quantum potential
And since we know the oscillator (on account of tunneling)  can penetrate beyond the:

 –A < x < A limits of the potential,  then we need to use:

yo (y) =  A exp [-y2/ 2]  where y = (mw/ h)1/2 x

To get the value of A the normalizing factor, we have:

ò ¥- ¥  y(y) 2  dy =  1

Then:

y(y) 2  =  [A exp (-y2/ 2)] 2  = A2 exp (-y2)

So that:   A2  ò ¥- ¥  exp (-y2) dy = 1

And we may use the well known integral:

ò ¥- ¥  exp (-y2) dy =  Öp

Simplifying:

A2  Öp  =   1  and A =   (1/Öp ) 1/2

We now modify the energy operator p^ = i h  ( /x).

Instead we use:  p y^ = p y ^2/ 2m=  (-i h Öa  /y) 2/ 2m

=  a ħ2 / 2m [2 /y 2 ]

(Note:  Öa  =  mw/ ħ )1/2

The appropriate operator equation is then:

H y ^   = (-i h Öa  /y) 2/ 2m + y (mw/ h)- 1/2

And Schrodinger’s equation becomes – in terms of y (y)  and a:

d2y/ dy2  +  (a2 – y2) y  = 0  

The acceptable solutions for the above are constrained by the condition:

ò¥- ¥  y(y) 2  dy =  1

Which shows that as y ® 0, y ® ¥

The properties of the given Schrodinger’s equation are such that the normalization condition can only be fulfilled if:

a       = 2n + 1 and n = 0, 1, 2, 3 ….

Since Öa  =  (mw/ ħ )1/2          and  w = 2pn

But E = ħ n  then:   a  =  2E/ ħ n

The energy levels of the harmonic oscillator are:

E n = (n + ½) ħ n     n = 0, 1, 2, 3

The energy is thereby quantized in steps of  ħ n  with the zero point energy:

E o =  ħ n/2

Because now :   a n  =  2 E n  / ħ n

We see that each choice of an a n  leads to a different wave function.  Each function then introduces a polynomial called a Hermite polynomial of the form: H n  (y) yielding either odd or even powers of y.  Thus, the Hermite polynomial must be introduced as part of any solutions for the wave function.

The form for any nth wave function then is:

yn  = (2mn / ħ ) ¼ (2n n!) -1/2 Hn (y) exp (-y2/2)

 

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