Quantum - Spherical Harmonics

Quantum Explanations – Issue II - Spherical Harmonics (Angular Momentum) Author: Ryan Jadrich Last Update: November 30, 2007 Introduction: Angular momentum on the macroscopic scale is really a trivial thing to discuss and model. Angular momentum has the exact same relations as linear momentum and is governed by strikingly similar equations. In the quantum world though something interesting happens with angular momentum. It becomes quantized like all other quantum things. You get certain energy levels and only certain states for the given quantum object which in this case will be a simple diatomic molecule. Working out this particular example is extremely complex and is possibly one of the most difficult examples to work out that can be solved exactly. After mastering this, the solution to the hydrogen atom is much less complex. Explanation: Part I. – First off we will consider what is called Legendre’s differential equation and its unique solutions. This will be integral to our concept of quantum angular momentum. The Legendre equation is stated below: (1  x 2 ) y ' '2 xy ' n(n  1) y  0

(1)

This equation can be solved using a power series method and we will try the general solution: 

y   ck x k

(2)

k 0

We can then take all the necessary derivatives of this trial function: 

y   ck x k k 0 

y '   c k k x k 1 k 1 

y ' '   c k k (k  1)x k  2 k 2

Copyright © Ryan Jadrich 2007

-1-

We can the plug in all of our functions into the Legendre equation (1) and get a new expression in terms of power series.       (1  x 2 )  c k k (k  1)x k 2   2 x  c k k x k 1   n(n  1)  c k x k   0  k 2   k 1   k 0 

(3)

We then simplify a little bit by multiplying through and get:     k 2  k k k c k ( k  1 ) x  c k ( k  1 ) x  2 c k x   k   k   k   n(n  1)c k x   0  k 2   k 2   k 1   k 0  Then we will take out necessary terms from each summation so that each summation starts at the same power of x which will be needed in order to combine power series.    2 c  6 c x  c k k (k  1)x k  2   ...  3  2 k 4        k ...   c k k (k  1)x   2c1 x   2c k k x k   ... k 2  k 2       ...  n(n  1)c0  n(n  1)c1 x   n(n  1)c k x k   0 k 2  

Then after reorganizing a bit and using a j substitution for k so that all of the power series begin with the same index number we end up with.

n(n  1)c0  2c2   (n  1)(n  2)c1  6c3 x   ( j  2)( j  1)c j 2  (n  j )(n  j  1)c j x j 

j2

For this expression to be true for all x it is necessary that all of the coefficients which are the bracketed parts be zero. This leads to:

n(n  1)c0  2c 2   0  c2  

n(n  1) c0 2!

(n  1)(n  2)c1  6c3 x  0  c3  

(n  1)(n  2) c1 3!

( j  2)( j  1)c  c j2  

j2



 ( n  j )(n  j  1)c j  0

(n  j )(n  j  1) cj ( j  2)( j  1)

Copyright © Ryan Jadrich 2007

for

j  2,3,4,...

-2-

0

The first few coefficients turn out to be: c0  c0 c1  c1 n( n  1) c0 2! (n  1)(n  2) c1  3! (n  2)(n  3) ( n  2)n(n  1)(n  3) c0 c2   43 4! (n  3)(n  4) ( n  3)(n  1)(n  2)(n  4) c3  c1  54 5! (n  4)(n  5) (n  4)(n  2)n(n  1)(n  3)(n  5) c4   c0  65 6! (n  5)(n  6) (n  5)(n  3)(n  1)(n  2)(n  4)(n  6) c5   c1  76 7!

c2   c3 c4 c5 c6 c7

Now we can form two linearly independent solutions from these coefficients and the form of a power series given as equation (2).  n(n  1) 2 (n  2)n(n  1)(n  3) 4 (n  4)(n  2)n(n  1)(n  3)(n  5) 6  y1  c 0 1  x  x  x  ... 2! 4! 6!   (n  1)(n  2) 3 (n  3)(n  1)(n  2)(n  4) 5   x  x  ... x  3! 5! y1  c1   ...  (n  5)(n  3)(n  1)(n  2)(n  4)(n  6) x 7  ...    7! The neat thing about this solution is that for an even integer for n the first series will terminate, and for an odd integer the second series will terminate. For this reason we take are solution to the equation as whichever series will terminate, and we opt to chose are coefficients in such a manner as to make them look uniform. These Legendre polynomials are the only well behaved solutions to this equation which is needed in quantum mechanics. 1  3  ...  (n  1) 2  4  ...  n 1  3  ...  n c1  (1) ( n 1) / 2 2  4  ...  (n  1)

c0  (1) n / 2

Copyright © Ryan Jadrich 2007

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With these coefficients we get solutions called Legendre polynomials which are commonly denoted by P. P0  1 P1  x 1 (3 x 2  1) 2 1 P3  (5 x 3  3x) 2 1 P4  (35 x 4  30 x 2  3) 8 1 P5  (63 x 5  70 x 3  15 x) 8 P2 

One way to generate these polynomials is called that of the Rodrigues formula as stated below without proof. l

Pl 





l 1 d  2  x 1 l  2 l!  dx 

This equation will be important to us later on were it will be restated and used to derive various relationships. These Legendre polynomials will be important in the solution to the quantum nature of angular momentum. Part II. – The second portion of this text will discuss the Schrödinger equation for that of angular momentum. It is assumed that knowledge of traditional angular momentum is present. The equation we are considering is given below:

2 2I

 1     1  2   sin      E  0   sin 2   2   sin   

(4)

Where I represents the moment of inertia of a diatomic molecule spinning about its center of mass with the bond length held constant.

Copyright © Ryan Jadrich 2007

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Now we just carry out the necessary differentiation and rearrange a bit to get: 2 2I

 1   sin 

   2  cos   sin    2 

 1  2     E  0 2 2   sin   

 cos    2 1  2  2 I    E  0  2 sin 2   2   2  sin      2 1  2 2 I  cos   sin    E sin   0   2 sin   2  2

(4)

(5)

Now we can use separation of variables in an attempt to find a solution to this differential equation. We will assume,

 ( ,  )  ( ) ( )

(6)

And substitute this into (5) to get,  ( )  2 () 1  2 () 2 I cos   sin    2 E sin  ()  0  sin   2  2    2 1  2 2I  cos    sin     2 E sin  ()  0 2 2  sin    

(7 )

We then divide equation seven by six and multiply by sine of theta to get, cos 

 1  2 1 1  2  1 2I  sin    E sin   0    2  sin   2   2

 sin  cos 

 1  2  1  2  1 2I  sin 2    E sin 2   0    2   2   2

(8)

Then we assume, 2I E  l (l  1) 2

For reasons which will become apparent. With this assumption our equation is now:

  1  2 1  2 1 2 2   l (l  1) sin    0  sin  sin  cos  2    2     

Copyright © Ryan Jadrich 2007

(9)

-5-

Since there are two separate parts to this equation dependent one separate variable, each piece must be equal to a constant. sin  cos 

 1  2 1  sin 2   l (l  1) sin 2   m 2 2    

 2 1  m 2 2  

(10)

(11)

Eleven is easily solved and the solution is quite simple.   c1e im  c 2 e im

(12)

Solving (10) though is an extremely difficult task, much more so than (11). First off we will assume: x  cos 

Then by the chain rule we can manipulate (10) as so,

  x  1  2 1 2 sin  cos   l (l  1) sin 2   m 2    sin  2 x      

(11)

Getting the relationship for the second derivative with x is a bit trickier than the first derivative. For this we will try something a little awkward.  

2   x       x    2   2   x   2 x  2       x  x  2  2

(12)

Next we get some simple relationships, x  cos  x   sin   x 2   x sin  2x   cos   2

(13) (14) (15)

Copyright © Ryan Jadrich 2007

-6-

Then we can plug (12) into (11) to get:

sin  cos 

2   x  1 x   2 x  1 2     sin      2   l (l  1) sin 2   m 2    x      x  x   

Then making all necessary substitutions using (13-15) we are left with,

 sin 2  cos 

  2 1   1 cos    l (l  1) sin 2   m 2  sin 2   2 sin 2   x x   x 

Then we shall divide through by sine squared and upon rearrangement get: sin 2 

  2 m2  2 cos   (  1 )   0 l l x x 2 sin 2 

(16)

Then by using our assumption that, x  cos 

The equation that we are left to solve after the separation of variables is: (1  x 2 )

 2  m2  2 x  l ( l  1 )   0 x x 2 (1  x 2 )

(16)

I assure you that this is a formidable problem and probably one of the most difficult ones you will encounter in quantum mechanics. The equation sure looks similar to Legendre’s equation which is: (1  x 2 )

d 2 d  2x  l (l  1)  0 2 dx dx

This equation while similar in looks is not half as hard to solve as the previous one which is referred to as the associated Legendre equation. The regular Legendre equation has relatively simple solutions which are referred to as Legendre polynomials. One way to attempt to solve the associated Legendre equation is to try and get it into a form that might look like the normal version. The first step in doing this is making the necessary substitution: ( x)  T (1  x 2 ) m / 2 C ( x)

Copyright © Ryan Jadrich 2007

-7-

Where T is just an arbitrary constant and C(x) is some unknown function. Do not attempt to understand why and how anyone even began to guess such a solution. It must have been some crazy smart math guy. After substituting in this function we defined we will get a new differential equation in terms of C(x) which actually looks more confusing than the original but oh well. (1  x 2 )

d 2C dC  2(m  1) x  (l  m)(l  m  1)C  0 2 dx dx

(17)

Under the convenient circumstances of m=0 our new equation becomes the original Legendre equation. (1  x 2 )

d 2C dC  2x  l (l  1)C  0 2 dx dx

for

m0

Now we can replace our unknown function C with the Legendre polynomial solutions P we found in the previous section. (1  x 2 )

d 2 Pl dP  2 x l  l (l  1) Pl  0 2 dx dx

(18)

Our next step is to differentiate this entire expression (18) m times and we will get yet another equation but it will end up looking very similar to (17).

dm dx m

2  dm 2 d Pl  ( 1 )  x    dx 2  dx m 

dm  dPl  Pl   0 2  (  1 ) x l l  dx  dx m  

Now these derivatives are rather difficult to work out and require the recognition of patterns for the coefficients of various terms. Don’t not worry too much about them as the pattern recognition is quite difficult, but if you feel so inclined work it out for yourself. It might be fun! m 2 2  Pl d m1 Pl d m Pl  dm  2 d Pl  2 d 2 (1  x ) 2   (1  x ) m  2  2mx m1  ( m  m) m  dx m  dx   dx dx dx 

dm dx m

m 1 d m Pl   dPl   d Pl 2 x  2 x  2 m   dx   dx m1 dx m    

d m Pl dm  P   l dx m dx m

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-8-

Now after all of the necessary substitutions we get: m2  Pl d m 1 Pl d m Pl   d m1 Pl d m Pl  d m Pl 2 d 2 x mx m m x m l l ( 1  )  2  (   )  2  2  (  1 ) 0     dx m 2 dx m 1 dx m   dx m1 dx m  dx m 

Then with some simplifications we get: m2  Pl d m 1 Pl d m Pl   d m 1 Pl d m Pl  d m Pl 2 d 2  2mx  ( m  m)  2m 0 (1  x )   2 x   l (l  1) dx m  2 dx m 1 dx m   dx m 1 dx m  dx m 

 (1  x 2 )

d m  2 Pl  d m 1 Pl d m 1 Pl   d m Pl d m Pl d m Pl  2 mx x m m m l l  2  2  (   )  2  (  1 )    0 dx m  2  dx m 1 dx m 1   dx m dx m dx m 

d m  2 Pl d m 1 Pl d m Pl 2  (1  x )  2(m  1) x  l (l  1)  ( m  m)  2m 0 dx m  2 dx m 1 dx m



2

 (1  x 2 )



d m  2 Pl d m 1 Pl d m Pl 2 m x l l m m  2 (  1 )  (  1 )   0 dx m  2 dx m 1 dx m





Now the last step is to change around the look of the coefficient in front of the last term of the last equation.

l (l  1)  m 2  m  l 2  l  m2  m  (l  m)(l  m  1) With the final form substituted into our equation we are left with. (1  x 2 )

d m 2 Pl d m1 Pl d m Pl  2 ( m  1 ) x   ( l  m )( l  m  1 )  0 dx m 2 dx m1 dx m

Whoa, this look just like (17) the original equation we set out to solve with our substituted function. For comparison purposes both equations are listed again below. (1  x 2 )

d 2C dC  2(m  1) x  (l  m)(l  m  1)C  0 2 dx dx

origional

d m 2 Pl d m 1 Pl d m Pl (1  x ) m 2  2(m  1) x m1  (l  m)(l  m  1) m  0 dx dx dx 2

Copyright © Ryan Jadrich 2007

-9-

What does this mean; well it means that the Legendre polynomials differentiated m times is our C function. Part III. – Now we have to remember in this section jut what it is that we are looking for after taking all of these detours. So this section will give a recap of all results and finally the solution to Schrödinger’s equation for angular momentum. First off the equation we are trying to solve was,

2 2I

 1     1  2    sin      E  0   sin 2   2   sin   

(4)

Where we assumed a separable solution of the form,

 ( ,  )  ( ) ( )

(6)

We found a solution readily for the phi function which was,   c1e im  c 2 e im

(12)

On the other hand for the other portion we were faced with much more difficulty. First off we assumed that the function was of the form, ( x)  T (1  x 2 ) m / 2 C ( x) Which gave us a new differential equation, (1  x 2 )

d 2C dC  2(m  1) x  (l  m)(l  m  1)C  0 2 dx dx

(17)

And had solutions of the form that we just found, C ( x) 

 m Pl x m

Now we are going to define new functions that solve the associated Legendre equation (16) in part which is, m

( x)  Pl  T (1  x 2 ) m / 2 C ( x) m

 Pl  (1) m (1  x 2 ) m / 2

 m Pl x m

Copyright © Ryan Jadrich 2007

(18)

- 10 -

Where we have chosen our constant T that is best suitable for our purposes. Now we are going to introduce with out proof something called the Rodrigues formula which gives a simple recursion relationship for the Legendre polynomials in terms of repeat derivatives as stated below. l

Pl 





l 1 d  2  x 1 l  2 l!  dx 

(19)

We can now insert this expression into (18), (1) m d  Pl  l (1  x 2 ) m / 2   2 l!  dx  m

l m

x

2



1

l

(20)

Now with Equations (18) and (20) we are going to prove some very important relationships involving these associated Legendre polynomials (which are not always actually polynomials ha!) The first relationship is to see if these functions are indeed orthogonal on the interval -1 to 1, which is good enough of an interval as when we substitute our cosine function in for x this ranges over a range of π radians which is the range that our theta value can assume. This orthognality relationship is shown below. 1

P

m

l

m

Pl ' dx  0

1

In order to prove this we will substitute equation (18) in for each of the associated Legendre polynomials. 1

P

l

m

m

Pl ' dx

1

1  d m P  d m P    (1  x 2 ) m  ml   ml '  dx  dx   dx  1

Copyright © Ryan Jadrich 2007

- 11 -

Now we take the final expression and integrate by parts once getting the expressions,  d mP  U  (1  x 2 ) m  ml   dx  m  d  2 m  d Pl  dU  (1  x )  m   dx dx   dx    d mP  dV   ml '  dx  dx  V 

d m 1Pl ' dx m 1

Then with integration by parts, 1

1 m m   m1 Pl '   m1 Pl ' d  2 m  d Pl  d 2 m  d Pl   d     Pl Pl ' dx  (1  x )  dx m   dx m1   1 dx (1  x )  dx m   dx m1 dx 1     1 1

m

m

It is clear that the portion outside of the integral equals zero because of the one minus x squared term which is zero at both limits. If we perform this repeated partial integration m times the portion outside the integral will always go to zero at the limits because it will be a polynomial with a factor of one minus x squared. After m integrations we get: m  dm  2 m  d Pl  P P dx ( 1 x )    Pl ' dx   l l '   m m  dx   1 1 dx  1

m

m

1

Now the portion in the integral, m  dm  2 m  d Pl  ( 1 ) x    m  m dx   dx  

Will be an even or odd polynomial of power l. For simplicity and with no loss of generality we assume l to be less than l’ because one of them must be smaller than the other if they are not the same. Now since this value is an even or odd polynomial of order l it can be expressed as a summation of even or odd Legendre polynomials of order l or less. See below. 1

P

l

1

m

1

Pl ' dx   c1 Pl  c 2 Pl  2  ...Pl ' dx m

1

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Since there is no Legendre polynomial of order l’ because its order is to large this integral vanishes because the Legendre polynomials are orthogonal on the interval -1 to 1. See below. 1

P

l

Pl ' dx 0

1

(Note: This orthognality is not proved for simplicities sake, but if you are interested look up what a Sturm-Liouville problem is and the theory and you will readily find this justification.) The next thing we need is the relationship when l = l’. This turns out to be: 1

P

m

l

m

Pl dx 

1

2 (l  m)! 2l  1 (l  m)!

To prove this relationship we are going to substitute in equation (20) this time which gives: 1

 Pl Pl dx  m

1

m

1 m l ml 1 2 m d 2 l  d 2 l   ( 1 ) ( 1 ) x x  m l   ml (1  x )  dx l 2  (2 l!) 1  dx   dx 

Guess what; more integration by parts. This is done as so,  d m l  U  (1  x 2 ) m  m l (1  x 2 ) l   dx  ml  d  2 m d dU  (1  x 2 ) l  dx (1  x ) ml dx  dx  d m l (1  x 2 ) l dx dx m l d ml 1 V  ml 1 (1  x 2 ) l dx dV 

After plugging in our relationships we end up with, 1 ml m  l 1   2 m d 2 l d 2 l   (1  x )  ml (1  x )  m l 1 (1  x )  ... 1 dx dx   1 m m   1  1 Pl Pl dx  (2 l l!) 2  1 ml 1 m l d   2 l d  2 m d 2 l ...   dx ml 1 (1  x ) dx (1  x ) dx m l (1  x )  dx     1 

Copyright © Ryan Jadrich 2007

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Once again it is clear that the portion outside the integral will become zero because it is a polynomial with a factor of one minus x squared. If we repeat this partial integration m+l times the outside portion will always be an even or odd polynomial with a factor of one minus x squared which will cause the expression to be equal to zero at both limits. After m+l integrations we are left with, 1

 Pl Pl dx  m

m

1

(1) ml (2 l l!) 2

1

2 l  (1  x )

1

d m l dx m l

m l   2 m d  ( 1 ) (1  x 2 ) l  dx x  m l dx  

Now since m ≤ l this allows us to simplify are expression like so,

(1) ml

d m l dx m l

 2 m d m  l 2l  d m l  x dx x   dx m l dx ml  

m l   2 m d  ( 1 ) (1  x 2 ) l  x  m l dx  

Why should this be so you might ask? Well first off d m l (1  x 2 ) l m l dx Is just an even or odd polynomial of order l-m which is easily seen. So it can be written generically as, d m l (1  x 2 ) l  (c1 x l  m  c 2 x1m 2  c3 x1m 4 ...) m l dx Then upon substitution,



d m l (1  x 2 ) m (c1 x l m  c 2 x1 m 2  c3 x 1 m 4 ...) m l dx



Now the expression in the brackets is just an even or odd polynomial of order m+l, and can be written generically as,



d m l (c1 x l  m  c 2 x 1 m 2  c3 x1 m 4 ...) m l dx



And since we are differentiating a polynomial of order l+m , l+m times the only thing that matters is the highest power term. Therefore we can make such a simplification without a loss of information because all other terms will just drop out.

Copyright © Ryan Jadrich 2007

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Are simplified equation is then, 1

(1) m l  P P dx  l l (2 l l!) 2 1 m

m

1

2 l ml  (1  x ) (1)

1

d m l dx m l

 2 m d m  l 2l  x  dx x dx ml  

(21)

The negative one to the power of (m+l) arises from the expansion of the polynomials. It’s not a big deal but you can check for yourself if you would like to. This new simplified expression can further be simplified into just constants, like so.

(1)

ml

(l  m)! d m  l  2 m d m l 2l  x   (1) m l (2l )! x ml  m l (l  m)! dx dx  

(22)

To see this lets first perform, d m l 2l d m l 1 d m l  2 2 l 1 x lx  2  2l (2l  1) x 2l  2  2l (2l  1)  ...  (l  m  1)x l  m m l m  l 1 m l  2 dx dx dx (2l )! l  m x  (l  m)!

Upon substitution, (1) ml

d m l dx ml

 (1) m l

ml  2 m d m l 2 l  ml d x x  (  1 )   dx m l dx m l  

 2 m (2l )! l m   x (l  m)! x   

 

d m l  (2l )! m l  (2l )! d ml m l m l x x  (  1 )   (l  m)! dx ml dx m l  (l  m)! 

Now let’s perform,

 

d m  l m l d m l 1 d m l  2 m  l 1 x  ( m  l ) x  (m  l )(m  l  1) x m l  2  (m  l )! dx m  l dx m l 1 dx m l 2 Upon substitution into the previous equation we get,

(1)

ml

 

(2l )! d m l ml (2l )!(m  l )! x  (1) ml m l (l  m)! dx (l  m)!

Which agrees with (22). With this expression inserted into (21) we have, (1) m l P P dx   l l (2 l l!) 2 1 1

m

m

1

 (1  x

) (1) m l

2 l

1

Copyright © Ryan Jadrich 2007

(2l )!(m  l )! dx (l  m)!

(23)

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Now all of this new stuff is just a bunch of constants which can be removed from the integral leaving us with, 1

 Pl Pl dx  m

m

1

1

(2l )!(m  l )! 1 (1  x 2 ) l dx 2  l (l  m)! (2 l!) 1

(23)

Now we can factor the expression inside the integral, 1

1

1

1

2 l l l  (1  x ) dx   (1  x) (1  x) dx

Then by performing partial integration on the integral we get (leaving out the preliminary steps), 1

(1  x) l 1  (1  x) (1  x) dx  (1  x) (l  1) 1 l

l

l

1

1

1 d  (1  x) l 1 (1  x) l dx  (l  1) 1 dx 1

It is again clear that the portion outside the integral will vanish because there is a factor that is zero at both 1 and -1. If we repeat this partial integration l times we will end up with, 1

l l  (1  x) (1  x) dx  

1

1

l (1) l 2l d x ( 1  ) (1  x) l dx l  (l  1)(l  2)(l  3)  ...  (2l ) 1 dx

But since, dl d l 1 d l 2 l l 1 (1  x)  l 1 l (1  x)  l  2 l (l  1)(1  x) l 2  (1) l l! l dx dx dx After substitution, 1

l l  (1  x) (1  x) dx 

1

 (1  x) 2l 1 l!   (l  1)(l  2)(l  3)  ...  (2l )  (2l  1)  l! 2 2l 1   (l  1)(l  2)(l  3)  ...  (2l ) (2l  1) 

2 l! 2 l 1

1

l! (1  x) 2l dx  (l  1)(l  2)(l  3)  ...  (2l ) 1   1  

1

2

(2l  1)!

(24)

Copyright © Ryan Jadrich 2007

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After substituting (24) back into (23) we get, 1

 Pl Pl dx  m

1

m





(2l )!(m  l )! 1 2 2l 1 l!  (l  m)! (2 l l!) 2 (2l  1)!

2

2(2l!) (m  l )!  (2l  1)! (l  m)! 2 (m  l )!   2l  1 (l  m)! 

Which is what we set out to prove. So finally to recap all of this with our new definitions we can write the solutions to equation (4) which is what we wanted as,

 ( ,  )  ( ) ( )   ( ,  )  (c1e im  c 2 e im )

(2l  1)(l  m)! m Pl 2(m  l )!

Because we want the function to be normalized. Now we set the constant c one equal to zero and c two equal to the square root of the reciprocal of 2π leading to,

 ( ,  )  Yl m ( ,  ) 

(2l  1)(l  m)! m Pl (cos  )e im 4 (m  l )!

(24)

The factor of (1/2π)^(1/2) is to normalize the phi portion. Now with all of these conditions our functions are referred to as spherical harmonics, and they solve the differential equation describing angular momentum in quantum mechanics.

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The first few spherical harmonics are, 1

0

Y0  0

Y1 

2  1 3 cos  2  1 3 (sin  )e i 2 2

1

Y1   0

Y2 





1 5 3 cos 2   1 4 

Y2

1



1 15 sin  cos  e i 2 2

Y2

2



1 15 sin 2  e  2i 4 2





The spherical harmonics with the appropriate normalizations given in equation (24) form an orthonormal set that follows, 2



 d  Y

l

0

m

( ,  )Yl ' ( ,  )sin  d   ll ' mm ' m'

0

The last final bit to be discussed is the energy levels of our little system. Obviously it is quantized, and the quantization comes from what we changed in equation (9), with the assumption that, 2I E  l (l  1) 2

And with rearranging the energy levels are obviously, E  l (l  1)

2 2I

(25)

As can be seen the energy levels are only dependent on the l quantity and is completely independent of the m quantity. This m quantity is called the magnetic quantum number and can have values of, l  m  l

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This restriction can be seen from equation (20) which is restated below with absolute values this time, m

(1) m /2 d  Pl  l (1  x 2 )   2 l!  dx  m

l m

x

2



1

l

(20)

Where m must be less than or equal to l, or greater than or equal to –l otherwise we just get the 0 solution which can’t be a valid solution in quantum mechanics unless we don’t have a particle ha! This is valid as in equation (16) m is squared and therefore only truly depends on the absolute value of the number, so we might make the adjustment to (24) to be,

 ( ,  )  Yl m ( ,  )   ( ,  )  Yl m ( ,  ) 

(2l  1)(l  m)! m Pl (cos  )e im 4 (m  l )! (2l  1)(l  m )! 4 ( m  l )!

Pl (cos  )e im m

positive negative

m m

(24) (25)

But we don’t make the change in the exponential term as it really doesn’t even matter, especially since there is an unreal number in the exponent. I hope you see what’s meant by this. The l quantity is referred to as the azithmul quantum number, and can have integer values of 0, 1, 2, 3, 4 ... These solutions will play an important role in the solution of the hydrogen atom.

Sources: Silbey, Robert J., Robert A. Alberty, and Moungi G. Bawendi. Physical Chemistry. 4th ed. Wiley, 2005. "Spherical Harmonic." Wolfram Mathworld. 02/17/05. Wolfram Mathworld. 25 Nov 2007 . "Department of Physics ." University of Utah. University of Utah. 26 Nov 2007 .

Copyright © Ryan Jadrich 2007

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