Methods of Mathematical Physics – Spring 2010-11
Week 7 – Friday 13 May 2011
Dr. E.J. Zita
Quantum Mechanics Ch.4: 3D
You have seen much of this already, in the context of modern physics and Boas. We’ll apply the
technique of separation of variables to solve the Schrödinger Eqn.
We saw in Boas on Tuesday:
Schrödinger’s equation (SE) 
Recall that i
2
2m
 2   V ( x)  i

t

 E  in quantum mechanics, where E is the energy.
t
SE has space derivatives and time derivatives, so guess a solution of the form    ( x, y, z )T (t ) .
We substituted    T into the PDE, then divided by    T to separate variables, and got:

2
1
2m 
2 ( x, y, z )  V ( x, y, z )  i
T (t )
 const
t
What separation constant did we choose? Always start with the easy ODE
We got two ordinary diffeqs to solve:
dT (t )
i
 E T (t )
dt

2
2m
 2 (r )  V (r ) (r )  E (r )
We found T (t ) eiEt . Then the general Schrödinger’s Eqn. has a solution of the
 iE t
form  n   n (r )e n , where the space-dependent part satisfies

2
2m
 2 (r )  V (r ) (r )  E (r )
The solution for  (r ) will depend on the geometry, and on the form of V ( r ) (including BC).
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4.1.1: Spherical Coordinates are natural for central forces, e.g. electrostatics (H atom):
V (r )  V (r )
2 
1   2  
1
 

r
 2
 sin 
2
r r  r  r sin   

  2 
1


 2 2  2
 r sin    
 Guess what kind of solutions we’ll get?
What are the variables in the PDE?
Try a solution of what form?
 (r , ,  )  R(r ) ( ) ( )  R(r ) Y ( ,  )
As usual, substitute = ______ into the PDE, then divide by  = ______ to separate variables:

2
2m
 2 (r , ,  )  V (r ) (r , ,  )  E (r , ,  )
 1   2 R  2mr 2

1  1  
Y ( ,  ) 
1  2Y ( ,  ) 
r

V
(
r
)

E

sin





0




2 
  sin 2   2 
 R r  r 
 Y ( ,  )  sin   
Which of these terms is constant?
 must be a function of only one variable
 that variable must not appear elsewhere in the equation
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 1   2 R  2mr 2


r
  2 V (r )  E   const : The solution to this equation will depend on V(r).
 R r  r 

 1  
Y ( ,  ) 
1  2Y ( ,  ) 

   const :
 sin 

Y ( ,  )  sin   
  sin 2   2 
This equation needs further separation.
1
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Solve
1 2Z
 K2.
2
Z z
(using K instead of k for reasons that will become clear…)
Now we have a second PDE in just 2 variables (what are they? _________________)
1 1   R  1 1  2
 K2  0
r

2
2
R r r  r   r 
Multiply by r 2 to separate variables. New PDE:
r   R  1  2
 K 2r 2  0
r

2
R r  r   
Now we have two ordinary differential eqns to solve:
1 d 2
 n 2
2
 d
r d  dR 
2 2
2
r
 K r  n
R dr  dr 
Solve the easiest one first (see why we chose the constant = -n2?)
r d  dR 
2 2
2
r
 K r n  0
R dr  dr 
p.640: “This is a Bessel equation with solutions J n ( Kr ) and Nn ( Kr ) .”
Does the last ODE look familiar?
Only the J n ( Kr ) solution is finite at the origin, so the radial solutions are R(r )  J n ( Kr )
Apply BC: temperature = 0 on the cylinder’s surface: u = 0 when r = a, for all  and z, so
R(a)  0  J n ( Ka)  J n (k ) (where k  Ka = zeroes of the Bessel function).
Then the general radial solution is R(r )  J n (kr a) , and the total solution is
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u (r ,  , z )  R(r ) ( ) Z ( z ) 
 J n (kr a)sin n e kz a

 kz a
 J n (kr a) cos n e
 HW: 13.5 #1 or #2, p.642
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