
This "prove", however, is not quite
satisfactory. It is not perfectly clear if solutions could exist that do not
obey Blochs theorem, and the meaning of the index k is left open. In
fact, we could have dropped the index without losing anything at this
stage. 


It does, however, give an idea about
the power of the symmetry considerations. 

A very similar prove is contained in
AlonsoFinn. It
uses a slightly different approach in arguing about symmetries. 


Again, we consider the onedimensional case, i.e.
V(x) = V(x + a) with a =
lattice constant. 


But now we argue that the probability of finding an electron at
x, i.e. y(x
)^{2}, must be the same at any indistinguishable position, i.e.



[y(x)]^{2} 
= 
[y(x + a)]^{2} 




This implies 


y(x) 
= 
C · y(x + a)




C^{2} 
= 
1 




We thus can express C as: 





for all a and k. At
this point k is an arbitrary parameter (with dimension
1/m). This insures that C = exp (ika) · exp
–(ika) =1 


We thus have 


y(x) 
= 
exp(ika) · y(x + a) 




And this is already a very general form of Blochs
theorem as shown below. 

Writing it straight forward for the
threedimensional case we obtain the general version of Blochs theorem 


y_{k}(r + T)

= 
exp (ik · T) · y_{k}(r) 




with T = translation vector of the
lattice and r = arbitrary vector in space. 


The index k now symbolizes that we are discussing that
particular solution of the Schrödinger equation that goes with the wave
vector k 

The generalization to three dimensions is not
really justified, but a rigorous mathematical treatment yields the same result.
The more common form of the Bloch theorem with the modulation function
u(k) can be obtained from the (onedimensional) form of the Bloch
theorem given above as follows: 


Multiplying y(x) = exp(ika) · y(x + a) with exp(ikx)
yields 


exp – (ikx) · y(x) 
= 
exp – (ikx) · exp( ika) · y(x + a) 
= 
exp (ik · [x + a]) · y(x + a) 



This requires unambiguously that
exp(ikx) · y(x) must be
periodic with the periodicity of the lattice, or exp(ikx) ·
y(x) = u(x) with
u(x) being periodic as before. 


This again gives Blochs theorem: 


y(x) 
= 
u(x) · exp – (ikx) 



Once more, no index k at y or u is required. We also did not require
specific boundary conditions. The meaning of k, however, is left
unspecified. Of course, the plan wave part of the equation makes it clear that
k has the function of a wave vector, but it has not been
explicitly introduced as such. 

