How do i go about showing that capital psi 2s and capital psi 2px are
normalized?  Orthogonal?

PART 1: Normalizing
A vector is normalized if it has a magnitude of 1.
Calculate the magnitude by taking the square root 
of the sum of the squares of each component  
   Magnitude = sqrt(psi_x*psi_x + psi_y*psi_y + psi_z*psi_z )
This formula for magnitude can be applied to each of your 
vectors to calculate each of their magnitudes.
If this magnitude already has a value of 1 then the vector
has been shown to be normalized.  

If the magnitude is not one, then you can do the following 
to make it normalized:
To normalize a vector just divide the original vector by 
the magnitude (notice that if it is already normalized, then
dividing by 1 won't change the vector)
   normalized x component of psi= psi_x / Magnitude
   normalized y component of psi= psi_y / Magnitude
   normalized z component of psi= psi_z / Magnitude
This method of normalizing can be applied to each of your
vectors if they are not already normalized.

PART 1: Showing Orthogonal
If two vectors are orthogonal, then their dot product will
be zero.
To make the writing here easier I will just show how to
calculate orthogonality on two vectors A and B (you can
substitute in your two psi vectors instead).  Each
vector has an x, y, and z component.  This can be
written as A_x, A_y, and A_z are the three components
of A and that B_x, B_y, and B_z are the three components
of B.
A and B are orthogonal if and only if
A_x*B_x + A_y*B_y + A_z*B_z = 0

If you calculate the left side and you get zero, then you
have shown that the two vectors are orthogonal.
If you don't get zero, then they are not orthogonal.

Since this is a Physics rather than a Math question, I guessed the
Customer may be asking about electronic orbitals rather than
three-dimensional vectors.  But your definitions for normalization and
orthogonality certainly are correct for the latter context.

-- mathtalk-ga

Oops, they were probably asking about orbitals.
But there is still a lot of math behind the answer.
If I assume you are dealing with electron orbitals
of a hydrogen atom, the derived eigenfunctions are
expressed in terms of Laguerre Polynomials.  Showing
the orthogonality and normalization of these is 
a rather painful exercise compared to the simple 
vectors that I was originally thinking about.
I hope somebody gives you more details on this proof.

Actually if you know two functions/vectors are
eigenfunctions/eigenvectors of a self-adjoint operator/symmetric
matrix, corresponding to distinct eigenvalues, then the orthogonality
property is easily established.

Let's sketch the proof for (real) symmetric matrices.  Let u,v be
eigenvalues of a symmetric matrix A:

A = A'

corresponding to distinct eigenvalues:

Au = bu, scalar b

Av = cv, scalar c

with b not equal to c.

Then:

b(u'v) = (u'A)v = u'(Av) = c(u'v)

(b-c)(u'v) = 0

Now since b - c is nonzero, this implies

u dot v = u'v = 0

so u,v are at least orthogonal.

The same idea carries over with minor changes to the eigenfunctions of
self-adjoint operators.  One slight embellishment is the use of
complex arithmetic in the (non-relativistic) Schrödinger equation (a
PDE).

regards, mathtalk-ga


regards, mathtalk-ga

Hi,

     I assume that you have the forms of the energy eigenfunctions for
the 2-s and 2-px orbitals. If you dont, these can be obtained by
either looking up some standard book or searching Google :)

     The 2-s orbital has a dependence on the radial coordinate alone
whereas the 2-px orbital depends on the radial coordinate (r) and the
angle theta. The general form of the function is

         psi-2-s  = (constant)*(2-t)*exp(-t/2)
         psi-2-px = (constant)*t*exp(-t/2)*cos(theta)

Normalized or not:
         We know that the abs(psi)-squared gives the probability of
locating an electron at a location. So, if the energy eigenfunction is
normalized, we must have :

           Integral(psi*psi) over all volume = 1.0

Orthogonality:

           Integral(psi1*psi2) over all volume = 0 if (psi1 != psi2)
           that is, integral of product of psi-2s and psi-2px over all volume is 0.

     Hope that helps.


Saikiran

Hi, Saikiran_r-ga:

Thanks, I think we've lost our original poster, but your Comments were
of interest to me!

regards, mathtalk-ga