# Adiabatic approxmation, estimate of d\phi/dx

1. Jun 16, 2011

### Derivator

Hi,

Messiah writes in his Quantum Mechanics book in chapter XVIII, section 14 (pages 786-789):

http://img827.imageshack.us/img827/4605/screenshot3si.png [Broken]

What does he mean, when he says "an increment of the order of 'a' is necessary to transform the function phi into a function that is orthogonal to it"? Why do we obtain an orthogonal function, when we increase the atomic distance?

Especially, why should d Phi/ dX be roughly equal to the orthogonal function divided by 'a'?

--
derivator

Last edited by a moderator: May 5, 2017
2. Jun 17, 2011

### DrDu

I suppose a is the Bohr radius. The overlapp of two atomic orbitals falls of exponentially for distances >>a. Hence they are rapidly becoming orthogonal to each other to any approximation when the distance supercedes a.

3. Jul 23, 2011

### Derivator

Thank's for your answer. I still don' t see, why d Phi/ dX should be roughly equal to the orthogonal function divided by 'a'. Could you explain this, please?

I mean, the fact that d Phi/ dX is orthogonal to Phi, is ok. One can justify this by integration by parts, for example. I don't see, why we have to divide the orthogonal function by 'a' in order to obtain roughly d Phi/ dX

4. Jul 24, 2011

### DrDu

No, it isn´t in general, e.g. take exp(ikX).
From the localization of the atomic orbitals it is clear that an orbital phi(x) and it´s shifted copy phi(x+y) become nearly orthogonal once y is of the order a. Now phi(X+y)=exp(y d/dX) phi(X)=phi(X)+(exp(y d/dX)-1) phi(X). As long as y is smaller than a, exp(y d/dX) can be replaced by 1+y d/dX (at least as far as it´s action in phi(X) ). Assuming phi to be normalized to one, <phi(X)|phi(X+y)>= 1 +y <phi(X) | d/dX phi(X)>. We know this approximation to become bad when y=a as then the left hand side is approximately 0. Hence a<phi(X) | d/dX phi(X)> is of order unity.

5. Jul 24, 2011

### Derivator

ok, for y=a:

0=<phi(X)|phi(X+a)>= 1 +a <phi(X) | d/dX phi(X)>

then

<phi(X) | d/dX phi(X)> = -1/a

But I still don't get it, thus there are two questions:

1)
why can we conclude from the last equation
<d/dX phi(X) | d/dX phi(X)> = 1/a^2 ?

2)
why can we conclude
d/dX phi(X) = phi(X+a)/a (at least roughly)

Last edited: Jul 24, 2011
6. Jul 25, 2011

### DrDu

I did show you why y has to be of the order of a for the two functions ( the shifted and the unshifted orbital) to become approximately orthogonal and how this is related to the derivative of the orbital with respect to X. I also would not buy the first part of the last sentence of Messiah. Rather
d/dX phi(X) = (phi(X+a)-phi(X))/a (approximately).
Now take the norm of both sides taking into account the orthogonality of the two functions on the right and you will get the last equation of Messiah up to an unimportant factor 2.

7. Jul 25, 2011

### Derivator

ah, I see, you also get this factor 2. I thought this couldn't be correct.

From
d/dX phi(X) = (phi(X+y)-phi(X))/y

I can conclude:
<d/dX phi(X)|d/dX phi(X)>
= 1/y^2 <phi(X+y)-phi(X)|phi(X+y)-phi(X)>
= 1/y^2 (<phi(X+y)|phi(X+y)> - <phi(X+y)|phi(X)> + <phi(X)|phi(X)> - <phi(X)|phi(X+y)>)
= 1/y^2 (1-0+0+1)
= 2/y^2