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### Topic: integral of a function belonging to a Irreducible rappresentation  (Read 3187 times)

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##### integral of a function belonging to a Irreducible rappresentation
« on: October 24, 2015, 04:47:27 AM »
Hi!!

Yesterday the Professor said:

" The integral of a Irreducible rappresentation is always ZERO,except when the base function  belongs  to A_1   IR (in other wolrds belong to the totalsymmetric rappresentation).

Then he did some examples about this fact  saying that "ns" orbitals have an integral =\=  0.
The "np" orbitals has an integral = 0  instead....Because they have two lobes symmetric  with an opposite sign

But now i have a doubt: consider the C2v group and the orbital 2p:  the pz orbital has A1 symmetry so his integral would be =/= 0  ...
So the Professeor makes a mistake?? Not all np orbitals have an integral equal to zero (...even if 2pz,2px and 2py has the same wave function...is only "traslate" in the space for each of them)

thanks!!

#### Corribus

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##### Re: integral of a function belonging to a Irreducible rappresentation
« Reply #1 on: October 24, 2015, 11:51:10 AM »
I'm not really following. It is not common to take the integral of a single representation - which would be akin to taking the integral of a single wavefunction. Integrating over the interaction between two or more representations is a common way of quickly assessing whether there is nonzero overlap (for example) between two orbitals. You bring up C2v - the z, x, and y oriented p-orbitals have linear bases of A1, B1 and B2 symmetry, respectively. The product of (x,x), (z,z) and (y,y) all yield A1 symmetry, indicating that each orbital overlaps with itself - and therefore a non-vanishing overlap integral. On the other hand, none of the products (x,y), (x,z), or (y,z) yield A1 symmetry, indicating that none of the orbitals overlap with each other - meaning they cannot combine to form new orbitals. This is especially important when determining whether orbitals on different nuclei can overlap to form molecular orbitals. Taking the integral of a single representation doesn't make a whole lot of sense - what is the meaning of an overlap between an orbital and nothing at all?
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##### Re: integral of a function belonging to a Irreducible rappresentation
« Reply #2 on: October 24, 2015, 01:17:09 PM »
I'm not really following. It is not common to take the integral of a single representation - which would be akin to taking the integral of a single wavefunction. Integrating over the interaction between two or more representations is a common way of quickly assessing whether there is nonzero overlap (for example) between two orbitals. You bring up C2v - the z, x, and y oriented p-orbitals have linear bases of A1, B1 and B2 symmetry, respectively. The product of (x,x), (z,z) and (y,y) all yield A1 symmetry, indicating that each orbital overlaps with itself - and therefore a non-vanishing overlap integral. On the other hand, none of the products (x,y), (x,z), or (y,z) yield A1 symmetry, indicating that none of the orbitals overlap with each other - meaning they cannot combine to form new orbitals. This is especially important when determining whether orbitals on different nuclei can overlap to form molecular orbitals. Taking the integral of a single representation doesn't make a whole lot of sense - what is the meaning of an overlap between an orbital and nothing at all?

HI

So it would be better to imagine always the integral between two functions,at least...

But why   only A1 meanings  that there is overlap??
Thanks

#### Corribus

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##### Re: integral of a function belonging to a Irreducible rappresentation
« Reply #3 on: October 24, 2015, 07:33:02 PM »
Are you familiar with the concept of even and odd functions? Even function is one where f(x) = f(-x) and odd function is one where f(x) = - f(-x). It should be fairly easy to see that these have different symmetries in a 2D coordinate system. And integration of an even function across all space (or from -x to x) is nonzero (unless it is accidentally zero), and an integration of an odd function is necessarily zero. It is straightforward to show that a product of two function is only even if the product of their symmetry representations belong to (or contains a component of) the A1 symmetry representation.

Most physical chemistry textbooks have some kind of formal proof of this concept, which is one of the reasons that we bother with symmetry at all - it helps us quickly determine whether certain integrals vanish without actually having to compute them.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman