# Help with a problem from Atkins Molecular Quantum Chemistry

• August 3rd 2007, 08:24 AM
MasterAir
Help with a problem from Atkins Molecular Quantum Chemistry
Hi all,

Problem
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Show that if $( /Omega f )* = - /Omega f*$, then $< /Omega > = 0$ for any real function $f$

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This is probably disgustingly simple, but it has me stumped :(

Sorry for the presentation, can anyone show me how to use TeX equations in my post? I am new to the forums.
• August 3rd 2007, 08:30 AM
topsquark
Quote:

Originally Posted by MasterAir
Hi all,

Problem
------------------------------------------------------------------------------------------------------

Show that if $( /Omega f )* = - /Omega f*$, then $< /Omega > = 0$ for any real function $f$

------------------------------------------------------------------------------------------------------

This is probably disgustingly simple, but it has me stumped :(

Sorry for the presentation, can anyone show me how to use TeX equations in my post? I am new to the forums.

Could you define your terms? I am assuming the < > are the standard expectation value brackets, but what are $\Omega$ and f?

See there first few posts in this thread for LaTeX.

-Dan
• August 3rd 2007, 08:36 AM
MasterAir
Quote:

Could you define your terms? I am assuming the < > are the standard expectation value brackets, but what are http://www.mathhelpforum.com/math-he...d3e90e31-1.gif and f?

See there first few posts in this thread for LaTeX.

-Dan
Omega is an operator, (I think it is Hermitian, but I am unsure.) f is any real function, $<\Omega>$ Is the expectation value for the operator Omega.

In words - if I have understood correctly - if the complex conjugate of the result of an (hermitian) operator acting on a function is equal to minus the effect of the same (hermitian) operator acting on the complex conjugate of the same real function, then the expectation value of the (hermitian) operator is zero for any real function f.
• August 3rd 2007, 09:06 AM
topsquark
Quote:

Originally Posted by MasterAir
Omega is an operator, (I think it is Hermitian, but I am unsure.) f is any real function, $<\Omega>$ Is the expectation value for the operator Omega.

In words - if I have understood correctly - if the complex conjugate of the result of an (hermitian) operator acting on a function is equal to minus the effect of the same (hermitian) operator acting on the complex conjugate of the same real function, then the expectation value of the (hermitian) operator is zero for any real function f.

Okay, so the complex conjugate of $\Omega f$ is equal to $-\Omega(f^*)$?

I'm still confused on one point. According to the definitions I know (Quantum Mechanics in general, not specifically Molecular Quantum Chemistry)
$\left < \Omega \right > = \int_V \psi ^* \Omega \psi \ d^3x$
where $\psi (x, y, z)$ is your wavefunction.

Where is f coming into this?

-Dan
• August 3rd 2007, 10:56 AM
topsquark
Okay, let me try this and you can correct me if I'm wrong on any point. (This would be easier if we were to use "bra-ket" notation. If you'd like me to rewrite it, just let me know.)

$\Omega$ is a Hermitian operator, so
$\left < \Omega \right > = \int_V \psi ^* \Omega \psi \ d^3x$ is a real number, where $\psi$ is a wavefunction.

So we know that
$\left < \Omega \right >^* = \left < \Omega \right >$

From here on I am taking $\psi = f$ as a real valued function. And we know that $( \Omega f)^* = - \Omega (f ^*) = - \Omega f$. (Given.)

Now
$\left < \Omega \right > ^* = \left ( \int_V f ^* \Omega f \ d^3x \right ) ^* = \left ( \int_V f \Omega f \ d^3x \right ) ^*$<-- Since f is a real valued function
$= \int_V f^* ( \Omega f )^* \ d^3x = \int_V f^* (- \Omega f ) \ d^3x = - \int_V f^* \Omega f \ d^3x = - \left < \Omega \right >$

But $\left < \Omega \right >^* = \left < \Omega \right >$ since the expectation value is real.

Thus
$\left < \Omega \right > = - \left < \Omega \right >$

Thus
$\left < \Omega \right > = 0$

-Dan
• August 4th 2007, 02:44 AM
MasterAir
Yes, I think it would be easier in bra-ket notation. However, thanks very much, I follow your answer.

I think the main reason I was stumped is that I failed to realise that $f=f*$ for a real function. Some, sort of mental block, I don't quite know.

• August 4th 2007, 07:06 AM
topsquark
Quote:

Originally Posted by MasterAir
Yes, I think it would be easier in bra-ket notation. However, thanks very much, I follow your answer.

I think the main reason I was stumped is that I failed to realise that $f=f*$ for a real function. Some, sort of mental block, I don't quite know.