# [Q] parity question.

1. Nov 16, 2008

### good_phy

Hi, i separated parity question from my question posted just below

Parity operator reveal parity of function such that $Pf = \pm f$

But In fact, in order to get parity of function f, we should already know parity of function

In that way, Why should we know Parity operator ? we already know parity of function!

Second, since partiy operator commute with Hamilitonian, expectation value of Parity is conserved in time.

What does it means? Why is it so important even author assign 5 page to make it clear.

2. Nov 16, 2008

### malawi_glenn

This is just the formal way to write it, the parity operator is that you change x-> -x etc.

You could argue the same for "why should we know the momentum operator, we already know the derivative of a function". This formalism becomes advantegous when dealing with Dirac notation, bras and kets (this is my impression).

That a quantity commutes with the hamiltonian in classical mechanics (the commutator is then the poisson bracket) means that the quantity is constant of motion, i.e it is conserved. The same holds for QM, if something commutes with the Hamiltonian H_1, it is a constant of motion, i.e holds its ititial value as long as the systems are subject to the hamiltonain H_1.

Which author and book are you studying at the moment?

3. Nov 16, 2008

### good_phy

I'm studying QM with Liboff fourth edition, it contains amazing contents

But i can not agree your answer because We should know momentum operator to obtain

corresponding eigenstate and eigenvalue which is used to predict experiment.

But for parity opeator, we can not evalute corresponding eigenstate and eigenvalue becuase infinity possible exists!

4. Nov 16, 2008

### malawi_glenn

As I said, this is just to introduce you to more abstract formalism, such when dealing with dirac formalism. And also for completness, in QM - everything is operators.

Do you A PRIORI know the parity of the function, lets say: Sin(x) over the interval -L < x < L ?? Well, just because it is "easier" to perform coordinate transformation than taking derivative doesn't mean that you are not doing anything. You always do something, noting is / should be a priori.

Now I KNOW from PREVIOUS (a posteriori) calculations that sin(-x) = -sin(x) over that intervall. In the same way I know from previous calculations what derivatives of functions is, so that $e^{-ax}$ is an eigenfunction to the derivative operator...

My point is, stressing this again, that you always do something. So saying that it is meaningless to perform the parity operation since one first has to know the parity of the function is totaly wrong - you always perform some action.

For the parity operator, you mirror all coordinates in your wavefunction. There are only two possible eigenvalues, +1 or -1.

There are infinite numbers of eigenvalues and eigenstates for momentum operator aswell, so what is the point you want to make?

Parity is a very important concept in physics, you require physics to be symmetric with respect to coordinate transformations such as parity and rotation.

5. Nov 17, 2008

### good_phy

So, u means even that i previously know parity of function is just interpreted as operating parity operator to the function?

Is it right? Is it essential in QM?

6. Nov 17, 2008

### malawi_glenn

Yes, one has to be formal! Even if i know the derivative of function, I have to write that I perform an operation with the p-operator.

How do you A PRIORI know the parity of a function/state if you don't perform the coordinate transformation?

And even though the parity operation is quite trivial, one has to be formal.