# Parity

• Apr 13th 2008, 10:29 PM
DiscreteW
Parity
With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

If $\displaystyle P$ is the Parity Operator, determine the parity of the functions below:

$\displaystyle P\sin{x} = ?$
$\displaystyle Pe^{-x} = ?$
$\displaystyle P(e^{x} + e^{-x}) = ?$

Now, if $\displaystyle H$ is the Hamiltonian Operator with $\displaystyle V(-x) = V(x)$, find the parity of

$\displaystyle PHe^{-x} = ?$

------

So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.
• Apr 14th 2008, 12:20 AM
Isomorphism
Quote:

Originally Posted by DiscreteW
With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

If $\displaystyle P$ is the Parity Operator, determine the parity of the functions below:

$\displaystyle P\sin{x} = ?$
$\displaystyle Pe^{-x} = ?$
$\displaystyle P(e^{x} + e^{-x}) = ?$

Now, if $\displaystyle H$ is the Hamiltonian Operator with $\displaystyle V(-x) = V(x)$, find the parity of

$\displaystyle PHe^{-x} = ?$

------

So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.

Is this related to some Physics concepts?
I am guessing that we need to define a function called P that will map like this P(f(x)) = f(-x).

$\displaystyle P\sin{x} = -\sin{x} \Rightarrow P(x) = -x$

$\displaystyle Pe^{-x} = $$\displaystyle e^{x}\Rightarrow P(x) = \frac1{x} \displaystyle P(e^{x} + e^{-x}) =$$\displaystyle e^{-x} + e^{x} \Rightarrow P(x) = x$

Hopefully Topsquark, our physics maestro(Cool), will help us out with the last one. I have no clue what you are talking about. What does $\displaystyle Hy$ do?
• Apr 14th 2008, 04:01 AM
topsquark
Quote:

Originally Posted by DiscreteW
With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

If $\displaystyle P$ is the Parity Operator, determine the parity of the functions below:

$\displaystyle P\sin{x} = ?$
$\displaystyle Pe^{-x} = ?$
$\displaystyle P(e^{x} + e^{-x}) = ?$

Now, if $\displaystyle H$ is the Hamiltonian Operator with $\displaystyle V(-x) = V(x)$, find the parity of

$\displaystyle PHe^{-x} = ?$

------

So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.

Quote:

Originally Posted by Isomorphism
Is this related to some Physics concepts?
I am guessing that we need to define a function called P that will map like this P(f(x)) = f(-x).

$\displaystyle P\sin{x} = -\sin{x} \Rightarrow P(x) = -x$

$\displaystyle Pe^{-x} = $$\displaystyle e^{x}\Rightarrow P(x) = \frac1{x} \displaystyle P(e^{x} + e^{-x}) =$$\displaystyle e^{-x} + e^{x} \Rightarrow P(x) = x$

Hopefully Topsquark, our physics maestro(Cool), will help us out with the last one. I have no clue what you are talking about. What does $\displaystyle Hy$ do?

The parity operator in 1 - D, as Isomorphism suggests, maps f(x) to f(-x). In 3-D it maps the function f(x, y, z) to f(-x, -y, -z).

I think all that is being asked for here is what the parity of the function is. So
$\displaystyle Psin(x) = sin(-x) = -sin(x)$, so sin(x) has an odd parity, or -1.

$\displaystyle Pe^{-x} = e^x \neq \pm e^{-x}$ so $\displaystyle e^{-x}$ is not a parity eigenstate.

$\displaystyle P(e^x + e^{-x}) = e^{-x} + e^x = +(e^x + e^{-x})$ so $\displaystyle e^x + e^{-x}$ has an even parity, or +1.

H is the Hamiltonian operator: $\displaystyle H = -\frac{\hbar ^2}{2m} \frac{d^2}{dx^2} + V(x)$

We can do $\displaystyle PHe^{-x}$ in two ways: Apply H to $\displaystyle e^{-x}$ then apply P, or we can apply P to $\displaystyle He^{-x}$ directly as P is a linear operator. I think at this level it is more instructive to take the former route, so
$\displaystyle PHe^{-x} = P \left ( -\frac{\hbar ^2}{2m}e^{-x} + V(x)e^{-x} \right ) = -\frac{\hbar ^2}{2m}e^x + V(-x)e^x$

We know that V(x) has an even parity so this is
$\displaystyle PHe^{-x} = -\frac{\hbar ^2}{2m}e^x + V(x)e^x = He^x \neq \pm He^{-x}$
so this is not a parity eigenstate.

-Dan