# Parity of wavefunctions

1. Apr 11, 2013

### ibysaiyan

Hi,
1. The problem statement, all variables and given/known data
A quantum harmonic oscillator is in a superposition of states(below):
$\Psi(x,t)$ = 1/$\sqrt{2}$ ($\Psi_{0}(x,t) + \Psi_{1}(x,t)$

$\Psi_{0}(x,t)$ = $\Phi(x) * e^{-iwt/2}$ and $\Psi_{1}(x,t)$ = $\Phi_{1}(x) * e^{-i3wt/2}$

Show that <x> = C cos(wt) ...

2. Relevant equations

Negative parity: f(-x) = -f(x)
Positive parity : f(x) = f(-x)

3. The attempt at a solution
On the mark schemes it shows that x|ψ(x)^2| = -(-x|ψ(-x)^2) are asymmetric functions,so when I expand the integral they should vanish,however, I thought that normalization condition allows the integral to be definite i.e = 1.

I have a very basic understanding of parity, it would be great if someone could explain to me just what role it plays.

P.S I have looked around for resources,and so far I have gathered that there are two distinct eigen-values when parity operator works(+-1).

Thanks.

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Last edited: Apr 11, 2013
2. Apr 11, 2013

### MisterX

Would you be able to show at least some work? Write out the terms in $\left\langle\psi \left|x\right| \psi \right>$.

if $f(x) = - f(-x)$, then what is
$\int_{-\infty}^\infty f(x) dx$ ?

if $f(x) = - f(-x)$ and $g(x) = g(-x)$, what is
$\int_{-\infty}^\infty f(x)g(x) dx$ ?

3. Apr 11, 2013

### vela

Staff Emeritus
You're comparing two different integrals. The normalization condition is
$$\int_{-\infty}^\infty \lvert \psi(x) \rvert^2\,dx = 1$$ while the integral from the expectation value is
$$\int_{-\infty}^\infty x\lvert \psi(x) \rvert^2\,dx.$$ Why would you think the first integral precludes the second one from vanishing? They have different integrands.

4. Apr 12, 2013

### ibysaiyan

I was in a rush hence my lack of input.

Here is what I get after expansion:

$\Psi(x,t)$ = 1/$\sqrt{2}$ ($\Psi_{0}(x,t) + \Psi_{1}(x,t)$

$\Psi_{0}(x,t)$ = $\Phi(x) e^{-iwt/2}$ and $\Psi_{1}(x,t)$ = $\Phi_{1}(x) e^{-i3wt/2}$

$\left\langle\psi \left|x\right| \psi \right> = 1/\sqrt{2} \int_{-\infty}^\infty dx [ x (\Phi_{0}\Phi_{1} + \Phi_{0}\Phi_{1}e^{-i\omega t} + \Phi_{1}\Phi_{0}e^{i\omega t} + \Phi_{1}\Phi_{1}$]

I understand that the integrand of a product of even-odd function is zero,but I still don't see why the two integrals above vanish,are they assumed to be asymmetric, is there any justification?

5. Apr 12, 2013

### ibysaiyan

Actually... I think I get it now...
(below is what I think happens}

Is it because $\Phi(x)$ = $-\Phi(-x)$(odd) and f(x) = f(-x) (even)...
Integral of odd*even = integral of odd = zero .
Also for the other two integrals(with exponential) they are defined as "symmetric limited function", something I am not familiar with.

I appreciate the responses.

EDIT: I was just restating the properties of odd/even functions,don't know why I had integral put in(fixed),sorry.

Last edited: Apr 12, 2013
6. Apr 12, 2013

### vela

Staff Emeritus
What is that notation even supposed to mean?

7. Apr 12, 2013

### ibysaiyan

Hi vela
I have fixed few typos so it should be clear now.

Last edited: Apr 12, 2013
8. Jun 25, 2013

### zeesun

Hi
Good thinking but not so much & clear definition.