What is the role of parity in quantum mechanics?

[ibysaiyan]

Hi,

Homework Statement

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) ...

Homework Equations

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

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.

 

[MisterX]

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

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 ?

 

[vela]

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.

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.

 

[ibysaiyan]

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.

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

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 ?

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&gt; = 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?

 

[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.

 

[vela]

What is that notation even supposed to mean?

 

[ibysaiyan]

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

 

[zeesun]

Hi
Good thinking but not so much & clear definition.