ibysaiyan
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Hi,
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) ...
Negative parity: f(-x) = -f(x)
Positive parity : f(x) = f(-x)
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.
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.
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