Parity Conservation

[Ed Quanta]

Ok,so check this situation out.

We have a one-dimensional box with walls at (-a/2,a/2). We know that the particle is in a state with energy probabilities

P(E1)=1/3, P(E2)=1/3, and P(E3)=1/3 while P(En)=0 for all n not equal to 1,2,3.

The parity is measured ideally and -1 is found. If some time later E is measured, what value is found? What is the answer if the original measurement found the parity to be 1?

I don't understand how if the parity of the state is measured ideally that -1 is found, since
we know Psi(x,0)= (square root of (2/3a))(cos(n(pi)x/a) + sin(2n(pi)x/a) + cos(3n(pi)x/a)

And when we take P, where P is the parity operator of our function, we note that only the sin function will become negative.

solving for <P> then, <P>= 1/3 + (-1/3) + 1/3=1/3, correct? This does not equal 1.


Anyway, given the initial state, now I am pretty sure that
Psi(x,t)= (square root of (2/3a))(cos(n(pi)x/a)e^-iE1t/hbar +
sin(2n(pi)x/a)e^-iE2t/hbar + cos(3n(pi)x/a)e^-iE3t/hbar)

How do I determine energy from this?
And then how will the energy change if the original parity happened to be -1?

[Tom Mattson]

And when we take P, where P is the parity operator of our function, we note that only the sin function will become negative.


Bingo. You measure a negative parity, so you must have found the particle in the only allowed basis state that has negative parity.


solving for <P> then, <P>= 1/3 + (-1/3) + 1/3=1/3, correct? This does not equal 1.


But they didn't say that you obtained an expectation value of -1, they said that you measured the particle with a parity of -1. That means that you found it in a negative parity eigenstate, and there's only one of those available, so....

Furthermore, if you find the parity to be 1, then you could be in either of the two even parity eigenstates.