- #1
dsr39
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We have to apply Ehrenfest's theorem and I don't think it was ever explained well to us. I have read that expectation values of measurable quantities behave according to classical physics equations
ie.
[tex]M\frac{d\left<x(t)\right>}{dt} = \left<p(t)\right>[/tex]
I think I must be applying this idea wrong because I don't see how this result works out for the calculation of <E(t)> for an infinite square well in a particular energy eigenstate.
For an infinite square well in a stationary state.
[tex]\left<x(t)\right>=const \to M\frac{d\left<x(t)\right>}{dt} = 0 [/tex]
That quantity is equal to momentum so <p> is zero, and does that also mean <E(t)> is zero since
[tex] \left<E(t)\right> = \frac{\left<p(t)\right>^2}{2m}[/tex]
But <E(t)> shouldn't be zero because the energies in stationary states for the infinite square well are very clearly defined by the boundary conditions and which stationary state you're in. How am I wrongly applying Ehrenfest's theorem... your help is greatly appreciated. Thank you.
ie.
[tex]M\frac{d\left<x(t)\right>}{dt} = \left<p(t)\right>[/tex]
I think I must be applying this idea wrong because I don't see how this result works out for the calculation of <E(t)> for an infinite square well in a particular energy eigenstate.
For an infinite square well in a stationary state.
[tex]\left<x(t)\right>=const \to M\frac{d\left<x(t)\right>}{dt} = 0 [/tex]
That quantity is equal to momentum so <p> is zero, and does that also mean <E(t)> is zero since
[tex] \left<E(t)\right> = \frac{\left<p(t)\right>^2}{2m}[/tex]
But <E(t)> shouldn't be zero because the energies in stationary states for the infinite square well are very clearly defined by the boundary conditions and which stationary state you're in. How am I wrongly applying Ehrenfest's theorem... your help is greatly appreciated. Thank you.