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## Homework Statement

Consider the nonstationary state:

[itex]\Psi = \sqrt{\frac{1}{3}}\Psi_{22-1} + \sqrt{\frac{2}{3}}\Psi_{110}[/itex]

Where [itex]\Psi_{22-1}[/itex] and [itex]\Psi_{110}[/itex] are normalized, orthogonal and stationary states of some radial potential. Is [itex]\Psi[/itex] properly normalized? Calculate the expectation value of [itex]L^{2}[/itex] and the uncertainty in [itex]L^{2}[/itex] for a particle in this state.

## Homework Equations

[itex]|a_{n_{1}l_{1}m_{1}}|^{2} + |a_{n_{2}l_{2}m_{2}}|^{2} = 1[/itex] (1)

[itex]<L^{2}> = \int_{all{}\Omega}\Psi^{*}(-\hbar^{2}\Lambda^{2})\Psi d\tau[/itex] (2)

[itex]\Delta L^{2} = \sqrt{<(L^{2})^{2}> - <L^{2}>^{2}}[/itex] (3)

## The Attempt at a Solution

For the first part, I think it's safe in this situation to use equation (1) where the [itex]a_{nlm}[/itex] terms are the two coefficients in [itex]\Psi[/itex], but I'm not sure if it applies when the quantum numbers aren't the same for the two stationary states.

For the second part, still not positive, but I think if I use equation (2), the [itex]L^{2}[/itex] operator results in [itex]\hbar^{2}l(l+1)[/itex] coming outside the integral, and because the two states are orthogonal, the integral and therefore [itex]L^{2}[/itex] goes to 0.

The last part has me tripped up the most because I don't know whether [itex]L^{2}[/itex] being zero automatically means the uncertainty is zero, but something tells me it's not that simple.

I'm just fuzzy on a lot of the concepts surrounding this topic so any clarification/confirmation will be much, much appreciated.

Thank you very much for your time, and if these answers are correct, sorry I wasted it :s

-Mike