Non-Stationary State Wavefunction - Normalized? <L^2>? Uncertainty on L^2?

[mkosmos2]

Homework Statement

Consider the nonstationary state:

\Psi = \sqrt{\frac{1}{3}}\Psi_{22-1} + \sqrt{\frac{2}{3}}\Psi_{110}

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

Homework Equations

|a_{n_{1}l_{1}m_{1}}|^{2} + |a_{n_{2}l_{2}m_{2}}|^{2} = 1 (1)

&lt;L^{2}&gt; = \int_{all{}\Omega}\Psi^{*}(-\hbar^{2}\Lambda^{2})\Psi d\tau (2)

\Delta L^{2} = \sqrt{&lt;(L^{2})^{2}&gt; - &lt;L^{2}&gt;^{2}} (3)

The Attempt at a Solution

For the first part, I think it's safe in this situation to use equation (1) where the a_{nlm} terms are the two coefficients in \Psi, 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 L^{2} operator results in \hbar^{2}l(l+1) coming outside the integral, and because the two states are orthogonal, the integral and therefore L^{2} goes to 0.

The last part has me tripped up the most because I don't know whether L^{2} 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

 

[sean_mp]

Ok, for the first part, all you need to know is that the states are orthogonal and normalized. This means
\big&lt; \Psi_{n&#039;l&#039;m&#039;} \big\lvert \, \Psi_{nlm} \big&gt; =\big&lt;n&#039;l&#039;m&#039; \big\lvert \,nlm\big&gt; = 0~~ \text{if}~~nlm=n&#039;l&#039;m&#039;;~1~~\text{if}~~nlm \ne n&#039;l&#039;m&#039;
So for your state, we have
\big&lt; \Psi \, \big\lvert \, \Psi \big&gt; =\bigg( \sqrt{ \frac{1}{3}} \Psi_{22-1}^*+ \sqrt{ \frac{2}{3}} \Psi_{110}^* \bigg)\bigg( \sqrt{ \frac{1}{3}} \Psi_{22-1}+ \sqrt{ \frac{2}{3}} \Psi_{110} \bigg)= \frac{1}{3}(1)+ \frac{2}{3}(1) + \frac{2}{6}(0)\frac{2}{6}(0)= \frac{3}{3} =1
So yes, it is normalized.
For the second part,
L^2 \big\lvert \,nlm \big&gt; = \hbar^2 l(l+1) \big\lvert \,nlm \big&gt;
or
\big&lt;n&#039;l&#039;m&#039; \big\lvert L^2 \big\lvert \,nlm \big&gt; = \big&lt;n&#039;l&#039;m&#039;\big\lvert \hbar^2 l(l+1) \big\lvert \,nlm \big&gt; = \hbar^2 l(l+1)~~\text{if}~ l=l&#039;;~~0~~ \text{if} ~~l \ne l&#039;
so,
\begin{multline}\big&lt;L^2 \big&gt; = \frac{1}{3} \big&lt;22-1 \big\lvert \,\hbar^2 2(2+1) \big\lvert 22-1 \big&gt; +\frac{2}{3} \big&lt;110 \big\lvert \,\hbar^2 1(1+1) \big\lvert 110 \big&gt;\\ = \frac{1}{3} \hbar^2 2(2+1) \big&lt;22-1 \big\lvert \ 22-1 \big&gt; +\frac{2}{3} \hbar^2 1(1+1) \big&lt;110 \big\lvert \, 110 \big&gt; = \frac{1}{3} \hbar^2 2(2+1) +\frac{2}{3} \hbar^2 1(1+1) \end{multline}

and etc. You get the idea. By the way, you need to do NO integration on this problem. That's the beauty of orthonormality.