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

  • Thread starter mkosmos2
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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
 

Answers and Replies

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

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