Hydrogen atom. state after L_z measurement

[Peeter]

Homework Statement

An initial state is given:

$${\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{3}}} \left( {\lvert {100} \rangle} + {\lvert {210} \rangle} + {\lvert {211} \rangle}\right)$$

An $L_z$ measurement is performed with outcome 0 at time t_0. What is the appropriate form for the ket ${\lvert {\psi_{\text{after}}(t_0)} \rangle}$ right after the measurement?

Homework Equations

$$t_0 = \frac{4 \pi \hbar }{E_I}$$
$$E_n = - E_I/n^2$$

The Attempt at a Solution

For the evolved state at this time $t_0$ (specially picked so that the numbers work out nicely) I get:

$${\lvert {\psi(t_0)} \rangle} = \frac{1}{{\sqrt{3}}} \left( e^{-i2\pi i/3} {\lvert {100} \rangle} + e^{i\pi i/3} ({\lvert {210} \rangle} + {\lvert {211} \rangle} )\right).$$

A zero measurement means that we've measured an $L_z$ eigenvalue of $m \hbar = 0$, or $m = 0$, so we can only have states

$${\lvert {\psi_{\text{after}}(t_0)} \rangle} = a {\lvert {100} \rangle} + b {\lvert {210} \rangle}$$

There are multiple m=0 states in the t_0 pre-$L_z$ ket, so I don't think that we can know any more about the distribution of those states after the $L_z$ measurement. Other than ${\left\lvert{a}\right\rvert}^2 + {\left\lvert{b}\right\rvert}^2 = 1$, can we say anything more about this post $L_z$ measurement state?

 

[diazona]

Sure you can. The ratios of the coefficients of the two m=0 states remain the same, they just get "re-normalized" to satisfy |a|² + |b|² = 1.

If you prefer to think about it a little more quantitatively, the effect of this measurement on your quantum state is equivalent to applying the projection operator
$$P = \sum_{n,l}\lvert nl0\rangle\langle nl0\rvert$$

 

[Peeter]

Thanks a lot! Measurement as a projection operator is a helpful way to think about this type of question (have state, measure alternate operator). That also verifies an answer that I arrived at mostly by intuition on a different practice question.