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1) How vwoulHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? go about writing the time dependent wave function given the wavefunction at $t=0$?

Suppose How would I go about writing the time dependent wave function given the wavefunction at $t=0$? an electron is in a superposition state in the hydrogen atom at $t=0$ with normalized wavefunction: How would How would I go about writing the time dependent wave function given the wavefunction at $t=0$? I go about writing the time dependent wave function given the wavefunction at $t=0$?

$$\psi(r,\theta,\phi) ~=~ A(2R_{10}Y_{00}+4R_{21}Y_{1,-1}).$$

2) What woHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? uld the tiHow wouHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? ld I go about writing the time dependent wave function given the wavefunction at $t=0$? me dependent wave function look like?

3) Also how couHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? ld I tell if the probabilities of measuring a certain energy, or angular momentum arHow would I go about writing the time dependent wave function given the wavefunction at $t=0$? e time dependent?

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1) In general, $\psi(\vec{r},t) = {\sf U}(t,0) \psi(\vec{r},0)$, where ${\sf U}(t,0)$ is the time-evolution operator (a unitary matrix).

2) Given your superposition state at initial time, after time $t$ the wave function would look like

$$\psi(r,\theta,\phi,t) = A \left( 2R_{10}Y_{00} e^{-iE_1 t/\hbar} + 4 R_{21}Y_{1,-1} e^{-iE_2 t/\hbar} \right)$$ where $E_1$ and $E_2$ are the ground and first excited energy eigenvalues of the Hydrogen atom (in atomic units, $E_1 = -1/2$ and $E_2 = -1/8$). By the way, your wave function is equivalent if you divide it by $2$ (you include $2$ in the normalization factor $A$).

3) The probability of measuring $E_1$ and $E_2$ would remain constant: $P_1 = |\langle R_{10} Y_{00} | \psi(r,\theta,\phi,t) \rangle |^2 = 4A^2 = 1/5$; $P_2 = 1 - P_1 = 4/5$.

You should by now understand that those are the same (constant) probabilities of measuring $L^2 = 0$ and $L^2 = 2\hbar^2$.

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