Exercise 620 the wave function of a hydrogen like atom at...

Exercise 6.20 The wave function of a hydrogen-like atom at time $t = 0$ is $$ \Psi(\vec{r}, 0) = \frac{1}{\sqrt{11}}\left[\sqrt{3}\psi_{2,1,-1}(\vec{r}) - \psi_{2,1,0}(\vec{r}) + \sqrt{5}\psi_{2,1,1}(\vec{r}) + \sqrt{2}\psi_{3,1,1}(\vec{r})\right], $$ where $\psi_{nlm}(\vec{r})$ is a normalized eigenfunction (i.e., $\psi_{nlm}(\vec{r}) = R_{nl}(r)Y_{lm}(\theta, \varphi)$). (a) What is the time-dependent wave function? (b) If a measurement of energy is made, what values could be found and with what probabilities? (c) What is the probability for a measurement of $\hat{L}_z$ which yields $-1\hbar$?

Transcript

00:01 In this problem, you are given the initial wave function for a hydrogen atom.

00:08 They want to know what that will be at some later time.

00:11 Now, how do we go about that? it's actually a very simple procedure, but let me take you through how it all arises.

00:19 Let's say we look at the time dependent, not the time dependent, but time independent, which is under equation.

00:26 And we assume we say we've solved it.

00:29 So we know all the energy eigenstates, you know all the energy, energy eigenvalues.

00:34 And we will expand out, as we've done above, our wave function at time equals zero in terms of those eigenstates.

00:48 So that's the equivalent of what we've given.

00:52 And now we want to look at what it would be at some later time.

00:57 So same type of expansion.

00:59 Now the coefficients carry time dependence.

01:03 And let's look at the time dependent.

01:05 So in his equation, h -minus, side t is equal to zero now when we act on when we act on our wave function here h acting on it will only act on the eigenstates so this becomes a n t e n now a partial derivative will act only on a as i h bar a and dot t and we still have the eigenstates outside now i wrote it like this because i'm going to make use of the fact that these are linearly independent.

01:54 That means the coefficients on each end must be zero.

01:59 Otherwise, it wouldn't be linearly independent.

02:01 So i get from that very simple partial differential equation, i -h -bar, a -n dot, z equal to e -n, a -n, and let me put the t dependence on this.

02:14 And that gives me for easily solved by integration, a -n -t is equal to a -n -0, e -m -i -e -n -t over h -barr.

02:30 And so i can write my expansion out now at any time, a -n -0, e -m -i -i -e -n -t over h -b -h -b -h -b -r -f -n -n.

02:46 I'll compare this.

02:48 With this, what changes? the only thing is that each original coefficient is multiplied by an exponential.

02:57 Exponential that depends on the energy eigenvalue for the particular state.

03:02 That's all there is to it.

03:04 It's a very simple procedure.

03:06 I've written it out here.

03:07 Notice each term now has an exponential, corresponding to the particular end value of that term, 2, 2, 2, and 3 here.

03:16 That's the wave function at time t.

03:24 Part b wants to know what energy values you could obtain and what probability.

03:32 Now notice there's only e2 and e3 in this.

03:34 Anything else, zero probability.

03:37 Only e2 and e3.

03:41 E2.

03:43 Probability of getting e2.

03:47 You can get e2 though by being in three different states.

03:53 So when that's the case, three possible ways of getting e2, you add up the probabilities.

04:02 P2 .1 minus 1 plus p210 plus p210 plus p2111.

04:11 That's how you get it.

04:13 Because you have three different ways, three different states you could be in and say you're in e2.

04:22 So you have that as the probability.

04:26 Now, i'll get back to that.

04:28 How do you calculate one of these? well, you do the integration, psi 2, 1 minus 1, star, psi, d3x, where are that? you're basically, you're taking the, this is a complex conjugate, odd to the square, so that, you know, z squared, z star, z.

04:58 So this is equal to, this is equal now...

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