25. The unperturbed Hamiltonian of a two-state system is...

25. The unperturbed Hamiltonian of a two-state system is represented by [ H_{0}=left(egin{array}{cc} E_{1}^{0} & 0 \ 0 & E_{2}^{0} end{array} ight) . ] There is, in addition, a time-dependent perturbation [ V(t)=left(egin{array}{cc} 0 & lambda cos omega t \ lambda cos omega t & 0 end{array} ight) quad(lambda ext { real }) . ] a. At ( t=0 ) the system is known to be in the first state, represented by [ left(egin{array}{l} 1 \ 0 end{array} ight) ext {. } ] Using time-dependent perturbation theory and assuming that ( E_{1}^{0}- ) ( E_{2}^{0} ) is not close to ( pm hbar omega ), derive an expression for the probability that the system be found in the second state represented by [ left(egin{array}{l} 0 \ 1 end{array} ight) ] as a function of ( t(t>0) ). b. Why is this procedure not valid when ( E_{1}^{0}-E_{2}^{0} ) is close to ( pm hbar omega ) ?

Transcript

00:01 In this problem, we're going to do time -dependent perturbation theory on a two -level system.

00:07 So in this problem, we have a hamiltonian, which has two eigenvalues, as denoted here, two energy states.

00:16 And we're going to perturb that with a time -dependent potential listed here.

00:23 It's going to be changing cosine of omega -t.

00:25 And we're going to assume that our wave function at time equals zero starts out in the e1 state.

00:35 And what we want to do here is to calculate the probability that at some later time, si is in the zero one state.

00:46 So let's go through this.

00:49 So what we know is that in general, xxi is going to equal some coefficient.

00:59 Let me call this c1, actually, c1.

01:05 And since this is a time, or we're perturbing this with a time -dependent part, this is going to attend on time, and this will be for the one -zero state.

01:16 And then we're also going to have a c2 for the zero -one state.

01:22 Right.

01:23 So the only thing we know right, now is that c1 of 0 is equal to 1 right and what we're going to do now is try to solve for c2 of t or approximate it by using time -dependent perturbation theory okay so what time -dependent perturbation theory tells us to the first order is that c2 of t is going to be approximately negative i over h -bar an integral like so, 0 of t, the v of t is going to be in here.

02:07 One of the factors of v of t at least, being one of these, but they're both the same.

02:15 So, okay, and then it's going to be, oh, and i wanna make this t prime, just a different shape, each of the two t's here, and then e to the i, e1, minus e2, zero, 0 divided by h bar t prime d t prime okay so this is the integral that we need to get through and so let's do that so in order to solve this you will use this identity that cosine of omega t is equal to e to the i omega t prime plus e to the i omega t prime plus e to the the minus i omega t prime divide by two.

03:16 So what this will do for us is allow us to express this integral as two exponentials, essentially.

03:28 So this is what this will become, integral from zero to t.

03:38 Let me factor off that two to the front here.

03:42 And then we're gonna get, let's see.

03:44 So what i want to do actually also, is i want to define omega not here as this e2 to the 0 divided by h bar.

03:56 So let's do that.

03:58 It'll make things a little bit easier to look at here, because then this will become e to the i, omega plus omega knot, t, plus e to the i, omega, not, t, plus e to the i, omega, not minus omega t, dt, and these should all have primes on it, these t's here.

04:20 Okay.

04:21 So this is the integral.

04:23 It's a pretty easy integral to solve.

04:26 This becomes minus i lambda to h bar, e to the i, amega, not plus omega, to the t minus one divided by omega not plus omega.

04:51 I realize that we're dividing by i here, and so this i actually is going to go away and plus e to the i omega not my anis omega t minus 1 divided by the difference between these omegas all right so this is what c2 of t is and if we want the probability of being in the state 2 as a function of time it's just going to be this modulus squared.

05:35 So you can go ahead and kill the way what that is.

05:38 It's going to be lambda squared, 2h bar, and then all of this squared on the inside, which i can do right away, i suppose.

05:53 So e to the i omega -not plus omega t minus 1, all that modulus squared divided by omega, not plus omega, all squared.

06:13 Okay, and then a very similar term here, t minus one, that's squared, divide by the difference.

06:27 Okay, and we want to square that, okay, and one more term here, it's going to be minus, or is it going to be minus? no, it should not be minus, it should be, plus 2 times e to the i omega not plus omega t minus 1 times an e to the i omega 0 times then e to the i omega minus omega not t minus 1 divided by and this is going to be um omega not squared minus omega squared okay brilliant and now we just got to do a little bit more out we should have our answer.

07:34 Okay, so there's going to be 1 minus 2 cosine of omega not plus omega.

07:59 So i'm kind of skipping some steps here because what this is going to be is right this times its complex conjugate and so this is going to be the summation of in fact, this isn't going to be a two here.

08:15 Maybe i shouldn't skip as many steps.

08:19 Yeah, let's just write this out.

08:24 Omega -0 plus omega -t, t minus 1.

08:29 E to the minus omega -not plus omega -t, t minus 1.

08:35 Okay.

08:37 Yeah, so we're going to get a 1 out of that, which i have right here for the two e's multiplied by each other.

08:46 And then we're going to have e to the i times negative 1, and the e of the minus i times negative 1.

08:54 And so that's going to give us a minus cosine, and there actually will be a 2 there, right? since it's actually will be 2 times cosine...

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