2. Time Development of Free Particle Superposition of...

2. Time Development of Free Particle Superposition of Momentum Eigenstates Consider an infinite domain free particle, of mass $m$, which enjoys a linear superposition of 1D momentum eigenstates, $|k\rangle$, as $|\psi\rangle = \frac{1+i}{2}|k_1\rangle + iC|k_2\rangle + \frac{1-i}{2}|k_3\rangle$. Recall that this type of infinite domain momentum eigenstates are $\langle x|k\rangle = e^{ikx}/\sqrt{2\pi}$, which are eigenstates of momentum, as $\hat{p}|k\rangle = \hbar k|k\rangle$; however the wave functions are not normalizable in the usual finite sense, but expectation values of any operator, $\hat{A}$, can be obtained as a ratio of the usual expectation value (as an inner product) with the normalization (inner product), where $\langle A\rangle = \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle}$, so that any wave function can be effectively normalized, as $1 = \frac{\langle\psi|\psi\rangle}{\langle k|k\rangle}$. Nevertheless, the momentum eigenstates, $|k\rangle$, form a generalized orthonormal basis, where the Dirac delta function inner product is $\langle k|k'\rangle = \delta(k-k')$, so that $\langle k|k\rangle = \delta(0) = \frac{1}{2\pi}\int^{\infty}_{-\infty} dx \to \infty$, where this infinity cancels when computing the normalization as well as any expectation value. a) Determine the $C$ coefficient to normalize the wave function, and write down the time-evolved normalized wave function, $\psi(x,t)$, using only the parameters given, $m,\hbar,k_1,k_2,k_3$, that is, not using the abbreviated notation of modes written as $|k\rangle$. b) Determine the expectation value of momentum, $\langle p\rangle$. c) If upon energy measurement, the energy value of $E = \frac{(\hbar k)^2}{2m}$ results, write down the resultant time-dependent collapsed wave function $\psi_{collapsed}(x,t)$.

Transcript

00:01 In this problem, you have multiple parts.

00:04 Due to time constraints placed on me, i cannot answer all parts.

00:13 Certainly resubmit if you need anything that i don't answer.

00:19 I'll start with the largest one, which is finding the eigenvalues and eigenvectors of s .y.

00:25 And then i'll see how my time is.

00:31 Tried it, it's almost double the time.

00:35 So let's look at this.

00:39 What we need to do is when we do looking for eigenvalues, we set up a determinant.

00:46 Well, here, though, we've got to be a little careful because we have this h -bower over 2 sitting out there.

00:52 So we have to, normally you would have your, whatever your symbol use, lambda or omega is what i use.

01:03 For your eigenvalue, you would be sitting here.

01:05 That's fine as long as there wasn't this h -bar over 2.

01:08 So what we have to do is call the eigenvalue, omega, h -bar -lamda over 2.

01:19 So we're looking for lambda, but lambda is not the eigenvalues.

01:24 Omega is the eigen -are the eigenvalues, or gives the eigenvalues.

01:32 So whatever lambda is, multiply by h -bar over 2, that gives you the eigen values.

01:37 1 .0 .01.

01:40 Now i can combine these matrices to have a single quantity.

01:47 Otherwise, how could i combine? i got h bar over 2.

01:51 So, h bar over 2 minus lambda, i, minus i, minus lambda.

02:02 My square brackets are for a matrix, and the up and down vertical ones indicate determine it.

02:12 So i get from this, h bar over two squared, determinant lambda i minus i minus lambda is equal to zero.

02:22 So the h bar over two squared doesn't mean anything.

02:26 So this becomes lambda squared minus minus i times i zero.

02:34 So this is lambda squared minus one is equal to zero.

02:38 So lambda is plus or minus 1.

02:42 So that gives me my eigenvalues, omega, are plus or minus h bar over 2.

02:50 Those are my eigenvalues.

02:52 All, now let's talk about the eigenvalues before i get into the eigenvectors.

02:56 All sx, s, y, s, c, all have eigenvalues plus or minus h bar over 2.

03:04 They all have that.

03:08 But the base, the eigenstates, the eigenbasis from each one are not the same.

03:18 Each one's eigenvector can be written in terms of the, is written in terms of the, say, the eigenstates, eigen basis of z.

03:29 So you don't, you don't, there's no simultaneous eigenvectors.

03:34 So it's like having, you have a vector for h bar over two.

03:39 As we're going to see for s .y plus h -bri -over -2, that can be written in terms of the plus h -bar -over -2 and the minus h -bri -over -2 that correspond to the s -z.

03:54 That is eigenvectors of s -z.

03:57 So it's like having a vector and be writing it into components of the other components.

04:04 That's really what's going on here.

04:07 It's like really what's going on is it's like having three sets of two unit vectors that each set can be written in terms of any of the other sets.

04:23 That's what we're doing.

04:24 That's really what we're doing.

04:26 But remember, each of those unit vectors are actually eigenvectors of a particular operator, but not of all.

04:35 Not of all? okay.

04:40 So we're going to do for, vamd equals 1.

04:44 This is h -bar over 2, eigenvector.

04:48 So h -bar over 2, minus 1, i, minus i, minus 1, x1, x2, is equal to 0, which you can just, a lot of times you just write 0, or you can write the column vector with 0 -0 in it, it's up to you.

05:09 Okay, so let's expand, the h -bar over 2 goes away, so let's expand this out.

05:15 This becomes minus x1, minus ix2 is equal to zero, and this is i x1, minus x2 is equal to zero.

05:24 Now you might say two equations, two unknowns, it's wonderful.

05:27 Well, it's not two equations.

05:30 If i were to multiply a second equation by i, i times i is minus one.

05:35 So i get minus x1, that's like the first equation, minus ix2.

05:40 They're the same equation.

05:43 They're the same equation.

05:44 They give us the same relationship.

05:45 So this gives me that x1 is equal to minus i x2.

05:54 Now let's look at some other further properties.

05:57 X1, the complex conjugate, is i, x2 complex conjugate.

06:03 So if i were to do x1, get the modulus square root of x1, which is x1 star times x1.

06:13 So that gives me modulus squared.

06:17 That would give me minus i times i, which is one times x2 star x2.

06:27 So the modulus square to both is the same.

06:35 And we're going to need that in a minute.

06:43 Okay.

06:45 So right now what we have plus one half, and this indicates that it's a y eigenstate.

06:58 It's an eigenstate of x.

07:00 Or z, that's xxx z, it's only for s1.

07:05 This is equal to, we have, we have, however you wanna write it, we can put minus i, we can leave, well actually let me just leave, x1, and then, or actually, let me do it this way, minus i, x2, and then just x2.

07:31 Just doing that, because when we do normalization, as we're going to do now, because that's what's going on here.

07:38 Why we have only one equation? because we're free to normalize.

07:42 We have to have that freedom.

07:44 If these were set by the equations, where's the freedom to normalize? so this is just putting in my result here.

07:59 Okay, so now let's normalize.

08:04 So we generate the ket, or we generate the bra from the ket, i should say.

08:12 What is that when it's a matrix, form.

08:15 You transpose, take complex conjugate.

08:19 So this is plus one -half y plus one -half y.

08:27 That's going to be one.

08:31 And so we transpose, which turns a column vector into a row vector, and take the complex conjugate.

08:38 So i becomes minus i, so this becomes just i, x2, and then just an x2.

08:47 Star and then x2 star times minus i x2 x2 x2 so i'm going to find x2 out of this i could have done the other one doesn't matter so expanding this out i times i is minus 1 times minus 1 it's 1 so it's x2 star x2 plus x2 star x2 that's the modulus squared twice.

09:19 2 times the modulus of x2 square equal 1...

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