2 Qubit Evolution For the mathematical description of qubit...

2 Qubit Evolution For the mathematical description of qubit control or its time evolution, the Pauli operators are a common choice. Let us consider a Hamiltonian $H = -\frac{\hbar\omega}{2}\sigma_y$ with the Pauli-y operator $\sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix}.$ Suppose the initial state of the system is $|\psi(0)\rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix}.$ (a) Find the wave function $|\psi(t)\rangle$ at time $t > 0$. (b) Find the probability of the system to have the outcome +1 upon measurement of $\sigma_z$ at $t > 0$. (c) Find the probability for the system to have the outcome +1 upon measurement of $\sigma_z$ at $t > 0$. (d) Replace the operator $\sigma_y$ with the Pauli-z operator $\sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix},$ and find the wave function $|\psi(t)\rangle$ at time $t > 0$. (e) Verify that the phase of the qubit cannot be determined by $\sigma_z$ measurement for this particular state $|\psi(t)\rangle$ and explain why. (f) Now let us generalize the qubit evolution. Consider a Hamiltonian $H = -\frac{\hbar\omega}{2}\mathbf{n} \cdot \sigma$, where n is a unit vector in $\mathbb{R}^3$ and $\sigma = [\sigma_x, \sigma_y, \sigma_z]$. Show that the time-evolution operator can be written as $U(t) = \cos\frac{\omega t}{2} \mathbb{1} + i(\mathbf{n} \cdot \sigma)\sin\frac{\omega t}{2}$, and calculate the wave function $|\psi(t)\rangle$ at time $t > 0$.

Transcript

00:01 So i've provided an illustration of the situation in this case.

00:04 So we have particles that are incident from the left.

00:08 And so we're going to have to divide our problem up into two regions.

00:12 So in region one, there will not be a potential, but there will be in region two.

00:17 This means that we're going to have two forms of the wave function.

00:20 So region one, the wave function will have two terms.

00:27 And that means a transmitted or, well, region.

00:30 One will have an incident, i should say, incident wave function.

00:35 That's this first term that you see right here.

00:39 And the second term will be a reflected term.

00:43 And in region 2, you'll simply have a transmitted term.

00:48 So these are the, this corresponds to particles that actually make it through the potential or make it through and are transmitted despite the fact that there is a non -zero potential.

00:58 So if we have a reflection probability, then we have to think in terms of b and a, the coefficients b and a.

01:09 Because after all, this term right here corresponds to reflected particles, because you see this minus sign right here, e to the negative ik1x.

01:21 That means the particles move to the left.

01:24 They move in the opposite direction as the incident particles, e to the i, k1, x, which move to the right, which are incident.

01:34 So we would, of course, have our reflection coefficient defined in terms of the ratio b over a.

01:41 So that reflection coefficient will be determined in this problem, and it will actually be the square of that ratio, b over a.

01:52 So what we have to do to get to that point is look at the boundary condition at x equals zero.

01:58 There's actually two such conditions.

02:01 The condition that at x equals zero, the wave functions, si -1 and si -2 are equal, and that their derivatives are equal to each other.

02:09 So d -si -1 d x and d -si -2 d x are equal to each other.

02:14 So we're going to evaluate those equalities in this problem, and we're going to arrive at r equal to b squared over a squared in terms of wave vectors, which are defined as well.

02:27 And before we get there, though, we need to go ahead and evaluate the wave vectors in terms of the energies.

02:34 So we look at the schrodinger equation in regions 1 and 2.

02:37 So we'll start with region 1 and get the wave vector in terms of energy e1.

02:42 This way, when we're given numerical values for the energies, we can get values for the k vectors ultimately.

02:51 So we go ahead, we substitute in the wave function for region one into the schrodinger equation knowing that the potential is zero in this region.

03:02 And we proceed as follows.

03:04 We take the derivative twice, which is what the second derivative does, and we go ahead and we get a relation from this.

03:14 So we get negative k1 squared on the terms of the wave function on the left hand side of the equation.

03:35 And now we factor k1 squared out.

03:38 The minus signs go away on the left side because we had negative h -bar -squared over 2m times negative k -1 squared.

03:46 And so we have h -bar -squared k -1 -squared over 2m multiplying the wave function on the left, and we have e -1 multiplying the wave function on the right.

03:56 So we get the energy, e -1, equalling h -bar -squared, k -1 -squared over 2m.

04:06 And so we'll go ahead, scroll down a bit, and you can see that relation, and you can see how these two constants multiply the wave function on both sides.

04:16 And so proceeding as follows, we go ahead and we get the wave vector of k in terms of the energy.

04:28 Now we look at region two, where there is a potential u, and we do a similar calculation.

04:35 And so i'll kind of go a bit faster through this, because you see that it's the same method.

04:42 So you would just go ahead, you substitute in your wave function, you take a second derivative.

04:47 This time, though, you do have a potential term.

04:50 And to deal with the potential term, since our potential is a constant, we don't have to worry about it being a function, you would simply subtract it over to e.

05:02 And we'll go ahead and move down right here.

05:06 And, you know, as you take the second derivative, you can go ahead and move u over to the right.

05:11 Hand side by subtraction and you'd have e minus u which is another constant instead of e you just have e minus you it's really e2 but we'll go what we can go ahead and label that as e2 and now you have once you get h bar squared k squared or k2 squared over 2m you have you'll have e2 minus you and then you can take a square root and you get that k2 is equal to the square root of 2m minus e2 minus u over h bar.

05:46 So now we look at our boundary condition at x equals zero.

05:51 And i say boundary condition because it literally is the boundary at x equal zero between the two regions.

05:57 So, si -1 equals psi -2.

06:00 At b equal, this means that at this point, a plus b is equal to c at the boundary.

06:08 And we also have the other condition that the, uh, the, derivatives have to be equal...

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