Recent content by cwill53

Okay, so basically you’re saying we impose the first equation in (A.4) and that’s what yields the second one? The reason that it’s confusing to me is that when the passage asks us to take the inner product, it makes it seems as if we know something about the perturbative corrections to the state...

I'm currently reading this passage to review perturbation theory. Just before Equation (A.4), this passage tells me to take the inner product of the proposed eigenstate ##|\psi _j\rangle## with itself. Writing this out, I got: $$1=\left \langle \psi _j| \psi _j\right \rangle=\left ( |\psi^0...

TL;DR Summary: Imagine you have infinite funds and cooperation to build a very ideal curriculum and student organization, both dedicated to quantum information science, at a university. What does this look like pedagogically? How would this be structured? What resources do students need access...

I've figured out the state of the system at the first barrier. It's $$|q_2q_1q_0\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|011\rangle)$$ I'm stuck trying to find what happens to the state of the system and what's read onto the classical register as a result of the measurement of the state in the...

I'm still pretty confused as to what exactly I need to do for this problem. I'm trying to figure out what the state of the entire system should be after Step 1a, but it doesn't quite make sense to me. I have $$ |\psi _{1a}\rangle=(I\otimes_K I)U_{\textup{CNOT}}(H|0 \rangle\otimes_K|0 \rangle...

I have done part A so far below, but I'm a bit behind on my reading, so I don't quite understand the action of the controlled-NOT gate on a single qubit. What I have so written so far for part B is: Let ##\mathcal{H}=(\mathbb{C}^2)^{\otimes 3}##. Let ##|\psi _{q_i}\rangle_k## , ##(i\in\left...

In physics there is a notation ##\nabla_i U## to refer to the gradient of the scalar function ##U## with respect to the coordinates of the ##i##-th particle, or whatever the case may be. A question asks me to prove that $$\nabla_1U(\mathbf{r}_1- \mathbf{r}_2 )=-\nabla_2U(\mathbf{r}_1-...

Right right. I remember that problem we did. I will readjust my notes to this answer then. Thanks a lot; your help has been very valuable.

Can you help me understand why exactly we need to be careful with the boundaries? I was able to complete the problem by using another substitution. In the picture, the z-component integral was (as you said) $$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi }\int_{0}^{\pi...

I see now that my vector was defined incorrectly. Thanks a lot.

Essentially the problem is this integral and its result: $$ \int_{0}^{\pi }\frac{\cos\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta=-\frac{(R^2+z^2)\log\left ( \frac{(R-z)^2}{(R+z)^2} \right )+4Rz}{4z^2}$$

That makes perfect sense now. Thanks. So I'm guessing now that my approach in my second reply was not correct then.

So for a sanity check I decided to use the following approach. Since I don't know how to TeX the fancy r used in Griffiths' book, I'll use ##\eta##. $$\mathbf{E}_{\textup{average}}=\frac{1}{4\pi R^2}\oint_{\textup{sphere}}^{}\mathbf{E}\, dS=\frac{1}{4\pi R^2}\oint_{\textup{sphere}}^{}\left (...

The picture above shows the integral that needs to be evaluated, and the associated picture ## \cos\alpha ## can be obtained via the law of cosines. I'm simply confused as to where the ##\cos\alpha ## comes from in the first place. I just don't see why ##\cos\alpha ## is necessary in this...

Right, right, thanks a ton.