Recent content by jumi

Thanks for the replies. So what exactly should I do? Are the formulas tms posted correct?

Homework Statement Using 5 different ammeters, you get the following data (all measured in Amps): I_{A} = 128 ± 2 I_{B} = 121 ± 1 I_{C} = 114 ± 8 I_{D} = 120 ± 3 I_{E} = 122 ± 4 Calculate the mean current and the standard deviation of the mean. Homework Equations Standard...

Ok, I have this circuit, and I'm trying to devise a way to create an 8x8 matrix for any given combination of target qubit and controlled qubit. (I is the identity matrix, and X is the CNOT operator) I want it to also include the qubit that passes through untouched because I'm trying to...

Homework Statement What is the 4x4 unitary matrix for the circuit in the computational basis. Homework Equations We were given the following relationship in our notes: . By letting A = H and B = I, the answer is supposedly supposed to be: simply by inspection (where H is the...

I don't understand how I could just go from (1) and (2) to (4)... Or why the numerator doesn't have a vector magnitude, whereas the demoninator does... I was, however, able to get the correct answer, just not using the full notation from equation (1). So I start with: \vec{E} (...

So there's this situation going on: http://imageshack.us/a/img826/7398/physicsforums.png Going from the definition of an electric field: (1) \vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} ∫ \frac{\vec{x} - \vec{x'}}{| \vec{x} - \vec{x'} | ^3} ρ( \vec{x'}) d^3x' (2) The...

I don't follow how this expression makes it easier: \langle x_{\pm} | = | x_{\pm} \rangle^{ \dagger } = \left( \frac{|0\rangle \pm |1\rangle}{\sqrt{2}} \right)^{ \dagger } = \frac{|0\rangle^{ \dagger } \pm |1\rangle^{ \dagger }}{\sqrt{2}} = \frac{\langle 0 | \pm \langle 1 |}{\sqrt{2}}

Thanks so much! I really appreciate all the help.

Oh, alright, I get it now. Disregard that second statement. I knew the complex conjugate showed up in some capacity, but you answered it in your first sentence. So all in all (sorry for continually pestering you...), I would take \left|x_{+}\right\rangle = \frac{\left|0\right\rangle +...

Ok, I had a chance to review the math a bit. So knowing, P(\alpha) = | \langle x_{\alpha} | \psi \rangle |^2 = \langle x_{\alpha} | \psi \rangle ^* \langle x_{\alpha} | \psi \rangle = \langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle, I can simply multiply the row vector...

Ok thanks, it's starting to piece together a little better now. I'll definitely give that book a read. So we can't say say |\alpha_{00}|^2 and |\alpha_{01}|^2 sum up to equal 1 since they were only 2 of the 4 options in the original function. But we CAN say that for the second measurement...

I'm actually an Engineering Physics major. Never taken a QM or probability course, either. Currently a junior. We're using "Quantum Computation and Quantum Information" by Nielsen and Chuang. I understood mostly everything in your first post. However, it seems like you just took a leap to...

Awesome reply, thanks. I don't have time to really sit down and mess around with the math tonight, but I'll take a look at it tomorrow and see what I can come up with. I'll make another reply if I'm still having trouble. Otherwise, I really appreciate the help. Thanks.

Ok, thanks. Yeah, I can see and reproduce the pattern, but I was wondering "why". No worries, though; I appreciate your help.

Ok, I understand that. However, in class, we did something like this: Say we have the same 2-qubit state we've been talking about, \left|ψ\right\rangle = \alpha_{00}\left|00\right\rangle + \alpha_{01}\left|01\right\rangle + \alpha_{10}\left|10\right\rangle +...