Recent content by PhysicsKush

Interessting, but how does that help me find the matrix generator of rotation ?

I first computed the operator ##\hat{T}## in the ##a,b,c## basis (assuming ##a = (1 \ 0 \ 0 )^{T} , b = (0 \ 1 \ 0)^{T}## and ##c = (0 \ 0 \ 1)^{T}##) and found $$ \hat{T} = \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}.$$ The eigenvalues and eigenvectors corresponding to this matrix...

Sorry , I am not sure to understand the setup. Are they just bouncing off the nodes?

I answered the first part of the question where I estimate the radius of ##O_{2}## is ##\approx 1.5 \times 10^{-10} \ \text{m}##: $$ p = \frac{KT}{l 4 \pi r^{2}} = \frac{(20+273.15)(1.38\times 10^{-23})}{(0.1)(4\pi)(1.5 \times 10^{-10})^{2}} = 0.143 \ \text{Pa}.$$ The confusion arises on the...

So I digged further and found out that by superposition the total electric field between the two wires should be $$ \frac{\lambda}{2\pi \epsilon_{0}R} + \frac{\lambda}{2\pi \epsilon_{0} (d-R)} = \frac{\lambda}{2 \pi \epsilon_{0}}\left( \frac{d}{R(d-R)}\right),$$ since ##R \in (0,d)##, it...

It is not stated, but I can only assume that is the logical case.

Right , the electric field can not be near the center because the electric field inside the conductor is ##0## ! I conclude the maximal electric field occurs right over the surface of the wire ?

You're right, well I think ##r## is the Gaussian radius which can be extended over the real radius, so essentially it is the distance from the center of the wire. Also, well since both wire produce an electric field that decays with ##\propto \hat{r}/r##, oh wait I realize my answer makes no...

For the first part, since $$ E(r) \propto \frac{1}{r} \hat{r}$$ by the principle of superposition the maximal electric field should be halfway in between the two wires. Then I'm not sure how to go about the second part of the question. I understand that the total potential due to the two wires...

Wow thank you so much ! Indeed, I completely dropped the absolute values and I did not think of setting ##\alpha \ \text{or} \ \beta = \frac{e^{i\delta_{\pm}}}{\sqrt{2}}##. Cheers!

Hello, all we saw in class is that for 2-level systems \begin{gather*} \ket{\pm x} = \frac{1}{\sqrt{2}}(\ket{+z} \pm \ket{-z}) \\ \ket{\pm y} = \frac{1}{\sqrt{2}}(\ket{+z} \pm i \ket{-z}) \end{gather*} and more generally, $$ \ket{\varphi} = \alpha \ket{+z} + \beta \ket{-z}$$ From the...

Right, that's a mistake. So ## \langle S_{x} \rangle = \lvert \langle+x | \varphi \rangle\rvert^{2}##, but I don't know what ##\varphi## is. Is it simply ##\langle S_{x} \rangle = \lvert \langle +x |+x\rangle \rvert^{2}## ?

By definition , ##\ket{+x} = \alpha \ket{+z} + \beta \ket{-z}.## Therefore we proceed , \begin{align*} \langle S_{x} \rangle &= \lvert \alpha \rvert^{2} \left(\frac{\hbar}{2}\right) + \lvert \beta\rvert^{2} \left(-\frac{\hbar}{2}\right) = (\alpha^{2} - \beta^{2})\left(\frac{\hbar}{2}\right).\\...

WOW ! I think I understand now ,all the pieces are connecting together ! Thank you so much for having persisted with me throughtout this. Couldn't thank you enough sir.

Well, I can not submit this picture since I don't understand how we arrived at this conclusion. I shall keep on searching for an alternate answer. Thank you again for your help !