Ok, if I understood correctly from here and also what I was taught back in my 3rd year QM and in susskind's, the direction of the vector on the bloch sphere corresponds to the direction I must align my detector in order to get 100% detection of the electron. In addition, the bloch sphere maps a ray in hilbert space to a vector on the sphere. Therefore the angle that appeared as the parameters on the Bloch sphere
are half of that for the corresponding ray in the hilbert space. Now my confusion:
The "spin vector" is defined to be $\langle \vec{S}\rangle =(\langle \hat{S}_x\rangle \langle \hat{S}_y\rangle \langle \hat{S}_z\rangle)$ where $S_i$ are the spin operators given by the pauli matrices under the standard basis. Because of the uncertainty principle, only one component is ever found with certainty thus we cannot speak of the "angle between two spin vectors" Now,
@BernardoMeurer This time is different, I am actually asking a question, I just have not tag acuriousmind in yet because I had not finished writing the question
suppose on the bloch sphere I have plotted the states $\lvert 1\rangle$ and $\frac{1}{\sqrt{2}}(\lvert 0\rangle+\lvert 1\rangle)$, therefore in the bloch sphere, the angle substended by these two vectors is $\frac{\pi}{2}$
I sometimes see $U(n)\subset SO(2n)$ stated casually
But the determinant is not always going to be one. What's going on here?
On the other hand, I guess elements of $U(n)$ preserve orientation and are orthogonal w.r.t. the inner product induced on $\Bbb R^{2n}$ by identifying it with $\Bbb C^n$, so they should be in $SO(2n)$.
@ACuriousMind Since in the original hilbert space where the spin states live, angles between rays are not well defined due to the spin vector can only ever have one component without uncertainty, how should I correctly interpret the physical meaning of the angle $\theta = \frac{\pi}{2}$ substended by the two aforementioned vectors on the bloch sphere?
@BernardoMeurer If you have two vectors $v_1,v_2\in\Bbb R^4$, how would you go about constructing a $4\times 4$ matrix $A$ with $\ker A=\vee\{v_1,v_2\}$?
I would use Gram-Schmidt to orthonormalize $v_1,v_2$, then construct a projector onto the orthogonal subspace.
Could someone provide an example of a manifold that is not smooth? All manifolds that come to mind are smooth! By a manifold, I mean a hausdorff, second countable locally euclidean space.
@Danu I think the point is you can approximate C^k diffeomorphisms by C^k+1 diffeomorphisms for all k > 0. Should follow from some sort of smooth approximation.
Let $\alpha$ be a $C^r$ differential structure on a manifold $M$, $r\ge 1$. Then for every $s>r$, there exists a compatible $C^s$ differential structure $\beta\subset\alpha$, and $\beta$ is unique up to $C^s$ diffeomorphism.
The proof seems to require more analysis than is healthy
I know of several reasons why the long line can't be a covering space for the circle, but I'm more curious in what exactly goes wrong with the following covering map.
Let $L$ be the long line and define $p: L \rightarrow \mathbb S^1$ by wrapping each segment of the line around the unit circle o...
@Slereah Give me a quick hand, I'm trying to translate my MIPS ASM into C to do some testing of my implementation
I have this assembly code:
SUBI R3,R0,#1
SUBI R6,R0,#-1
SUB R1,R3,R6
SUBI R4,R0,#3
B.EQ R4,R0,#7
SUB R2,R3,R6
SUB R3,R0,R6
SUB R6,R0,R2
SUBI R4,R4,#-1
ADD R1,R1,R2
B.NE R4,R0,#-5
SUB R2,R0,R1
B #0
@EmilioPisanty I have a question about PR boxes vs entanglement. I understood that PR boxes have stronger correlations than entanglement because the CHSH correlation function of PR boxes can go up to 4, beyond the Tsirelson bound of bell states. But physically speaking, how are they more nonlocal than entanglement, like what extra kind of correlation they can do that entanglement cannot?
In all likely hood, bound states do exist. If you were to look at the Ground state energy as a continuous function of the internuclear distance, starting from $R_ to \infinity$ you would need a length scale where the ground state energy vanishes
Could someone provide an example of a manifold that is not smooth? All manifolds that come to mind are smooth! By a manifold, I mean a hausdorff, second countable locally euclidean space.
