Does anyone know how to write the proof for "Is a graph with 8 vertices and 6 edges connected". I would think it's a no, because my intuition is that the longest connected graph I think is a linear tree, and such a graph with n vertices has n-1 edges ... so I would expect 7 vertices and 6 edges ... then 1 vertex will be left out
Apparently if you take all the numbers from $1$ to $10^{10}$, write them out as words, and order them alphabetically, the first odd number is just over $8$ million
I'll assume that it's a projective measurement, i.e. once you've measured $A$ and gotten $\lambda_i$ as the outcome, it should be the case that measuring $A$ again should still give the same outcome.
(On the other hand, if you measured A and then made a different measurement B, then A again, you wouldn't necessarily get the same result for A twice.)
anyways. If we assume $\langle \phi_i|\phi_j\rangle = 1$ if $i=j$ and 0 otherwise, then the above collapses to $\langle \psi |\psi\rangle = \sum_{i=1}^n |c_i|^2$
To make the link with what you wrote earlier, suppose we instead computed $\langle \phi_i | \psi\rangle=\sum_{j=1}^n c_j \langle \phi_i |\phi_j\rangle = c_i$
so our probabilities can be conveniently written as $p_i = \frac{|\langle \phi_i |\psi\rangle|^2}{\langle \psi|\psi \rangle}$
To make life easier, one usually goes a step further and scales $|\psi\rangle $ to have unit norm i.e. $\langle \psi|\psi\rangle = 1$
It's kind of like if measuring the vector $(3,4)$ had a $9/25$ chance of turning it into $\hat\imath$ and a $16/25$ chance of turning it into $\hat\jmath$
@AkivaWeinberger And if you placed a horizontal filter after this one, only 9/25 of the light passing through the first filter would pass through the second
i don't have a great sense of quantum teleportation tbh
@AkivaWeinberger yeah
the action of a polarizer is derived from electromagnetic wave theory
so the idea that you can make it work photon-by-photon is bizarre from that point of view
the main nice thing about the polarizer analogy imo is that you can have the following
Suppose you set up a horizontal polarizer and a vertical polarizer in sequence. Then no light should pass through both of them
So you measure polarization, selecting for horizontal polarization, and then measure again, selecting for vertical polarization. You end up with no light going through
For instance, suppose you send a beam of electrons through a Stern-Gerlach device. then you'll end up with two beams of electrons, deflected along two different directions
this is all prohibited subject matter I demand the SF ban both just as I was for my awesome response to will kunting being a banana anding also loving said fruit
So you won't typically see the Schrodinger equation invoked, at least not in the sense of $-\partial_x^2 \psi +V \psi = E\psi$
(You can invoke the Schrodinger equation in that sense, but you'll need to take $\psi$ to be itself a 2-vector and $V$ some matrix. most people don't bother going down that route and just worry about the spin degree of freedom rather than the space degrees)
I guess that's a big thing to appreciate about QM: So much of what's going on is to do with what information you treat as relevant and what information you discard as irrelevant
there's some subtlety attached to it, since the outcomes are continuous rather than discrete. but it is an observable.
yeah
You get into the weirdness of position measurements when you think about how an actual position detector would work, i.e. a realistic detector is a large but finite set of pixels with some local sensitivity
$\hat{S}_z$ is the operator corresponding to the observable $S_z$. you measure it by sending an electron through a stern-gerlach device (oriented in the $\hat{z}$ direction) and observing the deflection direction (up or down); we interpret it as the z-component of the electron's spin angular momentum
( angular momentum * angle has the same units as linear momentum * position)
Yeah. And then you'd have to relate that to the magnetic moment of an electron
So the linkage is: A spinning sphere of charge acts like a miniature current loop i.e. a magnetic dipole
And a magnetic dipole, passing through a magnetic field gradient, experiences a force depending on its orientation
the electron acts like it is spinning on its axis with angular momentum $\hbar/2$ and therefore has a certain magnetic moment, which means it'll get deflected by the magnetic field gradient
for that detector orientation, you'd get probabilities $\cos^2(\theta)$ for spin-up, $\sin^2(\theta)$ for spin-down (assuming you're sending in electrons which came out spin-up from an SG device oriented in the +z sense)
to be more pricse: Suppose you had two electron beams and you fed them into two Stern-Gerlach devices. One is oriented in the positive $z$ direction, whereas the other is flipped upside down.
In both cases, you'll get electrons deflected in the same way.
However, if you send the spin-up beam from the first one into the second one, you'll now find that they're all deflected down
(to get a sense of why that's an interesting case, note that the vectors for those three directions add to zero. up to rotation, that's actually the only way to add three unit vectors together and get zero)
in the CHSH setup, you'd have one observer who can measure Sx, Sy, while the other observer can measure along Sx',Sy'. specifically, you'd end up picking x'=(x+y)/sqrt(2) and y'=(-x+y)/sqrt(2)
sure. if I measure positive deflection, then the probability of you measuring negative deflection will be 1/4 iirc
To make this more quantitative, let your observables be $S_{a1},S_{a2},S_{a3}$ and mine be $S_{b1},S_{b2},S_{b3}$
If we measure along the same direction, we definitely get opposite results. For convenience, let me label these outcomes as $\pm 1$ rather than $\pm \hbar/2$
so, some relevant statistics: you'll always have $S_{a1}^2=1$, so $\langle S_{a1}^2\rangle =1$ and the same for the rest of our observables
Yeah. Equivalently, I could take my results and guess that your result is the opposite of mine
Which will work perfectly when we happen to pick the same direction
For now, though, let's leave them as written
In which case we always have $S_{a1}S_{b1}=-1\implies \langle S_{a1}S_{b1}\rangle =-1$
$\langle A\rangle$ is the usual physics way of writing the expected value of the observable $A$. that arises from the fact that $\langle A\rangle =\frac{\langle \psi | A|\psi\rangle}{\langle \psi|\psi\rangle}$, so it's handy notation
anyways. my statement above implies that $\langle S_{a1}S_{b2}\rangle = \frac14 (-1)+\frac34 (+1)=\frac12$
and that remains true if I replace 1,2 with another other pair of distinct indices
@user76284 I'm more used to writing r.v.s as capital letters and their values as lower case, e.g. $p(a)=\text{Pr}(A=a)$ for a discrete r.v. $A$
so I was mixing myself up
anyways. in the present case, we've got $p(a)=A^{-1/2}\varphi(a/\sqrt{A})$ and $p(b|a)=B^{-1/2}\varphi((b-a)/\sqrt{B})$ where $\varphi(z)$ is the standard normal pdf
If $f: I--->\mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x \in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$
(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f||...
so we're interested in $$p(b)=\int_{-\infty}^\infty \frac{1}{\sqrt{A}}\varphi\left(\frac{a}{\sqrt{A}}\right)\frac{1}{\sqrt{B}} \varphi \left(\frac{b-a}{\sqrt{B}}\right)\,da$$
