I am confused...the answers for this question seem to conflate hermitian operators with observables. or perhaps i am misunderstanding something. i don't think it makes sense to say that $|\langle \phi \lvert \rangle|^2$ for arbitrary $\lvert \phi \rangle, \lvert \psi \rangle \in \mathcal{H}$ corresponds to the probability that $\lvert \psi \rangle$ will be found in state $\lvert \phi \rangle$ upon "measuring".
In the proof of Wigner's theorem, the crucial role is played by the quantity $|\langle\phi|\psi\rangle|^2$ where $|\psi\rangle$ and $|\phi\rangle$ represent two distinct physical states (or more precisely, two distinct rays) of a system. Expanding the states in the orthonormal eigenbasis $\{|a_n\...
@SillyGoose No, such overlap integrals do mean the probability that $\left|\psi\right>$ will be found in state $\left|\phi\right>$, if you can find a measurement operator that can measure either state directly.
hm i see so e.g. for spin-1/2 $1/\sqrt{2}(\lvert 0 \rangle + \lvert 1 \rangle)$ i suppose in the context of measuring $S_z$ you could compute the probability of getting either spin value by projecting as in other peoples's answers
I know you linked to ACM's post that there are superselection rules. But those superselection rules can be taken backwards to mean that some certain states in the Hilbert space cannot be realised. Then there is no point talking about them
@SillyGoose It is easier to think of photon polarisations. You can get any angle's linear polarisation, and you can get circular polarisation. That should give you access to measure basically any state you want
and it is nice because you can not just measure any probability you like, but that you can also use the same setup to generate any state you want
of course, you will never get fermionic photons this way, but that is a situation that we would not have intended to scrutinise anyway
If I have some more time, I would directly answer the OP's question by saying the above. The rest of the answers are mired in the minutiae and are not actually answering the crux of his question, as you are concerned about
hm well im still confused because under the assumptions of textbook quantum mechanics any state post-measurement will be in an eigenstate of the observable you measured. so the observable to be measured in this case would have to have eigenstate $\lvert \phi \rangle$, which is what i am caught up on
i am not familiar with the photon polarization example
because what i am thinking is: is an arbitrary superposition of say energy eigenstates $\Sigma_n c_n \lvert E_n \rangle$ really an eigenstate of some observable? but idk .-.
@SillyGoose Ok, if that is the case, then you should consider measuring the spin of an electron in an arbitrary direction, and realise that, any $\left|\phi\right>$ that you write down, corresponds to spin up in some certain direction
Hm but something bugs me. The spin-1/2 system is almost trivial because it is a dimension two system so for any arbitrary state it is immediate that a post-measurement state will be one of the two eigenstates. And so the understanding of the amplitude squared as given in the current answers follows
But youre saying this is generally true for other systems
Hm well i think its true for orbital ang because i do recall linear combos of spherical harmonics are eigenstates of a rotated orbital ang operator
I guess i have to convince myself for non rotation related things :P
I would just argue that the formalism of QM is internally consistent. Even though it is only trivial to do all these measurements on angular momentum stuff, the formalism is the same, which means that it is in principle possible to do the same kind of stuff with any measurement operator you want.
sorry, any state you want
as long as it is not forbidden by some superselection rule
Then, does it mean that the corresponding magnetic moment is anti-parallel with the magnetic field direction?
Since the spin and the corresponding magnetic moment are anti parallel ?
$\mu_z=-2m_s\mu_b$ and since $E=-\mu_zB_z=2m_sm_b B$. So for $m_s=1/2$ you'd have a positive energy value for spin aligning with the magnetic field. While $m_s=-1/2$ you'd have a negative value for when spin is anti parallel to the magnetic field, so this implies that you get the smallest energy, for when the spin is opposite to the magnetic field, because the corresponding magnetic moment aligns with the B field. But isn't this wrong?
@PrateekMourya The question is worded very poorly. I think what it wants you to do is to compute the probability to measure the momentum of the particle as $2\hbar k$ at a time $t$ for a particle whose initial wavefunction at $t=0$ is the $\psi(x)$ from the question.
hi -- i have a question. :D somebody asked me recently why $v^ie^i$ is not correct notationally in GR. my answer was that vectors need to have contravariant and covariant components such that the overall vector is preserved and this is not how we write such components -- should be $v_ie^i$ or $v^ie_i$ depending on the basis, but then they asked what $v^ie^i$ would actually represent and i was unsure
i have another question about the lorentz transformation. it says in carroll that the velocity in reference in this coordinate transformation is $\frac{x}{t}$ where $x,t$ are the original coordinates. then it says the transformation is $t' = \gamma (t - vx)$ and $x' = \gamma (x - vt)$ then it says that in this transformation, the light cones stay the same.
i think i am misunderstanding smth because the line $x = t$ seems to become the origin in the new frame so ? how is that the light cones staying the same
@Relativisticcucumber I don't think that's the definition of $v$
but in any case even if it was, then setting $x' = 0$ (the origin in the primed frame) in your transformation is simply equivalent to $\gamma(x-vt) = 0\implies x = vt$.
i think im still not understanding this notion of $v$. on the one hand i was told this is the velocity of the one frame wrt the other frame, but then in this book it says $v = \frac{x}{t}$ and i am not sure how these are the same thing. the later one is slightly more sensible to me
@Mad the expectation value is linear, $\langle A + B\rangle = \langle A + B \rangle$ for any two operators $A,B$ and $\langle cA\rangle = c\langle A\rangle$ for any number $c$ and operator $A$
It is absloutely ridicioulous to the extent, that when the book says "the proof is trivial" or "left as an excercise" The professors literally write the same thing in his lectures
I guess I am missing something very trivial. If for a ferromagnetic material, the magnetisation v/s temperature relation is given by $\frac{M_s(T)}{M_s{0)}=tanh(\frac{T}{T_c}$, then why is this tan hyperbolic curve going down?? (Figure from Kittel)
hello -- for the vector transformation rule $V^{\mu^{\prime}} = \frac{\partial x^{\mu^{\prime}}}{\partial x^{\mu}}V^{\mu}$, does this solely refer to the components of the vector? it doesn't make reference to the bases?
bleh i feel i need to better understand whats important xD there should be 4 relevant transformation rules for vectors, right? 2 components and 2 basis? and one co and one contra for each?
I don't think about these transformation rules as something one should learn by heart. Every time I need to know how something transforms, I just re-derive it from the idea that vectors have $\partial_\mu$ as their basis and covectors $\mathrm{d}x^\mu$ as their basis
so a vector is $v = v^\mu \partial_\mu$ and I know the full expression doesn't change under coordinate change, so the chain rule gives how $\partial_\mu$ transforms and hence the $v^\mu$ transforms in the opposite way
@ShikharChamoli People are free to help or not help. If you want people to help you, you have to try to find some way so that people are more motivated to help you. Spamming the chat with calls for help will only make things worse.
It seems that you commonly have questions that are really out of the blue, especially when people are hotly talking about something else, leading to your questions being buried quickly. And quite often the questions you are asking seem to require too much work, and it does not seem like you put in enough effort to answer them yourself. For example, this question from Kittel that you brought in, is very well-treated in Kittel if you would just read it. The graph is obtained from the tanh
Ok, I'll wait for a better moment from the next time onwards.
@naturallyInconsistent This is the relation, given in Wahab. Honestly I was reading from multiple books. This is what's plotted and the relation is of tan hyperbolic, isn't it? I have thought enough but couldn't understand what's happening
@ACuriousMind i had a follow up to this. sorry this might be pedantic, but you wrote $x \rightarrow f(x)$ but in a simple case of polar coordinates, we would do something like $x \rightarrow r\text{cos}\theta$, right? so notationally, this seems to not match up? am i missing smth
@Relativisticcucumber What do you mean by "a means of doing that"? Spherical to Cartesian is just the map $(r,\theta,\phi) \mapsto (r\sin(\theta)\cos(\phi), r\sin(\theta)\sin(\phi), r\cos(\theta))$.
you are right that $x = r\cos(\theta)$, but I am talking about the map $f$ that sends a tuple of Cartesian coordinates $(x,y)$ to the corresponding tuple of polar coordinates $(r,\theta) = (\sqrt{x^2 + y^2}, \mathrm{atan2}(x/y))$
@Qmechanic They're aware that with the new buttons it can be hard to tell if you've voted, especially on meta sites, with their grey scale themes. Allegedly, they've made some improvements... meta.stackexchange.com/a/389771/334566
@Amit It's still very hard to see on my phone. But Samsung & its browser do some odd things in Dark mode, and I can't blame SE for that.
Eg, in an SVG with a filled rectangle & a filled polygon using identical colours, in dark mode they become not identical. :( It's rather annoying. But the Dark theme is much less harsh on my old eyes, so I (mostly) tolerate its quirks.
Yeah, I get that. I used to use dark mode too, but gave up because of issues like that. Instead I set the color balance to be slightly "warmer" so that white looks more yellow'ish
What I am saying is about my PC however, not phone
