1:03 AM
@Claudio I read your question, and the first thing that came to mind was that, originally you had a purely real-valued integral that is well-defined; why did you choose for the branch cut to be exactly along the line that defined the initial integral? Don't you think that would cause trouble?
@Pizza That is purely geometry and the definition of trigonometric functions; nothing to do with physics. We know the answer, and can help you, but you need to know what kind of answer you would accept as an answer.
Nothing, everybody is entitled to their opinion.
@imbAF I read the entire exchange you had with ACM and Claudio. I think you are incredibly confused and ain't getting what it is that they are trying to tell you at all. You seem to have many misconceptions and I will try to dispel some of them in the following.
You are doing Griffiths, which means at the this point that you have arrived at the Dirac Delta potential, almost end of Chapter 2, you must also have seen infinite square well = particle in a box, QHO, free particle, Dirac Delta potential, and I know you have also seen the Finite square well = particle in a box (i.e. there are two particle in a box and so you should be careful to state which one you are considering)
In the infinite square well and in the QHO cases, you have a discrete spectrum. This means that when you have a quantum state with fixed energy E, then you can always find an $\exists\varepsilon>0$ such that the interval $(E-\varepsilon,E+\varepsilon)$ has only just that one E energy state available. (When later you see degeneracies, that changes only to that one eigenvalue E exists, uniquely, no others, but that one eigenvalue may have many possible quantum states.)
Consider the QHO with eigenstates $\left|n\right>$ The point of what we call the "complete orthonormal basis" being so important and so useful, is that it forms what we can call the identity operator $\hat{\mathbb I}=\sum_n\left|n\right>\left<n\right|$, and this is something that isnt really expressible in the most basic introductions to QM, for the reason that they tend to avoid bra-ket notation.
Again, the bra-ket notation is necessary to state some stuff that we want to write down in QM. Consider the continous analogue that you should have seen in the free particle case, we want $\delta(x^\prime-x)=\left<x^\prime|x\right>$ (which is thus not a function, not normalisable for either of $\left<x^\prime\right|$ nor for $\left|x\right>$) but we want something that satisfies such a definition because we want the quantum state $\left|\psi\right>$ to be expressible by
$\left|\psi\right>=\hat{\mathbb I}\left|\psi\right>=\int\left|x\right>\mathrm dx\left<x|\psi\right>$ where the connection to the old notation is that $\psi(x)=\left<x|\psi\right>$
Only in this notation do you see that the quantum state $\left|\psi\right>$ requires you to state the entire function $\psi(x)$ over the whole space, not just one point $\psi(x_0)$.
> why we only have 1 bound state when the function depends upon position
(I know you stated that ACM beat it out of you, but this is part of the answer.)
2:10 AM
In particular, every energy eigenstate should spread out in space; there is never a position eigen"state" that is also an energy eigenstate because all physical Hamiltonians have momentum operators in them.
Now, when you first receive a quantum system to consider, you get to write down its Hamiltonian operator $\hat{\mathcal H}$
you only have that; you don't really yet know its Hilbert space, only that you know that some Hilbert space exists for it, because it is a quantum system, by postulating.
This means that the potential $\hat V$ is also given to you. You don't know what E values are the eigenvalues of $\hat{\mathcal H}$
This is very important, because you dont even know if you have bound states and/or scattering states. You don't know how many of each you have. You don't know which regions of energy corresponds to which type of solutions you might get. In particular, it is not so simple that E>0 is scattering and E<0 is bound; we prefer to have this nice separation, but it might not exist, e.g. QHO. This ties into your question about $E+V_0>0$
Oh, I forgot to point out that the identity operator and so forth that I had already mentioned above, answers your "outer product" question: Notice that the whole argument only makes sense in bra-ket notation, whereas what you were trying to write down in mixed notation is very muddied and you didn't actually make clear if you are doing outer product or inner product.
Anyway, before you can have $E<0$ be bound and $E>0$ be scattering / free, you have to consider, from the $V$ that you get inside the Hamiltonian operator $\hat{\mathcal H}$, the quantity $\min V(x\to\infty)$, because it is this that splits the energy eigenvalues.
If this quantity exists as a finite value at all, then your spectrum can have a discrete part and a continous part. If, like the QHO and the infinite square well, $\min V(x\to\infty)=\infty$, then you only have bound states and no more.
If the global $\min V(x)=\min V(x\to\infty)$, then every state is free.
There can be no energy eigenstate with an energy value smaller than $\min V$. This is proved as a theorem simply by arguing from the Schrödinger equation; If you tried a solution that way, you will find that there is no way to find a normalisable solution, and on top of that, it cannot be a free solution, i.e. you cannot find the free state's normalisation for it either. It is just unphysical.
This $\min V$ is your $V_0$ in the finite square well case. This should explain why it is necessary for $E+V_0>0$
The $\min V(x\to\infty)=0$ in the finite square well case also explains why the cut-off is $E<0$ for bound states and $E>0$ for free states; and thus there is only a finite number of bound states in the finite square well.
Only when $\min V(x\to\infty)$ is a finite number, does it make sense to shift by adding a constant to the potential, to shift this to zero, and then you have the nice separation of $E<0$ v.s. $E>0$
Now, for any system whose energy spectrum has both the discrete and continous parts, then the complete basis covers both. That is $\hat{\mathbb I}=\sum_n\left|n\right>\!\left<n\right|+\int\left|k\right>\frac{\mathrm dk}{2\pi\hbar}\left<k\right|$; when we write just the left part, it is really a lie, a lie that professors cannot help, because at the point in time introducing to students, students had yet known about this second part.
This, in particular, means that if you wanted to resolve $\delta(x^\prime-x)$, then, strictly speaking, you also need the scattering states integral part too.
This is of importance because the H atom wavefunctions are of this type, infinitely many bound states, and then Coulomb wavefunction free states.
Anyway, I dont think you should be reading Griffiths. You are too weird (for wanting mathematical rigour and completeness) and impatient. I don't know how you are meant to learn anything, but maybe you can get more out of Brian Hall's Quantum Theory for Mathematicians.
@user85795 starred for wiki that it is actually a fallacy. Not many people recognise it as a fallacy
3:27 AM
The tides causing a pet's water bowl to overflow is a bit of a stretch. — CPlus 4 hours ago
But that's what tides do! They stretch things! HEYOOOO!
3:45 AM
@rob I thought you were linking to mmesser's answer... Hoomans really love to mystify the tides and the moon~

3 hours later…
6:38 AM
hi
6:59 AM
@RyderRude @naturallyInconsistent Thanks so much for the replies, I solved the problem anyway :)

2 hours later…
9:01 AM
@naturallyInconsistent why, run around with pitchforks of course
9:34 AM
@Jakobian R O A R ~

1 hour later…
10:39 AM
I here by question the units in the title of this room.
h = h
11:24 AM
hi
@user85795 oh no, I am so going to be haunted by this for the next decade.
@user85795 they differ by 2pi
@RyderRude for someone always sending youtube links here you are one of the least qualified to not watch them before commenting.
11:57 AM
@naturallyInconsistent i saw it but missed the timestamp
now having seen it, i would say it's interesting. but also, a radian is dimensionless
e.g. u can exponentiate something carrying a radian unit

2 hours later…
1:36 PM
@Slereah, this may be up your alley:
https://thonyc.wordpress.com/from-%CF%84%E1%BD%B0-%CF%86%CF%85%CF%83%CE%B9%CE%BA%CE%AC-ta-physika-to-physics/

3 hours later…
4:48 PM
A cannon C is aimed at the sphere of mass m placed at a height h = 3m from ground level and at a horizontal distance d = 1.5 m, initially at rest. At time t = 0 the cannon fires and the sphere m is dropped. Determine the initial velocity v0 of the projectile for it to hit m at ground level.
My idea was to find when $m$ hit the ground, so $-h = v_{0y} t -1/2gt^2$ , but how do I know how much $v_{0y}$ of the mass Is?
Should I assume it's 0?
@Pizza both "initially at rest" and "dropped" imply that it is zero.
@naturallyInconsistent Ah ok, I was confused for a moment, so I can find from here the time at which the mass reaches the ground
It is said in a text that the "natural" integral measure on $\mathbb{C}P^1$ is $\lambda d \lambda$ where $\lambda$ is a left handed 2-spinor whose two components parameterizes the Riemann sphere $\mathbb{C}P^1$ by acting as homogeneous coordinates. How is the integral measure natural... I mean why consider $\lambda d\lambda$ as an integral measure on $\mathbb{C}P^1$?
I am not giving any context for now because I suspect that the naturalness is independent of the context in which I encountered this, but in case someone says that it might be not then I would be happy to share a snapshot of where I encountered this...
@naturallyInconsistent Then I substitute $x(t) = v0x \cdot t \to 1.5 = v0x \cdot 0.782 \to v0x = 1.91$
Now should I find v0y and then do the Pythagorean theorem?
5:19 PM
You can, but you dont have to. It is actually much easier than you think.
@naturallyInconsistent Could you give me some advice?
Im not sure. You were just having difficulty with trigonometry
Doing as I said above I found v0 = 4.27
That is correct
@naturallyInconsistent Okay, I'll try to improve on that aspect anyway
5:25 PM
are there anyone who live tex-es their class notes? if so, how do u handle diagrams?
@naturallyInconsistent Thanks so much for the confirmation!
@nickbros123 I have seen some gods do that. Very very very impressive
I mean, live-tex-ing at all. diagrams would be way more impressive
@naturallyInconsistent I can "write" stuff down pretty fast to keep pace with the class, I did that my thermodynamics course. Diagrams are a beast though, even for a course like thermodynamics where the only diagrams u draw are PV graphs, or A box and a circle representing reservoirs and engines.
@Pizza The question said that it was originally aimed at the mass; this means that the relationship between $v_{0y}$ and $v_{0x}$ is the relationship between $h$ and $d$, so I immediately knew that the answer, once you found $v_{0x}$, is that $v_0=v_{0x}\sqrt5$
5:45 PM
@Sanjana The usual "natural" metric/measure on $\mathbb{C}P^n$ is the Fubini-Study metric, but I don't even know how the expression "$\lambda\mathrm{d}\lambda$" is supposed to be a measure on $\mathbb{C}P^1$ - the homogeneous coordinates are not a coordinate chart in the sense of differential geometry, how am I supposed to integrate this thing over the Riemann sphere?

1 hour later…
7:14 PM
Since m1 > m2, the two blocks are d=3m apart, How can I find the time necessary for the 2 bodies to be at the same height?
Has the image been deleted?
(I also know the acceleration)
The idea of writing: d - y = 1/2 at^2
y = 1/2at^2
Where y is the displacement to reach the same position
Or I could do:
So i know the acceleration, i can use that, vi=0, and d=1.5, solve for t
we use d=1.5 because is that the blocks must meet at this middle point since the string length is constant

2 hours later…
9:00 PM
I am confused about the proper notation of the following expression:

$\langle \psi|\psi\rangle=\int\psi^{*}(x)\psi(x)dx=\int|\psi(x)|^2dx=\int \rho(x)dx$

But one can also write: $\langle \psi|\psi\rangle=|\psi|^2=\rho(x)$

If both expressions are true than one can write $\rho(x)=\int\rho(x)dx=1$ which clearly is wrong. What am I mistaking here?
@naturallyInconsistent Thanks for the long insightful reply