I'm trying to solve $(-\frac{1}{2}\nabla^2 -\frac{1}{r} + Fr\cos(\theta))\lvert\Psi\rangle$ using $\lvert\Psi\rangle = c_1\lvert 1s \rangle + c_2\lvert 2p_z \rangle$ hence it involves solving the equation $H c = E c$, using the secular determinant. However, I get to a weird result.
The determinant, yields: $(\frac{-1}{2} - \lambda)(\frac{-1}{8} - \lambda) - F^2 * C^2$, with $C$ being a number. I think this should not be too bad since -1/2, -1/8 are the solutions without magnetic field.
If I do baskhara, it wont solve unless given an $F$. I think this is why the problem says use $(1+x)^\frac{1}{2}\approx 1 + \frac{x}{2}$, and expand in taylor series.
You are more correct than you think. We are not using n as the Landau level specification for the following: $$\tag1e^{ikL}=1\qquad\implies\qquad\exists n\in\mathbb Z\ |\quad ikL=2\pi in$$ $$\tag2k=\frac{2\pi}Ln\qquad\implies\qquad\Delta k=\frac{2\pi}L\Delta n$$ The implied RHS in Equation (2) is the reason why we say that $k$ is quantised in units of $2\pi/L$. Now, if you want to count the number of states, you flip this around: $$\tag3\mathcal N=\sum\Delta n=\sum\frac L{2\pi}\Delta k\approx\int\frac L{2\pi}\mathrm dk$$
I asked a question about quantum entanglement. I would expect that people commenting on the question would be familiar with this phenomenon and its philosophical issues. But someone named WillO made a comment that shows a complete lack of knowledge of what entanglement is. When I tried to clarify...
@ACuriousMind i think this is not correct. it doesnt have to do with the particles being identical. it's just that there exists no non-zero function $f(x_1,x_2)$ which gets multiplied by different complex numbers respectively when you exchange the arguments once and then once again
so there can be no state vector with this property. this is purely mathematical
@ACuriousMind in terms of operators, any operator that exchanges 1 and 2 also exchanges 2 and 1. so they both have to be the same operator mathematically
Newton's second law is sometimes stated as F=kma and k is later set to 1. What is the use of this approach? is this just how Newton phrased it originally?
it seems like a useless step
maybe it is to emphasize that we could've worked with other definitions of force too, with $k>0$ and $k\neq 1$
I was looking into the Landau's argument why L can't depend on x for free particle. and look, he doesn't like saying much and somehow, I got confused. so I got a good proof why L can't depend on x. - just exactly follow my logic and proof.
because the space is homogeneous in inertial frame, Lagrangians for position x+a and Lagrangian for position x must yield the same e.o.m. Because of this idea, it's easy to show the following: $L(x+a, x' , t) = L(x, x', t) + d/dt f(x,t)$. (adding total time derivative doesn't change the eom and this is what should be difference between our lagrangians of position x and x+a because we know due to space homogeneity in inertial frame, their eoms must be the same).
$L(x+a, x' , t) - L(x, x', t) = d/dt f(x,t)$ and from this, $a ∂L/∂x = x' ∂f/∂x + ∂f/∂t$. since on the left side, there's no proportion to $x'$, $∂f/∂x$ must be 0, hence f is only a function of t. and we got: $a ∂L/∂x = ∂f/∂t$. In order for this to work for any $a$, ∂L/∂x must be 0, i.e we showed that L can't depend on x.
before I move to my question, is it clear what I said all along ?
well, $L(x+a, x', t) - L(x, x', t)$ is the difference between the lagrangians. since everything else is the same and they are just different by only a, that's what we get. Imagine L1 = 10k(x+a)+x' while L2 = 10kx + x'. well use $a\frac{\partial L}{\partial x}$ and you will get $a 10k$ which is the same as if you subtract by hand
@RyderRude you are missing that he is pointing out a much stronger criterion, namely that this relation is true for arbitrarily large a and not just the infinitesimal version. Then all higher derivatives of L by x must also vanish.
in particular, it is already sufficient to show that L is independent of x.
I am not sure why you tried to consider the function f(x,t) and its total time derivative; I know that is the general prescription in Lagrangian mechanics, but it is awkward to have it in there for this context.
@naturallyInconsistent i am saying that $L(x+a,x',t)-L(x,x',t)=a\frac{\partial L}{\partial x}$ is a step in the derivation. this is not valid for arbitrary $L$
@naturallyInconsistent reason is before Landau says anything in his book, he first mentions that if you add total time derivative of function f(x, t) to Lagrangian, it produces the same result. and we also know that space is homogeneous in inertial frame, which means that lagrangian for x+a and Lagrangian for x, must give the same results. the only way for these Lagrangians to give the exact same results(e.o.m), they must only be different by d/dt f(x,t)
@GiorgiLagidze I have read Landau myself. I know this derivation. I do not think he is meaning to use it here, but if you managed to give a good proof, power to you
problem now is the logic and proof i showed didn't even account for free particle case. I mean, if you use the same proof for non-free particle, it's the same thing you will get. even if particle is not free, since we observe it in inertial frame, space is still homogeneous, so e.o.m's must still be the same at different positions. which means my proof just works fine for any case in inertial frame
BUT for non-free particle, we know L must depend on x, otherwise, it doesn't give newton's second law correctly. So in my logic proof, where am I making a mistake such as it fails for non-free particle ? i am only interested yet to find a flaw in my proof.
problem now is the logic and proof i showed didn't even account for free particle case. I mean, if you use the same proof for non-free particle, it's the same thing you will get. even if particle is not free, since we observe it in inertial frame, space is still homogeneous, so e.o.m's must still be the same at different positions. which means my proof just works fine for any case in inertial frame. BUT for non-free particle, we know L must depend on x, otherwise, it doesn't give newton's second law correctly. So in my logic proof, where am I making a mistake such as it fails for non-free p…
No you don't have to consider the general case where $L$ changes by a total time derivative because we are talking about the form of $L$ in any one frame, not comparing $L$'s at different points, we are saying at any point things have to be the same as regards $x$ translations therefore it can't depend on $x$
It is a very simple point, it doesn't need anything more than this
@GiorgiLagidze In the non-free case, really, it is just that it is two or more particles interacting. Then L=L(x,y,x',y',t), and L(x+a,y+a,x',y',t)=L; here, the system doesnt care that you have translated by a overall, but y-x is going to be a variable that ignores a but yet can affect the dynamics.
In particular, L can depend non-trivially upon y-x and thus have (relative) position dependence. This is what we really meant.
This fact shows that increasing the dimensionality of a space is a tremendously scary increase in potential complexity.
if i try approaching the same way, we got: $L(x+a, y+a, x', y', t) - L(x, y, x', y', t) = \frac{d}{dt} f(x, t)$ $a\frac{\partial L}{\partial x} + a\frac{\partial L}{\partial y} = \dot x \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t}$ because of $\dot x$ on right side, left side doesn't contain the same proportion, i.e $\frac{\partial f}{\partial x}$ must be 0.
@GiorgiLagidze The idea that "since on the left side, there's no proportion to $x'$" makes no sense - in general, $\partial L/\partial x$ has arbitrary dependence on $x'$
If you want to waffle in this manner, at best you can say that because $L(x)$ and $L(x+a)$ should give the same eom, we must have $L(x+a,x',t) = L(x,x',t) + \frac{\partial L}{\partial x}a + ... = L(x,x',t) + \frac{d}{dt} f$ so since the total derivative is irrelevant $L$ can always be chosen so that its Taylor expansion in $x$ can be set to zero after the first term i.e. $L(x+a) = L(x)$ i.e. $L$ does not depend on $x$
@GiorgiLagidze in the non-free case, you have more than one x variable. homogeneity implies $L(x_1+a, x_2+a, v_1+v_2)-L(x_1,x_2,v_1,v_2)=\frac{df}{dt}$. so u get $a(\frac{\partial L}{\partial x_1}+\frac{\partial L}{\partial x_2})=0$. this just implies that $\frac{\partial L}{\partial x_1}=-\frac{\partial L}{\partial x_2}$
@bolbteppa You don't even need to talk about a Taylor expansion, you can just say that obviously $L$ differs only by a total time derivative from a function that's constant in $x$ and so you can just choose $L$ to be constant
if we got 2 particles, total time derivative added to the Lagrangian that doesn't change the eom must be functions of both x and y ? are we sure about this ?
the silly thing here is to expect that one could somehow derive constancy of $L$ while simultaneous considering the pseudo-symmetry case of changing by a total time derivative; of course you can't, but what is true is that among all these pseudo-symmetric Lagrangians there is one that is constant and that's the one we pick
@naturallyInconsistent if we got L(x+a, y+a, x', y', t) = L(x, y, x', y', t) + d/dt f(x,t) - note f only function of x and t. In this case, eom for x coordinate won't change but for y, it will, right ?
@GiorgiLagidze since we are nitpicking, isn't the first step here wrong? Two Lagrangians giving the same EoMs does not imply they differ by a total derivative, they could be constant multiple of eachother
there was this question by Dawkins that why do physicists even bother to talk about the singularity at the Big Bang. they can instead just begin the universe at a latter point in time :P
@GiorgiLagidze this is the correct version of this proof : $L(x+a,x',t)-L(x,x',t)=\frac{df}{dt}$ (The RHS can only be a function of $t$ for the EoM to remain unchanged here). Since the RHS only depends on $t$, u can immediately conclude that $\frac{\partial L}{\partial x}=0$ from the LHS
@GiorgiLagidze the original proof considers x dependency of the RHS, which is y u have to use some steps like "terms proportional to x' ", which r invalid because of what @ACuriousMind said
I don't particularly want to open this can of worms but I feel I have to: You must be aware that both Dawkins and Peterson are rather controversial figures for a variety of reasons (Peterson in particular holds views that would get him quickly banned from SE under the Code of Conduct were he to try and promote them here).
Randomly bringing them up in an unrelated chat and then saying nothing of more substance than "he likes to think" or "he gives rational answers mostly" is an extremely strange move (and exactly what trolls or astroturfing fans would do). I would like for this room to be a place where we can discuss even controversial things but this is far below the standard of discourse I would expect. Please do better.
@GiorgiLagidze there is also some lack of generality left because of what @fqq said. to take that into account, u will have to start with $L(x+a,x',t)-\lambda L(x,x',t)=\frac{df}{dt}$, for some $\lambda$. i dont know how the proof will proceed in this case. but this will make for the most general proof
we have to recover $\lambda =1$ somehow from the general starting case
@GiorgiLagidze that question has actually left me wondering for long. it reminds me of that theory where the universe could've begun just now and we couldn't prove otherwise
so why do we want to begin the universe at the singularity anyway? we could just pick a later point as the beginning and it wouldnt make any difference to science (as in, all observables remain the same)
Most of the stuff scientists have observed so far will eventually lead us that far back. We don't have any direct evidence the universe started at any particular point in time, so we default to the most logical. Or at least, easiest to prove/disprove.
@Someone yes. you could say that there are only two privileged time points to pick : "the universe began just now" and "the universe began at the singularity"
picking anything else would be arbitrary
but the choice really doesnt make any difference to science
I once heard someone say that the past is an illusion made by memories we can never confirm, and that the future is an illusion created by our perception of time, which comes from memories.
@RyderRude It's related to the idea that the simplest answers tend to be correct.
So the simplest explanation of the universe is a brain floating in space hallucinating everything that has ever happened, is happening, or will ever happen.
@Someone idk why i said "universe MAY not have a global present" i was thinking of spacelike foliations. The universe DEFINITELY doesnt have an objective present moment
because it depends on the observer. so the theory that only present exists is meaningless unless you go to Solipsism @Someone
The Boltzmann brain idea is this: if you believe that the universe will end in heat death, then there is an eternity of heat death. No matter how unlikely a fluctuation, eventually it will occur. If you also believe that statistical fluctuations in heat death can include arbitarily complex structures, just exponentially more unlikely the more the structure is, then even our universe could arise, in an instant, as such a fluctuation.
And since there is an eternity in heat death, the present moment is more likely to be such a fluctuation than at any point before heat death
But that's not all: You only know that you exist, and so you could not distinguish being a brain inside such a universe fluctuation, or just being a brain imagining that it sees the universe. The system "brain hallucinating a universe" is much simpler than "brain inside an actual universe", and so much more likely to occur by our initial assumptions
so if you buy the premises of this argument, you must conclude that the most likely explanation for your existence is that you are such a hallucinating brain - a Boltzmann brain
@Someone one criticism can be that we are using the laws of physics that we have hallucinated according to this theory to deduce the fact that we are hallucinating
if these laws are also a dream, who's to say these are the laws of the true universe
the core elements here are that the universe eventually spends an eternity in heat death and that arbitrarily complex fluctuations are unlikely but not impossible
in physics we often deal with weird probability distributions that assert things are unlikely but not impossible but actually everyone knows they're impossible
if we assume our universe to not be a hallucination, then this argument can be used to argue that there will be hallucinating brains in our far future
but using this to argue that our own universe is a hallucination is a circular argument
the simulation hypothesis hierarchy universe are very similar to this. it does not require the heat death, but it argues that intelligent life can create simulations of intelligent life, which leads to hierarchial universes @Someone
so it is likely that our universe is a simulation according to this argument. but it has never been proven if universes can be simulated
@ACuriousMind and im perfectly happy having the Big Bang just be the initiating point of such a fluctuation. Life just is; enjoy it and don't care too much about the rest.
@ACuriousMind somewhere out there in the multiverse, there is an Earth that had the doomsday scenario in Oppenheimer, i.e. atmosphere destroyed by nukes.
@naturallyInconsistent I mean that part isn't the weird part about the Boltzmann brains, you're right that if we just look at the universe fluctuations this is just a multiverse theory where the universes are fluctuations in a heat-dead larger universe rather than any of the other weird constructions
@naturallyInconsistent It's the kind of speculation which I think is pretty fun as long as we all agree the answers are ultimately completely irrelevant
I just meant it's very unclear, what, if anything, we should learn from Galois' death we couldn't from a million others: Don't point weapons at each other :P
I was very shocked when I saw someone do those symbolic manipulations to get derivatives using matrices; when $\varepsilon$ is a matrix that squares to zero. Turns out to be really helpful for symbolic calculus.
it would not have any meaning of limits as per usual calculus presentations, purely just an algebraic trick, but it obviously works
@naturallyInconsistent that's probably some variant of non-standard analysis, it doesn't matter that you can choose the $\epsilon$ there to be a matrix, you just need the algebraic property to square to 0
@Relativisticcucumber difficult to say. I very much like Primosale, which a type of Pecorino. Then, I love Grana Padano, Parmigiano Reggiano (very similar, but don't tell that to people from Emilia-Romagna). Last but not least, my beloved Gorgonzola
Now I have to decide what to do about the other elective course. I've put an experimental particle physics course as a placeholder but I already have another one. I'm not doing two hep-exp courses lol
