and for $p = 0$ we have $(-1)^1 = - 1 = - \mathrm{Re}[(1+i)] = - 1$
In fact, another proof of this insane identity is to note both sides change by the same factor under $p \to p+2$ so you only need to consider the $p=0,1$ cases
Yeah $p$ must be an integer
It's just the whole adding the $p$ even and $p$ odd cases together thing
I don't see any $\sqrt{AB} = \sqrt{A}\sqrt{B}$ kind of thing there
are articles in Quanta Magazine generally believable? I think the authors of these articles aren't scientists in the fields of the articles they write, so do they accurately convey the ideas in the original papers which they base to write these articles?
Hey guys, could someone explain why the probability that an electron is found between $r$ and $r+dr$ in the ground state is $\vert R_{10}\vert^2r^2dr$? I am actually only confused about the $r^2$ factor; I was expecting just to work with $dr$. I guess it has something to do with the fact that we consider a whole shell, and not just a tiny ‘line’ element $dr$, but not sure how to find sources on this?
Guys, why does the Hamiltonian commute with the displacement operator $Df(x)=f(x+a)$, if we are given a periodic potential $V(x+a)=V(x)$? I understand that $HDf(x)=Hf(x)$, but we need $DHf(x)=HDf(x)$, no?
Actually, I shouldn't say that $HDf(x)=Hf(x)$; it's just that $H$ works the same on $f(x+a)$ as on $f(x)$
As all community members know I am just a beginner of physics. Mistakes often happens from everyone .Why can't everyone give me a second chance to blot out my mistakes and this how I can improve my knowledge in physics .So I request community members to withdraw my ban.
@Slereah not sure what you mean by discrete version. but how can the kinetic term be invariant under translations? $\frac{d^2\psi(x+a)}{dx^2}$ surely isn't equal to $\frac{d^2\psi(x)}{dx^2}$?
Even the statement you can just take your spinor to be a Majorana spinor for simplicity for the 2-D superstring action is unbelievable, since you can only do that in $D = 2, 4 \mod 8$ for $D$ even...
I guess you just ignore it, I mean you can't even define $\mathrm{Spin}(1,D-1)$ as anything other than the set of operators $\exp (\frac{1}{8}[\gamma_m,\gamma_n])$ or something equivalent to this
Seems like you just define the Dirac conjugate with indices raised by definition
Guys, if we approximate a crystal potential by the Dirac comb, we need to do something about the fact that our solid doesn’t have infinite size. So Griffiths suggests to impose the following boundary condition: $\psi(x+Na)=\psi(x)$, where $N$ is the number of periods (of size $a$).
I don’t really understand why we have to do that; why can’t we just pretend that the crystal goes on forever? Or rather, why do we speficially impose these boundary conditions, and not something like $\psi(0)=\psi(Na)=0$?
@bolbteppa well, real crystals are infinite to all intents and purposes i.e. their size is so much greater than the lattice spacing that it is effectively infinite.
@ShaVuklia these are wrap around boundary conditions - there's probably a posh name for them but i don't know it.
In effect when you leave the crystal at one face you re-enter it at the opposite face.
I think it makes the crystal a 3-torus
The part of the crystal you are considering is only a small part of an effectively infinite crystal, so you don't want to impose restrictions like $\psi = 0$ at the edges because that's not physically reasonable.
I think the reason you impose periodic boundary conditions is because you can effectively ignore the real boundary conditions because as John said you can basically treat them as infinite and you barely expect the boundaries to affect anything
@ShaVuklia if you imposed $\psi=0$ at the edges of your region you'd turn it into a 3D box and you couldn't have any plane waves passing through it - only standing waves.
Since Bloch waves are going to be used to model electron states, and since Bloch waves are essentially plane waves, this would be a big problem.
I think it's a way to let you simply solve the equation in one interval, $0 < x < a$ for the Dirac comb and letting Bloch give you the solution for all intervals
right, I don't really understand what the work function is, but I'm imagining that if we have a crystal in empty space, then the potential outside of the crystal will rapidly decrease to zero
@JohnRennie in any case, this is the biggest reassurance for me that it's useful:p so thanks for the explanation
Guys, I don’t really understand why after $n=N-1$, we get no new solutions. Griffiths explains it as that the Bloch factor $e^{iKa}$ recycles, but I’m not sure why that means that there are only $N$ solutions possible. So say we have solutions $\psi$ and $\psi'$, belonging to $K$ and $K'$ respectively, and assume $e^{iKa}=e^{iK'a}$. Since it holds that $\psi(x+a)=e^{iKa}\psi(x)$, we know that $\psi(x+a)=e^{iK'a}\psi(x)$. But I should show that $\psi(x)=\psi'(a)$, right? How can I do this?
(if I haven't provided enough context, let me know)
$j$ comes from the total angular momentum which is the sum of the orbital angular momentum and the intrinsic angular momentum $\mathbf{J} = \mathbf{L}+\mathbf{S}$
And $s$ is the quantum numer associated with the spin and $l$ the quantum number associted with the orbital angular momentum
Given $l$ and $s$ I know that $j$ is restricted to the values $|l-s|, |l-s|+1,..., l+s-1,l+s$ but I don't know how I can determine $s$ from $j$ and $l$
Angular momentum addition in quantum mechanics is a little more complicated. It's less straightforward than classical mech. See these notes: www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf
I would like to have this question reopened.
In contrast to the users who expressed their apparent confusion, I found the question to be very clear. It also seems like one that would get a whole lot of Google hits.
Both complained that one could not know the parameters, but there's only three a...
The gauge covariant derivative is a variation of the covariant derivative used in general relativity. If a theory has gauge transformations, it means that some physical properties of certain equations are preserved under those transformations. Likewise, the gauge covariant derivative is the ordinary derivative modified in such a way as to make it behave like a true vector operator, so that equations written using the covariant derivative preserve their physical properties under gauge transformations.
== Overview ==
There are many ways in which to understand the gauge covariant derivative....
I don't fully get this beast yet
It's always bugged me how it differs from a normal covariant derivative you get from e.g. differentiating a vector field and it's basis
I think the gauge covariant derivative operates on some sort of fiber bundles that are different than the tangent bundle.....I think...
In other words the covariant derivative works on the tangent space to a manifold (space-time), but the gauge covariant derivative works on an arbitrary space that's attached to your manifold in a fiber bundle way.
...mmmmhm...hopefully I'm not just talking nonsense LOL