@OliverGregory See Wikipedia for the area of a circle on a surface of constant curvature

and, if I recall correctly, the area of an small disk of radius $r$ centred at a point $x$ on a general 2D Riemannian manifold is given by $\pi r^2 - \frac{\pi R(x) r^4}{24} + O(r^4)$, where $R(x)$ is the Ricci scalar at that point

@Biophysicist I do hope you’re dry and otherwise safe.

@Nobodyrecognizeable they aren't always positive, e.g. see the first paragraph here

@Nobodyrecognizeable if there is a potential energy term then the sign of the energy eigenvalues becomes arbitrary since we can take any value we want for the zero of the potential energy. For example the energy eigenvalues of the hydrogen atom are negative because we take the potential energy of the unbound electron to be zero.

If there is no potential energy term, e.g. a rigid rotor, then the energy eigenvalues are positive because they represent only kinetic energy and that is always positive.

Can someone help me with matlab?

t=[0:0.001:10]
vs=4*sin(pi * t)
for i =1:length(vs)
    if(vs(i)<=0.7)
        v(i)=0;
    else
        v(i)=vs(i)-0.7;
    end
end

i wanted plot v vs t, i.e plot (t,v)

however, plot doesnt work because apparently the lengths of v and t are unequal

length of t: 1001, length of vs=1001, but length of v= 1000001

How is this happening? I don't see why the lengths v and vs can differ (and as a result those of v and t)

Hi, I recently noticed something about spin 1 particles that seems strange. If I consider an eigenstate of the spin operator in an arbitrary direction (u_x, u_y, u_z), in some basis (say, in the eigenbasis of spin Sz), such an eigenstate can be written in terms of (u_x, u_y, u_z) -- which are 3 real parameters constrained by one equation. On the other hand, the generic state of a spin 1 particle can be parameterized using 4 real numbers.

This means that there are spin 1 particle states that cannot be seen as eigenstates of the spin operator in any direction.

For spin-1/2, this is not the case. There, you can parameterize a generic state using u_x, u_y, u_z themselves.

So, each spin 1/2 state can be seen as an eigenstate of spin operator in some direction.

I mean it is what it is but spin 1/2 looks much more intuitive in this respect, I'd usually not expect spin 1 to have weird properties.

@DvijD.C. that's because you have three different values of $m$ for a fixed axis for spin 1

for spin 1/2, you only have two choices of $m$ and you can associate them to the orientation of the axis

but if you do that for spin-1, (associate $m=1$ to one orientation of the axis and $m=-1$ for the other) then you haven't given any way to talk about the $m=0$ eigenstate associated to this axis

so that's your "missing" parameter

Hmm, just to clarify tho, it's not necessary that there exists a spin operator $\vec{u}\cdot\vec{\hat{S}}$ for which a given generic spin 1 state is an eigenstate (even after considering all three $m$ values), right?

I don't know if that's the case, I'd have to try to work it out

@DvijD.C. After thinking about it for a bit, I think it's not only not necessary, but also not true. The reason that you can write any state as an eigenstate of $\vec n\cdot \vec S$ for spin-1/2 is that in the 2d Hilbert space, the Pauli matrices (together with the identity) from a basis for the space of 2-by-2 matrices. So given a 2d vector $v$, you can pick any matrix $A$ it is an eigenvector of, write $A = \alpha I + \beta \vec n\cdot \vec S$ and there's your axis $\vec n$.

in the 3d Hilbert space of spin 1, it is no longer true that the spin operators form a basis for the space of operators

Yes, exactly! The other operators that complete the basis can be given by the three operators $S_iS_j + S_jS_i$, one operator $S_x^2 - S_y^2$, and one operator $-1/\sqrt{3}(S_x^2+S_y^2-2S_z^2)$, as mentioned in here on pp.4 in the context of tomography.

I remember seeing these combinations of operators somewhere in the context of representation theory but I can't seem to remember exactly where.

@JohnRennie i have an off topic question

Is the evil john rennie guy your alt

I can assure you that that's not the case

@Euler2 No :-)

Hey! Any DFT people in the house?

@Euler2 The picture is of the John Rennie who is an editor at Scientific American. As far as I know he is no relation.

@MoreAnonymous density functional theory?

@NiharKarve Yup

There are many users on Matter Modelling SE who are extremely knowledgeable about it (and friendly!)

Yea I posted there only :P

Somehoe I have one view and one upvote

Never seen that before

Wonder if that means a mod upvoted it

@OliverGregory On a surface of constant curvature K, geodesic triangles with angle a, b, c have area ((a + b + c) - pi)/K.

In general you need to use the Gauss-Bonnett formula with boundary term.

Note that I assume K is nonzero above, in the flat case you cannot say anything just by knowing the angles

There is a cosine law for triangles on a spacetime

It's a pretty cool proof

Although the letter $\phi$ does a lot of heavy lifting here

Because it's a pretty complicated term

Hi, everybody.

hello

Cosine laws in Riemannian geometry is, like, comparison inequalities. You get these by variating geodesics along Jacobi fields, or so is my understanding

I have never seen 129

What's the book?

@DanielSank howdy!

@Nobodyrecognizeable as John Rennie said, whther they are taken to be positive or not is essentially arbitrary. The important point is that the spectrum is bounded from below, so they can always taken to be positive by adding a constant. This is known as quantum stability. It's an important result of quantum mechanics that for reasonable potentials, $H=\frac{p^2}{2m}+V(x)$ has a spectrum bounded from below, even when $V$ is not bounded (explaining why the hydrogen atom is stable)