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2:49 AM
do stackoverflow have a chatroom?
btw, I am wondering if I write a function (for scientific computations), is it a good idea or bad idea to use any previously written functions? (either by me or others)
 
3:01 AM
@Shing several of them, but they're hosted on chat.stackoverflow.com
 
3:21 AM
lol what is the new profile page UI
 
 
4 hours later…
7:27 AM
@Shing depends on your goals in writing the function :P
If you're trying to figure out how things work, it can be worthwhile to try to write everything yourself. But if you just want correct software, it is usually better to rely on well-tested (!) functionality by others - don't reinvent the wheel.
 
 
1 hour later…
8:33 AM
It's pretty hard to write a function better than a standard package function
I mean you can do it, they're not written to be specifically the best in your specific context, but it's one of those things that could take weeks of work
unless that's your job it's not really worth it
 
always depends on what "better" means to you - sometimes "better" is just "doesn't pull in this one annoying dependency I don't want to have to deal with"
but yeah, don't be the guy writing their own cryptosystem after reading the Wiki article on cryptography :P
 
8:50 AM
alot of what you're paying for with eg. matlab, isn't just the implementation of the function. but also the testing/verification and optimisation
well, perhaps less so verification in the sense of formal verification
i agree with @ACuriousMind it really depends what your goals are
 
 
2 hours later…
10:54 AM
If you want to do it for fun go ahead
It's usually not too hard to write more efficient functions than the existing ones since standard functions will usually be filled with tests and special cases, since they have to work in a wide variety of cases
but you're gonna be shaving off nanoseconds
Also depending on the precision you require you can shave off time there too
Most standard functions will try to be precise up to the least significant bit
 
11:46 AM
yeah
 
 
2 hours later…
2:01 PM
0
A: Does Heisenberg's uncertainty principle imply discretization of position and momentum?

Matt TobiasHeisenberg's original paper actually does treat both space and momentum as quantized to derive the uncertainty principle: https://adammunich.com/heisenbergs-uncertainty-principal-the-actual-content-of-quantum-theoretical-kinematics-and-mechanics/ I'm not sure that specific derivation is too meani...

Would you look at this answer by Matt Tobias! If you haven't read Heisenberg's original paper you will be surprised
 
What a terrible month!
Nothing going right
months of investments and investments damned in a single day
 
 
2 hours later…
3:51 PM
@NiharKarve Removing the "Member since" and "Last seen" data is frankly ridiculous, IMHO, and a lot of people seem to share my opinion meta.stackexchange.com/q/368285/334566
 
4:36 PM
I wonder how they worked up the motivation to make these changes
Practically nobody in the community has taken issue with any of the old features
 
I see. Thanks, guys!
 
And it wasn't even trialed on a smaller scale?!
 
@NiharKarve Nope. No trial, or prior announcement. They don't do that anymore on trivial changes. Of course, they get to decide on what counts as trivial...
But here's a script that retrieves the "Member since" & "Last seen" data: meta.stackoverflow.com/a/410627/4014959 I guess that script may stop working in the near future, if they block those fields in the database.
@Shing What Slereah said. There's an art to writing reliable mathematical functions. The techniques that a pure mathematician might choose may not lead to the most efficient algorithm, or may not behave well with limited precision floating-point numbers. What kinds of functions are you talking about? Can you give some examples?
 
5:36 PM
Why aren't all unitary groups symmetries?
The inner product is invariant under all unitary operators
Sorry, I meant why aren't all unitary representations of groups represented by a unitary operator on a state, a symmetry
 
Not all wavefunctions transform under an irrep of a given group?
They will all transform under some rep of a given group, since the trivial representation is always a thing, but that's not saying much
 
What does it mean for a wavefunction to transform under an irrep of a given group?
If I have a group G, the action on a vector space V is just given by GV = V'
any vector fullfils that
 
Well no, because a group by itself does not act on a vector space
 
sorry I meant a vector V in a vector space
 
It's a mathematical object that isn't fundamentally related to vector spaces
That's what the representations are for
You're not gonna have a representation of $SU(3)$ that acts on $\mathbb{R}$ non-trivially
 
5:50 PM
But what dictates if a wavefunction lives in R or R^n?
 
Whatever the quantum theory you're using for it is!
Also it's usually either $\mathbb{C}^n$ or some variant of $L^2(\mathbb{R}^n)$
 
But what from the quantum theory dictates the symmetry?
I under stand a dynamical symmetry commutes with the Hamiltonian
With internal symmetries, am I just looking for a group where the Lagrangian transforms as a scalar?
 
Well, one thing to remember is that the quantum theory isn't just a Hilbert space
You have a Hilbert space at every point in time
for every $t$, $\psi(t)$ belongs to a Hilbert space
If your symmetry operator doesn't commute with the Hamiltonian, while it may be unitary at a given time, it may not later on
 
@DIRAC1930 What does "internal symmetry" mean in a quantum theory?
A symmetry is a unitary operator that commutes with the Hamiltonian
 
6:06 PM
Is that always the case even in more complicated field theories?
My guess is yes
 
That's the case in general since that's true for theories involving Hilbert spaces
 
you can be slightly more general and allow for time-dependent symmetries $Q(t)$, which are symmetries if $\partial_t Q + [Q,H] = 0$
 
Which is about all quantum theories
 
this is the case e.g. for Lorentz or conformal symmetry - the Lorentz boosts of course don't commute with the Hamiltonian since it transforms as the 0-th component of momentum, but it's still a symmetry of the theory
 
So should everything always commute with the energy-momentum 4 vector operator?
 
6:09 PM
whether or not you find an implementation of a gauge symmetry in QM depends on your quantization process - the "finished" quantum theory has a gauge-invariant space of states and hence only trivial "gauge symmetry" operators, but you might have intermediate spaces where you have non-trivial actions associated with gauge symmetries
@DIRAC1930 why "everything"?
 
Sorry I meant every symmetry
I think my confusion is over more basic things
 
it should commute with the Hamiltonian,
"symmetry" means "there is a conserved charge"
the conserved charge is precisely the generator of the symmetry that commutes with the Hamiltonian
if the charge doesn't commute with it, it's not conserved (since the Hamiltonian represents time evolution!), so it's not a symmetry
 
Ah okay thanks
So if I start with a Lagrangian that has an SU(2) symmetry, the states must transform under an irrep of that group so there must be 2 basis vectors?
 
I'm so confused about everything
 
6:15 PM
the hydrogen atom transforms under $\mathrm{SO}(3)\approx\mathrm{SU}(2)$ after all and it has infinitely many basis states!
but its state of space decomposes into the direct sum of irreps of SO(3), of course - the spherical harmonics
 
Sorry, I meant that the wavefucntion must have 2 components
 
not necessarily - again the hydrogen atom has SO(3) symmetry even if you ignore spin
the components of the wavefunction are usually associated with spin, not with symmetries in general
 
So why in the spin basis does a wavefunction have to have 2 components?
 
well, a particle with spin is essentially the combination of two quantum systems - of a stationary particle with spin, and that of a spinless particle that can move
the stationary particle is well-known to have state space $\mathbb{C}^{2s+1}$, where $s$ is its spin
the spinless moving particle has $L^2(\mathbb{R}^n,\mathbb{C})$ as its state space
we combine state spaces via the tensor product, and so we get $L^2(\mathbb{R}^n,\mathbb{C})\otimes \mathbb{C}^{2s+1}$, which is the same as $L^2(\mathbb{R}^n,\mathbb{C}^{2s+1})$, i.e. $\mathbb{C}^{2s+1}$-valued wavefunctions
there are other ways to think about it, but this way extends seamlessly to other degrees of freedom like weak or strong charges
you can gather up all the internal symmetries of your theory into some (possibly large) finite dimensional space $V$, and then the wavefunction of a particle charged under these symmetries is just a $V$-valued wavefunction
 
So if I found that the wavefunction $\psi$ has another degree of freedom $n$ where $n$ can take values $1,2$ or $3$, i.e. $\psi(x, n)$, the wavefunction would have 3 components in the $n$ basis?
 
6:23 PM
people would usually write that as $\psi_n(x)$ to emphasize $n$ is discrete, but yes
 
So how did Pauli infer that $\pm$ spin has the symmetry group of $SU(2)$?
Just from the non-relatavistic Schrodinger equation
Why not $SL(2)$ or $U(2)$ etc.
 
@DIRAC1930 spin is rotation
the rotation algebra is $\mathfrak{so}(3)\cong\mathfrak{su}(2)$, and so we look at representations of $\mathrm{SU}(2)$
 
I dunno if that's what happened historically, but I'm not rooting through Pauli's papers to find out
 
eh, trying to figure out QM historically makes it seem even more confusing than it already does :P
 
Lol
 
6:36 PM
there's a lot of false starts and questionable (from todays viewpoint) arguments in early quantum theory that were good at the time but really contribute little to a modern understanding
 
So symmetries are represented on the Hilbert space by unitary representations of a certain group
is that correct
 
Yes
 
But when just acting on wavefunctions, I dont have to worry about the unitary representation?
 
Why not
They act on them
 
what do you mean by "worry"?
the wavefunction is an element of the Hilbert space, you don't know how to act on a wavefunction without knowing the representation
 
6:41 PM
So $SO(3)$ isn't unitary but I assume we can act with it on wavefunctions
 
I mean $SO(3)$ isn't an operator
Unitary or not doesn't apply here!
But there are unitary reps of $SO(3)$ yes
 
Is $SU(2)$ a unitary rep of $SO(3)$ up to a sign?
Is that why Pauli used $SU(2)$ and not $SO(3)$?
 
you need to differentiate a lot more carefully between the group and one of its representations
$\mathrm{SO}(3)$, abstractly, is a group isomorphic to the group of real special orthogonal matrices in 3d
you can ask "what are the unitary representations of SO(3)?"
 
$SO(3)$ acts on a vector
WHy cant I just have the wavefunction as a $3$ component vector
 
the answer to that is that there are infinitely many, and just taking the 3d matrices acting on complex 3d vectors instead of real 3d vectors is one of these
@DIRAC1930 all groups act on vectors when you have a representation, that's the definition of what a representation is
@DIRAC1930 the real reason why one has to look at su(2) is technical but comes down to how the fact that QM states are really rays in Hilbert space rather than single vectors interacts with representation theory, see physics.stackexchange.com/q/203944/50583
 
6:49 PM
So if I find a symmetry in real life, I need to consider the unitary projective representations
Just finding the unitary representations will miss out a lot of information
 
well, how much information you lose depends on the group
e.g. if you already start with SU(2), then you gain nothing by knowing about projective representations - all its projective representations are already linear ones
 
But if you started off with $SO(3)$ you miss all the information about spin
Sorry, I meant just considering the unitary representations of $SO(3)$
 
yes, you'd never understand where the half-integer spins come from
 
So when considering the projective unitary representations of the Lorentz group
Does complexifying the algebra of $so(1,3)$ help find the projective representations of $SO(1,3)$?
Is that the reason why we do it?
 
yes
the complex linear representations of an algebra are the same as that of its complexification
and if you complexify the lorentz algebra, you get that this is the same as the complexification of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$
and so we can reduce the (finite-dimensional!) representation theory of the Lorentz algebra to the representation theory of $\mathfrak{su}(2)$ we already know
 
7:00 PM
Do you know of any complete sources where I can learn all of this from? Preferably something that starts off simple?
 
Well, the math is all in math books on the rep theory of Lie groups
 
What about the complexification of the lorentz algebra?
All sources I can find either skip that and just state the results or they are written more for mathemateicians
 
yeah, that's a common problem - the mathematicians do it correctly but rarely in a way accessible to physicists, and the physicists usually either just skip the proofs or butcher the math :P
Qmechanic's answer here is quite nice
 
I feel like it's one of the most important points in the whole of theoretical physics
 
you can check this out : springer.com/gp/book/9783319134666
 
 
3 hours later…
9:41 PM
Is a real lie algebra a lie algebra that has all real elements or is it one that gives rise a real Lie group?
 
"real" or "complex" Lie algebra refers to whether you allow linear combinations with real or complex coefficients of the generators
it somewhat confusingly has nothing to do with whether the matrices of the algebra in any given representation are real or complex
e.g. $\mathfrak{su}(2)$ is a 3-dimensional real Lie algebra, even if its fundamental rep consists of complex 2-by-2 matrices
 
So a complexified lie algebra is a real lie algebra where the generators are written in the form $G_i = X_i \pm \imath Y_i$ for $X,Y \in G$?
 
sort of, but the better way to look at it is that it's the algebra with the same generators but with complex coefficients allowed
 
But wont complex coefficients allow other combinations also?
 
I'm not sure what you mean
given a set of generators $T_i$, all elements of a real Lie algebra can be written as $\sum_i a_i T_i, a_i\in\mathbb{R}$
the complexification of this algebra is the algebra where all elements can be written as $\sum_i a_i T_i,a_i\in\mathbb{C}$
if you now look at this algebra again as a real Lie algebra (sounds confusing, I know), it is generated by the $T_i$ and the $\mathrm{i}T_i$ as a real Lie algebra, since all elements can be written as $\sum_i b_i T_i + \sum_i c_i \mathrm{i}T_i, b_i\in\mathbb{R}, c_i\in\mathbb{R}$
 
9:51 PM
Oh, i thought you meant that when you exponentiate $e^{\imath A_i T_i}$, if $A_i$ is real its a real lie algebra and if it can be complex its a complex lie algebra
Wait isn't that the same thing
 
yes, that's the same thing written differently
 
So is redifining the generators as $G_i = X_i \pm \imath Y_i$ just a convention?
 
well, you're always free to choose any basis of the algebra as your generators
it is often the case that complexifying one algebra and switching to the generators $T^{\pm}_i = T_i\pm\mathrm{i}T_i$ tells you something about a different algebra this algebra is also the complexification of
 
Is this different algebra the algebra of the projective representation when considering so(1,3)?
 
I'm not sure what that means
in the case of the Lorentz algebra you find that the complexification of $\mathfrak{so}(1,3)$ is in fact the same as the complexification of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$
and since we already know the representation theory of $\mathfrak{su}(2)$ from non-relativistic rotation, we use that to classify the Lorentz reps by pairs of spins $(s_1,s_2)$
 
10:06 PM
So the lie group formed through $\mathfrac{su}(2) \oplus \mathfrac{su}(2)$ is a projective representation of the lie group formed through $\mathfrac{so}(1,3)$?
 
no, the complexification is not related to the projective representation issue :/
we cover the "projective" issue already by only looking at the algebras
since except in exotic cases like the Witt algebra, all projective representations are linear representations of the universal cover, which in turn are linear representation of the algebra
the complexification is just because we're looking at complex representations - the state space in QM is a complex vector space, after all
it should be obvious that complex representations of a real algebra are the same as complex representations of its complex version (this might sound trivial, but becomes actually potent by different real algebras having the same complexifications)
 
How is the projective issue fixed just by looking at the algebras?
Sorry I missed your 3rd line
So $Spin(3)$ is the universal covering group of $SO(3)$. Since it is a universal covering group, it's linear representations from a projective representation of $SO(3)$. The Lie algebra of $Spin(3)$ is isomorphic to $su(2)$ therefore there exists a bijective map from $Spin(3)$ to $SU(2)$ so we can just consider the representations of $SU(2)$
Is this correct?
 
10:24 PM
the step where you conclude $\mathrm{Spin}(3)\cong\mathrm{SU}(2)$ is not correct
the Lie algebra of $\mathrm{SO}(3)$ itself is also isomorphic to $\mathfrak{su}(2)$ - all Lie groups with the same universal cover have the same algebra
 
What is the definition of a universal covering group?
 
in order to show that it's really $\mathrm{SU}(2)$ that is the universal cover you need to show that 1. $\mathrm{SU}(2)$ is simply-connected and 2. there is a surjective Lie homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$ with discrete fibers
it's instructive to do this some time, but you don't actually need to do that in this case - it 's enough that $\mathfrak{so}(3)\cong\mathfrak{su}(2)$
 
And then as a final step do you just complexify the lie algebra
Okay I think this is too complicated for the amount of maths I know
 

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