The h Bar

7:31 AM

There is two meanings of gauge transformation

Suppose a motorcycle starts at $v_i$ and then skids over a distance of $d$, ending the skid at $v_f$ .Find ${v_f}^2 - {v_i}^2$ in terms of the distance $d$ and the deceleration $a$.
How do I start about this problem?
If $d$ was displacement then I get $v^2_f-v^2_i=ad$.

Slereah

one is from the point of view of the principal bundle, the other is a more general automorphisms on fields that do not change the action

those things may be equivalent under some very broad understanding of things

BannedUser

Distance and displacement are not same in this condition to me...

ACuriousMind

@BannedUser I see you've already posted this in the PSS room, which is the more appropriate place for that - please don't post the same question in several rooms simultaneously, as people in one room might try to answer it when it has already been answered in the other.

BannedUser

and to skid it means acceleration is constant right?

@ACuriousMind Ah ok

B. Brekke

7:46 AM

The Schödinger equation is a linear differential equation, hence there is no wonder that a superposition of two solutions, is also a solution. However, in some cases, e.g. interacting many-particle systems, the equation is non-linear. How can I explain superposition in such a case?

Slereah

The equation is still linear even if there are multiple particles

You are still just applying operators to the tensor product of two wavefunctions

a very linear thing

B. Brekke

Even in the case $V=V(\psi)$?

Slereah

That's never the case in normal QM

I think it happens in the Newton-Schrodinger equation, but it's a semi-classical equation

B. Brekke

How about the case of an atom with multiple electrons with electron-electron Coulomb interaction?

Slereah

still the same thing

You have a complicated Hilbert space that is the product of many Hilbert spaces

ACuriousMind

7:50 AM

@B.Brekke The generic Schrödinger equation is $\mathrm{i}\partial_t\psi = H\psi$, where $H$ is the Hamiltonian operator (a linear operator!) on state space - it can't be non-linear by its very definition

Slereah

but within that space, a state is still just one vector, and the Schrodinger equation is still just $H\psi$

and you can still superpose states

B. Brekke

Is the state space just the tensor product of non-interacting state space or something more complicated?

Slereah

Some people have attempted non-linear versions of QM but it didn't really seem to bring much of interest

@B.Brekke Depends on the theory

ACuriousMind

@B.Brekke unless you're doing QFT and are interested in Haag's theorem, there is no such thing as an "interacting" or "non-interacting" state space, there's just a state space

Slereah

Sometimes it's the product, sometimes it's the symmetrized product, sometimes it's an entirely new space that doesn't have any obvious connection

ACuriousMind

7:53 AM

the state space of an $N$-particle system is just the $N$-fold tensor product of the 1-particle system, (anti-)symmetrized if you're doing (fermions) bosons

whether the particles interact or not is entirely irrelevant for this

the interaction changes what $H$ is, not what the state space is

B. Brekke

Okay, I see

And $H$ is a linear operator by definition. However, I can't reconcile this with the procedure where one does density functional theory self-consistently for example.

ACuriousMind

I don't really know much about DFT so I can't tell what problem you see there

B. Brekke

8:15 AM

My problem is that $H = K +V$, and in order to determine $V$, I need to know the density or the position of the electrons. In order to know their positions, I need to solve the Schrödinger equation. To me, the potential $V$ does not sound like a linear operator for interacting systems.

bolbteppa

$V$ is an operator built from operators

It satisfies $\hat{V}(a \psi_1 + b \psi_2) = a \hat{V} \psi_1 + b \hat{V} \psi_2$ as you'd need if the general wave function expands in a basis of eigenfunctions

ACuriousMind

@B.Brekke the positions are just operators

Slereah

Yes, you need $x$, not $\psi(x)$

Schrödinger–Newton equation

The Schrödinger–Newton equation, sometimes referred to as the Newton–Schrödinger or Schrödinger–Poisson equation, is a nonlinear modification of the Schrödinger equation with a Newtonian gravitational potential, where the gravitational potential emerges from the treatment of the wave function as a mass density, including a term that represents interaction of a particle with its own gravitational field. The inclusion of a self-interaction term represents a fundamental alteration of quantum mechanics. It can be written either as a single integro-differential equation or as a coupled system of a...

ACuriousMind

just like for one particle you have the position operator $x$, for $N$ particles you have position operators $x_1,\dots, x_N$. In both cases $V$ is just a function of $x$ resp. $x_i$, not anything to do with the actual state

Slereah

This is an example of a nonlinear system of QM

B. Brekke

8:31 AM

Ah, that clarifies it, thanks!

antimony

11:27 AM

whoever came up with that equation is bonkers (in a good way)

geocalc33

11:41 AM

so are the fields of gravity, electromagnetism, weak force and strong force sort of mixed together ?

im trying to visualize how they interact with each other

Nihar Karve

11:59 AM

why would they be mixed together, and what does it mean for forces to "interact with each other"?

geocalc33

the fields interact?

I mean they all vibrate and cause particles

where the vibrations are strong enough

Slereah

12:14 PM

I mean at high energy, yes

Two photons can produce an electron positron pair, which can produce a Z0 boson, etc etc

But those are like second or third order QFT effects

Very small

geocalc33

is there a mathematical structure that encodes how fields interact?

ACuriousMind

same as in classical physics - the Lagrangian/action

and then all the rest of QFT really is little more but a mathematical structure - any analogies and visualizations (like "fields vibrate", which sounds nice but really tells you very little about what to expect) are usually rather useless when it comes to actually trying to predict what will happen in a specific situation

(obligatory caveat: Unfortunately while QFT is mathematical, it often isn't rigorous in the sense actual mathematicians would want it to be)

geocalc33

12:36 PM

you can do this with fibre bundles as well right?

instead of using the lagrangian?

ACuriousMind

that's rather orthogonal

there are formulations that use bundle language but whether or not you use bundles has little to do with whether or not you have Lagrangians

geocalc33

okay

ACuriousMind

there are some cases where you don't have a Lagrangian mostly relating to conformal and special supersymmetric field theories, but "usual" QFT always starts from a classical action with a classical Lagrangian

geocalc33

can you have a fibre bundle of a 3-manifold whose projection to the base manifold is Minkowski 1+1?

not sure if that makes perfect sense yet

ACuriousMind

I don't really understand the question

you can have arbitrary bundles with fiber $E$ and base $B$ just by taking the trivial bundle $E\times B\to B$

geocalc33

12:52 PM

@ACuriousMind I mean you break up a 3-manifold M=(0,1)^3 into "threads" (a fibration) running from (0,1,1) to (1,0,0). I want the these "threads" to be compatible with the fibration on the base space (0,1)^2. I have that every face of the boundary cube bounding M, to be a component of Minkowski space. s.t. the boundary is flat space

it's mainly just an extension problem - flat space on the boundary - how do you extend to 3d

ACuriousMind

since $(0,1)\cong \mathbb{R}$ as manifolds, this sounds just like the standard fibration $\mathbb{R}\to\mathbb{R}^3\to\mathbb{R}^2$

I don't understand what the problem is

geocalc33

well the hard part is proving that M is a lorentzian manifold

I also tried assuming that slicing M with planes orthogonal to the faces of the boundary cube yields Minkowski (1+1)

do you have any ideas to proceed?

ACuriousMind

I don't really understand what you're trying to do

$\mathbb{R}^3$ is a manifold. If you endow it with the Lorentzian metric where $z$ is the time coordinate, then either the projection along any axis in the $x-y$ plane is a projection onto Lorentzian $\mathbb{R}^2$ with fibers $\mathbb{R}$

geocalc33

okay

ACuriousMind

what does this cube have to do with anything?

geocalc33

1:07 PM

I don't think the cube is completely necessary

Slereah

1:50 PM

I think he may be thinking about the hyperquadrics that people use to define (A)dS maybe?

BannedUser

2:35 PM

nyeees

Tim Crosby

5:44 PM

I'm required to solve the wave function for the potential V(x) = -α(δ(x+a) + δ(x-a)). I have got the wavefunctions Be^(-kx) , Ce^(-kx) + De^(kx) and Ae^kx for the three regions . I dont understand how he equates B = A and C =D.

Daniel Underwood

6:16 PM

I think it's typically by matching the wave functions at the interfaces

You could also say that you need symmetry under $x \mapsto -x$, but that may only work on the outsides

Actually I guess you can use it on the inside too if you're talking about the even case

Semiclassical

6:31 PM

@glS well, i'm thinking of a comment they make to the effect of "Bob's non-selective measurement on a given basis, is equivalent to a random unitary transformation which is diagonal in that basis"

and this limits the unitaries which Bob can apply. (i.e., an arbitrary unitary on Bob's subsystem won't be diagonal in any of the bases)

glS

6:51 PM

@Semiclassical "random unitary transformation diagonal in that basis"... as in, measuring with projections $\{P_1+P_2,P_3\}$ (thinking of a three-dimensional system) amounts to performing a random unitary of the form $U(2)\oplus U(1)$?

Semiclassical

7:01 PM

i think so? honestly i'm a bit fuzzy

glS

I probably just need to read the paper lol what was it again?

Semiclassical

arxiv.org/pdf/1211.7132.pdf

"We note in passing that making a non-selective measurement in basis b is equivalent to performing a random unitary transformation which is diagonal in the b basis"

with citation to a Schwinger textbook

glS

@Semiclassical thanks. Thing is, I don't understand what the analogy is supposed to mean. Is equivalent to doing a random unitary... and then what? Ignoring that degree of freedom?

now that I think about it, there was a paper about Nielsen that talked about intertwining operators and used an equivalence between random unitaries and some channels... let me see if I can find it, maybe it's relevant

@Semiclassical it was arxiv.org/abs/quant-ph/0205035, but it's not actually super relevant, only also used averages over unitaries. The point might be to observe that $\int dU (U\rho U^\dagger)$ probably corresponds to applying a completely dephasing channel to $\rho$. Starting with $\rho$ pure and averaging over unitaries with a certain block structure, I can see why this would correspond to "non-selective measurement"

Semiclassical

7:35 PM

@glS yeah, it's an interesting and plausible-enough claim

it'd be nice if they more more explicit, though

Semiclassical

8:10 PM

@gls the Schwinger reference seems pretty helpful, actually

on one page he looks at $|\langle c|M(B)|a\rangle|^2$ and asks how to apply it to three different scenarios

which is sensible

2) Indiscriminate measurement: $M(B)=1$. then $|\langle c|a\rangle|^2$ is the probability to go from $a$ to $c$. the measurement of B makes no difference

"The non-selective B measurement; the B measuring apparatus functions but no selection of b' atoms is performed. In the m.m. example, this means that the up and down beams are physically separated, but then the two beams are run along together to the next stage." (m.m. = magnetic moment?)

in which case the probability of going from $a$ to $c$ is the probability of going via $a\to b_1\to c$, plus that of going via $a\to b_2\to c$, plus etc

a few pages later he asks what the symbol $M(B)$ would be, and comes up with $M(B)=\sum_b e^{i\phi(b')}|b'\rangle\langle b'|$

where each $\phi(b')$ has to be some random angle to get the right statistics

if you don't do this averaging, then you'll apparently get interference terms when computing $|\langle c|M(B)|a \rangle|^2$

Semiclassical

10:13 PM

(Googling comes up with “pre-measurement” as a synonym for “non-selective measurement”, btw)

Transcript for