The h Bar

A vector space is a set of things which can be added together to get another one of those things, and which can be multiplied by a scalar to get another of those things.

So, for example, the set of all velocities in 3D space is a vector space.

You can add two velocity vectors to get another velocity vector.

Bernard Meurer

Ah, I see, makes sense

DanielSank

Here's another example: the set of all functions of a real number.

That's a vector space because given two elements $f$ and $g$ I can get another element by $h$ defined as $h(x) \equiv f(x) + g(x)$.

Bernard Meurer

But what if I throw in $f(x) = xi$ ?

Wouldn't that yield a complex? Or is that cheating?

DanielSank

@BernardMeurer No problem. I said function of a real number.

Bernard Meurer

3:01 AM

Ah, got it

Alright, jolly good so far

DanielSank

If we have $f: x \rightarrow ix$ and $g: x \rightarrow x^2$ then $h(x): x \rightarrow x^2 + ix$ is still a function of one real number.

@BernardMeurer Ok, so now you know what a vector is.

Bernard Meurer

Okay, so is a vector a kind of tensor?

DanielSank

Hold on.

Getting to that.

Bernard Meurer

Alright

DanielSank

Given a vector space, you can consider the set of linear functions on that vector space which return a real number.

For example, suppose our vector space is the set of pairs of two real numbers.

So, for example, $|v\rangle = (3,4)$ and $|w\rangle = (-1,2)$.

Ok?

Bernard Meurer

3:04 AM

Yeah, sounds fair

DanielSank

Note that $|v\rangle$ is just notation meaning "the vector named $v$".

Bernard Meurer

Kets are just vectors, no reason to be scared

DanielSank

Ok, well, we can cook up one of these linear functions by defining it like this:

$f: (x,y) \rightarrow (x \times -1, y \times 2)$.

In other words, $f: \vec{v} \rightarrow \vec{w} \cdot \vec{v}$.

Or in other words, $f: |v\rangle \rightarrow \langle w | v \rangle$.

Bernard Meurer

Wait wait

AH

DanielSank

I'm intentionally showing you many ways to write the same thing, so you're not surprised by new notation later in your life.

Bernard Meurer

3:06 AM

Got it

okay

DanielSank

Okay, so we have discovered a linear function of vectors in our vector space, namely the one defined by the inner product with $|w\rangle$.

Bernard Meurer

so a bra-ket is just a dot product?

DanielSank

@BernardMeurer Correct.

Bernard Meurer

inner product == dot product?

DanielSank

@BernardMeurer More or less.

Any inner product can be expressed as a dot product. I don't want to get into that yet.

Bernard Meurer

3:08 AM

Alright, that's okay

DanielSank

For our purposes right now inner product and dot product mean the same thing.

Oh, and I want to point out another example of these linear functions on a vector space.

Remember our vector space of functions?

Bernard Meurer

Yeah?

can you make a vector space of vector spaces?

like, a vector-space²

DanielSank

Here's a linear function on it: $f \rightarrow \int f(x) g(x) \, dx$ for some vector (i.e. function) $g$.

It's linear because $\int (f + h)(x) g(x) \, dx = \int f(x) g(x) \, dx + \int h(x) g(x) \, dx$.

Ok?

0celo7

@DanielSank Let $W$ be an $R$ module, $R$ a commutative ring

Bernard Meurer

Wait I'm processing

DanielSank

3:12 AM

@0celo7 Busy. Sorry.

Bernard Meurer

@DanielSank Alright, got it

DanielSank

@BernardMeurer Excellent. When you have a vector space, these linear functions of those vectors are called "co-vectors".

The co-vectors themselves form a vector space.

0celo7

@DanielSank wait a moment

what about the linear functions OF the linear functions

what about those

DanielSank

@0celo7 Please stop.

0celo7

?

DanielSank

3:16 AM

@BernardMeurer do you see that the co-vectors are a vector space? Try going through one of our examples above.

Bernard Meurer

@DanielSank Alright I'm following. Why are they special enough to get a name?

DanielSank

You'll see in a few minutes.

Bernard Meurer

Okay, so far I'm with you

DanielSank

Perhaps it will be more obvious if we take a common example. Ever heard of a gradient?

Bernard Meurer

@DanielSank The word isn't strange to me, not sure I recall the meaning

0celo7

3:17 AM

@DanielSank Yes

DanielSank

@0celo7 Please stop.

0celo7

?

DanielSank

@BernardMeurer I mean, the gradient of a function.

If you don't know about that it's fine, we can move on without it.

Bernard Meurer

@DanielSank Isn't that like differentiation but more general? That's all I remember

0celo7

@DanielSank I can explain things goodly, I think I got my girlfriend to understand what a homeomorphism is

DanielSank

3:19 AM

::0celo7 is awarded the physics pedagogy prize::

0celo7

>physics

the fuck?

DanielSank

@BernardMeurer Ok forget the gradient.

0celo7

@DanielSank since when is topology physics

Bernard Meurer

@DanielSank Alright, forgot it

DanielSank

@0celo7 Please stop.

0celo7

3:20 AM

Ok, what the fuck is it this time?

DanielSank

@BernardMeurer Ok, well, a tensor is simply a linear function which takes as arguments some number of co-vectors and vectors, and returns a number.

So, a co-vector is a tensor because it eats one vector to produce a number.

Similarly, a vector is a tensor because it eats one co-vector to produce a number.

A matrix is a tensor that eats one co-vector and one vector to produce a number.

@BernardMeurer ok?

Bernard Meurer

Wait wait, let the Pentium III here process this

DanielSank

@BernardMeurer ok

I'll type a few more things while you process.

Bernard Meurer

1.4GHz of power

DanielSank

The inner product is a tensor: you can think of it as the tensor which eats two vectors and returns a number.

Bernard Meurer

3:23 AM

Okay I think I got this

0celo7

@BernardMeurer Note that DS's construction works more generally with modules instead of vector spaces

DanielSank

@0celo7 He's in high school.

0celo7

@DanielSank I knew that in high school

Bernard Meurer

@DanielSank I'm done with high school, it's my one victory don't take it away from me :p

DanielSank

Why do you always come here and interrupt discussions by pointing out mathematical generalizations that you know the other participants don't know about?

Bernard Meurer

3:24 AM

@DanielSank Okay, I got the inner product being a tensor part

DanielSank

I perceive it as selfish, @0celo7. At least ask the participants if they know the constructs you're mentioning.

0celo7

@DanielSank extreme autism, probably

DanielSank

Ah, yes, the ignore feature, my old friend.

0celo7

Damn, foiled again

DanielSank

@BernardMeurer Great!

Now here's the cool part.

Well actually first some vocabulary...

Bernard Meurer

3:27 AM

@DanielSank :D

DanielSank

Suppose our tensor eats $k$ vectors and $l$ co-vectors. We will call this a $(k, l)$ tensor or a "tensor of rank $(k, l)$".

Bernard Meurer

Alright

0celo7

Hmm, I think it's the other way around

DanielSank

Ok, now the cool part.

Bernard Meurer

Okay I'm getting excited

DanielSank

3:28 AM

Any tensor can be constructed by summing together several pieces, each of which is formed by gluing together vectors and co-vectors. This "glue" is the tensor product, denoted $\otimes$.

0celo7

@BernardMeurer Yeah, a (k,l) tensor takes l vectors and k covectors.

DanielSank

Let's try the inner product as an example.

Bernard Meurer

So any given tensor can be divided into a series of tensor products of vector and co-vectors?

DanielSank

@BernardMeurer Yes.

Let's do an example.

Bernard Meurer

Alright, inner product

DanielSank

3:31 AM

Given two vectors $(a_0, a_1,\ldots a_n)$ and $(b_0, b_1,\ldots b_n)$, the inner product ought to be $\sum_i a_i b_i$, right?

Bernard Meurer

Wait, what's $i$?

DanielSank

An index that runs from 0 to $n$.

0celo7

@BernardMeurer an index taking the values 0,1,...,n

Bernard Meurer

Oh alright, silly question, got it, yeah

DanielSank

@BernardMeurer Alright. So let's see if we can construct this by gluing together vectors and covectors.

Let's write it out:

$a_0 b_0 + a_1 b_1 + \cdots + a_n b_n$.

I'm claiming that this thing can be broken down into a sum of glued vectors and co-vectors.

So, let's focus on the first term: $a_0 b_0$.

What does that term mean?

Bernard Meurer

3:35 AM

Is that rhetoric?

DanielSank

It means that if we feed the tensor a covector $a$ and a vector $b$, then the tensor multiplies their $0^\text{th}$ components together.

Bernard Meurer

AH, yes, it does mean that

DanielSank

Yes :)

Ok well, is there a co-vector whose action is to just grab the $0^\text{th}$ component of a vector?

Yes there is.

In fact, the act of "grabbing the $0^\text{th}$ component of a vector is a co-vector by definition.

$(a_0, a_1, \ldots, a_n) \rightarrow a_0$ is a linear function.

So it's a covector.

Bernard Meurer

:O

DanielSank

Still with me?

Bernard Meurer

3:37 AM

Cool!

DanielSank

Now, notice this: This particular co-vector has the property that it returns 0 on any basis vector except the $0^\text{th}$ one.

Bernard Meurer

Okay, makes sense

DanielSank

Suppose we have an orthogonal basis of our vector space $\{ e_i \} = \{e_0, e_1, \ldots e_n \}$.

You now see that each basis vector has a uniquely associated covector, which we denote $f_i$ for now, with the special property that $f_i$ acts on the various $e$'s and returns 1 only for $e_i$ and zero for all the others.

In your mind, think of the $e_i$ as columns of numbers and the $f_i$ as rows of numbers. They're transposes.

Bernard Meurer

Well every vector in existence has a covector right?

DanielSank

@BernardMeurer Yes. This is why they're called "covectors".

Bernard Meurer

3:41 AM

@DanielSank Yeah, exactly, so the basis vectors must do too; neat

DanielSank

@BernardMeurer Yeah.

Bernard Meurer

Okay I can visualize the column x row diagram

DanielSank

Good.

0celo7

@BernardMeurer What do you mean by that?

DanielSank

Now let's define $\otimes$.

Suppose I have two vectors $v$ and $w$.

We can make a new tensor called $v \otimes w$ with the property that:

Bernard Meurer

3:42 AM

or $|v\rangle$ and $|w\rangle$!

DanielSank

$(x \otimes y)(u, v) = x(u) y(v)$.

Bernard Meurer

Okay that one boggled me a bit

DanielSank

Ok, here $u$ and $v$ are co-vectors.

Oh crap.

Bernard Meurer

you used the same letters and you shouldnt?

DanielSank

@BernardMeurer Yeah, fixed.

Bernard Meurer

3:44 AM

Ah, yeah now it makes more sense hahaha

DanielSank

$x$ and $y$ are vectors, $u$ and $v$ are covectors.

Bernard Meurer

Okay I get that

DanielSank

Yeah, so that's what a tensor product is.

Now we see that the inner product is just this:

$f_0 \otimes e_0 + f_1 \otimes e_1 + \cdots + f_n \otimes e_n$.

Bernard Meurer

but wait, how are you defining $\otimes$ by using $\otimes$?

DanielSank

@BernardMeurer I defined $\otimes$ above without using itself.

Bernard Meurer

3:48 AM

Yeah and that part I got

but now you said that $\otimes$ is just $f_0 \otimes e_0 + f_1 \otimes e_1 + \cdots + f_n \otimes e_n$ and that I don't get

DanielSank

@BernardMeurer inner product :)

Bernard Meurer

AAAH

My bad, my bad

Okay, following ya

DanielSank

@BernardMeurer So that's kinda it.

Tensor product is a way to glue two smaller tensors together to get a new one.

The inner product is a really common case of a $(1, 1)$ tensor.

Bernard Meurer

@DanielSank Damn man thanks!

DanielSank

In a sense, any matrix is a (1, 1) tensor.

You feed it a row (covector) on the left and a column (vector) on the right, and it returns a number.

Oh, one more thing:

If you have a $(k, n)$ tensor and you feed it one vector, you get a $(k-1, n)$ tensor.

Bernard Meurer

3:52 AM

Why's that?

DanielSank

This is kind of how if you multiply a matrix on the right by a column you get a column back.

0celo7

sigh, other way around

DanielSank

@BernardMeurer Well, because what you're left with is a thing that needs 1 less covector to eat before it gives a numbers.

@0celo7 Actually, I think I got it right.

0celo7

@DanielSank $(k,n)$ has $k$ upper indices and $n$ lower.

DanielSank

@0celo7 Oh I did the whole thing backwards? Ok whatever.

@BernardMeurer We may have swapped the $k$ and $l$ in $(k, l)$. Whatever.

Bernard Meurer

3:54 AM

@DanielSank It's okay I got it still

DanielSank

Yeah none of this is that complicated.

Bernard Meurer

so in $(k, l)$ $k$ should be the vectors and $l$ covectors?

0celo7

@BernardMeurer Yes.

DanielSank

You'll hear people say that a tensor is something that transforms a certain way. That's a different but equivalent construction and in my opinion it's considerably less easy to understand.

0celo7

@DanielSank Well that's really only something that makes sense in the geometrical setting with tensor fields.

DanielSank

3:56 AM

You will constantly encounter physicists who mix up tensors with their representations in a particular basis.

0celo7

@Slereah take note ^

Bernard Meurer

So if I have a $(k,l)$ tensor and feed it one vector I'll get a $(k,l-1)$ tensor?

0celo7

@BernardMeurer Yes.

Bernard Meurer

Hmm, cool stuff

0celo7

@BernardMeurer I take it you don't want the module-theoretic definition of the tensor product?

Bernard Meurer

4:00 AM

@DanielSank and now to the initial question, why does the guy in this article (page 4) defines the 4 states of a 2 CBit system as being tensor products of ket pairs as $|0\rangle \otimes |1\rangle$ and so on?

@0celo7 You can say it, it'll make you sound smart and you know I won't understand it, so hey, why not?

0celo7

@BernardMeurer Sounds good

@DanielSank n00b move

DanielSank

@BernardMeurer Well, a quantum system is mathematically represented by a vector space.

The vectors are the states of the system, often written $|\Psi\rangle$.

@MAFIA36790 o/

user116211

@DanielSank hallo!

DanielSank

Dude, someone is flagging a lot this evening.

user116211

@DanielSank o.O

0celo7

4:03 AM

@DanielSank Holy shit Super Metroid is available on the new 3DS

DanielSank

@BernardMeurer The system is also characterized by it's Hamiltonian.

The Hamiltonian is a matrix (i.e. a (1,1) tensor).

Bernard Meurer

@DanielSank What's a hamiltonian?

DanielSank

@BernardMeurer Ah.

Ok, let's say this another way.

0celo7

@DanielSank That's not strictly true for infinite-dimensional vector spaces.

Bernard Meurer

I'm still a little confused when you define a matrix as a (1,1) tensor

DanielSank

4:05 AM

@BernardMeurer Well, it is a $(1, 1)$ tensor, all the time.

0celo7

Not in...fuck it

DanielSank

@0celo7 Bah!

@0celo7 It's the best video game ever made. Buy it. Play it.

0celo7

@DanielSank Yes, my new 3DS is arriving tomorrow.

I'll be putting that game on it

3075

@0celo7 I bought a ps vita instead of a 3ds and the games suck.

0celo7

Does the vita even have any games?

3075

4:06 AM

@0celo7 for some reason it's difficult to imagine you playing with a 3ds.

Bernard Meurer

@DanielSank I can see that, it just seems like such a dull statement :p

3075

@0celo7 yeah, most are jrpgs, kind of lame.

0celo7

@DanielSank I'm not trying to be pedantic...partially because I'm not sure what good thinking of matrices as (1,1) tensors is, honestly.

DanielSank

@0celo7 Oh, well, whatever. It's useful some times.

0celo7

@3075 I like those

DanielSank

4:07 AM

@BernardMeurer It is.

Let's go about this a different way.

If you have a quantum system, it's described by a set of state vectors $|\Psi\rangle$.

If you have a second quantum system it's described by a different set of state vectors $|\Phi\rangle$.

0celo7

@DanielSank Do quantum computing people use $\Psi$ instead of $\psi$?

DanielSank

If you consider both of those systems together, then the combined system is described by all the possible tensor products, i.e. $|\Psi\rangle \otimes |\Phi\rangle$

Bernard Meurer

@DanielSank State vectors? 'state' of the QBits?

DanielSank

@BernardMeurer Yes.

@0celo7 I dunno. Is there a standard in any field?

0celo7

@DanielSank You're the first person I've seen besides Weinberg to use $\Psi$

Bernard Meurer

4:10 AM

@DanielSank Alright, that makes good sense

DanielSank

@BernardMeurer It will make more sense if you ever have to use it.

3075

@0celo7 I remember I played this game called "lost dimension" and it was really good.

I bought a ps vita because of it.

Bernard Meurer

@DanielSank With your help hopefully I will someday :p

DanielSank

@BernardMeurer Come to SB and live in my apt. I need a housemate.

Bernard Meurer

@DanielSank Aren't you getting a wife?

0celo7

4:12 AM

@3075 believe it

Bernard Meurer

I'll GLADLY be your housemate

DanielSank

@BernardMeurer Yes but she lives across the country.

And will for about another year :(

She's the best.

Bernard Meurer

@DanielSank What, I could have swore she did research at SB

user116211

@yuggib: I've completed dihedral subgroups, symmetric subgroups, matrix subgroups and now about to enter homomorphisms and isomorphisms after quaternion groups.... I would let you know about any problem if I face.

DanielSank

@BernardMeurer Nope.

user116211

4:13 AM

@ManishEarth: o/

Bernard Meurer

@DanielSank I don't doubt that :)

@DanielSank but yeah, in two days we find out if I'm getting there :)

DanielSank

@BernardMeurer Oh, great.

Bernard Meurer

Just need the scholarship money,

DanielSank

Let me know.

I gotta go for now guys.

user116211

@BernardMeurer you'll :)

Bernard Meurer

4:14 AM

@MAFIA36790 Hoepfully

@DanielSank Will do

user116211

@DanielSank o/

Bernard Meurer

@DanielSank See you around man! Thanks for the class!

DanielSank

@MAFIA36790 \o

@BernardMeurer ciao

@0celo7 toodles

user116211

I meant to say groups ;_;

Danu

6:12 AM

Ah yes, @ACuriousMind of course that's what I was talking about; thanks. And sorry for the confusion @0celo7

user507974

7:01 AM

hey guys, anybody here do a lot of optics simulation (quantum effects included)

yuggib

8:05 AM

@MAFIA36790 I see

Physics Meta

9:04 AM

1

Q: Answering old questions and getting noticed

When I browse through old questions I usually find myself reading the question and maybe the top few answers. Same thing when there is an old question appearing on the front page. From what I've gathered other people must be proceeding similarly. Even when there is a good question (many upvotes)...

Secret

10:14 AM

(To be placed on PSE soon after polished) Why do we postulate the electromagnetic field (not electromagnetic wave, as otherwise the momentum density can be directly observed as doppler shift of the photons in the wave) having a momentum density (which is known to be unable to be directly observed and must be inferred by charges being moved around as if under some forces) when performing the experiment of two moving charges moving pass each other by travelling in opposite direction, and noting how the force resolved by Lorentz force law do not form an action reaction pair hence apparent viol…

ACuriousMind

10:50 AM

@Secret We don't "postulate" the field to have momentum. Maxwell's equations are translation invariant. So is the Lagrangian of electromagnetism. Apply Noether's theorem to translation invariance and the field momentum just falls out.

Secret

Ah I see, so that's my knowledge gap in the puzzle...

ACuriousMind

Well, you're right that it's often presented in the way "We want momentum conservation, so we define some field momentum".

But, to me, that's just the wrong way - you begin with translation invariance, and then use Noether's theorem. Conservation laws are not about what we "want", they are consequences of symmetries.

Secret

I agree, the usual notion of "we want A so we define A" will raise the question on how we will ever see any genuine violation behaviour if we are being preconditioned (is thsi the correct psychological term?) to just interpret the observations and make mdoels such that it conforms to our expectations

whereas the fundamental approach you highlighted above gives a justification on why it is sensible

0celo7

12:18 PM

@ACuriousMind Strange thing, I dreamt in German today...

ACuriousMind

12:34 PM

@0celo7 Must have been a nightmare, then

0celo7

@ACuriousMind No, people just started speaking in German!

Secret

Wikipedia
"In mathematics, especially in geometry and group theory, a lattice in $\mathbb{R}^n$ is a subgroup of $\mathbb{R}^n$ which is isomorphic to $\mathbb{Z}^n$, and which spans the real vector space $\mathbb{R}^n$."

how could it span $\mathbb{R}^n$ as the linear combination only has integer coefficients thus will definitely miss out points like $(\pi,0,0,...,0)$?

ACuriousMind

@0celo7 People who can't actually speak it?

@Secret "span" is meant in the sense of linear algebra.

0celo7

@ACuriousMind Well it started out with this one asshole I knew from grade school, who was German

ACuriousMind

I.e the "span" are all linear combinations with real coefficients.

0celo7

12:42 PM

but then everyone was speaking German!

ACuriousMind

@0celo7 My professional psychoanalytical opinion is that you are clearly afraid of said asshole and fear he might influence people in your life ;)

Secret

"In other words, for any basis of $\mathbb{R}^n$, the subgroup of all linear combinations with integer coefficients of the basis vectors forms a lattice."

It is explicitly stated here to be integers?

Lattice (group)

In mathematics, especially in geometry and group theory, a lattice in is a subgroup of which is isomorphic to , and which spans the real vector space . In other words, for any basis of , the subgroup of all linear combinations with integer coefficients of the basis vectors forms a lattice. A lattice may be viewed as a regular tiling of a space by a primitive cell. Lattices have many significant applications in pure mathematics, particularly in connection to Lie algebras, number theory and group theory. They also arise in applied mathematics in connection with coding theory, in cryptography because...

0celo7

@ACuriousMind he's an asshole because he gave me a Platzwunde

ACuriousMind

@Secret The lattice is not the span of its vectors.

0celo7

Not something I need to be afraid of now, Dr. Freud Jr.

Secret

12:45 PM

Ah yes, good ol's common span(S)=S misconception...

continues reading...

0celo7

@Secret what

Secret

(At least for my uni) most people who first learned linear algebra tend to make the mistake of using span(S) and S interchangeably

0celo7

what is S??

Secret

especially in the context of the span of a basis set

S is just a set of vectors

0celo7

well those are obviously not the same thing

Secret

12:48 PM

they are not the same, still we tend to make this mistake until we made it enough times then it started to die away

0celo7

@ACuriousMind do you like 2 Chainz?

ACuriousMind

@0celo7 no idea who or what that is

0celo7

So sad...

Birthday Song?

I Luv Dem Strippers?

I'm Different?

Black Unicorn?

ACuriousMind

Never heard of any of those

0celo7

1:25 PM

@ACuriousMind I'm shocked 😧

Qmechanic

1:35 PM

0

Q: Klein Gordon, Dirac, Proca

How do we know these equations : Klein-Gordon, Dirac, Proca is for spin 0, spin 1/2 , spin 1? How did people find Klein Gordon doesn't work for spin 1/2?

quantum-field-theory quantum-spin quantum-electrodynamics quantum-chromodynamics dirac-equation

Too broad?

Secret

1:45 PM

IMO
The Klein gordon and dirac part of the question is answered by the link you referred rahaa to
However, why proca cannot be recovered from spin=/=1 is not answered, thus perhaps users can guide rahaa to narrow down this question to asking about why proca must be from spin=1 instead