The h Bar

ACuriousMind

14:01

Okay, the heuristic here is that he thinks of $\phi(\vec x,t)^\dagger\lvert 0 \rangle$ as a particle at $\vec x,t$. and then $\langle 0 \rvert \phi(\vec x, t\lvert \psi \rangle$ is the overlap of the state $\psi$ with such a particle, i.e. this would be the QFT analogon to $\langle x \vert \psi \rangle$ being the wavefunction for $\psi$ in non-rel. QM.

So the object you are asking about would be the wavefunction of the one-particle state with momentum $\vec k$.

However, this is really only a heuristic.

There's no position operator in QFT, so it doesn't mean much to begin with to speak of a particle at $\vec x, t$.

And talking about wavefunctions in QFT is not...useful. He's trying to motivate "relativistic quantum mechanics" as an approximation to QFT from this, I think, but, well, don't worry too much about it.

ACuriousMind

It's an exam rather than homework (it's urgent) and it's precessing but autocorrect said otherwise. — Brayden 5 hours ago

Yes, that's an excellent response to being asked to read our homework policy :P

In which world would someone refuse to answer a homework question, but gladly do so when it's an exam question instead!?

Sean

Ok, I need a sanity check, if anyone's on right now...

the tips of all the compasses in my room are all pointing towards the north poles of all the bar magnets in my room. Why on earth would that be happening?

Bastian Treichler

14:18

@ACuriousMind with "a particle at $\vec x,t$" you mean something like a 4-position eigenvector? like $\hat x|x\rangle=x|x\rangle$ in QM?

ACuriousMind

@BastianTreichler Yes. Except that a proper relativistic position operator doesn't exist, so it's only a heuristic.

@Sean Why wouldn't it? If the field of the magnet is stronger than the earth magnetic field, that should happen, right?

(Also, why do you have multiple compasses and bar magnets in your room? :D)

Sean

i'm a teacher

except a compass point is the north pole of the needle, right?

so it should point to a south magnetic pole

and the earth's north pole is actually a south magnetic pole

ACuriousMind

Oh

Hm

Right

Sean

so claims every textbook i've ever read

ACuriousMind

Maybe your magnets are mislabeled?

Bastian Treichler

14:21

@ACuriousMind OK, but wouldn't $a^\dagger(\vec{x},t)|0\rangle$ be a better notation for that? After all, to create a particle from $|0\rangle$, we need a creation operator and not a field? Or IS a field operator some kind of creation operator?

Sean

i've wondered about that

but i don't know how or why that would have happened to all of them

ACuriousMind

@BastianTreichler Well...what would $a^\dagger(\vec x,t)$ be? Also, you have to see that the field is like $\phi = \int a + a^\dagger$, but the $a$ does nothing on the vacuum, so on the vacuum, you might imagine the field to act as a creation operator, since only the $a^\dagger$ part in it does anything.

@Sean Can you use one of the magnets as "compass" and see how it aligns?

Sean

that's a thought

ACuriousMind

Take a small one and suspend it from a string or something like that, I don't know if that actually works

Sean

i might try that

Bastian Treichler

14:28

@ACuriousMind In Zee's book he talks of the creation operator $a^\dagger(\vec{k})$, which I interpreted as a "creation operator for a particle of momentum $\vec k$". So when you said "a particle at (x,t)" I imagined the position analogue, something like $a^\dagger(x)|0\rangle$ with a four-vector $x$.

ACuriousMind

@BastianTreichler Well, but how would you define $a^\dagger(x)$? The $a^\dagger(\vec k)$ is half of the Fourier modes of the field $\phi$. You could just transform it back, but generally, transforming just one of $a,a^\dagger$ to position space is rather useless. Just...don't think too hard about this, I guess

0celo7

I bombed that ODE test.

Crap.

Bastian Treichler

@ACuriousMind ok, I still have many questions, but maybe it's indeed better to read on and look at it again at a later time. The heuristics makes sense to me, thanks!

0celo7

I did partial fractions correctly on the last question...and then I wrote $1=\tfrac{1}{2}\cdot 1$

Jesus...

Sean

@ACuriousMind Ok well it's not perfect but I have a bar magnet hanging from the cieling in my room and it does appear to suggest that it's mislabeled

0celo7

14:40

On another I multiplied one side of the ODE by $x^2$ and didn't touch the other side.

Sean

because the south pole is pointing to geographic north

0celo7

And of course all of the integrals with inverse powers are wrong.

Sean

which it shouldn't

but why would all these bar magnets across various brands (although mostly Sargent Welch) be mislabeled?

is there some kind of global magnetic conspiracy by bar magnet manufacturers against hapless science teachers?

ACuriousMind

Perhaps the labeling the industry actually does is not the one the physics books think it uses?

0celo7

@Slereah Hawking-Ellis. Chap 5. First few sections.

ACuriousMind

14:41

Unfortunately, I have no bar magnets, else I'd test if this is a localized phenomenon

0celo7

How is "north" and "south" defined for a magnet, anyway?

Axiom of Choice?

ACuriousMind

North points north, south points south, duh

Except it doesn't, apparently :D

0celo7

Duh?

Dude

I integrated $1/x^3$ wrong.

Things are not obvious to me.

I took the derivative correctly!

Wait a moment...

I did integrate correctly!

What is this conspiracy

ACuriousMind

Just to be sure, the correct indefinite integral is $-\frac{1}{2x^2}$, yes?

0celo7

@ACuriousMind Suppose I have a family of curves $x^2y=k,k\in\mathbb{R}$. I want to find a family of curves orthogonal to this one. We know that slopes of the family are $y'=-2k/x^3$. Then the slopes of the orthogonal family are $\tilde y'=-(y')^{-1}=x^3/2k$. Integrating gives $\tilde y=x^4/8k+C$.

It's wrong because I just graphed it and it's wrong.

So I must be stupid.

@ACuriousMind Yes. But I didn't need that.

On this problem, I did need it on another.

Slereah

14:53

aw there's a fuzzy wuzzy dog

(I don't like him)

Dammit Fallout

A dog is nice but dragging an AI around can be a bit inconvenient

Sean

@ACuriousMind apparantly not

ACuriousMind

@0celo7 I'm not sure what finding a "family of curves orthogonal curves" means.

0celo7

Or do I have to differentiate $x^2y=k$ to get $2xy+x^2y'=0,y'=-2y/x$. Then $\tilde y'=x/2y$.

ACuriousMind

There are indefinitely many curves which intersect a given curve such that the tangents of the two are orthogonal, the problem is not well-posed.

Sean

@ACuriousMind that's possible. The high school textbook industry isn't exactly known for its unparalleled accuracy

at least not here in the US

0celo7

14:55

@ACuriousMind Find a one? two? parameter family of these

but $x^2y=k$ so $\tilde y'=x^3/2k$...

ACuriousMind

@0celo7 So...you only need to find some curves that do that? Just draw the straight lines that are orthogonal to the tangents!

0celo7

wtf

ACuriousMind

I really don't understand what this exercise is asking for.

0celo7

I get the same answer

I give up

ACuriousMind

@Sean: Do you have an electromagnet?

Sean

14:57

i can make one

but yes

0celo7

Orthogonal trajectory

In mathematics, orthogonal trajectories are a family of curves in the plane that intersect a given family of curves at right angles. The problem is classical, but is now understood by means of complex analysis; see for example harmonic conjugate. For a family of level curves described by , where is a constant, the orthogonal trajectories may be found as the level curves of a new function by solving the partial differential equation for . This is literally a statement that the gradients of the functions (which are perpendicular to the curves) are orthogonal. The partial differential equation...

this is crazy

wwhat did I do wrong

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