The h Bar

Tobias Fünke

17:01

are A and B functions of p?

If so, then no; it has infinitely many possible solutions?! if they are constant, then it is not well-defined to begin with, I am afraid

imbAF

Honestly, I don't know what to say about that

Tobias Fünke

ehm ok

imbAF

Assume that I have this particular solution to K.G equation: $\phi(\vec x,t)=Ae^{-ipx}+Be^{+ipx}$ where p and x are in four vector notation.
Now for the general one I have:

$\Phi(x^\mu)=\frac{1}{(2\pi)^3}(Ae^{-ipx}+Be^{+ipx})d^3p$

Now I want to find the values of A and B, so I try to normalize the expression:

$\int \Phi(x^\mu)*\Phi(x)d^3x=1$

When substituting Phi* and Phi* I used p and p', so the integral will have d^3p'd^3pd^3x.

Assuming everything I wrong is correct. How would you proceed? My aim here is to get the $\frac{1}{2E}$ term

So this is where I stand

Tobias Fünke

I don't have MathJax atm, so it is not possible for me to decipher that, sorry. Perhaps someone else can help

imbAF

I think is best to make a thread for this, cuz I am not going nowhere for hours now

Tobias Fünke

17:05

but I told you today already that the $E$ term is due to the mass-shell condition

you want a Lorentz invariant measure

imbAF

That I understand

But as you can see

my calculations initiate from the wave function solution

and I want to reach the fourier decomposition in QM, not QFT

which means instead of the ladder operators, you have complex numbers a and a*

Tobias Fünke

I don't get what you want. But yes, you can solve the classical KG equation

and write down the general solution

but why do you want to normalize?

imbAF

because

trick to explain but I will try

In what I wrote above, you did notice how I expressed the general solution

oh wait you don't have MathJax

I will post a pic

This is how I start

with a very general expression of the decomposition

Feynmate

It's really just a choice of normalization. The expansion always holds, it's just about absorbing that term into the modes or not

imbAF

If you substitute the expression for $\phi$ in the integral, you will have an expression followed by d^3p' d^3p d^3x

I believe, that one should be able, through pure calculation

reach the following point

Feynmate

17:15

(and of course what Tobias said about invariance)

imbAF

While the way is different

namely here, we start with integration in spacetime d^4x\delta(...)\theta()

the end result must be possible from both angles

The reason why I am trying, what I am, is because I am starting from the wave function solution and then leveling up to this point. While in here and in many texts, we start from the fourier decomposition

Feynmate

It's the same idea; the first expansion is manifestly invariant; that's actually the way we prove that the measure with energy is invariant

imbAF

of the solution. And in no instance is made clear how that expression, which contains 1/2E was EVER derived

Tobias Fünke

then you follow the wrong resources

this should be explained in any reasonable text

...

Feynmate

Tobias entering ACM mode

imbAF

17:18

like L& L

who clearly do not derive the decomposition from the wavefunction solution of the eq.

Tobias Fünke

I don't know L&L; I just know that in any basic course this should be derived

imbAF

I believe that I gave a comprehensive explanation as to what the issue is here and why in no book I see the solution to my problem

@TobiasFünke the fourier decomposition?

Tobias Fünke

and I honestly do not understand why you spent hours with this instead of checking 1-2 notes and going through the relevant computation step by step

imbAF

@TobiasFünke I mean,.....

ok

Tobias Fünke

@Feynmate Please help me out :d I am no expert in relativistic QFT. But this should be derived in any intro course, no?

At least we did it in ours

and now after checking two books this was done there, too (although this are math books)

imbAF

17:21

You derived the decomposition starting from a particular solution, i.e wave fuction?

Tobias Fünke

I mean how the Lorentz invariant measure comes into play

imbAF

that comes into play

Feynmate

@TobiasFünke I'm a little tired so I'm not sure I'm any more helpful in this condition, but it would help me to really understand what's the problem with the normalization

@TobiasFünke This?

imbAF

if you START with decomposition in spacetime

Tobias Fünke

@Feynmate at least this is what I understood

imbAF

17:22

Ok, I will make it as clear as I possibly can

Feynmate

Well, if you are in a relativistic setting, you expect that the nornalization of states does not change with Lorentz transformations, so that's where the invariant measure comes into play

In the field expansion the energy factor is only there because of a normalization choice

Really, there is no reason in principle other than normalization, nothing to derive. The field is a function and you perform a Fourier expansion, then proceed to "divide" the Fourier transform into positive and negative modes, right?

16 mins ago, by imbAF

imbAF

I am working on posting soemthing. So by reading it, you'll be able to judge whether what I am trying to do is even possible or not. Because exchanging messages hasn't been working for the past 5 hours

Feynmate

I have just come into the discussion, so you could at least try to listen to someone who's trying to help you

This is the expansion. Understood, $A$ and $B$ are $A_\vec{p}$ and $B_\vec{p}$, functions of momentum

imbAF

Ok, but when I wrote them down I didn;t think of them as functions of momenta

I mean, I have to argue for it, no?

Feynmate

Then your integral is badly divergent

imbAF

17:31

correct

So how do you argue the momentum dependency ?

Feynmate

No, you don't those coefficients are just the Fourier transforms of the field

Nothing to argue

imbAF

So, when you have the K.G equation and you consider a wave function solution

you write

$\phi=A(\vec p)e^{ipx}$ ?

Feynmate

$\Phi(\vec{x},0)=\int\frac{d^3\vec{p}}{(2\pi)^3}\tilde{\Phi}(p)e^{i\vec{p}\cdot\vec{x}}$

That's the first step

imbAF

Ok, so that is the first mistake in what I wrote

I only have A or B

Feynmate

This is the Fourier transform, we haven't split into modes yet

imbAF

17:35

Ok

that is the generalized solution ?

could you say that?

since you are integrating over momenta

Feynmate

Again, I'm just Fourier transforming a field T_T

imbAF

I know that. I am asking if it is wrong to consider it as the general solution to the equation

But ok

Then what do we do, to avoid the divergence ?

Tobias Fünke

A FT is a FT, not a priori any solution

imbAF

ok

Feynmate

Now we use the fact that the field satisfies the KG equation and write the equation in momentum space i.e. the FT EoM

imbAF

17:39

hmm

Feynmate

$[\partial^2/\partial t^2+ \omega^2_\vec{p}]\tilde{\Phi}(p)=0$

Do you recognize this equation?

imbAF

atm I don't

FT of eom

Feynmate

That a HO EoM

With frequency $\omega_p$

imbAF

Ok, I recall the equation

Tobias Fünke

imbAF, since you speak german (?), you can check the book "Tutorium Quantenfeldtheorie"...perhaps it helps

imbAF

17:41

I am not sure how you got here

@TobiasFünke muss kaufen

Tobias Fünke

most German universities have contracts with Springer; it should be possible to get a PDF for free, at least with high probability ;)

anyway, I won't bother you guys anymore. see you later

Feynmate

*Oh no they are speaking a foreign language, I must say something to sound like I know what they are talking about. Come on, Feynmate, chill out. They won't notice a thing.*

Guten morgen

@TobiasFünke TOBIAAAAAAAAAS T_T

imbAF

I know that you can write in momentum space $(\partial^2_t + p^2+m^2)\phi$

so $\vec p^2+m^2=E^2$

right?

So, how do we continue ?

Feynmate

Now, for a QHO you know that $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$

I'm just writing the analogue of $\hat{x}\propto(a+a^\dagger)$ from QM, using the oscillators of our problem, $\tilde{\Phi}$

imbAF

yes \phi plays the role of \vec x

Feynmate

17:50

And of course, the respective annihilations/creations are labeled by the $p$ of the mode

Got this?

imbAF

meaning $a(\vec p), a(\vec p)^\dagger$

?

Feynmate

Just like $\tilde{\Phi}(p)$ plays the role of $x$, they play the roles of $a$ and $a^\dagger$

But of course you have a different mode for each $p$, so you label them with $p$

imbAF

yes

Feynmate

Okay, finally, we use the EoM of creation and annihilation of each mode, which are the same as QM

First replace this in the FT

9 mins ago, by Feynmate

Now, for a QHO you know that $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$

imbAF

I am not sure I am following

Feynmate

17:58

Just put that equation into the FT

imbAF

this one $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$ ?

in $\Phi(\vec{x},0)=\int\frac{d^3\vec{p}}{(2\pi)^3}\tilde{\Phi}(p)e^{i\vec{p}\cdot\vec{x}}$ ?

Feynmate

Indeed

imbAF

Ok

Feynmate

Then we want to write $\Phi(x,t)$, the one before was at zero. The time dependence is of course encoded in the coefficients, so we only have to write $a\to a(t)$

imbAF

with the subscript \vec p i assume

but ok

Feynmate

18:01

$\Phi(\vec{x},t)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p(t)+a^\dagger_p(t))e^{i\vec{p}\cdot\vec{x}}$

imbAF

I see

But one thing only. You have root of 2omega

I have seen notations where it;s simply 1/2\omega

Feynmate

For the field?

That happens in other cases

The field expansion usually has this normalization

imbAF

here

oh, this is another case

so you have other expression ?

@Feynmate physics.stackexchange.com/questions/442936/…

bolbteppa

@imbAF This is not a stationary state, the stationary state solutions are $e^{ipx}$, 10.6 of L&L vol 4, they are normalized as in 10.16.

Because the $p = (E,\mathbf{p})$ in this is only constrained to satisfy $p^2 = m^2$, the $E$ is not independent it depends on $\mathbf{p}$, so the stationary states are indexed by $\mathbf{p}$, however the sign of $E$ is also independent so we sum over the signs of it as well, and the general Fourier expansion as a sum over all possible stationary states is then 11.1.

You can then re-interpret the stationary states of this equation as describing a system of two 'species' of particles as explained after 11.1, where both species are described by the same equation in the same system, and interpret them as anti-particles

Feynmate

@imbAF You can do that, it just changes the coefficient of the modes

bolbteppa

18:13

The sign of the energy term in the exponential is just telling you whether the associated quantum field operator is going to be describing particles or anti-particles, as explained below 11.1

imbAF

Ok so it is vol 4 10.6 until 11,1

I was naïve into thinking that going as I went

one could get the same result

but of course, I ended up dealing with an integral that diverges

bolbteppa

(12.1 then discusses the more natural case where there is only one species of particles, which is what people would probably guess is going on at first)

Just read section 10, 11,12, skim 1 and 2, especially the last paragraphs of 2 for more on the sign of the exponential

Tobias Fünke

18:31

@Feynmate hehe

User198

18:54

Is there a generalization of the Moyal equation $\dot W= \{\{H,W\}\}$ ?

We can go Schrödinger equation $\rightarrow$ LvN equation $\rightarrow$ (Wigner transform to) $\rightarrow$ Moyal equation $\rightarrow$ ?

Now we are in phase space and practically made a full circle from the Hamilton's equation that are in phase space back to QM that is again in phase space. Did people feel the need to generalize further or is this practically it? The full circle?

I presume that it is (as always) possible to go even to greater generalizations and abstractions

ACuriousMind

19:53

@User198 it's not clear what you mean by a "generalization" here

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