Quantum Field Theory for the Gifted Amateur: Fourier Transforms & Excitations

[anklimekruk]

Dear Sir,
P 25 in quantum field theory for the gifted amateur One makes Fourier transforms from the position to the frequency space for the system of linear chain of N atoms.
How can I see that in the frequency space the excitations are uncoupled .

I also don’t understand equation 2.50

 

[ddd123]

Dear Sir,
P 25 in quantum field theory for the gifted amateur One makes Fourier transforms from the position to the frequency space for the system of linear chain of N atoms.
How can I see that in the frequency space the excitations are uncoupled .

Write the equations of motion for the Hamiltonians 2.45 and 2.59.

I also don’t understand equation 2.50

If m=0 you are simply counting all the j's, which means it equals N. If m ≠ 0 summing all terms they cancel out symmetrically.

 

[anklimekruk]

1)I am sorry but I don't understand why the comparaison of 2.45 with 2.59 means uncoupled frequenties.
The book says to compare 2.66 with 2.41 where it becomes clear that the HO's are uncoupled.
My question nevertheless is : can you see that before making all this calculations?
I have the principle: first understand the concept then prove it mathematicaly.
2)I understand when k = 0 i am counting N .I thought if m is not 0 the kronicker delta gave 0.I don't see the symmetric canceling out.

 

[ddd123]

1)I am sorry but I don't understand why the comparaison of 2.45 with 2.59 means uncoupled frequenties.

Write the equations of motion which you've learned from Hamiltonian mechanics. 2.45 is coupled, 2.59 is decoupled. It means that the resulting differential equations are dependent on each other for 2.45 and independent for 2.59.

My question nevertheless is : can you see that before making all this calculations?

You don't have to solve the equations of motion. Just by writing them you see that for the coupled system they form a system of differential equations that is difficult to solve. Of course, if you've clever enough you don't even need to write the equations of motion but just look at the Hamiltonians, but clearly you're not seeing this so you have to write the equations of motion.

2)I understand when k = 0 i am counting N .I thought if m is not 0 the kronicker delta gave 0.I don't see the symmetric canceling out.

Edit: never mind on that, I'm missing something too.

 

[anklimekruk]

I am going to do what you suggest.It will take some time,but i hope to understand.
Thank's a lot.

 

[anklimekruk]

Q 2) became clear.
Q1) I have studied both volumes of Susskind's the theoretical minimum. I hope to find there how to write the equations of motion for both H's and see that the dif eq's are coupled for 2.45 and uncoupled for 2.59.

 

[ddd123]

Q1) I have studied both volumes of Susskind's the theoretical minimum. I hope to find there how to write the equations of motion for both H's and see that the dif eq's are coupled for 2.45 and uncoupled for 2.59.

Hamilton's equations are

$\dot{x_i} = \frac{\partial H}{\partial p_i}$
$\dot{p_i} = -\frac{\partial H}{\partial x_i}$

 

[anklimekruk]

https://www.physicsforums.com/cid:0BBDC2FB-5CF0-954B-85E1-F2214146A7C5.png
https://www.physicsforums.com/cid:6FE7B597-2587-FD41-89BF-3670D418254F.png
https://www.physicsforums.com/cid:0D0F0552-CB7C-4245-80C5-E5C7467E7251.png

 

[anklimekruk]

I have tried to take the partial derivatives of both hamiltonian's with respect to p and x according to hamilton's equations but failed.
I don't understand how to do this particular problem.I think it is because of the indices.
Something else: I notice that in 2.45 & 2.59 we have operators.Not in hamilton's equations.
I very much appreciate your help.

 

[ddd123]

I can't see the images.

I have tried to take the partial derivatives of both hamiltonian's with respect to p and x according to hamilton's equations but failed.
I don't understand how to do this particular problem.I think it is because of the indices.

The main point is that, in one case, to find the solution at i you need the solution at i-1, so you have all equations dependent one another. In the other case, you can have independent equations to be solved separately, because the variables have been decoupled.

I suggest you google linear coupled oscillators and find a treatment that suits you.

Something else: I notice that in 2.45 & 2.59 we have operators.Not in hamilton's equations.

That only complicates the maths. That same decoupling argument works in classical mechanics so you can trust that.

 

[anklimekruk]

I understand your explanation of the dependence of i on i-1
In Gilbert Strangs Linear Algebra p 319 one can find how to take derivatives of operators in the exponential
I go to google following your advice
Thanks a lot !

 

[anklimekruk]

In QFT for the gifted amateur P 26 eq 2.58 is treated similarly to 2.56
I suppose i should substitute p in 2.56 by {(x_j+1) - (x_j)} and get the RHS of the first line of 2.58.
But {(x_j+1) - (x_j)} contains two terms one with index j+1 and one with index j and so i don't understand how to do it

 

[ddd123]

In QFT for the gifted amateur P 26 eq 2.58 is treated similarly to 2.56
I suppose i should substitute p in 2.56 by {(x_j+1) - (x_j)} and get the RHS of the first line of 2.58.
But {(x_j+1) - (x_j)} contains two terms one with index j+1 and one with index j and so i don't understand how to do it

Similarly means using 2.46 instead of 2.47.

 

[anklimekruk]

In my calculation I will encounter a summation over the term 2(X_j+1)(X_j).
I think it is the same as 2 (X_j)^2.
Than I can make the calculation
Is it correct?

 

[ddd123]

In my calculation I will encounter a summation over the term 2(X_j+1)(X_j).
I think it is the same as 2 (X_j)^2.
Than I can make the calculation
Is it correct?

No.

$\sum_k \tilde{x}_k e^{i k (j+1) a} - \sum_m \tilde{x}_m e^{i m j a} = \sum_k \tilde{x}_k e^{i k j a} ( e^{i k a} - 1)$

 

[anklimekruk]

I think it's clear
Thank you !

 

[anklimekruk]

https://www.physicsforums.com/cid:17989CD4-99C1-5842-ADFA-0587E697EEF5.png
I don't understand why from the definitions we automatically have... (P26 QFT for the gifted amateur)

 

[anklimekruk]

Dear Sir,
I hope you understand what is in the in my last mail.
Many mathematical signs are missing on my keybord.
I tried to use the app OneNote where i can use an electronic pencil and then copy and past it on my mails to You but it failed.
Is there an other option?
Thank you

 

[vanhees71]

Yes, in the forum you can use LaTeX, which you have to learn anyway to be able to type readable papers :-). You find an intro to how to use it in the forums here:

https://www.physicsforums.com/help/latexhelp/

 

[anklimekruk]

Thanks a lot I am going to do it later or tomorrow
Was it impossible to understand my question? The book uses the argument of hermitian matrices to assert that p_k is = to p_k_
I don't understand it

 

[anklimekruk]

Page 26 it says "with the definition we automaticaly have $p\dagger_k=p_\-k$ " I do see a minus sign in the exponential of $p\dagger$ because of the Fourier transform.
Is this the reason for the - sign?
But then it says:because $p\dagger$ and p are hermitian which means that if I take the complex conjugate and transpose I get the same matrix.
Complex conjugation does not yield a - sign so I don't understand it.
I must notice that this text is confusing!

 

[vanhees71]

It is really impossible to answer your questions without enough context. I don't own the book. So I can't help.

 

[ddd123]

I must notice that this text is confusing!

It's not confusing... You just have to calmly sit down and do the algebra. You seem like in a rush (or, no offense, lazy to do the algebra). The definition it talks about is of course 2.49:

$\tilde{p}^\dagger_k=\frac {1} {\sqrt{N}}\sum_j p^\dagger_j e^{i k j a}=\frac {1} {\sqrt{N}}\sum_j p_j e^{i k j a}=\tilde{p}_{-k}$

because $\hat{p}_j$ is hermitian.

 

[anklimekruk]

It is really impossible to answer your questions without enough context. I don't own the book. So I can't help.

It will be my pleasure to offer you the book as a present.
I guess you just have to give me the address where i can tell amazone to send it

 

[anklimekruk]

I want to ask if i understand it well:
In the first equality the - sign becomes + because it is $p\tilde\dagger$. In the second equality $p\dagger$ becomes $p$ because $p$ is hermitian and the third equality is definition again.
Is it correct?

 

[ddd123]

Yes.

 

[anklimekruk]

Thanks a lot

 

[anklimekruk]

In QFT for the gifted amateur p 26
I square 2.62 and 2.63 then i plug these values into 2.59 to get 2.64
Is it correct?

 

[ddd123]

No. Your questions, except for the first that was legitimate, are all solved by trivial plugging in. I don't know why you keep getting stuck at that level. It's going to get a lot harder than this, you know.

 

[anklimekruk]

Indeed I see I can trivialy plug in.But then I have a product of $\rm{p_k}$ with $\rm{p_k}$ .
Thats why I said "I square 2.63 and plug it in 2.59".And the same for 2.62.
In my humble opinion this yields the same result.
As for the fact that it is going to get much harder,I don't intend to give up.
I am passionated by the subject and will do my best to understand.I have already overcome some difficult points with your help, and i am gratefull to you.