Anihilation operator expression: is there a typo here?

[Tio Barnabe]

For those experienced with this stuff,

Weinberg argues (Weinberg, QFT, Volume 1) that an expression for the anihilation operator acting on a state vector when all particles are either all bosons or all fermions is $$a(q) \Phi_{q_1 q_2 ... q_N} = \sum_{r=1}^{N} ( \pm 1)^{r+1} \delta (q - q_r) \Phi_{q_1...q_{r-1} q_{r+1}...q_N}$$ I carefully tried to reproduce the scalar product of this state vector with another state vector, but I have not found the above expression to satisfy their inner product, for one thing: the plus sign in the exponent in $( \pm 1)^{r+1}$. It seems that it only agrees with the expression for the inner product if the sign is "$-$". Am I missing something else or is that sign a typo?

 

[Tio Barnabe]

Just for note: I'm trying to understand this expression for about one week or so! Weinberg actually proves it in the text, but I do not understand his proof either. After all this time, I arrived at a possible "proof", but that doesn't agree with the one by Weinberg for the plus sign I mentioned earlier.

 

[strangerep]

If you gave a more precise reference, i.e., page number(s), equation number(s), it would be a lot easier for others to help you.

 

[Tio Barnabe]

If you gave a more precise reference, i.e., page number(s), equation number(s), it would be a lot easier for others to help you.

The above can be found in page $172$ and the pages that follows from it. The equations are those appearing in there. One will have a better understanding of the context if one looks at them together, so it would not be so good to give their book order numbers.

 

[strangerep]

OK. As I read it, the minus sign is for fermions only (and then only if there's an odd number of fermion interchanges). The plus sign is for bosons only (hence not really necessary in that case).

(Tbh, I'm not sure what was confusing you -- the bottom paragraph on p172 explains all this, doesn't it?)

As for the inner product question, sometimes these things become clearer if you try it on a smaller specific example. E.g., try it for M,N just 1, then 2, etc, until you see the pattern. (But if you want to work through this in detail on PF, it should probably go in the Advanced Physics Homework forum, where you will have to show your attempt.)

 

[Tio Barnabe]

As I read it, the minus sign is for fermions only (and then only if there's an odd number of fermion interchanges). The plus sign is for bosons only (hence not really necessary in that case).

You've got this wrong. I'm asking about the sign in the exponent, not the signals in front of the number one.

(Tbh, I'm not sure what was confusing you -- the bottom paragraph on p172 explains all this, doesn't it?)

Let's say when I substitute the given expression in the inner product, I do not get what I should, i.e. the expression for the inner product as given in page $172$.

E.g., try it for M,N just 1, then 2, etc, until you see the pattern.

I did. I have been doing this thing for over a week.

 

[strangerep]

You've got this wrong. I'm asking about the sign in the exponent, not the signals in front of the number one.

(Sigh.) OK, now I feel guilty, so I'll do it...

I get (for fermions) $(-1)^{r-1}$, -- which is of course the same as $(-1)^{r+1}$, but the latter is indeed confusing, imho.

 

[Tio Barnabe]

Ok. After so many days working hard, I think I can finally understand what he does. I have changed the notation, because I find it more easy to go with then his notation:

First of all, the inner product is defined between vectors in a same vector space. What this means in the current case is that they must have the same number of indices.
The inner product between two vectors of same number $n$ of indices can be written as $$\langle n| n \rangle = \sum_{j=1}^{n!} ( \pm 1)^{j-1} \prod_{i=1}^{n} \delta (P(i,j) - i')$$ where is crucial to assume that there exists a function like $P$ above that will satisfy the permutations required by the situation.

Next, I found, using the elements $\{1,2,3 \}$ and fixing the element $\{1 \}$, writing a general expression for it and going through it back to the above expression, which proves that it works for every number $n$ of elements, that the above inner product can also be written as $$\langle n| n \rangle = \sum_{r=1}^{n} \delta(q - r) \delta_r \sum_{j=1}^{(n-1)!} ( \pm 1)^{j-1} \prod_{i \neq r} \delta (\bar{P}(i,j) - i')$$ (1) where it's assumed that a function $\bar{P}$ can be found, such that $\bar{P}(r,j) \equiv q, \forall j$ and $q \in \{n \}$. Furthermore, $\delta_r$ could be defined $\delta_{r = 1} = 1$, $\delta_{r > 1} = -1$ if the particles are fermions, and $\delta_{\forall r} = 1$ in case of bosons.

Let our candidate expression be $a(q) |n \rangle = \sum_{r =1}^{n} ( \pm 1)^{r-1} \delta (q - r) |n-1 \rangle$ (2)

Knowing all of the above, simply substitute the expression given by Weinberg, as you pointed out, it will not matter in the end which sign is up there in the exponent.

$$\langle n' | a(q) |n \rangle = \sum_{r=1}^{n} ( \pm 1)^{r-1} \delta(q-r) \langle n-1 | n-1 \rangle$$ but the above inner product is $$\sum_{j=1}^{(n-1)!} ( \pm 1)^{j-1} \prod_{i=1}^{n-1} \delta ( \bar{P}(i,j) - i')$$ Comparing this with (1) above, we see that (2) immediately satisfy the inner product for bosons. For fermions, we could ignore $\delta_r$ and $( \pm 1)^{r-1}$, because there is already the term $(-1)^{j-1}$ in the sum over the $n!$ possible permutations, and so (2) also satisfy for fermions. But I do not like simply ignoring terms. What are your thoughts? (If you read all of this text, LoL).

 

[strangerep]

But I do not like simply ignoring terms. What are your thoughts? (If you read all of this text, LoL).

Well, I read it, but I'm not sure what the question is.

As Weinberg says near the bottom of p174, he's doing all this in the reverse order from most textbooks. I.e., he's using some physical arguments and the inner product to derive the canonical (anti-)commutation relations. The latter are more usually postulated up front.

This is yet another reason why Weinberg should probably not be one's first introduction to QFT.

 

[bhobba]

This is yet another reason why Weinberg should probably not be one's first introduction to QFT.

Indeed not.

I have the set and have read a number of books on QFT up to Zee- QFT In a Nutshell.

I don't understand Zee as well as I would like to - and until I do will not undertake Weinberg.

Thanks
Bill

 

[Tio Barnabe]

I got it now. Nevertheless thank you @strangerep for trying to help me

 

[vanhees71]

Well, I read it, but I'm not sure what the question is.

As Weinberg says near the bottom of p174, he's doing all this in the reverse order from most textbooks. I.e., he's using some physical arguments and the inner product to derive the canonical (anti-)commutation relations. The latter are more usually postulated up front.

This is yet another reason why Weinberg should probably not be one's first introduction to QFT.

Indeed, but reading it after having gotten used to some calculations in QFT with the question, why these calculations make sense from a fundamental point of view, Weinberg's treatment is a revelation!

 

[vanhees71]

Indeed not.

I have the set and have read a number of books on QFT up to Zee- QFT In a Nutshell.

I don't understand Zee as well as I would like to - and until I do will not undertake Weinberg.

Thanks
Bill

Zee is the worst textbook on the subject, I've seen yet (and I've seen many!). My favorite is Schwartz as an intro-QFT book (or, if you prefer the path-integral only approach, Bailin and Love).

 

[strangerep]

Indeed, but reading [Weinberg] after having gotten used to some calculations in QFT with the question, why these calculations make sense from a fundamental point of view, Weinberg's treatment is a revelation!

Yes, indeed, if one can successfully navigate Weinberg torturous way of "explaining" things. Sometimes I think he's deliberately constructing an obstacle course for the reader.

@bhobba: Similar to what Hendrik said,... don't let non-completion of Zee stop from moving on to another QFT text.

 

[bhobba]

Zee is the worst textbook on the subject, I've seen yet (and I've seen many!). My favorite is Schwartz as an intro-QFT book (or, if you prefer the path-integral only approach, Bailin and Love).

I don't like it either. Think your advice might be good - have done QFT For The Gifted Amateur and Student Friendly QFT. But wanted something in the middle between those and Weinberg. I will check it out.

Added Later:
Had a look in my library just to see if I had purchased it somewhere along the line. But found Banks:
https://www.amazon.com/dp/0521850827/?tag=pfamazon01-20

What your opinion of that book?

Thanks
Bill

 

[vanhees71]

This one I don't know yet. I think QFT for the giftet amateur is very good too. I also don't know Student Friendly QFT.

 

[vanhees71]

Yes, indeed, if one can successfully navigate Weinberg torturous way of "explaining" things. Sometimes I think he's deliberately constructing an obstacle course for the reader.

@bhobba: Similar to what Hendrik said,... don't let non-completion of Zee stop from moving on to another QFT text.

Well, the problem with Weinberg for the beginner is that he aims at explaining the most general case right away, and that of course makes it hard to get started. E.g., it is completely sufficient for the QFT I+II lectures to treat the kind of fields needed in the Standard Model, i.e., scalars, Dirac fields (+chiral fermions when it comes to the electroweak sector, but here $\gamma^5$ is your friend and you don't need to formalize it to Weyl spinors but can use Dirac spinors with the projection operators to definite chirality states), and massless as well as massive spin-1 fields.