How to understand the standard momentum introduced in Weinberg's QFT

  • #1

Main Question or Discussion Point

How to understand the "standard" momentum introduced in Weinberg's QFT

I'm reading Weinberg's QFT Volume I. In page 63, you can find a formula (2.5.3) which states that the new state vector obtained by a Lorentz transformation is a linear combination of a whole bunch of other vectors.

[itex]U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'} C_{\sigma'\sigma}(\Lambda,p)\Psi_{\Lambda p,\sigma'}[/itex]

But in page 64, he tells us that we can choose a so-called "standard" momentum [itex]k^\mu[/itex] and associated with it, a "standard" Lorentz transformation [itex]L^\mu_{\phantom{x}\nu}(p)[/itex] such that we can define

[itex]\Psi_{p,\sigma}=N(p)U(L(p))\Psi_{k,\sigma}[/itex]

with [itex]p^\mu=L^\mu_{\phantom{x}\nu}(p)k^\nu[/itex].

Then the state vector [itex]U(L(p))\Psi_{k,\sigma}[/itex] is no longer a linear combination of several different [itex]\Psi_{p,\sigma}[/itex]'s. Why can we do this?
 

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  • #2
Bill_K
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It's like constructing Euler angles - for each value of p he wants to choose a particular basis for the σ label.

Using Lorentz transformations, the values of p you can generate all have the same value of p2 = m2. Out of these he picks one, a "standard" four-momentum k. This is like choosing a particular rest frame, k = (m,0,0,0). Next, for every other p value, he picks a "standard" way of getting there from k, which he calls L(p). L(p) could be, for example, a simple boost in the k-p plane.

Now L(p) not only maps k into p, it also uniquely maps the basis for σ states at k into a set of σ states for p. That is, Eq(2.5.5). So if he uses this to define ψ he doesn't need the matrix Cσ'σ any more - he's diagonalized it.
 
  • #3


It's like constructing Euler angles - for each value of p he wants to choose a particular basis for the σ label.

Using Lorentz transformations, the values of p you can generate all have the same value of p2 = m2. Out of these he picks one, a "standard" four-momentum k. This is like choosing a particular rest frame, k = (m,0,0,0). Next, for every other p value, he picks a "standard" way of getting there from k, which he calls L(p). L(p) could be, for example, a simple boost in the k-p plane.

Now L(p) not only maps k into p, it also uniquely maps the basis for σ states at k into a set of σ states for p. That is, Eq(2.5.5). So if he uses this to define ψ he doesn't need the matrix Cσ'σ any more - he's diagonalized it.
Thanks for your reply! But the problem is how he knows it is possible to find such kind of standard momentum. What if there isn't the standard momentum? So we need the proof to this possibility.
 
  • #4
Bill_K
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I don't understand. k can be ANY four-vector at all with k2 = m2. Just pick one!
 
  • #5


I don't understand. k can be ANY four-vector at all with k2 = m2. Just pick one!
My purpose is simply to ask if someone can provide a proof to show that it is indeed possible to find a standard momentum, instead of merely saying we can...
 
  • #6
Bill_K
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All right, qinglong, if you are unable to write down k = (m, 0, 0, 0) I guess I can be of no further help to you.

All the four-vectors with p2 = m2 and p0 > 0 lie on a hyperboloid. It is a connected three-surface, and Lorentz transformations are transitive on it. Which means that any p on it can be obtained from any other p' on it by means of a Lorentz transformation, p = L(p,p') p'. (This is what the Lorentz group does, by definition!) Therefore, choose p' to be anything that's convenient, p' = k say. That's all you need. There are no other conditions to be satisfied.

For some reason you are making it out to be much more complicated than it really is.
 
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  • #7


All right, qinglong, if you are unable to write down k = (m, 0, 0, 0) I guess I can be of no further help to you.

All the four-vectors with p2 = m2 and p0 > 0 lie on a hyperboloid. It is a connected three-surface, and Lorentz transformations are transitive on it. Which means that any p on it can be obtained from any other p' on it by means of a Lorentz transformation, p = L(p,p') p'. Therefore, choose p' to be anything that's convenient, p' = k say. That's all you need. There are no other conditions to be satisfied.

For some reason you are making it out to be much more complicated than it really is.
Sorry, you misunderstand my question.

My question is actually why such kind of Lorentz transformation L(p) can only change the value of momentum and fix the value of [itex]\sigma[/itex]. Notice that, in the first equation, [itex]U(\Lambda)\Psi_{p,\sigma}[/itex] has been written as a linear combination of [itex]\Psi_{\Lambda p,\sigma'}[/itex]'s. But after you apply this standard transformation, you don't change [itex]\sigma[/itex] which is too special.

So I want to know why it is possible.
 
  • #8
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You confused yourself too much. It is merely a DEFINITION of the general states [itex]\Psi_p{}[/itex]. You can show that those states defined in this way are complete. Naturally, under this definition, the C matrix for U(L) is diagonalized. This is a nice way to simplify the analysis without loss of any generality.
 
  • #9


You confused yourself too much. It is merely a DEFINITION of the general states [itex]\Psi_p{}[/itex]. You can show that those states defined in this way are complete. Naturally, under this definition, the C matrix for U(L) is diagonalized. This is a nice way to simplify the analysis without loss of any generality.
Thanks for your reply! But it shouldn't be just a definition. When you give a definition, you should certify it. You should give a well-defined definition, instead of throwing a whole bunch of random chosen stuff into a box and saying that they are the same simply because you've thrown them in the same box.

Here, I'm not saying that Weinberg's definition is wrong, but want to know why we wouldn't lose generality in this way. Can you tell me explicitly? Thanks again!
 
  • #10
Bill_K
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The states only need to be linearly independent. Which they are.
 
  • #11


I am sort of convinced by my teacher. He reminded me of the fact that the common eigenstates of two commuting operators can be formed by direct product of the eigenstates for the two operators. In this way, we can linearly combine the eigenstates with fixed momentum but varying \sigma, since momentum and \sigma commute with each other. Then, we might be able to redefine the operator of \sigma such that U(L(p)) doesn't alter \sigma.

But still, there is a new problem, that is, can we really redefine the operator of \sigma? It appears to me that it's possible. I am not quite sure.
 

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