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I am looking into the section of the book by Peskin and Schroeder in which they connect the $S$-matrix to probabilities.

They start by considering the in state, which is a two-particle state \begin{equation} |\phi_{A}\phi_{B}\rangle_{in} \equiv \int \frac{d^3 k_A}{(2\pi)^3}\frac{1}{\sqrt{2E_{k_A}}}\phi_A(k_A) \int \frac{d^3 k_B}{(2\pi)^3}\frac{1}{\sqrt{2E_{k_B}}}\phi_B(k_B)|k_Ak_B\rangle_{in}. \tag{4.68} \end{equation} As one can see, it is not a product of momentum eigenstates, but it is smeared over momentum space with some functions.

Instead, the out state is considered to be the state with definite momenta of particles \begin{equation} {}_{out}\langle p_1p_2 \dots p_n|. \end{equation} Eventually, they compute the probability density for the scattering process as \begin{equation} {\cal P}({A} {B}\to 1,2,\dots, n)=\prod_{f=1}^n \frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f}|{}_{out}\langle p_1p_2 \dots p_n| \phi_{A}\phi_{B}\rangle_{in}|^2. \tag{4.74} \end{equation}

Question. Why is this formula so asymmetric with respect to the change of in and out particles? Why do in particles feature as wave packets, while out particles enter as eigenstates of momentum? P$\&$S explain this by, essentially, saying that this is how the design of a collider works. That is, the incoming particles are the beam that hits the target, so they are wave packets, while detectors can measure particles with definite momentum. Is this a real reason? What are the conditions on the wave packets then except that they are normalised? It seems that P$\&$S suggest that these are simultaneously localised in space, at least in the transverse direction and, at the same time, have a sharp peak at a given momentum. It seems that these two things are incompatible.

My guess. I guess, that in reality the problem that one encounters is that unless the in states are integrated against test functions, then (4.74) has a product of momentum conserving delta functions. This is why amplitudes where integrated against test functions. I guess, the same could have been done with out particles, just this is not necessary, as the issue with delta functions can be resolved smearing in states only.

What is the actual reason?

Highlighting the mathematical issue. What I've said in words and what may need a clarification is the following. If we just consider the scattering probability for eigenstates of momenta, we will end up having a factor \begin{equation} |{\cal M}|^2\Big(\delta^4(\sum p_{in}- \sum p_{out})\Big)^2. \end{equation} So we get an issue of $(\delta(x))^2$ type. Having wave packets instead of eigenstates of momentum resolves it as the formula above gets replaces with something like \begin{equation} \int |{\cal M}|^2\delta^4(\sum p_{in}- \sum p_{out})\delta^4(\sum p'_{in}- \sum p_{out}) dp_{in} dp'_{in}\phi(p_{in})\phi^*(p'_{in}) \end{equation} in the sense that this expression becomes finite and well-defined.

What I, probably, want. I think, it makes sense to expect something symmetric in ingoing and outgoing particles. For example, can one write a formula of the form \begin{equation} {\cal P}({A} {B}\to 1,2,\dots, n)=\prod_{f=1}^n \frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f} \frac{d^3p_A}{(2\pi)^3}\frac{1}{2E_A} \frac{d^3p_B}{(2\pi)^3}\frac{1}{2E_B} |{}_{out}\langle p_1p_2 \dots p_n| p_{A}p_{B}\rangle_{in}|^2 \times (\dots) . \end{equation} The last formula definitely does not exist because of the issue with the Dirac delta squared. Is there anything reminiscent of it with test functions involved?

Edit 1.

Ok, I think it is clear why one integrates over final states, but not over initial states. The reason is simply that the initial state is assumed to be known and fixed, while the final state is known only probabilistically and it is this probability that we compute. It is the probability density not the probability because the space of potential final states is continuous/infinite-dimensional. If one considers the reverse process (with $p_1,\dots, p_n$ incoming and measures $p_{\cal A}$ and $p_{\cal B}$) then the probability density computed needs to be integrated over $p_{\cal A}$ and $p_{\cal B}$. So, the whole asymmetry on that part comes from the fact that the problem is asymmetric with respect to in and out states.

Still, the question why we actually need to smear the initial states as in (4.68) is not entirely settled for me.

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    $\begingroup$ It is much easier to think about "somewhat localized" states because a thorough treatment of plane-wave-in to plane-wave-out necessarily leads to infinite results, both for mathematical as well as physical reasons. It's not just an artifact of poorly chosen measures. We are simply running up against the problem that, no matter how weak the interaction in the bulk, a beam of infinite volume will always experience an infinite number of scattering events. Some authors are trying to bypass that with these kind of approximations. You decide if that "does it for you". $\endgroup$ Commented Apr 4, 2023 at 17:14
  • $\begingroup$ yes. I understand this. Indeed, if we scatter only definite momenta states, then these states are all around the space and one cannot say that they are far from each other, hence, interactions cannot be ignored. So, this is the issue with defining what asymptotic states are. But this should equally refer both to in and out particles. Not mentioning that I would like to have something more precise than just words, e. g. a packet here, a packet there. Something that actually ensures that particles are separated. $\endgroup$
    – Dr.Yoma
    Commented Apr 4, 2023 at 17:38
  • $\begingroup$ besides that, it feels like the issue here is not so much about defining the asymptotic states. The thing is that if you just consider all states with definite momenta you get $\delta^4(p_{in}-p_{out})$ squared. This -- $(\delta(x))^2$ makes no sense. So, I guess, the prime reason to consider packets is to resolve this issue with distributions. $\endgroup$
    – Dr.Yoma
    Commented Apr 4, 2023 at 17:44
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    $\begingroup$ I believe, this is a commonly acknowledged thing that amplitudes/correlators etc are distributions and they only make sense when integrated against something. So, I guess, this procedure with wave packets is some form of implementation of a general procedure of dealing with distributions. $\endgroup$
    – Dr.Yoma
    Commented Apr 4, 2023 at 17:44
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    $\begingroup$ Maybe you could resolve the problem this way: imagine using a theory without interaction in most of the bulk and then turning on the interaction terms in a well defined volume of e.g. radius R (maybe use a Gaussian "bubble"?). So now there should be an asymptotic behavior of the scattering against R, which is irrelevant, while all the physics stays well defined and finite for any finite R. Would this be an acceptable physical "measure"? $\endgroup$ Commented Apr 4, 2023 at 17:52

2 Answers 2

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I do not have a copy of the Peskin and Schroeder at hand, but just from what you mentioned, I would answer the following:

  • While the expressions in (4.68) could be interpreted as wave packets in momentum space, they can also be seen as a statistical ensemble describing a beam in which particles of type $A$ appear in the beam in a momentum interval $(k_A, k_A + d k_A)$ with the probability $\phi_A(k_A) dk_A$ and analogously for type $B$ particles. Then $|\phi_A \phi_B \rangle_{\text{in}}$ is just the average inbound state, which is exactly what you need for calculating the cross section for some final state.
  • The probabilities for all final states you can detect add up, which is why in general one sums over all final states and averages over all initial states when calculating scattering cross sections. Notice that in (4.74) there appears the measure $\prod_{f=1}^n d^3 p_f$, so this equation describes, as you correctly mentioned, a probability density. If you wanted to get a probability, you would integrate ("sum infinitely many small contributions") over that measure, so the final state kind of implicitly contains an integral just like the initial one. Only the distribution functions are absent, which is due to the sum vs. average consideration.
  • From the first point there follows that the condition on $\phi_A$ and $\phi_B$ is just that they are valid probability distribution functions. To get meaningful results, they should also describe the real momentum distribution in the beam of the collider you are interested in. These are, to my knowledge, usually determined experimentally.

Remarks:

  • The result (4.74) depends on the full form of the probability densities $\phi_A$ and $\phi_B$. Namely, they enter as part of $|\phi_A\phi_B \rangle_{\text{in}}$, which is, mathematically speaking, a functional of these two functions.
  • Under "My guess." it is suggested in the OP that the probability densities are just a means of regularisation. If one wants to calculate a physical process happening in some particle collider, this is certainly not the case, because the densities are part of the modelling of physical reality. If instead an interaction of momentum eigenstates should be considered, there will of course be no probability densities as part of the physical modelling. Any arising singularities in this case will have to be considered carefully with respect to their physical meaning, so that it can be decided if regularisation is needed and by what means it should be done. An unmotivated ad-hoc procedure of arbitrarily adding some fixed densities will, as far as I can tell, likely lead to meaningless results.
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  • $\begingroup$ So, you, essentially, say that it is done this way, because this is the way the experiment is designed. This is what Pesking and Schroeder say as well. But I do not find this very convincing. One reason is that, eventually the result that they give does not depend on the profiles of packets -- the only thing that matters that they are localised around fixed momenta (no specifics). Another reason is that if you do the computation without this trick, you get the problem with $\delta^2(x)$. Thus, I get more and more convinced that this $\delta^2(x)$ is the actual reason $\endgroup$
    – Dr.Yoma
    Commented Apr 8, 2023 at 22:11
  • $\begingroup$ @Dr.Yoma My reply to the two reasons you stated would not fit in a comment, so I edited the answer instead, adding two remarks. Sorry if they do not read like a direct answer to your comment, but I thought the answer should be understandable for future readers without the comments. $\endgroup$
    – sim0
    Commented Apr 10, 2023 at 10:01
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In the following, I will discuss the necessity of wave packets from the standpoint of semi-axiomatic quantum field theory. (I will cite Duncan as a reference.)

In the following explanation, essentially the following requirements are used:

  1. Existence of the interacting vacuum $|\Omega\rangle$
  2. Existence of asymptotic states and asymptotic fields. Asymptotic states are the state in Fock space constructed by asymptotic fields.
  3. Asymptotic fields are free field and in-state and out-state is assumed to construct the same Fock space. (Sometimes the second and third requirements are called as the asymptotic completeness.)

More axiomic point of view, an uniqueness of the vacuum is also added to the requirement: $|\Omega\rangle=|0\rangle_{\mathrm{in}}= |0\rangle_{\mathrm{out}}, $ but here I don’t add it.

This answer become too long. So I summarize the point.

  1. To introduce wave packets and normalize states is not necessary, but sufficient to show asymptotic relation (see below):$$\langle\alpha|\phi_{\mathrm{in}/\mathrm{out}}(x)|\beta\rangle{=}\lim_{t\rightarrow\pm\infty}\frac{1}{\sqrt{Z}} \langle\alpha|\phi(x)|\beta\rangle.$$
  2. The asymptotic relation above is satisfied only in the sense of weak-limit. Especially, the operator relation $\phi_{\mathrm{in}/\mathrm{out}}(x){=}\lim_{t\rightarrow\pm\infty}\frac{\phi(x)}{\sqrt{Z}}$ leads us contradiction.(see below)
  3. If we don’t introduce the wave packets, strictly speaking there are two problems. One is a problem about weak-limit. Weak limit, which we introduce to construct the well-defined asymptotic relation (see-below), is defined through the normalized states, so it is sufficient to introduce wave packets. The second reason is that to use the Riemann-Lebesgue lemma, which plays an important role when we show the asymptotic relation above, wave packets is also important to make the integrand well-integrable.
  4. More rigorously, if we want to prove the LSZ in a mathematically correct way, we need to introduce wave packets for both the in- and out- states. Otherwise, our Fock space would not consist of normalized states.

Answer

Firstly, let us define asymptotic fields by a following relation: $$\phi_{\mathrm{in}/\mathrm{out}}(x)\overset{?}{=}\lim_{t\rightarrow\pm\infty}\frac{\phi(x)}{\sqrt{Z}},$$ where $Z$ is the wave-function renormalization constant and $\phi(x)$ at r.h.s. is the interacting scalar field, which satisfies the following equal-time commutation relation: $$[\phi _{\mathrm{in}/\mathrm{out}}(t,\mathbf{x}), \dot\phi _{\mathrm{in}/\mathrm{out}}(t, \mathbf{y})]=i\delta^3(\mathbf x-\mathbf y)$$

However this definition does not work. Let us show this. Let’s consider the following commutator: $$\lim_{t\rightarrow\pm\infty}[\phi(t,\mathbf{x}), \dot\phi(t, \mathbf{y})]=Z [\phi _{\mathrm{in}/\mathrm{out}}(t,\mathbf{x}), \dot\phi _{\mathrm{in}/\mathrm{out}}(t, \mathbf{y})]=Zi\delta^3(\mathbf x -\mathbf y),$$ but this must be equal to the second equation. But this is contradict to the fact that $Z<1$ in the interacting system. So the first equation is ill-defined definition.

In fact, let us impose the asymptotic relation (I mean the first equation) as a limit of following expectation value (this is weak limit in mathematical sense): $$\langle\alpha|\phi_{\mathrm{in}/\mathrm{out}}(x)|\beta\rangle{=}\lim_{t\rightarrow\pm\infty}\frac{1}{\sqrt{Z}} \langle\alpha|\phi(x)|\beta\rangle, $$ where $|\alpha\rangle$ and $|\beta\rangle$ are arbitrary normalized states.

Next, let’s introduce the field localized by wave packets: $$a_g^\dagger(t,\mathbf p)=-i\int d^3 x g_p(x)\overset{\leftrightarrow}{\partial_0}\phi(x),$$ $$a_{g,\ \mathrm{in}/\mathrm{out}}^\dagger(\mathbf p)=-i\int d^3 x g_p( x)\overset{\leftrightarrow}{\partial_0}\phi _{\mathrm{in}/\mathrm{out}}(x),$$ $$g_p(x)=\int \frac{d^3 p’}{(2\pi)^32E_\mathbf{p’}} \frac{1}{(\sqrt{(2\pi)}\ \delta)^{3/2}}e^{-\frac{(\mathbf p-\mathbf p’)^2}{(2\delta)^2}}e^{-i(E_\mathbf{p’} t-\mathbf p\cdot\mathbf x)},$$ where $g_p$ is a wave packet. To satisfy the asymptotic relation, the following relation may be sufficient: $$\langle\alpha| a_{g,\ \mathrm{in}/\mathrm{out}}^\dagger(\mathbf p) |\beta\rangle{=}\lim_{t\rightarrow\pm\infty}\frac{1}{\sqrt{Z}} \langle\alpha| a_{g}^\dagger(t,\mathbf p) |\beta\rangle$$

What we should do is to check the equation above is satisfied if we use the well-normalized state by wave packets. It is possible to show this relation exactly, but it is waste of a paper, so I give one example: inner product of single-particle and two-particle states. Of course, this amplitude must be vanish. We will check it and how wave packets work in this calculation.

Let us prepare a normalizable 2-particle state: $$|\beta\rangle:=\int d^3 P f(\mathbf P)|\mathbf P, \mathbf p\rangle,$$ where $\mathbf P$ is the CoM momentum, and $\mathbf p$ is the relative momentum. By simple calculation, we can show $$\langle\beta| a_g^\dagger(t,\mathbf p) |0\rangle=\int \frac{d^3 \mathbf P }{2E_\mathbf P}f^*(\mathbf P) \frac{1}{(\sqrt{(2\pi)}\ \delta)^{3/2}}e^{-\frac{(\mathbf p-\mathbf P)^2}{(2\delta)^2}}e^{i(P^0-\sqrt{\mathbf P^2+m^2})t}$$ $$\times(P^0+\sqrt{\mathbf P^2+m^2})\langle\mathbf P,\mathbf p|\phi(0)|0\rangle, $$ where $P^0=\sqrt{\mathbf{P}^2+(2m)^2}\geq \sqrt{\mathbf P^2+m^2}. $

Because our state vectors are all well-normalized so this integrand without $e^{i(P^0-\sqrt{\mathbf P^2+m^2})t}$ factor is integrable, then we can use Riemann-Lebesgue lemma, when taking the limit $t\rightarrow \pm\infty$, and we obtain $\langle\beta| a_g^\dagger(t,\mathbf p) |0\rangle \rightarrow 0$ as expected.

Here, wave packets plays important role because if we don’t introduce the wave packets, $a_g^\dagger(t,\mathbf p) |0\rangle$ is no more normalizable in mathematical sense and we can’t use the Riemann-Lebesgue lemma. If so, we cannot show the asymptotic relation $\langle\alpha|\phi_{\mathrm{in}/\mathrm{out}}(x)|\beta\rangle{=}\lim_{t\rightarrow\pm\infty}\frac{1}{\sqrt{Z}} \langle\alpha|\phi(x)|\beta\rangle$.

That’s why we introduce wave-packets.

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  • $\begingroup$ Thank you for a reply and for a reference. I am still digesting them :) $\endgroup$
    – Dr.Yoma
    Commented Apr 11, 2023 at 14:09

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