Asymptotic states in the Heisenberg and Schrödinger pictures

In summary: To sum up, the problem is that the asymptotic operators can be used to create momentum eigenstates, so why are they used to create non-momentum eigenstates?In summary, the quantity of interest in scattering theory is the amplitude for a system to evolve from a collection of momentum eigenstates to another collection of momentum eigenstates. This amplitude can be expressed in the Heisenberg picture as a product of creation and annihilation operators acting on the vacuum state. However, when examining the use of asymptotic operators in the literature, it becomes unclear why they are used to create non-momentum eigenstates instead of the expected momentum eigenstates. This contradiction requires further clarification.
  • #1
UnreliableObserver
4
4
In scattering theory, the quantity of interest is the amplitude for the system—initially prepared as a collection of (approximate) momentum eigenstates—to evolve into some other collection of momentum eigenstates. For example, for ##m\to n## scattering, the amplitude we're interested in is $$\bra{p_1\cdots p_n}U(t_f,t_i)\ket{k_1\cdots k_m},\tag*{(1)}$$ where ##t_i## is the time at which we prepare the state, ##t_f## is the time at which we observe the result of the scattering process, and ##U(t_2,t_1)## is the time evolution operator.

We can imagine finding operators ##a_p## capable of creating such states from the vacuum, in which case we could write ##(1)## as $$\bra{\Omega}a_{p_1}\cdots a_{p_n}U(t_f,t_i)a_{k_1}^\dagger\cdots a_{k_m}^\dagger\ket{\Omega},\tag*{(2)}$$ or, in the Heisenberg picture, $$\bra{\Omega}a_{p_1}(t_f)\cdots a_{p_n}(t_f) a_{k_1}^\dagger(t_i)\cdots a_{k_m}^\dagger(t_i)\ket{\Omega},\tag*{(3)}$$ where ##a_p(t)=U^\dagger (t, t_0)a_p U(t, t_0)##. I hope this is all rather uncontroversial so far.

Now, it can be shown (see e.g. Srednicki QFT) that the operators that create single-particle momentum eigenstates from the vacuum are ##(a_p^{\pm\infty})^\dagger=\lim_{t\to\pm\infty}a_p^\dagger(t)## where
$$a_p^\dagger(t)=\int{\rm d}^3x\,e^{-ipx}\left(\omega_{\boldsymbol{p}}\phi(x)-i\pi(x)\right).\tag*{(4)}$$ However after this point things start to fall apart for me. According to everything I've read, these operators are used to create the "initial" and "final" states ##\ket{i}=(a_{k_1}^{-\infty})^\dagger\cdots(a_{k_m}^{-\infty})^\dagger\ket{\Omega}## and ##\ket{f}=(a_{p_1}^{+\infty})^\dagger\cdots(a_{p_n}^{+\infty})^\dagger\ket{\Omega}##, with which the corresponding S-matrix element is given as ##\langle f|i\rangle##. But how can this be right? If these operators do indeed create single-particle momentum eigenstates, then isn't this inner product very trivial? If we refer back to the Heisenberg picture expression ##(3)##, we are free to write it as ##\langle f|i\rangle##, but these states are clearly not products of momentum eigenstates, but rather whatever complicated mess results from the action of ##a_p(t)## on the vacuum (as they must be to give a non-trivial amplitude). The products of eigenstates are those in ##(1)##, so shouldn't the operators ##(a_p^{\pm\infty})^\dagger## instead assume the role of those in ##(2)##?
 
Last edited:
Physics news on Phys.org
  • #2
First of all it should be
$$\hat{a}_{\text{H} p}(t)=\hat{U}(t,t_0) \hat{a}_{\text{S}p} \hat{U}^{\dagger}(t,t_0).$$
Then using ##\hat{U} |\Omega \rangle=|\Omega \rangle## you can see that indeed (2) and (3) are equivalent. Also note that the products are not trivial, because the annihilation operators, creating the final states out of the vacuum, are at time ##t_f \rightarrow \infty## and the creation operators, creating the initial states out of the vacuum, are at time ##t_i \rightarrow -\infty##.

It's indeed everything else than trivial to understand this subtle point of asymptotically free in and out states. Everything boils finally down to the socalled "LSZ-reduction formula" (named after Lehmann, Symanzik, and Zimmermann).

I think this point is best explained in the good old book by Bjorken and Drell, Quantum Field Theory (not the other one, named "relativistic quantum mechanics", which discusses a completely outdated version called "hole theory" a la Dirac; QFT is at the end more consistent and easier to understand).
 
  • Like
Likes dextercioby, PeroK and topsquark
  • #3
Thanks for the reply vanhees.
vanhees71 said:
First of all it should be
$$\hat{a}_{\text{H} p}(t)=\hat{U}(t,t_0) \hat{a}_{\text{S}p} \hat{U}^{\dagger}(t,t_0).$$
I don't think this is true. Using ##a_p(t)=U^\dagger (t, t_0)a_p U(t, t_0)##, ##(3)## can be written as $$\bra{\Omega}U^\dagger (t_f, t_0)a_{p_1} U(t_f, t_0)\cdots U^\dagger (t_f, t_0)a_{p_n}\underbrace{U(t_f, t_0) U^\dagger (t_i, t_0)}_{U(t_f, t_i)}a_{k_1}^\dagger U(t_i, t_0)\cdots U^\dagger (t_i, t_0)a_{k_m}^\dagger U(t_i, t_0)\ket{\Omega},$$ which recovers ##(2)##.

However, this isn't the part I'm concerned about.

Let me clarify. I'm perfectly happy with the progression from ##(1)##-##(3)##. That is, I accept that ##(2)## and ##(3)## are equivalent and are both non-trivial expressions. The issue I'm having is with reconciling them with how the asymptotic operators ##a_p^{\pm\infty}## are used in the literature. Here's the breakdown:
  1. In ##(2)##, the vectors ##a_p^\dagger|\Omega\rangle## are momentum eigenstates (by construction). The product of momentum eigenstates ##|k_1\cdots k_m\rangle## is evolved forward in time to ##t_f## and compared with ##|p_1\cdots p_n\rangle##. This inner product is clearly non-trivial.
  2. In ##(3)##, the vectors ##a_p(t)^\dagger|\Omega\rangle## are not momentum eigenstates, but in general some complicated states with a form that depends on the interactions. The inner product between ##a_{k_1}^\dagger(t_i)\cdots a_{k_m}^\dagger(t_i)|\Omega\rangle## and ##a_{p_1}^\dagger(t_f)\cdots a_{p_n}^\dagger(t_f)|\Omega\rangle## is clearly non-trivial and is equal to ##(2)##.
  3. The asymptotic operators ##(a_p^{\pm\infty})^\dagger## are the operators that create momentum eigenstates from the vacuum. That is, ##(a_p^{\pm\infty})^\dagger|\Omega\rangle## is a momentum eigenstate. The inner product between ##(a_{k_1}^{-\infty})^\dagger\cdots (a_{k_m}^{-\infty})^\dagger|\Omega\rangle=|k_1\cdots k_m\rangle## and ##(a_{p_1}^{+\infty})^\dagger\cdots (a_{p_n}^{+\infty})^\dagger|\Omega\rangle=|p_1\cdots p_n\rangle## therefore is trivial. The operators act like those in ##(2)## but are used like those in ##(3)##! What gives?
 
  • #4
As I said, it's not that simple. Look up the LSZ reduction formalism. As I said, it's most clearly explained in the QFT book by Bjorken and Drell.
 
  • #5
UnreliableObserver said:
... but these states are clearly not products of momentum eigenstates, but rather whatever complicated mess results ...
You should distinguish the final state before measurement, from the final state after the measurement. The final state before measurement is indeed a mess, as you said. This mess can be written as a messy superposition of momentum eigenstates, because the momentum eigenstates constitute a (Fock) basis. But when we measure, what we observe is a definite number of particles with some definite momenta. In other words, we measure in the Fock basis of momentum eigenstates, so the state after the measurement is one of those Fock states. Hence the state after the measurement is no longer a mess at all. But we do not know in advance which particular non-messy state will this be, so we compute the probability of ending up in any particular non-messy state. The S-matrix element is precisely the probability amplitude that the messy state before measurement will end up in a certain non-messy state after the measurement.

With equations, the final state before measurement is ##S|i\rangle##, which is a mess when written explicitly. Denoting the final state after the measurement with ##|f\rangle##, the probability amplitude that the final state after the measurement will be ##|f\rangle## is
$$\langle f|S|i\rangle$$

A note on notation: Sometimes people denote ##S|i\rangle## with ##|i(\infty)\rangle## or something like that, write the probability amplitude above as ##\langle f |i(\infty)\rangle##, and then further shorten the notation by writing it just as ##\langle f |i\rangle##. Such a shorter notation is rather confusing and misleading.
 
Last edited:
  • Like
Likes PeroK
  • #6
Hi Demystifier. I agree with everything you've said, however the states ##S|i\rangle## and ##|f\rangle## you refer to are the states ##U(t_f,t_i)|k_1\cdots k_m\rangle## and ##|p_1\cdots p_n\rangle## I wrote down in the Schrodinger-picture expression ##(1)##. These are different to the "messy" Heisenberg picture states, ##a_{k_1}^\dagger(t_i)\cdots a_{k_m}^\dagger(t_i)|\Omega\rangle## and ##a_{p_1}^\dagger(t_f)\cdots a_{p_n}^\dagger(t_f)|\Omega\rangle##, that I refer to in the final paragraph. My confusion concerns the fact that the asymptotic operators ##(a_p^{\pm\infty})^\dagger## are said to create momentum eigenstates, which would make the inner product between ##(a_{k_1}^{-\infty})^\dagger\cdots (a_{k_m}^{-\infty})^\dagger|\Omega\rangle=|k_1\cdots k_m\rangle## and ##(a_{p_1}^{+\infty})^\dagger\cdots (a_{p_n}^{+\infty})^\dagger|\Omega\rangle=|p_1\cdots p_n\rangle## simply an inner product between collections of momentum eigenstates, which is very trivial (a delta function). Yet, in every QFT text I've come across, this inner product is said to be the relevant ##S##-matrix element.

vanhees has suggested things are more subtle than this, so I will give Bjorken and Drell a read. From skimming ##S##-matrix chapter, it seems that I might be failing to distinguish between the states created by ##(a_p^{-\infty})^\dagger## (the "in" states) and those created by ##(a_p^{+\infty})^\dagger## (the "out" states). Even though the inner product between a pair of "in" states or a pair of "out" states is trivial in the sense I described, the inner product between an "in" state and and "out" state is (somehow) the non-trivial ##S##-matrix element.

I'll report back 🙂
 
  • Like
Likes vanhees71, Demystifier and PeroK
  • #7
UnreliableObserver said:
##\ket{i}=(a_{k_1}^{-\infty})^\dagger\cdots(a_{k_m}^{-\infty})^\dagger\ket{\Omega}## and ##\ket{f}=(a_{p_1}^{+\infty})^\dagger\cdots(a_{p_n}^{+\infty})^\dagger\ket{\Omega}##, with which the corresponding S-matrix element is given as ##\langle f|i\rangle##. But how can this be right? If these operators do indeed create single-particle momentum eigenstates, then isn't this inner product very trivial?
The point is that ##(a_{p_1}^{+\infty})^\dagger## do not create single-particle momentum eigenstates. Only ##(a_{p_1}^{-\infty})^\dagger## do, and ##a_{p_1}^{-\infty} \neq a_{p_1}^{+\infty}##.

But ##(a_{p_1}^{+\infty})^\dagger## and ##a_{p_1}^{+\infty}## satisfy the same creation and destruction algebra, as ##(a_{p_1}^{-\infty})^\dagger## and ##a_{p_1}^{-\infty}##. So what do ##(a_{p_1}^{+\infty})^\dagger## create? In the condensed matter language, I would say that they create "quasi-particles", rather than physical particles. What exactly is the difference? The world of these "quasi-particles" alone looks the same as the world of "physical particles" alone. What distinguishes them is their properties relative to our measuring apparatuses. Our measuring apparatuses are made such that only states created by ##(a_{p_1}^{-\infty})^\dagger## manifest as particles in the way we are used to. In principle one could construct a different measuring apparatus with which only states created by ##(a_{p_1}^{+\infty})^\dagger## manifest as particles in the way we are used to, but in practice we don't work with such measuring apparatuses. An example of a "different measuring apparatus" that sees "quasi-particles" as particles we are used to is an accelerating particle detector, this effect is known as Unruh effect.
 
  • #8
UnreliableObserver said:
##(a_{k_1}^{-\infty})^\dagger\cdots (a_{k_m}^{-\infty})^\dagger|\Omega\rangle=|k_1\cdots k_m\rangle## and ##(a_{p_1}^{+\infty})^\dagger\cdots (a_{p_n}^{+\infty})^\dagger|\Omega\rangle=|p_1\cdots p_n\rangle##
In view of my post above, the first formula is correct but the second isn't. Perhaps the second formula could be written more correctly as
$$(a_{p_1}^{+\infty})^\dagger\cdots (a_{p_n}^{+\infty})^\dagger|\Omega\rangle=|\bar{p}_1\cdots \bar{p}_n\rangle$$
where the bar denotes that these states are "quasi-particles", rather than physical particles.
 
Last edited:
  • #9
Finally getting back around to this :)

So it seems that was indeed it. The problem was that I was writing down all these momentum eigenstates without paying attention to which momentum operators they are actually eigenstates of!

The operators ##(a_p^{\pm\infty})^\dagger## don't just "create momentum eigenstates", they specifically create eigenstates of the asymptotic momentum operators ##\hat{p}(\pm\infty)=U^\dagger(\pm\infty,t_0)\,\hat{p}\,U(\pm\infty,t_0)##. Likewise, the ##a_p^\dagger## create eigenstates of the Schrödinger-picture operator ##\hat{p}##. My problem was implicitly thinking of ##|i\rangle## and ##|f\rangle## as collections of eigenstates of the same Schrödinger-picture operator ##\hat{p}##, when they are really eigenstates of two different operators (##\hat{p}(-\infty)## and ##\hat{p}(\infty)##), making their inner product very non-trivial indeed.

This also makes it clear how the fixed Heisenberg picture states ##|i\rangle## and ##|f\rangle## can be interpreted as particles at asymptotic times. It is obvious in the Schrödinger picture, where one explicitly evolves the state—initially in a product of eigenstates of the (fixed) operator ##\hat{p}## at ##t=-\infty##—to ##t=+\infty##. In the Heisenberg picture, it is the operators that undergo the evolution, and the (fixed) state is a product of eigenstates of the momentum operator only at the specific times ##t\pm\infty##. So the pictures are equivalent.

It seems quite simple in hindsight. Neverthless, do let me know whether or not this all makes sense!
 
  • Like
Likes vanhees71

Similar threads

  • Quantum Physics
Replies
2
Views
735
Replies
1
Views
1K
Replies
1
Views
761
  • Quantum Physics
Replies
2
Views
896
  • Quantum Physics
Replies
6
Views
872
Replies
16
Views
1K
Replies
1
Views
1K
  • Quantum Physics
Replies
1
Views
1K
  • Quantum Physics
Replies
5
Views
2K
Replies
10
Views
2K
Back
Top