• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

146 submissions , 123 unreviewed
3,953 questions , 1,403 unanswered
4,889 answers , 20,762 comments
1,470 users with positive rep
507 active unimported users
More ...

Why is there a time dependence in the Heisenberg states of the Haag-Ruelle scattering theory?

+ 7 like - 0 dislike

I'm reading R. Haag's famous book "Local Quantum Physics: Fields, Particles, Algebras", 2nd edition, and I'm very puzzled by the way he treats the Heisenberg picture in the Haag-Ruelle scattering theory. It begins in section "II.3 Physical Interpretation in Terms of Particles", where, on page 76, he clearly states "Our description is in the Heisenberg picture. So $\Psi_{i\alpha}$ describes the state "sub specie aeternitatis"; we may assign to it, as in (I.3.29), a wave function in space-time obeying the Klein-Gordon equation."

Then, on page 77, he says: "Suppose the state vectors $\Psi_1$, $\Psi_2$ describe states which at some particular time $t$ are localized in separated space regions $V_1$, $V_2$." From here on the whole construction begins.

I would very much appreciate it if an expert in Haag-Ruelle scattering or whoever knows the answer, would answer my question as to why a state vector in the Heisenberg picture like $\Psi_1$ and $\Psi_2$ above depends on time, when it's common knowledge that there is no time dependence assigned to the state vectors in the the Heisenberg picture?

EDIT 1: Up until recently I didn't even know how a scattering process might be described in the Heisenberg picture of QM, since once the initial state is prepared at $t_i = - \infty$ , this state will remain unchanged for all time and it will be the same for $t_f = + \infty$, and hence there could be no scattering (let alone particle production, 3-body scattering, rearangement collisions, etc.). How to solve this problem? Then I have discovered one of the most lucid presentations in the paper of H. Ekstein, "Scattering in field theory", http://link.springer.com/article/10.1007/BF02745471

The basic idea is the following: one prepares a state of the system at $t_i = -\infty$ by measuring a complete set of compatible observables represented by operators in the Heisenberg picture (i.e., time dependent), say $A(t_{i}), B(t_{i})$, etc. Obviously, this prepared state is a common eigenvector of these operators, say $|a,b,...; t_{i}\rangle$ corresponding to the eigenvalues (obtained in measurement) $a, b$,.... , i.e., $A(t_{i})|a,b,...; t_{i}\rangle = a|a,b,...; t_{i}\rangle, B(t_{i})|a,b,...; t_{i}\rangle = b|a,b,...;t_{i}\rangle$, etc.

Then, one lets the system evolve from $t_i = -\infty$ to $t_f = +\infty$. Obviously, the state vector of the system remains unchanged, namely $|a,b,...; t_{i}\rangle$ for any time $t$, with $t_i \leq t \leq t_f$, since we are in the Heisenberg picture, but the operators representing dynamical observables do change in time according to the Heisenberg equation of motion.

At time $t_f = +\infty$, one measures again the system choosing a complete set of compatible observables, say $C(t_{f}), D(t_{f})$,.... As a result of this measurement, the state of the system changes, at time $t = t_f$, from $|a,b,...; t_{i}\rangle$ to $|c,d,...; t_{f}\rangle$, where $|c,d,...; t_{f}\rangle$ is a common eigenvector of the operators $C(t_{f}), D(t_{f})$,..., corresponding to the eigenvalues $c, d,$... obtained in the measurement (at time $t = t_f$), i.e. $C(t_{f})|c,d,....; t_{f}\rangle = c|c,d,....; t_{f}\rangle, D(t_{f})|c,d,....; t_{f}\rangle = d|c,d,....; t_{f}\rangle$, etc.

The quantity of interest is the transition amplitude from the Heisenberg state $|a,b,...; t_{i}\rangle$ to the Heisenberg state $|c,d,...; t_{f}\rangle$, and this is given by the S-matrix element $S_{a,b,...; c,d,...} = \langle c,d,...; t_{f}| a,b,...; t_{i}\rangle$.

To summarize: the key to understanding scattering in either the Schrodinger or Heisenberg picture is to realize that it implies 2 experimental operations, namely preparation at $t = t_i$ and measurement at $t = t_f$.

A logical approach to solving a scattering problem in the Heisenberg picture (as presented by Ekstein) is the following:

  • H0) For any given observable solve the Heisenberg equation of motion to find its dependence on time, i.e. the operator $A(t)$.
  • H1) For any Heisenberg operator (representing an observable) $A(t)$ find the asymptotic values $A_i = \lim_{t \rightarrow -\infty} A(t)$ and $A_f = \lim_{t \rightarrow +\infty} A(t)$
  • H2) Solve the eigenvalue problem for the asymptotic operators $A_i$ and $A_f$. The eigenvectors are the corresponding asymptotic scattering states.
  • H3) Select a complete system of compatible observables (CSCO) that corresponds to state preparation at $t = t_i$, denoted generically by $A_i$. Select a CSCO that corresponds to measurement at $t = t_f$, denoted generically by $C_f$.
  • H4) Calculate matrix elements between eigenvectors determined in step H2), namely $\langle c, t_{f}| a, t_{i}\rangle$, where $|a, t_{i}\rangle$ is an eigenvector of $A_i = A(t_{i})$, and $|c, t_{f}\rangle$ is an eigenvector of $C_f = C(t_{f})$.

Regarding the Haag-Ruelle scattering, things are very confusing. The main argument is the same in all the books available. Instead of following the very logical steps H1)-H4) presented above, one starts by constructing a vector depending on a parameter $"t"$ and shows that this vector has limits when $|t|$ becomes infinite. I must say that this type of reasoning is reminiscent of the way one treats scattering in the Schrodinger picture (SP). In the SP, one starts with an arbitrary state vector $|\Psi (t)\rangle$ which is time dependent according to the SP and then must show that $|\Psi (t)\rangle$ has asymptotes when (the real time) $|t|$ becomes infinite.

I would be very grateful if you could help me with some answers to these questions:

  • 1) What is the relation between the parameter $"t"$ of H-R scattering and the real time, since when $"t"$ becomes infinite they claim to have obtained the asymptotic scattering states?
  • 2) What is the physical interpretation of the vectors $\psi_t$ in H-R scattering? Are they obtained as a result of a measurement? Are they in the Heisenberg picture or in the Schrodinger picture?
  • 3) Is there a CSCO such that the H-R asymptotic scattering states are the eigenvectors of this CSCO? If yes, is this CSCO the asymptotic limit of a finite time Heisenberg CSCO, as described in steps H1)-H4)?
  • 4) Can one obtain asymptotic scattering states for an ARBITRARY CSCO using the H-R method? This should be the case since one can prepare the initial state as one wants at $t = t_i$, and then can choose to measure what observable one wants at $t = t_f$, and hence the CSCOs corresponding to preparation and measurement must be arbitrary.

EDIT 2: @Pedro Ribeiro Your objections to Ekstein's construction are perhaps unfounded:

  • I chose a discrete spectrum for CSCOs in my presentation from EDIT 1 only to convey the general idea with minimum notation. In case of a continuous spectrum one can use spectral projection operators as per von Neumann's QM.
  • A Heisenberg operator $A(t)$ acts in the full Hilbert space, i.e. in the same Hilbert space on which the total Hamiltonian $H$ acts. The Haag theorem has to do with the fact that the free Hamiltonian $H_0$ and the full Hamiltonian $H$ act on 2 different Hilbert spaces. There is no connection between $A(t)$ and $H_0$ or its associated Hilbert space for any time $t$, finite or infinite. Hence, Haag's theorem has no bearing on $\lim_{t \rightarrow \pm\infty} A(t)$ and hence does not forbid the existence of this limit. Examples: If $A(t)$ commutes with $H$, then $A(t)$ is constant in time and the limit surely exists (see, e.g., the momentum operator). As a matter of fact, the whole LSZ idea is based on such limits!

It's only one way a state can depend on time $t$ in the Heisenberg picture. That time $t$ has to be a time at which some Heisenberg operator, say $A(t)$, is measured on the system, and as an effect the state becomes an eigenvector $|a,t\rangle$ of that operator. Otherwise, state vectors in the Heisenberg picture do not evolve dynamically in time! One can look at my post.

From your presentation is still not very clear if the parameter $"t"$ is the time at which one chooses to measure a CSCO on the system and obtains an eigenvector(?) $\psi_t$. For that one has to construct such a Heisenberg CSCO and show that $\psi_t$ is its eigenvector (corresponding to some eigenvalue) at time $t$. Can one show that?

In the meantime I've discovered some lecture notes by Haag published in Lectures in theoretical physics, Volume III, edited by Brittin and Downs, Interscience Publishers. Starting on page 343 Haag discusses his theory and in his own words says very clearly that the $\psi_t$ states are manifestly in the Schrodinger picture, and $t$ is regular time. Only the asymptotic limits of $\psi_t$ Haag considers to represent scattering states in the Heisenberg picture. But even that cannot work since $\psi_t$ has 2 limits, $\psi_{\pm} = \lim_{t\rightarrow\pm\infty}\psi_t$, and hence one needs 2 different Heisenberg pictures, one that coincides with the Schrodinger picture at $t = -\infty$, and a 2nd one, which coincides with the Schrodinger picture at $t = +\infty$. So, he doesn't stay all the time in the Heisenberg picture, but uses most of the time the Schrodinger picture, and in the end, apparently, 2 different Heisenberg pictures. However, it's well known that the Schrodinger picture does not exist in relativistic qft due to vacuum polarization effects!!! What is left of Haag-Ruelle theory, then???

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
asked Aug 16, 2016 in Theoretical Physics by Andrea Becker (35 points) [ no revision ]

3 Answers

+ 5 like - 0 dislike

The "Heisenberg picture" referred to in page 76 of Haag's book applies to the single-particle Hilbert space $\mathcal{H}^{(1)}$ and therefore to the "in" and "out" Hilbert spaces only, that is, at times $t\to\pm\infty$ respectively. The discussion on page 77, on its turn, refers to states in the interacting (Wightman-GNS) Hilbert space. In this respect, it must be remarked that the discussion on page 77 (particularly formulae (II.3.3) and (II.3.4)) is not very precise - what Haag really means is the content of Theorem 4.2.1, pp. 88, as (EDIT) I shall explain in more detail below.

The recent expansion of your question made clearer the issues which are troubling you. Firstly, there are a few points to address before assessing your questions 1)-4) in a conceptually rigorous way:

  • It seems you are dealing with observables with a pure point spectrum only. Most observables are not of this kind - points in the continuous part of the spectrum are not eigenvalues in the sense they have corresponding eigenvectors. They do have what is called corresponding "generalized eigenvectors", which strictly speaking are not contained in the Hilbert space.

  • In QFT, the limits $\lim_{t\to\pm\infty}A(t)$ usually do not exist in the operator sense for the relevant observables. This is mainly due to Haag's theorem, the same which tells us that there is no interaction picture in QFT. That is the technical reason why the time parameter $t$ must appear in state vectors, since the asymptotic limit can only be approached by applying $A(t)$ first to some state (namely, the vacuum state).

The above points show that making steps H1)-H2) rigorous (specially in the context of QFT) is quite problematic. H3)-H4), on the other hand, are not that far off.

Secondly, I want to stress a few conceptual points about Haag-Ruelle scattering theory. I do so at the risk of being a bit pedantic, but I want to set a precise context in a self-contained manner. Recall that the Haag-Ruelle theory is a scattering framework for quantum field theories. Regardless of whether you work with Wightman fields or a Haag-Kastler net of C*-algebras, this means that all (smeared) fields and all (local) observables are thought of as being localized in a certain region of space-time, in the sense of relativistic microcausality: observables localized in causally disjoint space-time regions should commute (for smeared fields, they either commute or anti-commute, depending on their spin). This is radically different from (non-relativistic) quantum mechanics. Particularly, any given local observable should be thought of being measured within a certain region of space and within a certain interval of time. A "sharp time" localization for observables is only possible for free fields, which of course have a trivial scattering theory. In other words, local observables and smeared fields in QFT are always in the Heisenberg picture, but their time localization is usually not "sharp".

Given any local observable or smeared field polynomial $A$ localized in a space-time region $\mathscr{O}$, the effect of time translations (using the unitary time evolution of the theory) simply has the effect of translating the localization region $\mathscr{O}$ of $A$ in time - more precisely, the localization region of $A(t)$ is $$\mathscr{O}_t=\{(x^0+t,x^1,x^2,x^3)\ |\ (x^0,x^1,x^2,x^3)\in\mathscr{O}\}\ .$$ More generally, if $U(x)$ is the unitary operator implementing the space-time translation by $x=(t,\mathbf{x})$, then $A_x=U(x)AU(x)^*$ is localized in $\mathscr{O}+x$ (so that $\mathscr{O}+(t,\mathbf{0})=\mathscr{O}_t$). This (I hope) answers your question 1).

However, the input and output of an scattering experiment are about large times and large distances away from the scattering center, so it is more appropriate to talk about momentum localization when dealing with scattering states. For constructing the latter, we need local observables or smeared field polynomials with a nonzero transition amplitude between the vacuum state and a one-particle subspace with mass (say) $m>0$, whose existence is one of the assumptions of the Haag-Ruelle theory. Such operators exist thanks to the Reeh-Schlieder theorem. One then localizes such an operator (let us call it $Q$) in an energy-momentum region $\widehat{K}$ disjoint from the remainder of the energy-momentum spectrum (recall that there is an open neighborhood $m^2-\epsilon<p^2<m^2+\epsilon$, $0<\epsilon<m^2$ in energy-momentum space whose only points $p$ belonging to the energy-momentum spectrum lie precisely in the mass shell $p^2=m^2$, by the mass gap assumption of the Haag-Ruelle theory) by smearing the operator-valued function $x\mapsto Q_x$ with a tempered test function $f$

$$Q_f=\int_{\mathbb{R}^4}f(x)Q_x\mathrm{d}^4 x$$

whose Fourier transform $\hat{f}$ is of the form $\hat{f}(p)=h(p^2)\tilde{f}(\mathbf{p})$, where $h$ is a smooth function on $\mathbb{R}$ supported in $(m^2-\epsilon,m^2+\epsilon)$ and $\text{supp}\tilde{f}$ is such that $\{(\sqrt{\mathbf{p}^2+m^2},\mathbf{p})\ |\ \mathbf{p}\in\text{supp}\tilde{f}\}\subset\widehat{K}$. One obtains that if $|\Omega\rangle$ is the vacuum vector, then $Q_f|\Omega\rangle$ is a one-particle state with momentum wave function $\tilde{f}(\mathbf{p})$. We then write $Q(t,f)=(Q_f)_t$ - since the one-particle subspace is invariant under the action of the translation group, $Q(t,f)|\Omega\rangle$ is still a one-particle state, with momentum wave function $e^{-it\sqrt{\mathbf{p}^2+m^2}}\tilde{f}(\mathbf{p})$. It should become clear at this point that the precise form of the local observable $Q$ is not important.

A way to think of the observable $Q(t,f)$ is as follows: applying $Q(t,f)$ to the vacuum state adds an "energy-momentum chunk" to it, localized in $\widehat{K}\cap\{p^2=m^2\}$. By the uncertainty principle, $Q_f$ cannot be a local observable, but it is "almost local" in the sense that the commutator with any observable localized in a causally disjoint region should be "negligible" at large distances, more or less like tempered test functions with non-compact support (e.g. Gaussians). The effect of the time translation by an amount $t$ is that the approximate localization center spatially disperses with displacements $t\mathbf{v}=t\mathbf{p}/m$, where $\mathbf{p}$ belongs to the support of $\tilde{f}$. Think of it as a dispersing bunch of classical particles of mass $m$ in free motion at speeds $\mathbf{v}=\mathbf{p}/m$. This intuitive picture can be made rigorous with the aid of the stationary phase method.

If one now considers an operator monomial $Q(t,f_1)\cdots Q(t,f_n)$, where $\hat{f}_j(p)=h(p^2)\tilde{f}_j(\mathbf{p})$, $j=1,\ldots,n$, one may think of it as adding $n$ "energy-momentum chunks" to the vacuum state. The key point is that if the supports of the $\tilde{f}_j$'s are all disjoint, the corresponding localization centers move away from each other so that their commutators become negligible at large times. So, in a sense, the "almost local" observables $Q(t,f_j)$ become "asymptotically compatible" and the above "energy-momentum chunks" become effectively non-interacting at large times, thus giving origin asymptotically to $n$-particle states. This is made precise by the statement that $$\psi_t=Q(t,f_1)\cdots Q(t,f_n)|\Omega\rangle$$ converges to $n$-particle states with momentum wave functions $\tilde{f}_j(\mathbf{p})$, $j=1,\ldots,n$ as $t\to\pm\infty$ for each $n$. At finite but large times, one may think of $\psi_t$ as a state which yields a nonzero response around time $t$ from a coincidence arrangement of $n$ detectors with "momentum detection windows" contained in the supports of the $\tilde{f}_j$'s and (approximate) space-time localization regions contained in those of the $Q(t,f_j)$'s, but yields a "negligible" response from a similar coincidence arrangement of $n+1$ detectors. This interpretation may even be used (somewhat tautologically) to provide an operational definition of what is a particle in QFT. This (I hope as well) answers your question 2).

Now we are as well in a position to address questions 3) and 4). In QFT the Hilbert space is generated by applying all smeared field operator polynomials or all local observables (not necessarily compatible!) to the vacuum state. In fact, by the Reeh-Schlieder theorem, such states are a total set in the Hilbert space even if we restrict to a single space-time region with nonvoid causal complement. The "in" and "out" Hilbert spaces in Haag-Ruelle theory, however, are obtained by applying to the vacuum state a special subset of "almost local" operations - namely, polynomials in the $Q(t,f_j)$'s for all $f_j$ as above - and taking respectively the asymptotic limits $t\to\pm\infty$. As discussed in the previous paragraph, the observables $Q(t,f_j)$ are only "asymptotically" compatible, but as I pointed at the beginning, this picture must be taken cum grano salis since the operator limits $\lim_{t\to\pm\infty}Q(t,f_j)$ usually do not exist. Nonetheless, since the "in" and "out" Hilbert spaces are obtained as subspaces of the interacting Hilbert space, any "in" state may be prepared with arbitrary precision by applying local operations (even in a single space-time region with nonvoid causal complement) to the vacuum state. This is the closest we can get to a positive answer to your question 3). As for question 4), this is related to whether the "in" and "out" Hilbert spaces coincide with the whole interacting Hilbert space, that is, whether our field theory is asymptotically complete. This is usually an additional assumption, which has never been proven except in trivial (i.e. free) cases. We do know, however, that whenever a model has bound states, solitons, etc., then asymptotic completeness fails.

Finally, I must point out that the treatment of Haag-Ruelle scattering theory in Haag's book is almost telegraphic at parts (such as these) and not really a good first place to learn this topic. Better references are Section XI.16, pp. 317-331 Volume III (Scattering Theory) of the book Methods of Modern Mathematical Physics by Michael Reed and Barry Simon (Academic Press, 1979) and Chapter 5 of the book Mathematical Theory of Quantum Fields by Huzihiro Araki (Oxford University Press, 1999), particularly in the above order - Reed and Simon introduce the pedagogically simplifying assumption that the field operator itself interpolates between the vacuum and the one-particle Hilbert space (physically, the particles appearing in the asymptotic states are not "composite" with respect to the field). As discussed above, this assumption can be circumvented with the help of the Reeh-Schlieder theorem.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Pedro Lauridsen Ribeiro
answered Aug 16, 2016 by Pedro Lauridsen Ribeiro (580 points) [ no revision ]
Thanks for the references, but I'm not concerned at this time with math manipulations, but with the physical way of stating the scattering problem in the Heisenberg picture and the interpretation of results. How can one really state/describe scattering in the Heisenberg picture when the state vectors are constant in time?! Why are the state vectors time-dependent? Are we returning to the Schroedinger picture?

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
The scattering states are constructed by means of a time-dependent, almost local "one-particle creation operator" $q_{i,\alpha}(h_i,t)$ (see formula (II.4.15) in pp. 88), which are then applied to the vacuum vector as in formula (II.4.17) - that's why in Haag-Ruelle scattering one is really working in the Heisenberg picture. The resulting (non-factorized) states have the approximate localization stated in page 77 and are clearly time-dependent, but the time dependence is removed in the asymptotic limit $t\to\pm\infty$. Just by reading pages 76-77 alone one is not able to get the above picture.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Pedro Lauridsen Ribeiro
I know all that. I've read to the end, but there is still an INTERPRETATIONAL problem. In scattering theory formulated in Schrodinger picture one has to show that an arbitrary $|\Psi (t)\rangle$ converges to asymptotic states when $|t|$ becomes infinite. But in the Sch picture $|\Psi\rangle$ IS time dependent. In the book and everywhere in the H-R theory one constructs a time-dependent $|\Psi (t)\rangle$ and shows that it has limits, but what is the physical interpretation of this vector in the Heisenberg Picture? Moreover, is this time-dependent vector the most general it can be?

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
Your objections to H1) and H2) are unfounded. See EDIT 2 above.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
+ 2 like - 0 dislike

Unfortunately, I don't have precise references at the minute about the following argument, but only some notes taken during lectures of S. Doplicher.

The Haag-Ruelle scattering theory starts from the observation that observables cannot be used to construct asymptotic states from the vacuum, since they leave the superselection sectors invariant. Hence one needs to use field operators. Considerations on the Fourier transform lead to the conclusion that, given a field operator $B$, one has to construct a quasi-local operator $\tilde B$ out of localisation data for a single-particle state [the details should be contained in the original work of Haag-Ruelle]. A single-particle state is then constructed simply as $$\phi = B\Omega$$

We now construct the Heisenberg state. By this I mean a state that does not vary in time. This can be achieved by considering the continuity equation associated to the Klein-Gordon field equation, and in particular by considering the time-independent inner product that comes from it. To be concrete, take the one particle state $phi$ and set $$B_\phi(t)\Omega := \int_{\mathbb R^3}\overline{\phi(x)}\overset{\leftrightarrow}{\partial_0}U(x,I)B\Omega\ \text d^3\mathbf x,$$ where $U$ is a representation of the Poincaré group on Fock space. Observe that, in general, $B_\phi(t)$ will depend on time, but by construction $B_\phi(t)\Omega$ won't. Hence $$\psi:=B_\phi(t)\Omega = B_\phi(0)\Omega,\qquad\forall t\in\mathbb R$$ in practice, and this is how one can go about getting the asymptotic limit.

The construction of $n$-particle states is based on the choice of single-particle states with disjoint support in the momentum space. This is to guarantee that, in the asymptotic limit, the particles will be well separated (read far apart), in space and practically free, i.e. non-interacting. The state is then of the form $$\Psi^t := B_{1\phi_1}(t)\cdots B_{n\phi_n}(t)\Omega,$$ where $B_k$ and $\phi_k$ is a choice of quasi-local operators and solutions to the Klein-Gordon equations done as described above.

The property of clustering then shows that the above state has the form of a product of states, and therefore one can set $$\Psi^{\text{in}} = \psi_1\times^{\text{in}}\cdots\times^{\text{in}}\psi_n:=\lim_{t\to-\infty}\Psi^t$$ and similarly for the outgoing $n$-particle states.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Phoenix87
answered Aug 20, 2016 by Phoenix87 (40 points) [ no revision ]
Most voted comments show all comments
The $t$ in $\Psi^t$ is merely a label. As stated earlier, $B_\phi(t)\Omega = B_\phi(0)\Omega$. The time-dependence is only in the product of the quasi-local operators, but as soon as this product touches the vacuum vector it becomes time-independent by construction in the asymptotic limit by the clustering property.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Phoenix87
Not for any $t$ in the case of the $n$-particle state. You have to make use of the hypothesis on the support in momentum space of the single particle wave-functions and the clustering to then use the property that $B_\phi(t)\Omega = B_\phi(0)\Omega$ for any $t$.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Phoenix87
If it's only the asymptotic limit that is time independent, then for finite time $\Psi^{t}$ depends on $t$ and it cannot be in the Heisenberg picture.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
But the asymptotic states is what the HR theory is about, isn't it? This is what you use to then define the $S$ matrix as the unitary operator that rotates an "in" state to the corresponding "out" state (caveats: 1. some people use the opposite conventions; 2. asymptotic state completeness should be assumed).

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Phoenix87
It doesn't really matter, as the state associated to the vector $\Psi^t$ is constructed as $\omega_{\Psi^t}(A)=\frac{(\Psi^t,A\Psi^t)}{(\Psi^t,\Psi^t)}$.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Phoenix87
Most recent comments show all comments
One can choose a function $\phi$, indeed, but once chosen it cannot be changed and it evolves in time. So my question remains. Again, any proof would be greatly appreciated.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
I've found a sketch of the proof in Haag's book, and the norm is finite, indeed. I would like to thank you very much for all the help you've given me. I greatly appreciate it!

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user Andrea Becker
+ 0 like - 0 dislike

This is not an authoritative comment to you interpretation concerns. The Schrodinger properties of the time dependent state $\Psi^t$ are, actually, never used. As was commented above, the limit $\lim_{t\rightarrow\pm\infty}A(t)$ usually does not exist in the operator sense, thus one acts on a Heisenberg state to get a well defined quantity. Therefore, I would interpret $\Psi^t$ as a symbol that has to be substituted by the actual definition in the proof and thus stay in the Heisenberg picture.

This post imported from StackExchange Physics at 2017-11-23 19:27 (UTC), posted by SE-user dlont
answered Oct 30, 2016 by dlont (0 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights