Quantcast
  • 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.

News

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

Attributions

(propose a free ad)

Site Statistics

146 submissions , 123 unreviewed
3,961 questions , 1,408 unanswered
4,911 answers , 20,872 comments
1,470 users with positive rep
519 active unimported users
More ...

  In what picture of QM is the Haag-Ruelle scattering theory formulated?

+ 4 like - 0 dislike
354 views

I've started reading on scattering in AQFT and now I'm studying the Haag-Ruelle scattering theory. I've read the presentations in Bogoliubov's book on axiomatic qft, in Jost's book on qft, in Haag's book on qft and, also, that  which can be found in Encyclopedia of Mathematical Physics.

I've pretty much understood the idea and the math involved, but I still have some lingering questions about H-R scattering theory which I think are related to, or generated by a deeply ingrained prejudice of mine about axiomatic qft. It all started after I found out that Haag's theorem forbids the interaction picture in qft. The only 2 pictures left were Schrodinger and Heisenberg (for describing scattering and time evolution in general). Then I stumbled upon a paper by Dirac (Quantum electrodynamics without dead wood) published in Phys. Rev. in which he shows that the Schrodinger picture must be banned from qft since a state vector at finite time $t$, say $\psi (t)$, cannot exist (more exactly is infinite and hence not in the Hilbert space) due to vacuum polarization effects. The only picture left for studying qft rigorously was the Heisenberg picture.

My deeply ingrained prejudice is that any mathematical construction in scattering theory must not violate the main characteristics of the Heisenberg picture, namely that all the state vectors one works with, or constructs, even at an intermediary step, should be time independent. I had hopped that this should be the case with the Haag-Ruelle scattering theory. The Heisenberg picture is necessary to maintain Lorentz invariance and to avoid vacuum polarization effects.

From Bogoliubov's book, and also from Haag's book, I understood that the many-particle state vector one constructs in the formalism, namely $\psi_t$, is NOT time independent (as a Heisenberg picture vector should be) for finite time $t$, and hence one does not remain at this step of the proof in the Heisenberg picture. (When the infinite time limits of $\psi_t$  are taken one obtains the scattering states.) This comes in contradiction with my prejudice and I have some questions which I would be most thankful if you could help me with.

1) In what picture of QM should one consider the time-dependent many-particle state vector $\psi_t$ to be in? Is it correct to assume that it is in the Schrodinger picture? If yes, how does one cope with vacuum polarization effects?

2) Since $\psi_t$ is time-dependent it is not Lorentz-invariant. Is it a bad thing that the H-R construction is not Lorentz invariant from the beginning (at finite time $t$) to the end (at $t = -\infty$)?

3) If the answer to my question 1) is "the Schrodinger picture", can one regard the H-R scattering theory as a non-Lorentz-invariant Schrodinger picture type of scattering theory?  

4) Is my prejudice (described above - namely that one has to always remain in the Heisenberg picture while doing qft) justified from a physical point of view or it is just a bad way of thinking?

Thank you!

asked Aug 26, 2016 in Theoretical Physics by mhrt (55 points) [ revision history ]
retagged Aug 26, 2016 by mhrt

2 Answers

+ 3 like - 0 dislike

The Heisenberg picture is the only one where relativistic quantum field theory can be rigorously defined, e.g., through the Wightman axioms. The latter imply the existence of a 4-dimensional group of translations generated by the 4-momentum vector $p$. Once one select a future-like direction to determine time, one gets from $p$ the Hamiltonian $H=p_0$, and can use it to define time-dependent states $\psi(t)$ in the usual way. Together with the spatial part of the momentum, this gives a conventional, frame-dependent Schroedinger representation of the states.

Thus the Schroedinger picture exists but is frame dependent. You can regard any operator $A$ on the Hilbert space as a Schroedinger observable and find its expectation as time changes in the usual Schroedinger way, and translate it to the Heisenberg picture in the usual way such that $A$ becomes explicitly time-dependent and the state is fixed.

However, in 4-dimensional relativistic interacting quantum field theories, fields must be smeared in space and in time in order to produce densely defined operators (rather than distributions). Thus, unlike in the nonrelativistic case, the fixed-time spatial fields $\phi(x)=\Phi(t,x)|_{t=0}$, where $x$ is 3-dimensional, are not well-defined objects. 

Thus although the Schroedinger picture exists it does not represent local fields at a fixed time. The attempt to pretend it did leads to the problems mentioned in Dirac's paper. In this sense, the relativistic field theory cannot be made well-defined without transcending the Schroedinger picture. 

The above is completely independent of scattering theory. In scattering theory one has to construct the asymptotic Hilbert space. Unlike the interacting Hilbert space, the asymptotic Hilbert space is a Fock space and must therefore be defined by a limiting procedure. This is what is done in the Haag-Ruelle theory. Note that because of the Lorentz invariance of the future cone, the resulting asymptotic space does not depend on the choice of the time direction, as long as it points into the future cone. 

There are presentations of the Haag-Ruelle theory that are a little more in the spirit of the Heisenberg picture; see the comments at the end of p.379 of Volume 3 of Reed and Simon. One can probably work almost completely in the Heisenberg picture if one works out the algebra of asymptotic constants (in analogy with the nonrelativistic case treated in Section 3.4 of Volume 3 of Thirring), but to show that the corresponding asymptotic operators have a Fock representation one apparently needs to go through some Schroedinger-like computations.

The meaning of the S-matrix in the Heisenberg picture is the following: There are frame-dependent isometries $\Omega_\pm$ from the interacting Hilbert space to the asymptotic Hilbert space such that for any two states $\phi,\psi$ in the Hilbert space of the interacting theory, the asymptotic states $\phi_-=\Omega_-\phi$ and $\psi_+=\Omega_+\psi$ satisfy the limiting relation $\lim_{t\to\infty}\langle\phi|e^{-itH/\hbar}|\psi\rangle=\langle\phi_-|\psi_+\rangle=\langle\phi|S|\psi\rangle$, where $S=\Omega_-^*\Omega_+$ is the S-matrix. This makes no reference at all to Schroedinger states. The latter are used only to construct the unitary mappings. But of course one can rewrite the limiting relation also in the Schroedinger picture by splitting $t$ into a sum of two times that separately go to infinity, and then attach these two times to the state.

answered Aug 26, 2016 by Arnold Neumaier (12,640 points) [ revision history ]
edited Aug 26, 2016 by Arnold Neumaier
Most voted comments show all comments

@ArnoldNeumaier 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. How can one calculate transition probabilities using state vectors that belong to 2 different Heisenberg pictures? 

I added a paragraph to my answer that - I hope - explains better what happens.

@ArnoldNeumaier One can define a scattering S-matrix in the Heisenberg picture as you did, namely $S = \langle\phi_{-}|\psi_{+}\rangle$ only when $|\phi_{-}\rangle$ and $|\psi_{+}\rangle$ belong to the same Heisenberg picture, say a Heisenberg picture that at some time, say $t_{0}$, coincides with the Schrodinger picture. But in H-R scattering one has 2 different Heisenberg pictures (see my first comment) and I don't know as to how one can define an S-matrix in this case. This is my confusion about.     

The only way out of my dilemma is that the asymptotic scattering states $|\phi_{-}\rangle$ and $|\psi_{+}\rangle$ should also be considered/interpreted in the Schrodinger picture, and never in the Heisenberg picture, since otherwise one is led to 2 different Heisenberg pictures and a scattering S-matrix cannot be defined if it's based on 2 different Heisenberg pictures. The conclusion is that the Haag-Ruelle scattering theory is entirely formulated in, and based on the Schrodinger picture of QM, at any finite time $t$, and also at $t = \pm\infty$.

@mhrt: The physics is not in the scattering amplitudes but only in their absolute squares. These are expectations and hence picture independent. One can see this also in other ways:

1. The formula defines the scattering matrix, independent of the picture, and obviously agrees with experiment, as the success of QFT for the interpretation of high energy collision experiments shows. Thus all the physics is in this formula.

2. That this mathematical definition really gives the conventional scattering matrix can be seen most easily when viewed in the Schroedinger picture, by comparing with the nonrelativistic situation. This is probably the reason why most derivations choose the Schroedinger formulation. Note that - as I had mentioned in my answer - there is nothing intrinsically wrong with that formulation; the only defect is that one cannot define the fields at a fixed time (and hence has some difficulties with arguing about observables as Hermitian operators).

3. The physical interpretation of the matrix elements of interest in scattering experiments is also independent of the picture. Indeed, the physical information in the S-matrix consists of the absolute squares of the matrix elements between asymptotic momentum eigenstates. But momentum states have a trivial Schroedinger dynamics, affecting only the phase and hence not the absolute squares of the matrix elements. Thus Schroedinger and Heisenberg picture agree on the physical predictions.

If this still doesn't satisfy you then no amount of explanation will help, and you'll have to wait till you grow out of your prejudices.

@ArnoldNeumaier Now it's crystal clear. I would like to thank you very much for all your help, and to apologize to you for asking so many questions. Best wishes, mike.

@ArnoldNeumaier Yes, when one calculates expectation values of some operators, the pictures are irrelevant, indeed. But in a scattering problem one calculates transition amplitudes and not expectation values. If the scattering operator were known or if it could be defined in the Heisenberg picture in a meaningful way, then the pictures would be irrelevant, but in the Haag-Ruelle scattering there is no other way as far as I know to define the S-matrix other than the way you mentioned, via the asymptotic scattering states. Therefore, for H-R scattering one must state very clearly and openly in what picture of QM one works.

+ 0 like - 0 dislike

The Haag-Ruelle scattering theory, as presented in all the books there are, is formulated entirely (that is, from beginning to end) in the Heisenberg picture of QM. My question and my comments above were raised by me due to a poor understanding of the Heisenberg picture, which was in turn due to the systematic brainwashing that all the textbooks on QM superbly do when discussing the Heisenberg picture. Indeed, right from the outset the reader of a QM textbook is told that in the Heisenberg picture all the time dependence is assigned to the operators representing various observables, whereas all the state vectors are time-independent.

My humble opinion is that one should erase this phrase from every textbook on QM, and here is why. Let $|\psi\rangle$ be a state vector describing a state in the Heisenberg picture, and let $\hat{A}_{H}(t): \cal{H} \rightarrow {H}$ be a Heisenberg operator that acts on the Hilbert space of states $\cal{H}$. According to the superposition principle of QM, every properly normalized state vector of $\cal{H}$ represents a possible state of the system. Therefore, up to normalization, the time-dependent state vector $|\phi (t)\rangle =  \hat{A}_{H}(t) |\psi\rangle \in\cal{H} $ represents a possible state of the system in the Heisenberg picture of QM.  One can see that one can obtain time-dependent state vectors while remaining entirely in the Heisenberg picture, since the image of any range of a Heisenberg operator contains only time-dependent vectors. 

Rudolf Haag obviously knew better that the textbooks on QM and based his scattering theory entirely on such constructions, and hence his formulation is entirely within the Heisenberg picture of QM.

answered Sep 17, 2016 by mhrt (55 points) [ no revision ]

Your time-dependent state vector is a vector in the Hilbert space, but it does not describe a time-dependent state in the Heisenberg picture, since the latter is defined to have time-independent state vectors.

This means that in the Heisenberg picture, the temporal dynamics of the measurable expectations in the pure state $\psi$ is always given by $\langle\psi |A(t)|\psi\rangle$. This only holds for the time-independent state vector, and that characterizes the Heisenberg picture.

On the other hand, you can define arbitrary paths in the Hilbert space defining a $\psi(t)$ but these $t$-dependent states don't have a temporal meaning as a Heisenberg state - i.e., $\langle\psi(t) |A(t)|\psi(t)\rangle$ means nothing.

@ArnoldNeumaier For the physical interpretation of $\langle\psi (t)|\hat{C}(t)|\psi (t)\rangle$ you can look at Section II.4.3 (starting on page 92) of Haag's book on qft.

One can form this expression, and for some $C(t)$ also give it an interpretation, but still this doesn't give $\psi(t)$ the meaning of a state in a physical sense. The Hilbert space contains innumerable vectors, but in the Heisenberg picture only one is distinguished as the state of the system, and this one is time-independent.

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:
p$\hbar$ysicsOverf$\varnothing$ow
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
...