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???
Answer
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.
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