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Why do we assume unitary evolution?

  1. Jul 24, 2011 #1
    I think this is the right forum for this.. are there any physical reasons to assume the evolution of a quantum system is given by a group of unitaries rather than a semigroup of isometries (or, if you're in the Heisenberg picture, group of automorphisms rather than semigroup of endomorphisms)?

    In all the baby examples I've seen (particle in a box, field in a steady state universe..) the unitary formulation is natural, but looking at what happens near t=0 in our universe surely keeping a fully reversible evolution isn't option? In particular being able to reverse to t<0 seems a bit nonsensical? Would there be any drastic physical consequences if I assumed my evolution was isometric/endomorphic? _O

    (p.s. I'm a mathematician, so I may be missing the obvious here!!)
     
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  3. Jul 24, 2011 #2
    Evolution has to be unitary otherwise probabilities wouldn't add to equal one.

    So its a basic reality condition for quantum theory to actual be a physical theory.
     
  4. Jul 24, 2011 #3
    The problem is not with unitarity, as Finbar points out, that is a logical requirement. The problem you noticed is that of *time* --- in a cosmological setting, it is the existence/direction of time that is subtle, and requires some careful thinking. In the regime where gravity is classical (i.e. non-quantum) and the FRW solution is applicable, there is a preferred foliation of spacetime and therefore a *physical* time that is fiducial. As one goes further back, quantum effects should start kicking in, and the FRW solution will no longer be a good approximation; it is expected that *something* (i.e. we don't know what, but people have guesses) will modify the definition of time, and so there will not be a problem.
     
  5. Jul 24, 2011 #4
    OK, conserving probabilities is clearly necessary but I'm not sure exactly what you mean.. if [itex](V_t)[/itex] is a semigroup of isometries and [itex]\psi_t=V_t\psi_0[/itex] we obtain [itex]<\psi_t|\varphi_t>=<\psi_0|\varphi_0>[/itex], which is what I understand by conservation of probability

    The time aspect is a good point.. I also liked the use of the word 'fiducial'

    EDIT: thinking further I think I get what you mean - if we do not assume unitarity then at time t>0 some states will be impossible. Then I think things like [itex] \int{(dp/2\pi)|p><p|=1} [/itex] might not hold - but I'll have to check it..
     
    Last edited: Jul 24, 2011
  6. Jul 25, 2011 #5

    Fra

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    The question is answered, and I am quite sure this isn't what you asked, but a deeper version of your question why unitarity is: why do we assume that information is conserved? Clearly this is an assumption, based on assuming we have a "closed system", it's not a logical requirement.

    The scheme we have is: fixed timeless statespace, in which the state is confined as per some distribution at all times.

    But the idea of a timeless statespace is not something that's inferred in the general case. It's an assumption.

    This is why I personally see the unitarity condition as a rationality constraint on expectations. Ie. given lack of further information, the EXPECTATION is that information is conserved, simply because there is no rational ground for expecting something not reflected in current evidence state.

    IMHO, in this sense unitarity is then best thought of as a kind of "tangent space" to statespace. But as the observer participaces in actual interactions, the tangent space changes. So the unitarity is valid only as a expectation in between measurements.

    I consider this to be an important point, but I am quite sure this isn't what OP asked for. I thought it might be interesting to just point it out.

    /Fredrik
     
  7. Jul 25, 2011 #6
    Unitarity is a requirement that the sum of squares of all amplitudes add up to 1.
    If this does not hold then we cannot interpret the squares of the amplitudes as
    probabilities; we can't use them to make a prediction for an experiment.

    Whats your interpretation of a non-unitariry set of amplitudes?

    For me unitarity is a constraint on state-space corresponding to an actual physical theory.
     
  8. Jul 25, 2011 #7

    Fra

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    I'm not arguing for something foolish here, I'm just trying to add another subtle point.
    I interpret it as a lack of decidability, manifested as a violated expectation. This makes sense if you take a decision theoretic view of physics, rather than a descriptive view.

    For me a theory does not describe nature; it is an INTERACTION tool, that is evolving.

    If we presume that a physical theory must alwas be perfectly decidable about the future - ie to a "description" of the future that can be verified in retrospect as sequence of events, then I see your position (and your logic, and this is of course the standard logic, I think however this is flawed)
    For me the assumption of unitarity is a rationality condition, in a decision process, not a constraint on actual evolution. This is why what I think of as the "expected evolution" is always unitary! This is a logical constraint. And this is exactly what the schrödinger equation gives us.

    So yes, an expected non-unitarty evolution would be a contradiction.
    However, non-unitary evolution established in retrospect is not a contradiction - it's food for evolving the theory.

    But my key point is that the ACTUAL evolution, typically deviates from the a priori expected evolution unless the theory has reached an equilibrium point, which can of course happen, but it's not a valid presumption in my view. There are several reasons why in some cases the equilibrium point is never reached, and the theory just keeps evolving.

    In my view, this is not a problem, beucase predictions you mention are something completely different from DESCRIPTION. Predictions are a bona fide element of game perspective. Predictions only served as to rationally placed your bets. Your bets can be correct (as in rational) yet end up "wrong" in retrospect. But this is not a flaw, this is in my view natural elements of interactions.

    We need to distinguis between

    1) a game: where predictions make us place bets (action), and we then subsequently react on the feedbeack from the actions as to strengthen or weaken the theory in a rational way.

    2) a description: where we just "test" whether a sequence of observations matches (in retrospect!) what our prior expectations(predictions) was. Either the predictions was right or wrong. But either way, there is no mechanism for evolving the theory. IT's either corroboration or falsification.

    So for me unitarity is not a constraint on the description! It's a rationality expectation in the decision process. These two views coincide in many cases, but the difference is how to understand a theory/law that is not stable.

    /Fredrik
     
  9. Jul 25, 2011 #8
    So you want a description of the evolution of a system which is not necessarily unitary to be understood as a meaningful description of nature away from physical states for which measurements are possible?


    I think such parts of "state-space" probably correspond to states without any preferred foliation i.e. a parameter t. In which case maybe its impossible to even define unitary evolution anyway.
     
  10. Jul 25, 2011 #9

    Fra

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    Mmm No. Bearning in mind my distinction between descripions and decision; my main point is that we seek isn't a "description" at all. I seek a strategy for evolving a theory (related to selection in theory space; except of course the theory space is not static).

    The context of my thinking is in a picture where you have an evolving theory space, and where "renormalization" of theories are taken to a whole new level, where THEORIES are litterally interacting in time. And not that I don't mean a theory that describes interactions, I mean two theories that are interacting with each other, causing an evolution in theory space.

    My ambition is to seek a deeper understanding of this, as well as unification of interactions.

    The notion of a timeless state spaces, and unitary descriptions are in my picture always equilibrium points - corresponding to populations of theories where the expectations are stable, and thus there exists defined objective transformations laws between the theories (corresponding to observer-observer transformations).

    About measurements: you can see that in two ways; as verifications of a descirption, or as feedback into the evolution.

    So what we all know as objective physics, will be unitary as usual. It's just that I argue that there is a better way to understand the unitarity and origin of observer invariance, rather than as logical constraints. In my view they are more like equilibrium conditions, or expecations.


    /Fredrik
     
  11. Jul 25, 2011 #10
    Nice words! But I'm not sure what you mean. Theory space is the space spanned by all possible couplings in the theory i.e. coefficients of all possible terms that can be written down in the action. Surely if this space itself evolves you must be evolving the symmetries of the theory? Also with respect to what is the theory space evolving??

    I prefer the idea that a theory evolves in theory space which we assume static.

    Interactions in theory space is a nice idea. Could you conceive of a coupling between say two theories then? Would this in turn define some notion of measurement? In the sense that one system(theory) interacts with the other system(theory) and the nature of this interaction gives changes the first system which constitute a measurement of the second?

    But then maybe such an interaction is just a very special case of interactions between theories?

    A further question? Do you consider a path integral only to exist in equilibrium? I believe that the analogous statement is true in statistical physics; away from equilibrium there
    is no partition function. So what you want is some non-equllibrium version of quantum theory?
     
  12. Jul 26, 2011 #11

    Fra

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    A static theory space would certainly be perferrable if existing, I agree. IMHO, however there are reasons to think that this doesn't exist. There are several ways to argue for that, but ONE problem is that if you were about to write down the MOST general statespace, it would be infinitely infinitely large to the point where it would be unmeasurable. The idea is that just like in biology, and evolution, the state space itself IS also evolving and expanding along with evolution. This is in fact a key point for stability in my thinking.

    Aslo my ideas would not work with a static state space, simply because we would miss what I think is a key point in understanding interactions. The total integration space must always be bounded when we compute an expectations, or it just gets ill ldefined.
    This is all fuzzy nonstandad and something I've been and still is working on but...

    Yes, there are "couplings" between theories! But that's not to say that the space of coupling constants is static, it's not (in my picture that is).

    Each theory even have inertia, which is measured in therms of the AMOUNT of information that has generated the theory and thus corresponds to some confidence in it. A massive theory is not as easily deformed when interacting with a small one.

    There are also "distance" between theories, defined in various ways as a sort of information divergence. Ie. a theory are in the "same place in theory space" only if they are identical. So the interactions between the theories creates the theory space - a kind of "relational picture" in theory space; no interacting theories - no theory space. This means that theory space is not defined or known globally, all we have are locally defined patches where there are interacting theories.

    Locally interacting "clusters of theories" can define an equivalence class of theories; it's something like this that I picture will be on the form of what we are used to.
    Yes, any "interacting" between theories is like a measurement. But in the decision picture, not descriptive picture. A theory measures other theories simply as a way of survival. There is a variation and selection in this picture. So this is a version of "evolving law" but very different than smolins csn. I am aiming at a more detailed description of it.
    Maybe you put it like that. As I like to put it, I am seeking a generalisation of measuremt theory that is truly intrinsic; this is also the reason why we do loose some decidability. But this is not a flaw; it's a feature of nature. And acknowleding it, will (I think) help understand the workings of this.

    A path integral is a way to comptue and expectations, by combining evidence counts. So part of what I'm working on will aim to be generalisations of the path integral, that I think is more well defined, but also observer dependent. IF we are talking about somewhat objective path integrals (as opposed to the subjective ones off equilibrium) then they mightbe seen as some equilibirum.

    In particular the measure structure of the integration space (or generally everything that is "baggage" and not part of the initial state explicitly) is negotiated due to a local equilibirum.

    The nice part is that here the measure structure and size of the state space will yield observable consequences for the action of a system. Ie. you can tell from how it acts, what state space it "sees" from the inside.

    Observer invariant state spaces are equlibriu points, off equilibrium we have more like a "observer democracy" rather than "observer invariants constraint".

    /Fredrik
     
  13. Jul 26, 2011 #12

    Fra

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    It's evolving with respect to itself only. Ie. the future evolution, is expected (or described but I prefer to avoid that word) with respect to the prior state. So it's a purely relational and stepwise evolution, this is why there simply is no global evolution. The global evolution is unpredictable from the inside; but the global evolution as seen by one theory; CAN be EXPECT by another (bigger theory). This is somehow the exploit I want to use to make this predictive.

    Also, the expectation of this evolution can only be made relative to another theory! There is no external description. So, you can assess one theory with respect to another theory only; there is in my new NO meta theory that restores decidability.

    Instead, each observer only has an illusion of the "expected" sort of meta theory that is the result of truncating what it can encode, and this will constrain it's actions.

    So the thought behind this, is that I combine it with an action principle which I call "rational action", where the ACTION rather than beeing arbirarity parameterized by some ad hoc lagrangian, ad hoc string action or similar, is construction as random action seen from the inside. The non randomness arises from the fact that the ergodics and the "randomness" is observer dependent. This is like a genralization of brownian motion, but where you have further complications like non-commutative information.

    But I admit that there are some massive unsolved issues in this, I'm fully aware of it and working on it.

    /Fredrik
     
  14. Mar 21, 2012 #13
    Does someone have any proof ? I think we have to make other assumptions.
     
  15. Mar 22, 2012 #14
    Here are the conclusions I've reached about unitary evolution.
    I think that the simplest we to motivate the unitarity of the temporal evolution lies in the framework of special relativity, where space and time are "treated in the same way": as discussed e.g. by Weinberg in "the quantum theory of fields", vol.1, a physically reasonable way to express the Poincarè invariance of the the spacetime is to assume that the modulus of the transition amplitude between two arbitrary normalized states in an inertial reference frame is the same as the modulus of the transition amplitude between the two states "translated" into another inertial reference frame; in mathematical terms we have the following equation


    | [itex]\langle[/itex] [itex]\psi[/itex][itex]_{2}[/itex] | [itex]\psi[/itex][itex]_{1}[/itex] [itex]\rangle[/itex] | = | [itex]\langle[/itex] [itex]\chi[/itex][itex]_{2}[/itex] | [itex]\chi [/itex][itex]_{1}[/itex] [itex]\rangle[/itex] | [itex]\;[/itex] [itex]\;[/itex] [itex]\;[/itex] [itex]\;[/itex] [itex]\;[/itex] [itex]\;[/itex] [itex]\;[/itex] (1)


    where |[itex]\psi[/itex][itex]_{1}[/itex][itex]\rangle[/itex] and |[itex]\psi[/itex][itex]_{2}[/itex][itex]\rangle[/itex] are two arbitrary states in a given reference frame and |[itex]\chi[/itex][itex]_{1}[/itex][itex]\rangle[/itex] and |[itex]\chi[/itex][itex]_{2}[/itex][itex]\rangle[/itex] are the same states "seen" by another inertial observer. This means of course that two observers measure the same physics.
    Now, Wigner's theorem states (roughly speaking) that, under hypothesis (1), there exists a (anti)unitary operator which implements the transformation: given two inertial frame O and O', there exists an operator U(O,O') such that, for every state |[itex]\psi[/itex][itex]\rangle[/itex] in O, the state seen in O' is |[itex]\chi[/itex][itex]\rangle[/itex] =U(O,O')|[itex]\psi[/itex][itex]\rangle[/itex].

    The last step is to recognize that time translations are nothing but a particular Poincarè transformation: by definition the generator of time translations is the hamiltonian.

    This is, in my opinion, a physically reasonable motivation to impose the unitarity of the temporal evolution.

    Let me know what you think.

    Best,
    Francesco
     
    Last edited: Mar 22, 2012
  16. Mar 22, 2012 #15
    I think that Weinberg's derivation is misleading because he assumes a priori unitary evolution using the Heisenberg picture, in which states don't evolve in time, so the probability transitions are trivially constant.

    In the Schrodinger picture if we look at a system at time [itex]t[/itex] in a configuration described by a state vector [itex] \psi [/itex], then a frame of reference with the origin of time at [itex] -s[/itex] will simply look the same configuration at time [itex] t+s [/itex], described by the same vector [itex] \psi [/itex].
    So imho the reasonable motivation leads to the identity operator, not something like [itex] e^{-isH} [/itex].
     
  17. Mar 22, 2012 #16
    Dear naffin,
    I don't think so (if I've correctly understood what you said). In my opinion the situation is clearer if one sees space translations: in that case it is easier to imagine that two states in one given frame and the "translated" states must have the same transition probability (and this happens also in non relativistic quantum mechanics). Now I'm asking: what is the difference between space and time translations in special relativity?
    But maybe I'm just missing your point.
    Best,
    Francesco
     
  18. Mar 22, 2012 #17

    fzero

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    You don't have to assume unitary evolution, just that there is a complete set of orthonormal states at any time. Then we can expand the ith state at time t' in terms of the states at time t via:

    [tex] | i ; t' \rangle = \sum_j U_{ij}(t',t) |j; t\rangle, [/tex]

    where

    [tex] U_{ij}(t',t) = \langle j; t|i;t'\rangle[/tex]

    are c-numbers. If anything, this only assumes locality, nothing in particular about U has been assumed yet. Then orthonormality of the [itex]|i ; t' \rangle [/itex] implies that

    [tex] \delta_{ij} = \langle j, t' | i ; t' \rangle = \sum_{k,l} (U^*)_{jk}(t',t) U_{il}(t',t) \langle k, t | l ; t \rangle , [/tex]

    from which

    [tex] U_{ik}(t',t) (U^*)_{jk}(t',t) = \delta_{ij}.[/tex]

    So [itex]\hat{U}[/itex] must be unitary.
     
  19. Mar 22, 2012 #18
    A transformation between states (rays) is implementable by a unitary operator if and only if it is one-to-one and preserves transitions probabilites.
    Let's focus on the Schrodinger picture: at a certain time [itex] t[/itex] we describe a system configuration with a vector [itex] \psi [/itex]. The obvious thing to say is that, whatever frame you choose, you will look at the system in a configuration that is symmetric (i.e. implementable by a unitary transformation) to the previous one, at another certain time [itex] t' [/itex].
    But this is not the point of Weinberg because he assumes a priori unitary evolution using the Heisenberg picture (in each frame). Under this assumption the point is that all descriptions of a system are symmetric, and that is 'obvious' .
    Symmetry in pure time translations [itex] (\vec{x},t) \longmapsto ( \vec{x},t +s) [/itex] means exactly that at time [itex] t+s [/itex] the situation is symmetric to that at time [itex] t [/itex], it doesn't mean just that in the new reference frame there exist a time [itex] t' [/itex] at which you see a symmetric description.
    Try to justify the meaning of the time translational invariance (in the Schr. picture) : what is the vector [itex] U \psi (t)[/itex] ?


    Are you assuming that an orthonormal set evolves into another orthonormal set? Nice point of view about unitarity, but I wouldn't say that this is an obvious thing.
     
  20. Mar 22, 2012 #19

    This is not true, in my opinion: the starting point of Weinberg, as far as I have understood, is that two different inertial observers measure the same transition probabilities (chap. 2 of vol.1, if I remember correctly). In that case, one is just interested in building the Hilbert space of states, just assuming the existence of operators which satisfy the Poincarè algebra.


    It is true that U(t,t')ψ(t)≈ψ(t+t') and so the whole thing seems a tautology; the central point, in my opinion, is that space and time are treated in the same way in special relativity: then, nobody finds unreasonable that if we have a spacial translation (that is, the system is shifted in space) then it is conceivable to assume that this is a symmetry of the space-time; because of the very nature of special relativity, time is not "privileged" with respect to space and so, as it is reasonable to assume space translation as an invariance of the space-time, it is also reasonable to assume that time translation (that is, the system is shifted in time) is a symmetry of space time.

    I hope I've made my point of view clear; let me know if something is wrong, in your opinion: this is a very interesting topic.

    Best,
    Francesco
     
    Last edited: Mar 22, 2012
  21. Mar 23, 2012 #20
    The difference between space and time is that a system evolves with time, not with space.
    I think I've found a nice way to explain what I was trying to say: in a given frame of reference we describe a system with a trajectory in the space of states [itex] \psi(t) [/itex].
    It seems obvious to me that in any other frame you can describe the same system with a trajectory [itex] \psi'(t) [/itex] such that it is linked to [itex] \psi(t) [/itex] by a symmetry transformation [itex] U [/itex] possibly time-dependent. I'm just assuming that we can paremetrize the system with the same time, with the property that corresponding descriptions are symmetric.
    So in a space-translated frame we see a function [itex] \psi_{\vec{a}}(t) = U_{\vec{a}}(t) \psi(t)[/itex], in a time-translated frame [itex] \psi_{s}(t) = U_{s}(t) \psi(t)[/itex], in a rotated frame [itex] \psi_{\vec{\theta}}(t) = U_{\vec{\theta}}(t) \psi(t)[/itex], in a boosted frame [itex] \psi_{\vec{v}}(t) = U_{\vec{v}}(t) \psi(t)[/itex].

    Choosing [itex] \psi(t) = | \vec{x} \rangle[/itex] we want [itex] \psi_{\vec{a}}(t) = | \vec{x} + \vec{a} \rangle[/itex] and [itex] \psi_{\vec{\theta}}(t) = | R(\vec{\theta})\vec{x} \rangle[/itex], choosing [itex] \psi(t) = | \vec{p} \rangle[/itex] we want [itex] \psi_{\vec{v}}(t) = | \vec{p} - m\vec{v} \rangle [/itex]. But choosing whatever we want [itex] \psi(t) = | b \rangle [/itex] we expect [itex] \psi_{s}(t) = | b \rangle [/itex], so the reasonable symmetry transformation is just the identity.

    My point is that time translation unitarity is on a different level, it is a stronger assumption.
     
  22. Mar 23, 2012 #21
    This is false, see Weinberg 1 pag. 109 (this is the subtle and central point of the question, in my opinion).

    [itex] \psi_{s}(t) = | b \rangle _{t+s}[/itex]

    Let me ask a (much) more general question: let us forget about special relativity for a moment and let's turn to general relativity: is there any difference between space and time in that case? In my opinion no, because of the principle of general covariance: space and time are just labels that we give to identify a reference frame.
    In my opinion the same thing happens in special relativity; of course, as you have pointed, this is not a proof of the validity of the unitary evolution: it is just a physically reasonable assumption that can be imposed because in special relativity ("like in gr", very roughly speaking) space and time have the same role.
    So, if we impose space translation invariance, then it is also reasonable to impose time translation invariance.
    In classical mechanics (or, if you want, non relativistic quantum mechanics) this is not true because space and time have different roles: indeed unitarity must be imposed as an external principle which has nothing to do with the symmetries of the spacetime; in special relativity it can be imposed (not proved, as far as I know), but the motivation lies in the strict analogy between space and time (for my personal taste, this is much more convincing).
     
  23. Mar 23, 2012 #22
    Uhm, the operator [itex]U_{s}(t)[/itex] that I am using is different than Weinberg's, and I was assuming [itex]|b \rangle [/itex] as an arbitrary time-indipendent vector.
    Anyway I realize that using time-indipendent vectors in my previous example was obscuring the main idea, so I try to be more clear.
    In my reference of frame I describe a state with the function [itex] | \psi (t) \rangle [/itex]. I use the same time to parametrize the system in another frame of reference. For space translations we obtain [itex] | \psi_{ \vec{a}} (t) \rangle [/itex], such that

    [itex] \langle \vec{x} | \psi_{\vec{a}} (t) \rangle = \langle \vec{x} | U_{\vec{a}} (t) | \psi (t) \rangle = \langle \vec{x} + \vec{a} | \psi (t) \rangle [/itex].

    Analogously for rotations and boosts.
    For time translations we obtain however [itex] \langle b | \psi_{s} (t) \rangle = \langle b | U_{s} | \psi (t) \rangle = \langle b | \psi (t) \rangle [/itex], i.e. nothing new: a time translated observer measures exactly the same things, just at different times (translated of [itex] s[/itex], here [itex]t[/itex] is the time of the starting frame ).

    Translations, rotations and boosts are symmetries about ''contemporaneous'' observation (at the same time [itex]t[/itex]), while time translation symmetry is about different times. This one can't be justified with my argument (that I think it is the only one obvious).

    A 'classical' example to clarify: imagine a green particle that at a certain time becomes red.
    Symmetry assumption about space translations is nothing more than saying that in a translated frame when I observe the particle becoming green I have an equivalent (symmetric, unitary equivalent) configuration to that one seen from the starting frame (during the same transition from green to red).
    Symmetry assumption about time translations means instead that I observe an equivalent configuration at every time.

    I hope that now is clear what I was trying to say, and that there isn't any mistake.

    I can't see the difference (in this aspect) between the classical case and the relativistic one.
    I haven't studied general relativity yet, maybe I'm missing the link with Special relativty. Anyway I wouldn't say that time and space are perfectly equivalent, essentially because of the metric tensor, i.e. you can't exchange their labels.
     
  24. Mar 23, 2012 #23
    I've seen that conservation of transition probabilities is often interpreted as conservation of information or entropy.
    An interesting accessible paper for example is:
    http://arxiv.org/abs/quant-ph/0407118
     
  25. Mar 23, 2012 #24
    I agree with you if we are talking about non relativistic theory (and, indeed, one should impose the unitarity of the evolution). What you have said cannot be true in a relativistic theory simply because in a relativistic theory the momentum and the energy mix under a boost (or, alternatively, space and time mix under a boost): this is, once again, the central point and difference between non relativistic and relativistic physics (the role of time).

    I've understood your point about the difference between space and time: in your opinion the time is like a label and it is not important where its origin is; and, indeed, in your framework, I agree with you that imposing the conservation of probabilities for time translated states is a strong assumption (and, once again, this is what is done in usual nonrelativistic mechanics).

    The claim I stress (and stressed) is that all these difficulties can be overcome in special relativity because (apart from the metric tensor, as you have correctly emphasized) space and time are not "distinguishable". As far as I know, the metric tensor is not so important in determining the properties of the spacetime (and, indeed, Lorentz transformations can be deduced without any reference to the metric tensor). I've not got a precise idea of the relashionships between all these quantities.

    I again remark that mine is not a proof: it is just my point of view on the justification of treating space and time in the same way.

    In the end, I don't understand your "classic" example (and its quantum counterpart) : what is dynamics and what is observation? Can you make the problem clearer please?


    Things become harder and harder :(

    P.S. Maybe this is a good point to see where our opinions diverge: in Weinberg, if I remember correctly, there is a discussion to see how galilean invariance can be deduced as a low energy limit of Poincaré invariance.
     
  26. Mar 24, 2012 #25
    In my frame of reference I describe a particle with a function [itex] | \psi(t) \rangle [/itex].
    Let's focus on a precise time [itex] t_0 [/itex], for example when the particle changes its color; actually in quantum mechanics you see the color only by measurement, in this sense it is a 'classical' example: I'm not talking about measurements, but I think this is a good way to express the fact that a certain configuration of the system is objectively ''observable''.
    In any other frame of reference you will look at a certain time the particle changing its color, or in a more appropriate quantum language: at a certain time (when the particle becomes red) you describe the particle (i.e. its transition probabilities) in a symmetric/equivalent way to the description of the particle at the corresponding time in the starting frame. It is the basic point of my argument, let me know if this is not clear or if you don't agree with this property of the description.
    Talking about colors I think is just a nice way to express the fact that there is something objective, like the observation of a space-time event in classical relativistic mechanics.
    I'm just bypassing some difficulties in expressing this simple fact in a correct quantum vocabulary.

    Just like events in classical mechanics, in quantum mechanics you describe the same objective system configuration (the particle changing the color) in a manner which depends on your frame of reference. But an obvious thing is that the descriptions must be related by a symmetry transformation: when the particle change its color we all agree on the structure of the transition probabilities (altough we may disagree about what are the the possible ''outcomes''/transitions because our subjectively identical observables are different).

    My aim was to demonstrate that space translations, rotations and boosts symmetries are a directly consequence of this obvious fact, but (as you have pointed out) I was completely using only non relativistic mechanics. I don't know if this completely changes all the story, however I try to rephrase using a more general language.
    When the particle changes its color it appears to me in a state [itex] | \psi(t_0) \rangle [/itex].
    What happens in another frame of reference?
    In a space translated, rotated or boosted frame when the particle changes its color I observe a different configuration, i.e. I describe the system with a different state vector.
    In a time translated frame this is not the case. I observe instead exactly the same thing: the unitary operator is the identity.
    (What happens if we consider a time operator ? This would not be true...)

    In other words I assume we can parametrize the evolution of a system with a parameter [itex]b[/itex] (it's a sort of causality parameter, it could be the time of any frame: a particular value of [itex]b[/itex] could correspond to a ''change of color'' as intended above).
    Depending on the frame we choose we observe a different evolution in our set of states [itex] \psi (b)[/itex].
    The description in a space translated frame it is naturally connected with the previous by a unitary transformation [itex] | \psi_{\vec{a}}(b) \rangle = U_{\vec{a}} | \psi (b) \rangle [/itex], and so on with the other space-time transformation.
    So we obtain a representation of the Poincarè (or Galileo) group at each [itex] b [/itex].
    If we don't admit time operators the pure time translation would be the identity (how can we distinguish the same configuration of a system if we have just the clock backward?).
    In non relativistic quantum mechanics the remaining transformations are exactly those generated by P,Q and J because of the absolute property of time (if in our frame the parameter [itex] b [/itex] corresponds with [itex] t [/itex], then it is the same for the others, so this passive transformation is just an usual active transformation).
    In relativistic quantum mechanics boosts are surely an exception to what I've just said (because time is different in boosted frames), and I don't know if the transformation coming out from this whole argument is the same obtained by, say, Weinberg.

    These are the reasons why I think Weinberg assumes that the whole spacetime history of a state could be described by a fixed state vector (i.e. unitary evolution in the Heisenberg picture). With this assumption the question about transformations is: what is the state described by another frame?
    This is a different question than that I was asking me before: what is the state described by another frame at the same value of [itex]b[/itex] ?

    A simple question could be enlightening to me: you said that Weinberg doesn't assume a priori unitary evolution, then I'm asking you: if we use the Schrodinger picture and we are describing a system with a vector [itex] | \psi (t) \rangle[/itex], what is the physical meaning of the vector [itex]U(\Lambda,a)| \psi (t) \rangle[/itex] assuming that [itex]U(\Lambda,a)[/itex] is the transformation corresponding to a pure time translation in the passive point of view ? It is of course a vector in the new frame describing the same system, but at what time? At the same [itex] b [/itex] ?
     
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