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Natural Science Forum / Physics / Research / December 2007operator satisfying relations [H,T] = i hbar
Hi, I am reading John Baez's writing on energy-time uncertainty and have a question about his claim of the non-existance of any observable varaible T satisfying relations [H,T] = i hbar. In fact, for a classical harmonic oscillator, H=P if we use action-angle variables P and Q, and {P,Q}=1 as usual, i.e {H,Q}=1. So the Q would satisfy [H,Q] = i hbar after quantization. But this seems to create a paradox due to his following argument. I would appreciate any comments or refence material. Thanks a lot. In below, I quote John Baez's original argument from his website: ------------------------------------------------------------------------------------------------------------------------ The problem is, for physically realistic Hamiltonians H one can prove there is no operator T with [H,T] = i hbar In other words, there is no time observable! The reason is this: by the Stone-von Neumann uniqueness theorem, any pair of operators satisfying the canonical commutation relations [H,T] = i hbar can only be a slightly disguised version of the familiar operators p and q. These operators p and q are unbounded below - i.e., their spectra extend all the way down to negative infinity. But a physically realistic Hamiltonian must be bounded below! (Here I am glossing over some mathematical nuances: if you read the precise statement of the Stone-von Neumann theorem, you'll see how to fill in these details.) Crudely speaking, this theorem says that it's impossible to construct a clock that works perfectly no matter what its state is. That's not surprising - but it's sort of surprising that you can *prove* it, and it's sort of interesting to see what assumptions you need to prove it. This theorem is usually called Pauli's theorem - it first appeared in 1926. The simple proof can be found e.g. in section 2 of http://arxiv.org/abs/quant-ph/9908033 . IMO the clearest explanation for this theorem is in terms of Rovelli's distinction between partial and complete observables, http://arxiv.org/abs/gr-qc/0110035 . Partial observables can be measured, but only complete observables can be predicted and correspond to operators on the quantum level. A real experiment consists of two measurements of partial observables, the reading of a detector A and the reading of a clock t. Neither A nor t is predicted by the theory and subject to quantum fluctuations, only the correlation A(t). Therefore neither A nor t are operators, but only A(t). On Dec 10, 8:31 am, mingyum...@yahoo.com wrote: > Hi, > [quoted text clipped - 6 lines] > his following argument. I would appreciate any comments or refence > material. Thanks a lot. To make it clear that the Poisson bracket can't go over to a commutator for any physical system, and how it relates to the existence of a groundstate, suppose one has an Hermitian operator T with [H,T] = i hbar. Now define the operator U by U := exp[ i w T ] for some positive constant w (eg, the frequency of the oscillator for your particular case). By expanding U as a power series, it is easy to show that [H,U] = -hbar w U. (*) Further, since T is Hermitian, one has U* U = exp[ -i w T] exp[ i w T] = 1. (**) The idea now is to use (*) and (**) to obtain a contradiction, for any system that has a groundstate (i.e., for all physical systems). In particular, the Hamiltonian of any physical system has a groundstate |0>, corresponding to the lowest possible energy E0 of the system. Now consider the state |psi> defined by |psi> := U |0> . Using (*) above, one has H |psi> = H U |0> = [H,U] |0> + U H |0> = -hbar w U |0> + E0 U |0> = (E0 - hbar w) |psi> . This says that |psi> is an eigenstate of H with a LOWER energy than the groundstate energy! The only possibility that this can hold is if |psi>=0. But, using (**), one has <psi|psi> = <0| U* U |0> = <0|0>, where the righthand side cannot vanish, as it is the norm of the groundstate. QED. By the way, although one cannot define a Hermitian time operator an oscillator with frequency w, one can still define a perfectly good time observable using so-called POVMs (probability operator valued measures). In particular, denoting the n'th energy eigenstate by |n>, and the period 2pi/w by T, one defines the kets |t> by |t> = (1/sqrt{T}) sum_n exp{ -i nwt } |n> . It is easy to show that exp{ -i hbar H tau } = | t + tau >, and that int_0^T |t> <t| dt = 1. The probability of the "time" being measured to have the value t in [0, T) for state |psi> is then given by p(t|psi) = | <t|psi> |^2 . One can similarly define POVM time observables for the free particle and various other (but not all) physical systems (see, eg, "Probabilistic and Statistical Aspects of Quantum Theory" by A. S. Holevo). > [snip] > By the way, although one cannot define a Hermitian time operator an [quoted text clipped - 14 lines] > "Probabilistic and Statistical Aspects of Quantum Theory" by A. S. > Holevo). I think this is very misleading. The sum written to "define" |t> does not converge, so the reader has to guess how it might be translated into meaningful mathematics. My guess is as follows. The Hamiltonian H would presumably satisfy H|n> = n|n>. Since an unquoted part of the post specified that H be bounded below by 0, the sum should be over n >= 0. Taking w := 1 for notational simplicity, the Hilbert space with orthonormal basis the set of |n> could be realized as the set of square-integrable functions on [0, 2pi] with Fourier transforms in the span of functions t -> exp(int) with n >= 0. Such a function would be denoted |n>, and the action of the Hamiltonian on such a function (to assure H|n>=|n>) would be to differentiate it (and multiply by i). That is, the Hamiltonian is what is usually called the "momentum" operator, except that it operates on square-integrable functions on [0,2pi] instead of on functions defined on the whole real line. The operator T would then be multiplication by the coordinate function, f(t) -> tf(t), so as to assure HT-TH = i. Formally, the (generalized) eigenfunctions |t> for T would be as the poster stated (essentially, they are Dirac delta "functions"). Thus the "time" operator" would be what is usually called the "position" operator. I think any usual translation of the poster's undefined notation into meaningful mathematics would be essentially the same as the above, apart from notation. But if it were correct, it would contradict the result that HT-TH = i is impossible for H bounded below. What is wrong? First of all, the alleged operator T does not preserve the stated Hilbert space. That is, if a function f is a linear combination of |n>, then Tf need not be, so the mathematics turns out to be nonsense. It is also probably physical nonsense. Where in the world would one find a physical system, on which one could do actual experiments, whose "energy" turns out to be identical to its "momentum", and whose "time" turns out to be identical to its "position"? My intention is not to criticize the poster with username "a student". He probably is a student, and she probably got the above nonsense from some book or paper. There is a lot of such nonsense in the literature (check the literature on the "quantum phase" operator). The point of this post is that one has to read the physics literature with a great deal of caution, realizing that nonsense may even be more the rule than the exception. The physical meaning of formal algebraic calculations should always be carefully examined. > > [snip] > > By the way, although one cannot define a Hermitian time operator an [quoted text clipped - 32 lines] > except that it operates on square-integrable functions on [0,2pi] > instead of on functions defined on the whole real line. Essentially OK so far (but think of [0,2pi) as the circle, so that there are periodic boundary conditions). There are no convergence issues, at least, no more than in defining momentum kets by |p> = \int dx exp{ipx/hbar] |x> . > The operator T would then be multiplication by the coordinate function, > f(t) -> tf(t), so as to assure HT-TH = i. Formally, the (generalized) > eigenfunctions |t> for T would be as the poster stated (essentially, > they are Dirac delta "functions"). Thus the "time" operator" would be > what is usually called the "position" operator. No, I do not allege, nor define, any such 'operator' T. I only work with the POVM {|t><t|}. Note in particular that the states |t> and | t'> are NOT orthogonal for distinct t and t', and hence are not eigenstates of any Hermitian operator. Note that if one tries to define T by T = \int dt t |t> <t| , for example, then one finds that one does NOT have f(T) = \int dt f(t) |t> <t| . Hence I certainly do not claim there is some T with HT-TH = i (that would be contradictory to my earlier post showing there is not such a T !). > I think any usual translation of the poster's undefined notation into > meaningful mathematics would be essentially the same as the above, apart > from notation. But if it were correct, it would contradict the result > that HT-TH = i is impossible for H bounded below. What is wrong? No contradiction, as there is no T. Perhaps you are unfamiliar with POVM observables? Any book on quantum information theory will explain these - or, try Wikipedia at http://en.wikipedia.org/wiki/POVM - they are essential for the description of all possible measurements on a system, in terms of the system's Hilbert space. Briefly, a POVM (positive operator valued measure) is a set of positive operators which sum (or integrate) to the identity operator. For a given POVM {E_k}, the probability of result k is defined to be p(k|rho) = tr[rho E_k}, for a system described by density operator rho. This is equivalent to some Hermitian operator E acting on the system Hilbert space if and only if the E_k are orthogonal projection operators. However, there is still a fundamental connection with measurement and Hermitian operators: for a given POVM, one can always find an "apparatus" system, described by some fixed initial density operator sigma; a fixed unitary operator S acting on the composite Hilbert space of the system and apparatus; and an Hermitian operator A of the apparatus, such that for any rho, p(k|rho) = tr[ S rho x sigma S* A_k ], where A_k is the projection onto the k'th eigenspace of A (try googling Naimark extension theorem). Hence there is nothing inherently unphysical about POVMs - they are just as measurable, in principle, as any Hermitian operator is. > First of all, the alleged operator T does not preserve the stated > Hilbert space. That is, if a function f is a linear combination of |n>, > then Tf need not be, so the mathematics turns out to be nonsense. No such operator alleged, as previously mentioned. The POVM {|t><t|} is mathematically perfectly well defined. > It is also probably physical nonsense. Where in the world would one > find a physical system, on which one could do actual experiments, whose > "energy" turns out to be identical to its "momentum", and whose "time" > turns out to be identical to its "position"? Not sure what you mean. The position and momentum of an oscillator are defined in the usual way, as are the energy and number operators. None of these are equal to each other! Several representations of the "time" (or "phase") POVM above, in terms of measurement via interaction with an associated apparatus as per the Naimark extension theorem above, have been given. Hence this observable is, in principle, just as "physical" as any Hermitian operator A. > My intention is not to criticize the poster with username "a student". > He probably is a student, and she probably got the above nonsense from > some book or paper. There is a lot of such nonsense in the literature > (check the literature on the "quantum phase" operator). Aren't we all students?! True, there is some nonsense in the literature in this regard. There are also some valid papers (and books). I happen to have done the work necessary to sort out the good from the bad. Sorry if I have communicated some of the results badly. Some particular references have been given elsewhere in this thread. > The point of this post is that one has to read the physics literature > with a great deal of caution, realizing that nonsense may even be more > the rule than the exception. The physical meaning of formal algebraic > calculations should always be carefully examined. Couldn't agree more! > Hi, > [quoted text clipped - 6 lines] > his following argument. I would appreciate any comments or refence > material. Thanks a lot. Your construction of the supposed paradox is not clear to me, but I suspect that any apparent contradiction between your example and John Baez's assertion may be because your H may not be bounded below. His assertion was that for any H which *is* bounded below, an Hermition Q satisfying HQ - QH = i cannot exist. There is a sketch of an argument justifying the assertion in a review of "Quantum Paradoxes" by Y. Aharonov and D. Rohrlich at www.math.umb.edu/~sp/paradoxes.pdf . It is basically a formal calculation showing that if Q exists, then the spectral measure (hence the spectrum) of H must be invariant under translation (hence cannot be bounded below). The calculation is rigorous for bounded, Hermitian H, and Q. It is algebraically correct (e.g., satisfies the usual standard of mathematical rigor of physics texts) for any Hermition H, Q. I have the impression that the argument may be originally due to Wigner, but I'm not sure. Aharonov and Rohrlich seem to assume that for any H, Q will exist, which the argument shows is false. But that's just my interpretation. The book is so vaguely written that it's hard to tell what they are assuming. I would be interested to hear what others think of this book. There are two reviews of it on amazon.com, mine and one of Michael Steiner, who is a serious researcher. He liked the book, giving it a a five-star rating (the highest). I didn't, and gave it two stars (one step above the lowest). mingyuming wrote {e4cb1894-b4b7-43dd-a8c8-a1d00b0d05b1@b15g2000hsa.googlegroups.com} on Sun, 09 Dec 2007 21:31:04 +0000: > Hi, > [quoted text clipped - 8 lines] > > In below, I quote John Baez's original argument from his website: -------------------------------------------------------------------------- > The problem is, for physically realistic Hamiltonians H one can prove > there is no operator T with > > [H,T] = i hbar > > In other words, there is no time observable! One would start by understanding what H and T are and what are their limits of applicability. In non-relativistic quantum mechanics t is an evolution parameter. The t in a Schrodinger equation is basically Newton concept of time. Since t is not a observable, it has no sense to build an operator for it; T has no sense. Now turns to relativistic theory. There is two: relativistic quantum field theory and relativistic quantum many-body dynamics. In relativistic quantum field theory t is a parameter again. T has no sense. In relativistic quantum many-body dynamics t is an observable. Therefore associated a quantum operator T. but H is not the generator of time translations (energy H is valid for non-relativistic case and field theory only). The evolutor is the hamiltonian K. One equation of motion follows from known Stuckbelger/Horwitz/Piron theory, {PARTIAL |PSI> / PARTIAL TAU} = K |PSI> The Hamiltonian K is the covariant K = {p^a p_a / 2M} + V and wavefunction is PSI = PSI(x^a; TAU). Notice that x^0 = ct. Once you have quantum operator for time, you can study its spectrum, t_op = SUM |PSI> t <PSI| In [1] Horwitz computes quantum interference in time and gets separation between diffraction peaks of order 10^{−15} sec. Value in good agreement with the results obtained in a recent experiment (see literature on [1]). Another interesting aspect are unstable quantum systems. The Hamiltonian used by Baez is only valid for studying the dynamics of stable quantum systems. For unstable quantum systems H does not work. For unstable quantum systems an equation of evolution is [2] {PARTIAL RHO / PARTIAL T} = -iL RHO Where L is a superoperator with complex spectral decomposition. The complex modes z_v contain the decay rates of the unstable system (e.g. radiactive nuclei). Since the generator of the evolution is now a superoperator. It has no sense the search for a time operator T. One may search a time superoperator. This is done in "Explicit construction of a time superoperator for quantum unstable systems" [3]. [1] Found. of Phys 2007 37(4-5), 734. Lawrence, Horwitz [2] Chaos, Solitons and Fractals 2001 12, 2591. G. Ordonez, G; Petrosky T; Karpov, E; Prigogine, I. [3] Phys. Rev. A 2001 63, 052106. Ordonez G; Petrosky, T; Prigogine, I.  SignatureI follow http://canonicalscience.com/guidelines.txt Juan R. Gonzlez-lvarez wrote {fjm2q1$u67$1@aioe.org} on Tue, 11 Dec 2007 21:39:40 +0000: > mingyuming wrote > {e4cb1894-b4b7-43dd-a8c8-a1d00b0d05b1@b15g2000hsa.googlegroups.com} on > Sun, 09 Dec 2007 21:31:04 +0000: >> In below, I quote John Baez's original argument from his website: -------------------------------------------------------------------------- >> The problem is, for physically realistic Hamiltonians H one can prove >> there is no operator T with >> >> [H,T] = i hbar >> >> In other words, there is no time observable! I add several corrections to my post. > One would start by understanding what H and T are and what are their > limits of applicability. [quoted text clipped - 3 lines] > is not a observable, it has no sense to build an operator for it; T has > no sense. Since T has no sense (t is not an observable) the equation [H, T] = i hbar is meaningless. > Now turns to relativistic theory. There is two: relativistic quantum > field theory and relativistic quantum many-body dynamics. > > In relativistic quantum field theory t is a parameter again. T has no > sense. [H, T] = i hbar is meaningless again. > In relativistic quantum many-body dynamics t is an observable. Therefore > associated a quantum operator T. but H is not the generator of time > translations (energy H is valid for non-relativistic case and field > theory only). Since the generator of quantum evolution is not H (but K), one would not search [H, T] = i hbar even if now T can be defined. > The evolutor is the hamiltonian K. One equation of motion > follows from known Stuckbelger/Horwitz/Piron theory, [quoted text clipped - 10 lines] > > t_op = SUM |PSI> t <PSI| with t_op = T. > In [1] Horwitz computes quantum interference in time and gets separation > between diffraction peaks of order 10^{−15} sec. Value in good agreement [quoted text clipped - 4 lines] > used by Baez is only valid for studying the dynamics of stable quantum > systems. For unstable quantum systems H does not work. Eliminates "only". Above we saw that K is the generator for many-body systems. > For unstable quantum systems an equation of evolution is [2] > > {PARTIAL RHO / PARTIAL T} = -iL RHO typo, {PARTIAL RHO / PARTIAL t} = -iL RHO > Where L is a superoperator with complex spectral decomposition. The > complex modes z_v contain the decay rates of the unstable system (e.g. [quoted text clipped - 13 lines] > > [3] Phys. Rev. A 2001 63, 052106. Ordonez G; Petrosky, T; Prigogine, I. SignatureI follow http://canonicalscience.com/guidelines.txt On Dec 9, 4:31 pm, mingyum...@yahoo.com wrote: > Hi, > [quoted text clipped - 31 lines] > surprising - but it's sort of surprising that you can *prove* it, and > it's sort of interesting to see what assumptions you need to prove it. Thank you for all the answers. One thing that is still not clear to me is this: We can write down the classical (not quantum mechanical) dynamic equation for the oscillator in the so-called action-angle variables by the well know canonical transformation found in text books. Suppose we call P the action and Q the angle, then P and Q are well known functions of p (the usual momentum) and q (the usual coordinate) found in textbook. Further more, P and Q are real physical quantities. (it is also possible to express p and q in terms of P and Q). Here, we are not talking about quantum mechanics yet. In classical mechanics, we thus have {P,Q}=1 and H=P in the classical sense (here, no quantum mechanics is involved). So, {H,Q}=1 in classical sense. The above Possion brackets are in classical sense. This is a classical Hamiltonian system. And we can write down the classical equations in terms of P and Q (not in terms of p and q). Then, by standard procedure, if we quantize the above classical Hamiltonian system (written as P and Q), the classical variable will become quantum mechanical operators, and the Possion bracket become commutors. This means, for the oscillator, [P,Q]=ihbar and [H,Q]=ihbar, where P,Q,H are operators and [,] are commutors. Since P and Q (and H) are real physical quantities in classical limit, I would argue they are observables in quantum mechanics for the oscillator. We can just rename our Q to T is we want to. I would appreciate if someone can point out the mistakes in the logical steps. > Thank you for all the answers. One thing that is still not clear to me > is this: [quoted text clipped - 25 lines] > I would appreciate if someone can point out the mistakes in the > logical steps. The "mistakes" are all in the para beginning "Then, by the standard procedure", and in particular in your interpretation of what you are doing. You are indeed "quantising" an oscillator by your procedure, but it is not equivalent to the standard quantisation. In general this is because so-called Dirac quantisation is not actually well- defined - it is only an ad hoc prescription for trying to find the "right" quantum Hamiltonian operator for a given system (for which there are MANY candidates, all having the same classical limit !). Technically, the inequivalence arises from the fact that if you make a classical canonical transformation from (q,p) to (Q,P), the two sets of Dirac operators are not necessarily related by any unitary transformation. The case of the oscillator is simple enough to actually make a connection of sorts between the two inequivalent quantisations. Briefly, the usual quantisation gives a set of energy eigenstates |0>, |1>, |2>, .... . Now, compare this to a quantisation of a 1D rotator (i.e., a particle on a circle), with canonical coordinates (phi, J) - which are just the prototypical action-angle variables. We know that the quantisation gives in this case a set of angular momentum eigenstates ..., |-2>, |-1>, |0>, |1>, |2>, ... . You can see the inconsistency now already: the spectrum of an "action" variable has arbitrarily negative values - so the Hilbert space is too large compared to that of the standard quantisation. In particular, following your quantisation method will give negative energy eigenvalues, which are unphysical. For this reason we choose the first, inequivalent quantisation. To indicate the formal connection between the two, let's suppose we follow your quantisation through, so that the energy spectrum has negative and positive eigenvalues. Now consider the subspace corresponding to the span of just the zero and positive eigenvalues. Then, this subspace is clearly unitarily equivalent to the usual Hilbert space of the oscillator. In this sense, we can interpret the loss of a Hermitian time operator for the oscillator as being from having to work on only half the "full" Hilbert space of both positive and negative eigenvalues. For example, consider the operator V = sum |n> <n+1| with summation over positive and negative eigenvalues. This operator is unitary, and is essentially the same as exp[i phi]. However, if we truncate to nonnegative eigenvalues, we get the operator U = |0> <1| + |1> <2| + .... which is only semi-unitary (one has UU*=1, but not U*U=1). Let me know if you would like references. >Thank you for all the answers. One thing that is still not clear to me >is this: [quoted text clipped - 25 lines] >I would appreciate if someone can point out the mistakes in the >logical steps. Here's a partial answer, I think. One of the hypotheses of the theorem that says there can be no operator T with [H,T] = i hbar is that H be bounded below. The operator H = P is not bounded below, so the theorem doesn't apply. -Ted Signature[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] >[snip] > Then, by standard procedure, if we quantize the above classical [quoted text clipped - 8 lines] > I would appreciate if someone can point out the mistakes in the > logical steps. In essence, you seem to be saying that there exist operators P and Q satisfying [P,Q] := PQ-QP = i, and that if we rename P as H and call it a "Hamiltonian" then also [H,Q] = i. And we can rename Q as T and call it a "time" operator. Or from a more physical point of view, perhaps you are asserting that there exists a physical system whose Hamiltonian H is identical to P and such that if we measure the above Q, then we are really measuring time. That's all fine, but it misses the point. Your original post seemed to ask why the above construction does not contradict John Baez's assertion that for an operator H which is bounded below (as is traditionally assumed for physical Hamiltonians), there cannot exist an operator T with [H,T] = 1. THE ANSWER: The reason there is no contradiction between your construction and Baez's assertion is that your "Hamiltonian" is not bounded below. It is possible to arrange the formalism so that it may APPEAR that your Hamiltonian may be bounded below, but this is an illusion. If you try to write out the mathematics rigorously, you will find that it cannot be done. Trying to force your Hamiltonian to be bounded below causes a contradiction to appear elsewhere in the argument. A reply to "a student" (another poster in this thread) explains this more fully. On Dec 13, 3:59 pm, a student <of_1001_nig...@hotmail.com> wrote: > > Thank you for all the answers. One thing that is still not clear to me > > is this: [quoted text clipped - 72 lines] > > - Show quoted text - Thank you for the explaination. I'd like to have references for further reading. On Dec 13, 3:59 pm, a student <of_1001_nig...@hotmail.com> wrote: > > Thank you for all the answers. One thing that is still not clear to me > > is this: [quoted text clipped - 72 lines] > > - Show quoted text - I was always under the impression that canonical trasforms in classical mechanics corresponds to unitary tranforms in the quantum mechanics. But as you convienced me, this is not always true. If you can provide some references, I am interested in knowing under what condition it is true or not true. There has to be some relationship even when this simple coorespondence fails. On Dec 14, 6:00 pm, mingyum...@yahoo.com wrote: [snip] > I was always under the impression that canonical trasforms in > classical mechanics corresponds to unitary tranforms in the quantum > mechanics. But as you convienced me, this is not always true. If you > can provide some references, I am interested in knowing under what > condition it is true or not true. There has to be some relationship > even when this simple coorespondence fails. In regard to the oscillator, and the two different Hilbert spaces for (q,p) and (E,t), I was thinking of references such as [1] R.G. Newton, "Quantum action-angle variables for harmonic oscillators", Ann. Phys. 124 (1980) 327 (I remember this as very clear); [2] D.T. Pegg and S.M. Barnett, "Phase in quantum optics", J. Phys. A 19 (1986) 3849 (less clear). As to general references for inequivalent quantisations, I don't know much about this area. The abstract of the paper [3] Boya et al, "Exotic representations of the isotropic harmonic oscillator", Int. J. Mod. Phys. A 13 (1998) 3347, available at http://adsabs.harvard.edu/abs/1998IJMPA..13.3347B, looks interesting, but I can't get hold of the full paper. A google search shows there are plenty of refs for field theories, but not for finite configuration spaces such as the oscillator. The POVM description of oscillator time/phase was first given in [4] C.W. Helstrom, Int. J. Theoret. Phys. 11 (1974) 357, and his book [5] C.W. Helstrom, "Quantum Detection and Estimation Theory" (Academic Press, 1976) has a good intro to POVMs in general (he calls them POMs). An independent approach which foreshadowed the same idea is [6] J.M. Levy-Leblond, "Who is afraid of non-Hermitian operators? - a quantum description of angle and phase", Ann. Phys. 101 (1976) 319. A book that reviews all the different attempts to resolve the description of time/phase for the harmonic oscillator is [7] Perinova, Luks and Perina, "Phase in Quantum Optics" (World Scientific, 1998). > In below, I quote John Baez's original argument from his website: > > But a physically realistic Hamiltonian must be bounded below! Hmm. I once got that remark thrown at me about the Hamiltonian for the parametric oscillator, which being cubic, isn't bounded below. Nevertheless it _is_ bounded below in the region where the approximations used in its construction are valid; i.e. for low excitations. I consider that physically realistic enough for practical purposes. Signature---------------------------------+--------------------------------- Dr. Paul Kinsler Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714 Imperial College London, Dr.Paul.Kinsler@physics.org SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/ >> In below, I quote John Baez's original argument from his website: >> > But a physically realistic Hamiltonian must be bounded below! [quoted text clipped - 5 lines] > used in its construction are valid; i.e. for low excitations. I consider > that physically realistic enough for practical purposes. Similar remarks, about the boundedness from below of physically realistic Hamiltonians, are made often. It should be pointed out, though, that their validity strongly depends on what is meant by "physically realistic". Let me give an example. Think back to projectile calculations in introductory physics. The only force present is a uniform and constant acceleration due to gravity. This is none other than a linear potential. This potential has no minimum, and the total energy is unbounded from below. However, this observation doesn't seem to bother any of the introductory physics students or their instructors, and rightly so. The reason is that they are not looking for the minimum of the potential or for stable/periodic particle motion. Instead, they are looking for the trajectory of the projectile particle between leaving and hitting the ground. This trajectory is part of a larger trajectory where the ground is absent and the projectile comes from and escapes to infinity. While this trajectory itself is not physically realistic, parts of it are and there is no harm in studying the entire trajectory before deciding which part of it is relevant. Coming back to quantum mechanics, it's simple to formulate a Hamiltonian for the analogous problem. It is again not bounded from below. This implies that there is no ground state. However, there is still an entire L^2 space of states, just none of them can claim to be of minimal energy. The Hamiltonian can be extended to be self-adjoint and a unitary evolution operator U(t) = exp(-iHt) can be defined. So, even if one is considering a linear potential together with infinite potential walls (such as the surface of the Earth), this evolution operator may give a good approximation to the short time dynamics of a wave packet distant from any of the potential walls. While this may not be as simple as in classical mechanics, this use of U(t) is still possible and I think falls into a physically realistic scenario (if only as an approximation). A similar discussion can be made for the inverted harmonic oscillator. Igor |
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