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Consider the following quantum system: a particle in a one dimensional box (= infinite potential well). The energy eigenstates wave functions all vanish outside the box. But the position eigenstates wave functions don't all vanish outside the box. Each one of them is a delta function at a specific location, and some of these locations are outside the box. So it seems that there is no overlap between certain position eigenstates and all energy eigenstates. So the energy eigenstates don't span the whole Hilbert space! And these position states have zero probability for any energy outcome in a measurement!

Now, I know that when speaking about an infinite potential well, it is assumed the particle cannot be outside the well. But I don't see any reason to assume this from the postulates of quantum mechanics. Is there an implicit additional postulates that says: "The Hilbert space of the system is spanned by the Hamiltonian operator's eigenstates (and not other operators, such as position)"? Or is the infinite potential well just an ill defined system because it contains infinities (just like free particle...)?

Qmechanic
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Lior
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  • Apart from the truncation-to-a-box issue, it is essentially a duplicate of this Phys.SE post in the sense that position eigenstates are not part of the Hilbert space. – Qmechanic Sep 05 '13 at 18:14
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    @Qmechanic, I don't see why it is a duplicate. Can you explain please? – Lior Sep 05 '13 at 18:24
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    @Qmechanic I think Lior is right. 'Apart from the core issue of the question...'. ;). – Emilio Pisanty Sep 05 '13 at 21:26
  • P.S, I know my question might seem silly or not worth the discussion, because it is obvious what a working physicist must do when she encounters a particle in a box, but to me the fact that I need to do something that is not derived from postulates but rather based on "common sense" or "intuition" means that this model of reality (= physical theory) is incomplete. – Lior Sep 06 '13 at 12:03

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The problem is that you're assuming that the postulates of quantum mechanics automatically assign systems a full position representation... whereas some systems (like a particle with spin) do not have such a representation.

The solution, then, is to look carefully at the postulates of quantum mechanics. There are a bunch of abstract ones - states are rays in Hilbert space, observables are hermitian operators, existence of a hamiltonian, normal unitary evolution under it, probabilities are expectation values, what happens with measurements, and so on - but none of those tell you which Hilbert space to use for which physical system, or what hermitian operators to use for your particular physical observables.

For that, you first need a lot of physical intuition, and you follow a general recipe which goes more or less as

If the system has a classical representation which includes a canonical symplectic structure with position and momentum coordinates defined on the whole real line, and a Poisson bracket which satisfies $\{x,p\}=1$, then assign a Hilbert space tensor factor of $L_2(\mathbb R)$ to each space dimension with position as the $x$ operator and momentum as such and such a derivative.

and which is known as canonical quantization.

Note an important caveat in this recipe: it requires position to be defined on an unbounded interval. Because of von Neumann's representation theorem, postulating the canonical commutation relations $[x,p]=i\hbar$ automatically requires the spectrum of both to be $(-\infty,\infty)$.

This is a very tricky point, and even Dirac stumbled with it: he proposed a quantum theory for the phase of a harmonic oscillator (The quantum theory of the emission and absorption of radiation. P.A.M. Dirac. Proc. R. Soc. Lond. A 114 no. 767, pp. 243–65 (1927)) which eventually proved to be fundamentally flawed. (A good source for why is probably R. Lynch, Phys. Rep. 256, 367 (1995), but Elsevier seems to be down at the moment.)

The bottom line of this is that you need to look at your classical system before you decide how you're going to quantize it. For a particle in an infinite well, does the classical system include the positions outside the well? If so, what's the potential there? It must be "very large", because "infinite" is not a valid value of an operator (i.e. $\hat V|x\rangle=\infty|x\rangle \notin \mathcal H$)... and then you're back in a finite well.

If your classical system does not include positions outside that box, then you need to be careful with what you want your quantum system to be. You definitely can't ask your quantum system to do more than your classical one, so position states outside the box should not form part of your Hilbert space. In one strike, this fixes your problem: energy eigenstates will span all of Hilbert space.

You still need to decide what operators you need to use for momentum and energy, and physical intuition usually serves well there. However, if you want to know exactly why we do things like we do, then you should be looking at the classical system for guidance as to how to quantize. As it happens, the classical system is not completely trouble-free, and any troubles you have quantizing you might have seen coming just from looking at the classical system! For an interesting take on this, I recommend the paper

Classical symptoms of quantum illnesses. Chengjun Zhu and John R. Klauder. Am. J. Phys. 61 no. 7, p. 605 (1993).

This includes a discussion, at the end of section III, of precisely this problem.

Emilio Pisanty
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  • For a classical particle in a box, with finite but very high walls, there is no restriction at all for the particle to be outside the box. The outside-the-box states have equal rights as the in-the-box ones. Of course, a transition between them requires a lot of energy, as the classical equations of motion tell us. But these states in no way disappear in the limiting process of taking the potential to infinity. So by your reasoning, when I quantize the system I should leave these states. – Lior Sep 06 '13 at 07:47
  • @Lior If you keep those states after the limiting process, you also break classical mechanics. You have an undefined or infinite hamiltonian on huge parts of phase space. If you want your qm to be sound, you need working classical mechanics, and that means throwing away those states. – Emilio Pisanty Sep 06 '13 at 10:45
  • I disagree. For any finite value of the potential, even if enormous, the classical particle can be outside the box. Therefore when you take the limit it can still be there. (also, you can always shift all the potentials in the system by a constant). Anyway, what I understand from your explanation is that I should not look for a missing postulate but rather for some principal in quantizing the classical system. – Lior Sep 06 '13 at 11:54
  • @Lior Yes, your last comment is right: you need to complement the postulates with a quantization procedure (or at least a procedure to actually find the right space and operators). However, you are wrong in thinking that having some behaviour allowed at finite energies then it will persist in the limit. Try constructing such a mechanics! what do Hamilton's equations look like there? – Emilio Pisanty Sep 06 '13 at 12:35
  • Define a function e(V) in which V is the value of the potential, and in which e(V) equals 1 whenever the classical particle is aloud to exist outside the box, and equals 0 otherwise. For any value of V, e(V)=1, therefore e(V) is the constant function 1. By elementary calculus, lim e(V) = 1, where the limit is taken on V to infinity. – Lior Sep 06 '13 at 12:52
  • @Lior it is unfortunately not that simple, and you have no guarantee that e(V) would be continuous at infinity. (Try, for example, $f(x)=1$ if $x$ is finite and 0 if it is infinite.) If you could construct a meaningful hamiltonian mechanics in that region (do try it!), then you could talk about quantizing it. – Emilio Pisanty Sep 06 '13 at 13:03
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    @Lior: With infinte potential the Hamiltonian is not defined on the whole Hilbert space (as an operator! - you can write it down, but <s|H|s> is not defined for many vectors/states in the full Hilbert space which was OK for the system with finite potential). Point is: The limit process has no physical meaning: It's just a way to make educated guesses for the proper description of some systems. – M.E.L. Sep 06 '13 at 13:24
  • Suppose you have a free quantum particle, and then at time t you turn on a potential which is an infinite well. I am not assuming that the particle is inside the well prior to time t - it may or may not, or it can be in a superposition of states. What is the Hilbert space of this system? Does it include position eigenstates outside of the well? Alternatively, you can assume that within a finite time span T you continuously turn on the potential from zero to infinity. – Lior Sep 06 '13 at 13:52
  • If within a finite time $T$ you continuously turn on a potential that diverges to infinity, then Schrödinger's equation will cease to have solutions. But, again, this is also present in classical mechanics, as the corresponding classical system also fails to have global solutions for all initial conditions. – Emilio Pisanty Sep 06 '13 at 13:56
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    Probably the best way to put it: There is no V(x) with V being infinite outside of the box. You simply can not write it down in terms of standard analysis. So the limiting process has no meaning. The expression "infinite potential well" or seomthing like this has also no meaning inside the mathematical formalism. It's just an informal way to talk about a physical interpretation of the situation we want to describe. The limit process also has no meaning: In the limit the operators do no converge. And that's it: There is no question left to asked. – M.E.L. Sep 06 '13 at 15:06
  • (I hope you don't lose the patience with me...) So if the infinite well is not a result of a limiting process (i.e, I can't say that it models the situation of a potential well with the value of 10000^10000 eV), what could it model? And why do we require that the wave function is zero on the edge of the box, if it is meaningless to say that the potential is infinity over there? Remember that that's how we determine the discrete spectrum of the system... – Lior Sep 06 '13 at 16:35
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    @Lior: Have you ever seen an infinite potential well? It's just about making a completely imaginary model universe (where things only exist in a box) in the (justified) hope the behaviour described/calculated is somehow related to a group of more realistic systems you're interested in (e.g. deep potential wells). This hope stems mostly from (informal) physical interpretation of the mathematical model. I'm almost sure one could complete existing mathematical model (e.g.) by way of non standard analysis to describe the limiting process, but I don't know of anybody who has done that). – M.E.L. Sep 06 '13 at 19:26
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    I'm out of here now: It gets repetitive. But thanks for the interesting question. Made me think a bit about the process of creating physical models :-). – M.E.L. Sep 06 '13 at 19:28
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My understanding is that in various problems in quantum mechanics the final step is to restrict the Hilbert space to physically permissible states. In this problem, such a restriction requires that the state is supported exclusively on the spatial interval in which the potential is finite. This would imply the resolution of your paradox is that the position eigenstates outside of this interval are not in the Hilbert space.

This is not the only example of such a restriction. In the harmonic oscillator there is a similar restriction, that we limit our Hilbert space to states which can be eventually annihilated to the vacuum, and we reject those which can be lowered arbitrarily. Similarly when quantising the vector field we find the non-physical degrees of freedom allow for states of zero norm, in order to recover physicality, and a theory which obeys the appropriate gauge conditions, we reject these.

ComptonScattering
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  • This comment is more of a question (serious one). Why are you using quantum mechanics; presumably the ultimate theory of physical reality (at least at small scales), to discuss a "problem", which is all about a quite fictitious thing; to whit, a "one dimensional box, or infinite potential well". What purpose is there in discussing something never observed or known to not even exist ?? Like I said, it's a serious question. –  Sep 06 '13 at 19:43
  • I will try to answer though if you are unsatisfied it may be useful to discuss further. If a theory contains a logical inconsistency, then it is not valid, and definitely not a theory for everything. While it is true the proof is in the pudding and the ultimate value of a theory is in its predictive power, we should pay attention to problems as they arise as solving them may help guide developments, and may prevent us from wild goose chases. As for the physicality of the problem, I don't believe its an appalling approximation to certain physical systems, though these are of course finite. – ComptonScattering Sep 07 '13 at 00:09
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This is just a question of choosing a base for the vector (Hilbert) space: The set of position eigenstates is one base, the set of energy eigenstates is another. They can be expressed in terms of each other (vectors/states from base B1 can be expressed as a sum of vectors from base B2).

All in the theory of vector spaces...

OP said: "But the point of my question is that for the particle in the box, a certain position eigenstate (any one which has as it's wave function a Dirac delta outside the box) cannot be expressed in terms of energy eigenstates"

Ah, you're right, now I understand your question. I think the point is, that the well is infinite. For a pfinite potential well, you always have "unbound" higher energy states which can be used to sum up the eigenvectors outside of the box. In the case of infinite well, those states have an energy of infinite (their energy goes to infinite if you let the energy well move to infinity). This way they sort of "vanish" from the model. But: If you accept the same for the position eigenstates (just dripping thos outside the box) everything is OK again: You have "universe" inside a box (a universe where the Koordinate is by design limited to some intervall) and again all position eigenstates can be used to build energy eigenstates and vice versa.

M.E.L.
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  • But the point of my question is that for the particle in the box, a certain position eigenstate (any one which has as it's wave function a Dirac delta outside the box) cannot be expressed in terms of energy eigenstates. – Lior Sep 05 '13 at 18:07
  • @Lior: I've put my response into my answer. That you have to throw away a part of the psoition eigenstates is IMHO due to the fact that [H,x] would not be right any more. – M.E.L. Sep 05 '13 at 18:18
  • And why the down votes? – M.E.L. Sep 05 '13 at 18:20
  • I see how this limiting process of taking the potential to infinity makes the energy wave functions vanish outside the well, but I don't see how the postulates or the limiting process should make me abandon the position wave functions outside the well. – Lior Sep 05 '13 at 18:33
  • @Lior: Just to get the "normal" situation: A Hilbert space where the x eigenvetors and the E eigenvectors span up the whole space. I admit, it seems pretty ad hoc, to throw away a part of the state space. If we don't do it, then you're right, that the Hamiltonian eigenverctors become special. Then the situation is that there is a Hilbert space, but only a subspace corresponds to physical states and this subspace is spanned by the energy eigenvectors. This is perhaps as well - to accept that the Hamiltonian is indeed special in QM (and in any dynamical theory as well). – M.E.L. Sep 05 '13 at 19:44
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    Another ways to look at this: (1) The axioms don't say you can get your Hilbert space from the finite case by some continous transition. They state, that there is a Hilbert space which describes the state of the physical system. Inthis case ist's the space of all function that are zero outside of the box. – M.E.L. Sep 05 '13 at 19:57