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Momentum,discrete or continuous spectrum?

  1. Jan 3, 2013 #1

    ShayanJ

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    At first let's take a look at the eigenvalue problem for the momentum operator in x-representation.
    [itex]
    -i \hbar \frac{d}{dx} \psi(x)=p \psi(x) \Rightarrow \psi_p(x)=C e^{{ipx}/{\hbar}}
    [/itex]
    The orthogonality condition is:
    [itex]
    \langle p_i|p_j \rangle=C^2 \int_R e^{i(p_j-p_i)x/{\hbar}} dx=C^2 \delta(p_i-p_j)
    [/itex]
    And so the spectrum of the momentum operator is continuous regardless of the problem at hand.
    Now consider the particle in infinite well as an example.
    The energy levels are given by:
    [itex]
    E_n=\large{\frac{n^2 h^2}{8 m L^2}}
    [/itex]
    We can find the momentum levels :
    [itex]
    \large{\frac{p_n^2}{2m}}\small{=E_n} \Rightarrow \large{\frac{p_n^2}{2m}\small{=}\frac{n^2 h^2}{8 m L^2}} \small{\Rightarrow p_n^2}=\large{\frac{n^2 h^2}{4 L^2}} \small{\Rightarrow p_n}=\large{\frac{n h}{2L}}
    [/itex]
    Which means the momentum is quantized in this example.But we know that the spectrum of the momentum operator should always be continuous.
    What am I missing?
    Thanks
     
    Last edited: Jan 3, 2013
  2. jcsd
  3. Jan 3, 2013 #2
    the reason your orthogonality yields delta function relies on the integration domain being the entire real line, which is not true in bound states, in fact bound states all have discrete spectrum, and scattering states continuous spectrum
     
  4. Jan 3, 2013 #3

    vanhees71

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    Here some care is necessary. For the infinite well you don't have a well-defined momentum operator. In fact your Hilbert space is represented (in the position representation) by the wave functions in [itex]\mathrm{L}^2([0,L])[/itex] with the boundary conditions
    [tex]\psi(0)=\psi(d)=0.[/tex]
    Now you can consider the operator [itex]-\mathrm{i} \partial_x[/itex], which represents the momentum operator in free space. The important point, however is that this is a self-adjoint operator in free space. That's not the case in the case of the infinite well. Formally it seems all fine, because
    [tex]\int_0^L \mathrm{d} x \psi_1^*(x) (-\mathrm{i} \partial_x) \psi_2(x)=\int_0^L (-\mathrm{i} \partial_x \psi_1)^* \psi_2(x),[/tex]
    because you can integrate by parts and the boundary terms vanish due to the boundary conditions of the wave functions.

    However, [itex]-\mathrm{i} \partial_x[/itex] doesn't map any dense subspace of the Hilbert space into itself, and thus the operator is only Hermitean and not (essentially) self-adjoint! In consequence, there are no (generatlized) eigenfunctions and thus the spectrum of this operator is empty. Of course the eigen-value equation can be solved, and you get the exponential solutions as in the free case, but there is no solution, fulfilling the boundary conditions. So for the infinite well it doesn't make sense to consider this operator as an observable.

    This is not the case for the Hamiltonian, because
    [tex]\hat{H}=-\frac{\partial_x^2}{2m}[/tex]
    is self-adjoint. The eigenvalue equation is
    [tex]u_E''(x)=-2m E u_E(x).[/tex]
    The solutions of this differential equation, fulfilling the boundary conditions are given by
    [tex]u_E(x)=C \sin(k_E x) \quad \text{with} \quad k_E=\sqrt{2 m E}.[/tex]
    The boundary conditions imply
    [tex]k_E L:=k_n L= n \pi, \quad n \in \mathbb{N}=\{1,2,3,\ldots \}.[/tex]
    The energy eigenvalues are thus
    [tex]E_n=\frac{k_n^2}{2m}=\frac{\pi^2 n^2}{4 L^2}.[/tex]
    From known theorems about Fourier series we know that this is a complete orthonormal set of functions in the Hilbert space [itex]\mathrm{L}^2([0,L])[/itex] with the homogeneous boundary conditions suitable for the infinite square well potential.

    One should note that there is no problem with defining the momentum operator in a slightly different space, namely [itex]\mathrm{L}^2([0,L])[/itex] with periodic boundary conditions,
    [tex]\psi(0)=\psi(L).[/tex]
    Then the operator [itex]\hat{p}=-\mathrm{i} \partial_x[/itex] is Hermitian, because when doing the integration by parts again the boundary terms drop because of these boundary conditions. Further the eigenvalue equation again has the exponentials as solution, i.e.,
    [itex]u_p(x)=C \exp(\mathrm{i} p x), \quad p L=2 \pi n, \quad n \in \mathbb{Z}=\{0,\pm 1,\pm 2,\ldots \}.[/itex]
    These are obviously [itex]\mathrm{L}^2([0,d])[/itex] functions and a complete set of orthonormal Hilbert-space vectors. Thus the operator in this Hilbert space is self-adjoint and thus a valid description of an observable.

    Note that the spectrum of the corresponding Hamilton operator, which now can be written as
    [tex]\hat{H}=\frac{\hat{p}^2}{2m}[/tex]
    is different from the infinite-square-well case and that the energy eigenvalues are twofold degenerate (except the ground state for [itex]n=0[/itex] with energy eigenvalue 0), because the energy eigenvectors are given by the momentum eigenvectors with [itex]p=\pm \sqrt{2 m E}[/itex], where [itex]p[/itex] must take one of the discrete values [itex]p=2 \pi n/L[/itex], and for [itex]n \neq 0[/itex] for the corresponding energy eigenvalue [itex]E_n=(4 \pi n^2)/L^2[/itex] both momentum eigenstates for [itex]p=\pm 2 \pi n/L[/itex] lead to the same energy value. Note that for the infinite well the energy eigenvectors where non-degenerate.

    The deeper reason for all this are symmetries: The example of the Hilbert space for functions on a finite interval with periodic boundary conditions has to symmetries, which are absent for the infinte well: (a) The physics is invariant under translations by multiples of [itex]L[/itex] and (b) the physics is invariant under space reflections at the middle of the interval. The latter is responsible for the two-fold degeneracy of the non-zero energy-eigenvalues.

    This simple example shows, how important the subtleties of self-adjointness of the operators that represent observables indeed are! A more formal mathematically rigorous treatment can be found in the paper
    Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
    http://arxiv.org/abs/quant-ph/9907069
     
    Last edited: Jan 3, 2013
  5. Jan 4, 2013 #4

    ShayanJ

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    I didn't understand this part,could you clarify?
    Thanks guys
     
  6. Jan 4, 2013 #5

    vanhees71

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    What precisely haven't you understood?
     
  7. Jan 4, 2013 #6

    ShayanJ

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    At first you said momentum operator is not self adjoint and then you said it is in the example given.I don't understand why it's self adjoint in that example.
    I also don't understand what you mean by the first sentence of what I quoted.
     
  8. Jan 4, 2013 #7

    vanhees71

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    Perhaps I shouldn't have posted this second example. I wanted to make clear that for the self-adjointness the domain of the Hilbert space on which the operators are defined is important and that observables should be described by operators that are not only Hermitean but self-adjoint, i.e., they must in addition to be Hermtian also be defined on a dense subset of the Hilbert space and the image of this subset under the operator must belong to this subset again.

    In the first example, the Hilbert space was given as the functions that are square integrable functions on the interval [itex][0,L][/itex] and fulfill the boundary conditions
    [tex]\psi(0)=\psi(L)=0.[/tex]
    This the wave functions have to fulfill for the infinite-square-well potential.

    As I've showed in my previous posting, on this Hilbert space the operator [itex]-\mathrm{i} \partial_x[/itex] is Hermitean but not self-adjoint, and this leads to trouble with its possible interpretation as an operator representing an observable. In other words, in this space you cannot use this operator as representant of an observable, and in this sense there doesn't exist a momentum operator in this Hilbert space.

    In the second example, the Hilbert space was given as the square integrable functions on the interval [itex][0,L][/itex] but with periodic boundary conditions
    [tex]\psi(0)=\psi(L).[/tex]
    The difference is that here any values on the boundaries are allowed not only 0 but the function should still take the same value ath the boundaries of the interval. This is of course not representing the physics of a particle constrained to this interval by an infinite-square potential but more like a particle free to move along a circle.

    In this case, the operator [itex]-\mathrm{i} \partial_x[/itex] is self-adjoint and thus represents an observable. And thus you can define it as the momentum operator. In the picture of a particle along a circle it's the rotations around the center of the circle, i.e., it has more the characteristics of an angular momentum.
     
  9. Jan 4, 2013 #8

    jtbell

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    You know the wave function ##\Psi_n(x,t)## for the infinite square well in position space, right? Try calculating the momentum space wave function ##\Phi_n(p,t)## by using the Fourier transform and see what you get.

    $$\Phi_n(p,t) = \frac{1}{\sqrt{2 \pi \hbar}} \int^L_0 {\Psi_n(x,t) e^{-ipx/\hbar} dt}$$

    (The limits don't go to infinity because ##\Psi_n = 0## outside the well. That makes a big difference!)
     
    Last edited: Jan 4, 2013
  10. Jan 4, 2013 #9

    dextercioby

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    Part of the bolded part actually isn't really so. The underlined part of what you wrote means that one should seek for invariant (also densly defined) subsets of the operators' maximal domains, so that within these subsets the normal algebraic operations like addition, composition and combinations of the 2 (e.g. the commutator) would make sense. But if one indeed finds these subdomains, they, due to their inherent restriction, would make the operators, when considers acting on these subdomains, no longer self-adjoint, but essentially-self adjoint at most, thing which would be excellent, actually, as argued below.

    The issue regarding essential self-adjointness on invariant subdomains is good enough for two reasons:

    1. essentially self-adjoint operators have a unique self-adjoint extension, so we can safely take them as the true observables.
    2. invariant subdomains on which the operators are typically e.s.a. can be further/supplimentary topologized and with the additional topology one can build rigged Hilbert spaces which are by the Gelfand-Kostyuchenko-Maurin theorem the necessary* mathematical constructs to accommodate the standard formulations of Quantum Mechanics and Axiomatical Quantum Field Theory.

    *I think of them as necessary, as physicists love the bra-ket formulation of quantum physics which really can't be formulated à la von Neumann 1932.

    P.S. Sides, if the operator is hermitean (I use the word 'symmetric'), then it's automatically defined on a dense everywhere subset.
     
    Last edited: Jan 4, 2013
  11. Jan 4, 2013 #10

    strangerep

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    I'm not sure this is true for QFT...

    In QM, one typically works with a finite number of independent operators, and finds the intersection of their domains on which they're all well-defined everywhere, giving a rigged Hilbert space as you said.

    But... what if we have an infinite number of independent operators? How do we know that the intersection of their domains is something less trivial the empty set?? That would at least need a sophisticated argument in terms of some topology on the operators, or so I guess. This is one aspect of rigged Hilbert space theory that I haven't seen addressed thoroughly, and I'd dearly like to know the answer.
     
  12. Jan 4, 2013 #11

    dextercioby

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    Strangerep, indeed the problem you raised it quite important. As far as I know, the only two explicit constructions (actually using the proper name) of rigged Hilbert spaces within quantum field theory are the 1975 textbook by Bogolubov, Logunov, Todorov (also known as BLT*, Publisher: Benjamin) and the article by M. Gadella <A rigged Hilbert space for the free radiation field> (JMPhys, Vol.26, Iss.4., page 725, 1985).

    Due to the 'space-time' indirect (i.e. through test functions) functional dependence of each field/operator, one can say that they are indeed uncountably infinitely many operators. But I think this is misleading. I don't see this as an issue, the Schwartz space of test functions of 4 variables is an invariant essential self-adjointness domain for an arbitrary number of operators, and there's an (operator) algebra of finite number of generators in each space-time point. i.e. for each test function. The RHS over the (separable) Fock space should be the mathematical tool to describe free field theories.

    Perhaps I'm wrong, but the only source to treat QFT from RHS perspective is BLT*. Each mathematical physicists should own a copy of this book, like they should own copies of MTW or Wald.

    *As opposed to BLOT (Kluwer, 1990).
     
  13. Jan 4, 2013 #12

    strangerep

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    Have you checked the price of 2nd-hand copies? :eek:

    I read the Gadella paper some time ago, but imho it's insufficiently careful in extending its conclusions from finite tensor products to the infinite tensor product case. The BLT treatment seems similar on this point.

    It's a bit like the way some physicists see a well-defined n-dimensional integral $$\int dx_1 \dots dx_n ~ f(x_1, \dots , x_n) ~$$ and blithely assume it remains well-defined in the limit ##n\to\infty##.

    But you've got to prove that the subspace of the Fock space on which all those operators are defined everywhere is non-empty. Maybe an infinite tensor product of Schwartz spaces is ok though, since every member of the space is a function of rapid decrease. For ordinary inf-dim integrals, one usually adopts a Gaussian measure in place of the now-ill-defined translation-invariant countably-additive Lebesgue measure, so maybe if one is integrating only functions that fall off like Gaussians, all is well. But I don't know -- I'm only guessing. Hopefully Micromass or DarMM will stop by and set me straight properly.... :smile:
     
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