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Quantum mechanical operators.

  1. Jan 12, 2005 #1
    In quantum mechanic you always use operators. You can get the eigenvalues of functions (with this operators). But how can we derive this operators? (Can somebody me derive the operator of momentum oder the operator of place?).

    Thanks for every answer.
  2. jcsd
  3. Jan 12, 2005 #2


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    Rigorously, it cannot be derived from the postulates of QM.
    However, it is natural to assign the operator X (which corresponds to multiplication with x) to position and [itex]\frac{\hbar}{i}\frac{\partial}{\partial x}[/itex] momentum. (Considering 1D only).
    There are all kinds of properties that hint towards this.
    I`ll try to give a few. I`m sure others can supplement on this.

    Firstly, (important but not convincing), they both posses a continuous spectrum. Which means all real values are possible for the position and momentum values and experiment has shown that this is indeed the case.

    Second. By applying the following postulate (the fourth or so)
    -- which states in this case that the probability of measuring the position (momentum) between x and x+dx (p+dp) is:
    [tex]|\langle \phi_x | \psi \rangle |dx \qquad \left(\, |\langle \phi_p | \psi \rangle |\right dp \, )[/tex]
    where [itex]\phi_x \, (\phi_p)[/itex] is the eigenvector associated with the eigenvalue x (p) of the observable X (P)--
    we can derive the probabilistic interpretation of the wavefunction as well as that of its Fourier transform.

    The eigenvalue of the eigenstate [itex]\exp(ikx)[/itex] of the momentum operator has eigenvalue [itex]\hbar k[/itex].
    Remember that a wavefunction in this state has a well-defined momentum.
    According to the de Broglie relation: [itex]p=h/ \lambda[/itex] and the relation between wavenumber and wavelength: [itex]\lambda=2\pi/k[/itex] we find that the momentum is [itex]p=\hbar k[/itex]. Exactly that eigenvalue.

    It can also be 'derived' if we use the probabilistic interpretation of the wavefunction and postulate that [itex]\langle p \rangle = m\frac{d}{dt}\langle x \rangle [/itex]. This is done in Griffiths Intro. to QM. Although it may be better to use the above arguments and show that this operator leads to this nice classical result between the two observables.

    I hope that helps.
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3


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    Or, for a more abstract point of view, quantum mechanical operators simply form an algebra with some properties.

    The familiar operators acting on a space of wave-functions just happens to be a convenient representation of this algebra. That QM began with this representation is purely history.

    However, there's probably a theorem or something that says this type of representation is a good one.
  5. Jan 12, 2005 #4


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    If u thought Hurkyl's opinion was abstract,check this out. :tongue2:

    I'll try to give some rigurosity to the statements made by Galileo.

    The heart of the second/quantization postulate.How to quantize classical systems.Take the Hamiltonian observable and associate to it a densly defined self-adjoint linear operator.That's how you get quantum observables.As for the fundamental commutatio relations,they come from the canonical quantization rule (used in QFT as well):
    [tex] [A,B]^{*}\rightarrow \frac{1}{i\hbar} [\hat{A},\hat{B}]_{\pm} [/tex](1)
    ,which reads:graded Dirac bracket goes to graded commutator multiplied by (1/i\hbar).

    As for momentum and position operator,let's consider the simplest case of the general formula (1).Consider a classical dynamical system described at Hamiltonian level by 2 commutative/bosonic phase space coordinates (i assumed that the system was nondegenerate) "[itex]x[/itex]" and "[itex]p_{x}[/itex]".Then graded Dirac bracket goes to bosonic Poisson bracket which has the form
    [tex] [x,p_{x}]=1. [/tex](2)
    Applying the quantization postulate,we get:the densly defined unbounded selfadjoint linear operators [itex] \hat{x} [/itex] and [itex] \hat{p}_{x} [/itex] (i didn't want to place the index under the hat,should it should have been) which satistfy the fundamental commutation relation:
    [tex] [\hat{x},\hat{p}_{x}]_{-}=i\hbar \hat{1} [/tex] (3)

    So that's how u quantize a simple system.


    PS.U see that the quantization postulate has been applied to the constant "1" as well.It became the unit operator.
  6. Jan 12, 2005 #5


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    Historically, Bohr's Correspondence Principle was an important guide. It says that the values predicted by the use of quantum mechanical equations need to coincide with the classical values in the limit of large quantum numbers.

    There are weaknesses in that principle, however, and there are situations where it is not applicable.
  7. Jan 12, 2005 #6


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    I won't say anything new,this is done in most serious QM texts.I can pin point to one,specifically,so i'll try to use the course which has been taught to me at school and my memory.

    Consider one quantum system with 2 degrees of freedom (like in the prior post) described by the densly-defined unbounded selfadjoint linear operators [itex] \hat{x} [/itex] and [itex] \hat{p} [/tex].For simplicity,i didn't place the index "x",but it's understood.
    Consider the system in the pure state [itex] |\psi\rangle [/itex].I want to apply Dirac's formalism (which is found in the nonrelativistic QM in the Dirac/traditional/vectors and operators formulation in the textbook/Copenhagen interpretation) to compute this number:
    [tex]A=:\langle \psi|\hat{p}|\psi\rangle [/tex] (1)
    in the position representation and extract,in this representation,the form of the momentum operator,knowing the form of the state vector,which,for hystorical arguments,is called "wavefunction".

    We make use of both momentum and coordinate representations.Passing from the separable abstract Hilbert space in which the state vector [itex] |\psi\rangle [/itex] is found to the more familiar [itex] L^{2}(R) [/itex] is made by means of a set of generalized Dirac orthonormed continuous unbounded linear functionals [itex] \langle p| [/itex] and [itex]\langle x| [/itex] which satisfy both the orthormalization relations (nonimportant in this case),but also allow the construction of generalized orthogonal projectors which satisfy the completeness relations:
    [tex] \int dp |p\rangle \langle p|=\hat{1} [/tex] (2)
    and similar for the "x".Moving the momentum operator into the linear functional Hilbert space [itex] \tilde{H} [/itex] (in which we considered the linear functionals),it satisfies the spectral equation:

    [tex] \langle p|\hat{p}=p\langle p| [/tex](3)

    Passing into the original Hilbert space of states is made,of course,not formally (as encountered in the case of spectral equations which don't have solutions it the original Hilbert space) and it reads:
    [tex] \hat{p}|p \rangle =p |p\rangle [/tex] (4)

    We write the original quantity (average) inserting the completeness realation (2) and making use of (3):
    [tex] A=:\langle \psi|\hat_{p}|\psi\rangle =\int dp \langle \psi|p\rangle \langle p|\hat{p}|\psi\rangle [/tex] (5)

    and definig the momentum wave function
    [tex] \langle p|\psi\rangle =\tilde{\psi} (p) [/tex] (6)
    ,we get
    [tex] A=\int dp \tilde{\psi}^{*} (p) p\tilde{\psi}(p) [/tex] (7)

    We define the wave vector (component "x",of course)
    [tex] k=:\frac{p}{\hbar} [/tex] (8)
    (this is due to Louis de Broglie,Ph.D thesis,Nov.1924) ans we write (7) as
    [tex] A= \int \frac{dk}{2\pi} \tilde{\psi}^{*}(k)\hbar k\tilde{\psi} [/tex] (9)

    I remember being taught in my first seminar of QM the Parseval identity (apply it for 1D):
    [tex] \int \frac{dk}{2\pi} \tilde{f}^{*}(k) \tilde{g}(k)=\int dx f^{*}(x) g(x) [/tex] (10)

    Making the notations:
    [tex] \tilde{f}(k)\rightarrow \tilde{\psi}(k) [/tex] (11)
    [tex] \tilde{g}(k)\rightarrow \hbar k \tilde{\psi}(k) [/tex] (12)
    [tex] f(x)\rightarrow \psi(x) [/tex] (13)
    and using the Fourier transform and the Parseval identity (10)
    [tex] g(x)=\int\frac{dk}{2\pi} \hbar k\tilde{\psi}(k) e^{ikx} =-i\hbar \frac{d}{dx}\int \frac{dk}{2pi} \psi (k) e^{ikx} =-i\hbar \frac{d\psi(x)}{dx} [/tex] (14)

    Going with (14) into (9),by using again the Parseval identity,we find:
    [tex] A=\int dx \psi^{*}(x) (-i\hbar \frac{d}{dx}) \psi(x) [/tex](15)
    ,from which we extract the form of the operator of momentum in the coordinate representation:
    [tex] \hat{p}_{x}=-i\hbar\frac{d}{dx} \hat{1} [/tex] (16)


    PS.I couldn't come up with a simpler and at same time mathematically and QM rigurous approach... :tongue2:
  8. Jan 13, 2005 #7
    Thanks for the answers. Mhh, this hilbert space. I begun to learn working with the hilbert space a few days ago. This vector space is really as complex as the name is. Does someone know a site with some theory of this space? I ve only found some sites of universities and I dont study physics yet. I have to study your answers later (in 2 weeks, when I can calculate with this).
  9. Jan 13, 2005 #8


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    Before you get too deep, why don't you just restrict yourself to the simplest Hilbert space, the one the books use for introductory problems, the complex numbers. Then the operators ar just complex numbers too.

    The next step up would be to consider the cartesian product of a small number of copies of the complex numbers, say 3; this would be the space of vectors descibed by three components, each one a complex number. Then the operators would be matrices (3X3 ones in this example) that would act on the vectors.
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