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Prrof of momentum operator

  1. Nov 9, 2012 #1
    I know that the average momentum <p> is defined as m[itex]\frac{d}{dt}<x>[/itex]. But why is this also equal to :
    [itex]\int[/itex]ψ*[itex]\frac{h}{(2\pi)i}[/itex][itex]\frac{\partial ψ}{\partial x}[/itex]dx ?

    the integral goes from negative inf to inf, * indicates conjugate,ψ the wavefunction.

    Also, why is it in general that for any average value <q> there is an operator [q] such that
    <q> = [itex]\int[/itex]ψ*[q] ψ dx ?

    I cannot find simple explanation anywhere and my book just skips the derivation..

    Also, how come for the potential energy operator [U(x)]:
    [U(x)] = U([x]) = U(x) ??
    I understand that [x] = x, so U([x]) must equal U(x), but I dont understand the jump from [U(x)] to U([x])..
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    bhobba

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    Get a hold of Quantum Mechanics - A Modern Development by Ballentine and see Chapter 3 for a deeper understanding of the why of momentum operators etc

    But one of the axioms of QM is <A> = Tr(uA) = <u|A|u> if u is a pure state from which your formulas follow.

    Thanks
    Bill
     
    Last edited: Nov 9, 2012
  4. Nov 10, 2012 #3

    dextercioby

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    The average momentum in a pure state [itex] \psi[/itex] is not defined by [itex] m\frac{d}{dt} \langle \hat{x}\rangle_{\psi} [/itex], but by [itex] \langle\psi,\hat{p} \psi\rangle [/itex].
     
  5. Nov 10, 2012 #4
    Sorry Im having trouble understanding what pure state is. Is this same thing as 'stationary state'? My book defines a stationary state as one in which probabilities are constant in time and the solutions can be given in separable form.

    Also ive attached image of the page of my book that defines momentum like in my previous post. I am assuming this definition is for a stationary state since the book pretty much only deals with these states. But I don't know if this is the same thing as a pure state..
     

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  6. Nov 10, 2012 #5

    Jano L.

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    Hello Aziza,
    you have interesting questions.

    One possible way to see it is this. We write down the expression for average value <x>:

    [tex]
    \langle x\rangle =\int \psi^* x \psi dx,
    [/tex]

    which is due to Born (probabilistic understanding of [itex]\psi[/itex]). With this assumption, we calculate average momentum

    [tex]
    m \frac{d}{dt} \langle x \rangle = m \int \partial_t \psi^*x\psi + \psi^* x \partial_t \psi ~dx~~(*)
    [/tex]


    Now we assume that time change of [itex]\psi[/itex] is described by Schroediger's time dependent equation:

    [tex]
    \partial_t \psi = \frac{1}{i\hbar} \hat{H} \psi
    [/tex]

    with the simple Hamiltonian (more complicated ones can be used)

    [tex]
    \hat{H} = \frac{\hat{p}^2}{2m} + U(x),
    [/tex]

    where
    [tex]
    \hat{p} = -i\hbar \frac{\partial}{\partial x}
    [/tex]

    is assumed.

    Plugging these two equations to express the time derivatives in the equation (*) and using Hermitian property of H, we obtain


    [tex]
    m \frac{d}{dt} \langle x \rangle = \int \psi^* \frac{im}{\hbar}[\hat{H}x - x\hat{H}] \psi ~dx
    [/tex]

    It is easily checked that

    [tex]
    \frac{im}{\hbar}[\hat{H}x - x\hat{H}] = \hat{p}
    [/tex]

    Well, it works for x due to Born's rule and for p due to above derivation, so it is possible that it will work also also for other more complicated expressions based on them. But there could be some problems with operators containing higher derivatives than second and I think it is always good to check for every new operator whether <q> = ∫ψ*[q] ψ dx works.

    There are two aspects.

    First, the question is, why the operator U in Schroedigner's equation is the multiplication by the same function of coordinates as given by classical theory and not something more complicated?

    Simple answer is, this is what Schroedinger tried and it worked (mostly for explaining hydrogen and other spectra). There is presently no deduction of general Schroedinger's equation or the rule "take the classical expression for U to from H" from something more simple or more clear.

    Second, the question is, why average of potential energy is calculated by the formula

    [tex]
    \langle U\rangle = \int \psi^* U(x) \psi ~dx
    [/tex]

    where U(x) is just the standard potential energy function of coordinate of particle, and not with U(x) more general?

    This is probably what you refer to in
    This can be explained as consistency with or consequence of the classical theory.
    In probability theory, assuming [itex]d(x)[/itex] is the probability density that the particle is at x and assuming that the potential energy for particle at x is given correctly by classical theory as U(x), the average value is calculated by means of probability density as

    [tex]
    \langle U \rangle = \int d(x) U(x) dx.
    [/tex]

    In the theory using [itex]\psi[/itex], Born introduced the prescription that d(x) should be calculated from the [itex]\psi[/itex] function as

    [tex]
    d(x) = \psi^*(x)\psi(x),
    [/tex]

    which gives us

    [tex]
    \langle U\rangle = \int \psi^* U(x) \psi dx
    [/tex]

    with simple function U(x), so nothing more complicated than this seems to be necessary.
     
  7. Nov 10, 2012 #6

    bhobba

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    This is one of the problems with the more elementary texts - they don't really give the full story.

    They give the impression that quantum states are elements of a complex vector space - this is not entirely true - it is just one issue they gloss over - there are others as well. They are in fact positive operators of trace 1. |u><u| where u is any element of the vector space is such an operator and is known as a pure state. It can be proven that any state that is not a pure state is a convex sum of pure states - non pure states are called mixed states. Physically their interpretation is a statistical ensemble of pure states and is very important in decoherence theory as well as other areas of QM such as interpretational issues. To understand this more I really do recommend getting a hold of the book by Ballentine I mentioned before where it is explained properly.

    Stationary states are the eigenvectors of the observable you are considering. Because of that when the operator is applied to the stationary state its a number and you can take it to the outside of the average formula so it is always that value.

    However for now just simply accept that one of the foundational axioms of QM is the average of an observable A, <A>, is <A> = <u|A|u> and your formulas follow trivially from that axiom by expanding the inner product in terms of the eigenvectors of position.

    It is an unfortunate issue with QM that the textbooks giving a first brush with QM often leave important issues unanswered and if you are a genuinely thinking student can leave you scratching your head. Rest assured that more advanced texts like the excellent book by Ballentine (another plug - do give it a quick look - it will make things a lot clearer) fix them all up and in a mathematically very beautiful way. I well remember my first brush and its use of the damned Dirac Delta function and calling the vector space containing them Hilbert spaces - it isn't - I was pulling my hair out - other books like Von Neumann's classic Mathematical Foundations of QM - gave the correct mathematical theory that skirted such things and it took me ages to sort it out by delving into some very advanced mathematical tomes. Eventually I did but it took me ages. I wish I had read Ballentine where even that issue is explained properly.

    Thanks
    Bill
     
    Last edited: Nov 10, 2012
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