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Kinetic energy of matter

  1. Jun 13, 2009 #1
    Wikipedia's derivation of the Schrodinger equation apparently includes the premise that the energy of a particle is the product of Planck's constant and the particle's frequency.

    E = hf

    I have seen this equation before for photons but when applied to matter, I'm confused. If we assume the following

    [tex] E = \frac {p^2}{2m} [/tex]

    [tex] \lambda = \frac {h}{p} [/tex]

    [tex] p = mv [/tex]

    [tex] \lambda f = v [/tex]

    where

    E = energy
    p = momentum
    m = mass
    λ = wavelength
    f = frequency
    v = velocity

    simple algebraic manipulation yields E = hf/2 , not hf. The only relationship not mentioned in the derivation is the λf = v, but how could that not be true for a wave of any kind?
     
  2. jcsd
  3. Jun 13, 2009 #2
    One commonly accepted set of postulates for QM is:



    1. The state of a system is described by a vector belonging to a Hilbert space possessing an inner product.

    2. Observables are represented by hermitian operators (dynamical variables in classical mechanics) belonging to the same Hilbert space.
    The observed quantities are given by eingenvalues of this operator.
    Observation causes the system to assume the associated eingenvector (collapse).

    3. The evolution of the system is governed by the Schrodinger equation, derived from a Hamiltonian function which represents the distribution of the energy in the system.

    4. The probability amplitude of finding the particle in a given state is given by the inner product between this state and the original one (the probability is the square of the product's modulus).

    5. Two states can be represented by one state as their tensorial product.



    Treating the Schrodinger equation as the starting point is found for example in Griffiths' textbook.

    The kinetic energy appears when constructing the Hamiltonian function.
     
  4. Jun 14, 2009 #3

    clem

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    Science Advisor

    The p=mv equation should not be used. The v in your mv is not simply related to the v which is the velocity of the wave in f lambda=v. The wave function is
    exp[p.x-(p^2/2m)t/hbar]. You can find the wave and group velocity from that.
     
  5. Jun 14, 2009 #4
    This is a very good question which deserves a clear answer. Snoopies has done his calculations correctly and he is right to be confused. You get the wrong velocity for the particle if you take the formulas at face value.

    The phase velocity which Clem alludes to in post #3 is the apparent velocity of a single frequency. But for a pure frequency the particle is everywhere at once. To localize the particle, you need to take a mixture of frequencies spread closely around an average. This gives you a wave packet...and the velocity of the wave packet is altogether different from the phase velocity of the pure wave. The phase velocity is w/k, as expected. The group velocity turns out to be dw/dk.
     
  6. Jun 15, 2009 #5
    Phase velocity versus group velocity -- of course! Thanks conway, you have completely clarified this issue for me.

    Incidentally intervoxel, I just took out of copy of Herbert S. Green's Matrix Mechanics (1965) and it looks like a good introduction to what you were talking about, although for the moment I don't have a clear idea of how vectors and matrices are isomorphic to waves and functions of waves. But I am just beginning to study all this.
     
  7. Jun 15, 2009 #6

    diazona

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    Homework Helper

    It's no substitute for reading the book, but here's the gist: a function (which describes a wave - like the sine function, for example) is basically just a list of values, indexed by some number. f(0) = 0, f(1) = 2, f(2) = 5, f(3) = 1, etc. You can take those values and put them into a vector, [0 2 5 1 ...]. Linear operations on the function are equivalent to multiplying the corresponding vector by a matrix. And it turns out that even if your function is defined at all points on the real line, you can still treat its values as a vector, even though you can't list them out individually.
     
  8. Jun 15, 2009 #7
    Diazona has indeed given the gist of it; I'd like to expand on his explanation by giving a specific example of the matrix which corresponds to the differentiation operator. Take a matrix with all zeros except for +1 in the superdiagonal row and -1 in the subdiagonal row. Operating with this on a target vector gives a new vector which is the difference of (nearly) adjacent elements.

    This corresponds exactly to how you execute ordinary differentiation as a convolution integral with the derivative of a delta function.
     
  9. Jun 15, 2009 #8
    QM isn't trivial stuff anyway, but the journey is worthwhile (I'm a beginner too).

    Interestingly, while Griffiths goes from wave functions to vectors, Sakurai starts with vectors and eventually defines wavefunctions as projections of the state vectors on a suitable basis.

    Good luck, snoopies622.
     
  10. Jun 15, 2009 #9
    I have found that chapter XII in Schiff "Quantum Mechanics" on the relativistic wave equation is very good.
    For relativistic particles, including photons,
    E2 = (m0c2)2 + (pc)2
    where E is total energy, m0c2 is rest mass, and pc is momentum in energy units.
     
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