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I The postulate of Quantum Mechanics and Eigenvalue equation

  1. Apr 26, 2016 #1
    According to one of the postulates of quantum mechanics, every measured observable q is an eigenvalue of a corresponding linear Hermitian operator Q. Which means, that q must satisfy the equation Qψ = qψ. But according to Griffiths chapter 3, this equation can only be followed from σQ = 0. It makes no sense to me because not every observable has zero standard deviation. Can someone explain this?
     
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  3. Apr 26, 2016 #2

    Simon Bridge

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    The exact energy levels (ferinstance) are for idealized situations... real energy levels are best characterised by a line with some thickness appropriate to the uncertainty of the state - which leads to it's stability. So you are learning that the postulates are not exactly true.

    You also need to distinguish between uncertainties in the Heisenberg sense and measurement uncertainties arising from the way measurement equipment works.
    Real life is messy.
     
  4. Apr 26, 2016 #3
    I am only considering idealized theoretical situations. The position of harmonic oscillator, for example, according to the postulate, a measured position value q is an eigenvalue of the equation xψ = qψ. But from this equation we can deduce that σx = 0. This is not true since |ψ|2 is not a single valued function. I am so confused...
     
  5. Apr 26, 2016 #4

    Nugatory

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    The postulate says that if you measure a given observable, the result will be one of the eigenvalues of that observable. However, the system state is not necessarily an eigenstate before you measure; it may be a superposition of several different eigenstates with different eigenvalues and then your measurement may yield any of several different results. Formally, you prepare an ensemble of systems all in the same initial state and measure the observable on each instance. If the initial state is an eigenstate of the observable, you will get the corresponding eigenvalue on every measurement and ##\sigma## will be zero. However, if the initial state is a superposition you will get different results (all eigenvalues of one of the many eigenstates making up the superposition) on the different measurements and ##\sigma## will be non-zero.

    For a number of reasons, it is not possible to prepare a system in an exact eigenstate of the position operator, so ##\sigma_x## will always be non-zero.
     
  6. Apr 27, 2016 #5

    A. Neumaier

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    The standard deviation depends on the state. In an eigenstate of ##Q##, there is no uncertainty about the value, so the standard deviation is zero, as it should be.
     
  7. Apr 27, 2016 #6

    Simon Bridge

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    What they said... plus: what is the probability that a measurement of position will yeild the specific eigenvalue q? Explain how you worked it out so I can see your current understanding and I'll get back to you.

    Note: the position operator has continuous eigenvalues, as does momentum.
    The particular value for position is one of the allowed eigenvalues.
     
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