Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measurements in Quantum mechanics

  1. Mar 20, 2013 #1
    Ok so I'm currently revising my quantum theory course from this year and I've reached the section on the postulates for measurements in quantum mechanics. The one I'm having trouble with is "The only result of a precise measurement of some observable A is one of the eigenvalues of the corresponding operator [itex]\hat{A}[/itex]."

    My main conceptual problem with this is that suppose I'm measuring the energy of some particle, it has an infinite number of values of energy I could measure it at, which suggests that the associated operator has an infinite number of eigenvalues. Is this just a natural consequence of working in an infinitely-dimensional Hilbert space, in which case how is the fact that we have 3-spacial dimensions encoded; or am I misunderstanding the nature of continuous basis vectors?
  2. jcsd
  3. Mar 20, 2013 #2
    Well you don't measure something's energy at a certain value, you just measure its energy and you get a result. What the postulate basically means is that say you have a system in a superposition of n energy (or any other observable) eigenstates, when you perform a measurement the result will be the eigenvalue of one of the eigenstates. Systems can be a superposition of a finite number of energy eigenstates or an infinite number. Some operators do admit infinite eigenstates and hence eigenvalues while some do not, for example the spin operators admit only two eigenstates while the position operator admits infinitely many (Dirac delta functions shifted to any point in space).
    The three spatial dimensions in our world are represented by the three position operators, I'm not sure what you are asking in that last bit there. What do you mean by continuous basis vectors? We normally take eigenstates as our basi, which can be generated continuously or discretely by the respective operator, for example the energy eigenstates advance discretely in a harmonic oscillator.
  4. Mar 20, 2013 #3
    Yes I definitely phrased that badly, what I meant was there are an infinite number of results one could get from measuring the particle's energy. I'm working through a whole heap of misunderstanding here and suspect I'm getting rather too caught up on Hilbert space but bear with me:
    My understanding of operators are that they map a vector onto another vector, so is the picture I should have that: particles are represented by a vector in Hilbert space, when the particle moves, it's had the position operator applied to it to shift it a certain distance which has mapped that particle onto a new vector in Hilbert space. At this point I guess I'm kind of confused as to how eigenvectors and eigenvalues fit into this picture. I always sort of assumed there was one eigenvector per dimension for a given operator, so does the set of all eigenvectors for different operators span different sub-spaces? I suppose what I'm asking is do different operators have different dimensionalities?

    Sorry that this is somewhat rambly, as I say I'm very confused at the moment.
  5. Mar 20, 2013 #4
    I recommend grabbing a text book because there is only so much I can explain here on a forum, and there is alot to explain in this case.

    We are not really interested in what an observable operator does to a state but rather the eigenstates and eigenvalues which it admits, these eigenvalues are the results you would get if you measured this observable, and the eigenstates are states that have a definite outcome for that measurement, corresponding to its eigenvalue.

    When we say that a state is in a superposition of multiple energies, we mean that if you projected the state onto an energy eigenbasi (Which span the vector space, due to the fact that the energy operator, the hamiltonian is hermitian), you will get a sum (generally infinite) of energy eigenvectors.


    Where |E_n> is the nth energy eigenstate.

    However if the hamiltonian admits continuous eigenvalues, we would have to replace this by an integral. (As is in the case of the free particle)

    When we perform an energy measurement, the probability amplitude that we will get that energy is the inner product between the eigenstate with that energy and the state.

    <E_n|State>=Probability Amplitude

    However since we have expanded the state in the energy basi, the inner product really is just


    And since the states are orthogonal and normalized <E_n|E_j>=0 for all n=/=j, and 1 if n=j


    Now this is the probability amplitude, the probability is C*C=|C|^2.
    Last edited: Mar 20, 2013
  6. Mar 20, 2013 #5
    Yeah I'm delving back into my notes now, I mostly follow the specifics just keep losing sight of the big picture. Thanks for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook