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Sharp quantum states

  1. Sep 15, 2015 #1
    Hey all,
    I'm reading through an anecdotal work about the philosophical foundations of quantum field theory and the authors keep referring to states having the ability to be "sharp." As in it's possible for P to be sharp if the system is mixed, where P is some property of the system. Thanks! IR
     
  2. jcsd
  3. Sep 15, 2015 #2

    BvU

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    Hello,

    probably they refer to the Heisenberg uncertainty relation (which google).
     
  4. Sep 15, 2015 #3

    vanhees71

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    Well, philosophers tend to be not very sharp usually (some are, because they have a proper education in math, logics, and other hard sciences). That's why you cannot make sense of the phrase "a quantum state is sharp".

    I've a vague clue what they might mean. A state in quantum theory is represented by a normalized vector in Hilbert space (for the purpose of discussion we can be a bit sloppy here; more precisely a pure state is represented by a ray in Hilbert space and not by a single vector). Observables are represented by self-adjoint operators in Hilbert space. These have (generalized) eigenvectors and eigenvalues, and the eigenvalues are the possible outcomes of exact measurements of this observable. If you know the pure state of the quantum system (and you can only know it, if the system has been prepared such as to be in that state), then the probability (density) to find a possible eigenvalue ##a## of the observable ##A##, represented by the self-adjoint operator ##\hat{A}## is given by
    $$P(a)=|\langle a|\psi \rangle|^2,$$
    where I have assumed that the eigenvalues are not degenerate, i.e., that for each (generalized) eigenvalue ##a## there's exactly one (generalized) eigenvector ##|a \rangle##. This is known as "Born's rule" and is one of the basic postulates of quantum theory.

    What I mean with "generalized eigenvector and eigenvalue" is the following. Many operators/observables in quantum theory have eigenvalues which build a continuum. Examples are a component of a position or momentum vector of a massive, which have the entire real axis as a spectrum, i.e., generalized eigenvalues. Sometimes you also have observables which can only take discrete values. An example is one component of angular momentum like ##\hat{J}_z##, which can take only either half-integer or integer values (for half-integer and integer spin particles respectively) or you can have the case that an operator has both a discrete and a continuous spectrum like the Hamiltonian of a hydrogen atom.

    The generalized eigenvector for an eigenvalue in the continuous part of the spectrum is never normalizable to 1 and thus can never represent a state, but you can always normalized these generalized states in the sense of a ##\delta## distribution,
    $$\langle a |a' \rangle=\delta(a-a').$$
    After setting up this mathematics (which however is very rough and not really accurate; a mathematician would cry, when reading this ;-)) we can discuss, what might be meant by this phrase of a state being sharp.

    If you have an observable with a discrete spectrum. Then if you take an eigenvalue of the operator in this discrete part of the spectrum, you have a true normalizable eigenvector ##|a \rangle##, which you can normalize such that ##\langle a|a \rangle=1##, you can prepare your system to be in this state. For a theorist the most simple way to think of such a preparation procedure is that you measure somehow the observable in a way such that you can through all particles away which have not the value ##a## for this observable and keep those which have this value (this is known as a von Neumann filter measurement).

    Now there's a theorem saying that eigenvectors of a self-adjoint operator with different eigenvalues (let's for simplicity take the example for an operator that has only discrete nondegenerate eigenvalues) are orthogonal to each other and that they are complete, i.e., you can span the entire Hilbert space of states as a series
    $$|\psi \rangle=\sum_{a'} \psi_{a'} |a' \rangle.$$
    Then if the particle is prepared in the state ##|\psi \rangle=|a \rangle## the probability to measure the eigenvalue ##a'## is given by
    $$P(a')=|\langle a|a' \rangle|^2=\delta_{a,a'}.$$
    So you have ##P(a)=1## and ##P(a')=0## for all ##a' \neq a##. This means that you measure the value ##a## with certainty, if the particle is prepared in an eigenstate of the operator that represents this measured observable. Then you say that due to the preparation procedure the value of the observable is determined or you also say it has the "sharp value" ##a##. In this sense you can say that a true normalizable eigenstate is "sharp" with the meaning that then this value of the considered observable is "sharp".

    It is also immediately clear that this can never work with generalized eigenvalues in the continuous spectrum of the operator. Formally you can argue very much in the same way, but then you have ##\langle{a'}{a}=\delta(a-a')##, but this implies that you cannot square this to get a probability. The reason is that the generalized eigenvector ##|a \rangle## is not a proper state. You can only prepare the particle to take the value ##a## within a "small interval around ##a##" but not to be really sharply ##a##.

    This is all consistent due to the Heisenberg-Robertson uncertainty relation between incompatible observables, and there are always pairs of incompatible observables. This means the following: You can only prepare a particle in a state such that two observables ##A## and ##B## take always sharp values, if there is a complete set of (generalized) simultaneous eigenvectors of both operators, representing these observables. One can show that under this condition the two operators commute (and also the opposite is true, i.e., if the operators commute you always find a common complete set of (generalized) eigenvalues). So one calls observables compatible, if the representing operators commute. If they do not commute, you have the uncertainty relation, which states that for any (true normalizable) state ##|\psi \rangle## the standard deviations of the observables ##\Delta A## and ##\Delta B## fullfil
    $$\Delta A \Delta B \geq \frac{1}{2} |\langle{\psi}|[\hat{A},\hat{B}]\psi \rangle|.$$
    The most famous uncertainty relation is that between position and momentum components in the same direction,
    $$[\hat{x],\hat{p}_x]=\frac{\hbar}{2} \hat{1},$$
    which implies
    $$\Delta x \Delta p_x \geq{\hbar}{2}.$$
    This in turn implies that there's no proper (normalizable) state in which you can prepare the particle to have either a sharp position or a sharp momentum (because position and momenum operators have only a continuous spectrum). You can prepare the particle with as sharp a position as you like, i.e., you can make the standard deviation ##\Delta x## as small as you like (but not 0!), but then the Heisenberg-Robertson uncertainty relation tells you that the momentum must be quite undetermined, i.e., it's standard deviation ##\Delta p_x## becomes pretty large.
     
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