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Limited study of QM and all I know of operators

  1. Jul 17, 2004 #1
    So I've had limited study of QM and all I know of operators is that you use them to operate on the wave funtion, then integrate that over the valid range and *bam*, they pop out the expectation for a certain value such as momentum, position or any other function. I understand how expectation values work, but what I don't understand is why the operators take on such values. Doe they just make the math work out nicely or is there some mathematical/physical meaning I am missing?
  2. jcsd
  3. Jul 17, 2004 #2


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    Keep reading your QM, particularly Dirac's book, and learn aboout Sturm Liouville differential equations. It is big subject (functional analysis), sophisticated math, but I'm sure you'll get it with solid application.

    Reilly Atkinson
  4. Jul 18, 2004 #3


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    I'd suggest you to continue reading books on QM, and asking questions here :-) The short answer is that operators make the math work out nicely, but this is intimately related to the basic postulates of quantum mechanics.
    The story can be sketched as follows:
    you have to POSTULATE the possible outcomes (real numbers) of a measurement (for example, for position, it is just all real numbers ; for spin, it is, depending on the case, say, +1/2 and -1/2), and you have to postulate the states which correspond to these measurements exactly. It then follows that you can sum the projection operators on the "exact result states" multiplied with the respective outcomes, and that will fabricate your magical "operator". The operation you then apply to calculate expectation values is just a nice "trick" to weight the outcomes with their probabilities (that's a fundamental postulate of QM). Unfortunately, most QM texts take on the opposite appproach and introduce "magically" these hermitean operators.
    A book which does a rather good job at it is Sakurai, unfortunately, it is not a beginners' book.
    In the end it doesn't matter which approach you take, but the "sum over projection operators" is probably much clearer on an intuitive level.

  5. Jul 18, 2004 #4
    I myself have been trying to better understand the operators that represent observables. I have also studied QM in school, but we spent time doing complicated math and I didn't gain enough understanding.
    Studying on my own I have learnt much more.
    I agree with averything Vanesch said. The process of getting the expectation value consists of doing a weighted average. I doubt though it is necessary or convenient to get into projection operators to explain this.
    The way I see it, when you are in Hilbert space, state space, or whatever you want to call it, the variable on which you build your basis (set of base vectors) ceases to be a variable in the ordinary sense, and each of it's values becomes just a label. In the case of a wavefunction in position space, each of the possible positions of the particle becomes a base vector (a dimension of that space). So, when you are in Hilbert space, you are working with a different type of variable. Each of the possible positions ( in position space) becomes a variable of it's own, but the units of this variable are not length or volume but a dimensionless number (probability amplitude) related to the square root of a probability + a phase factor. As we are now (in Hilbert space) dealing with amplitudes and not the original variables, we kind of loose the ability to manipulate the original variable. A position of 3cm is not a value of the position variable anymore but just a label for a component of the state vector, and this individual position becomes a variable of it's own, but with units related to probability (dimensionless).
    Now, let's say you do a calculation in Hilbert space and get a particular vector. This vector just represents something like a distribution of probabilities, but you are missing the numbers with units of length you normally have in your classical variable.
    The operator takes you form this world of probabilities to the world of classical variables by converting the labels back to regular numbers that can be multiplied, added, etc. It adds these numbers in the form of eigenvalues, which occupy the diagonal of the operator's matrix when the basis is that that corresponds to the operator.
    About my explanation: I realize that I have been somewhat repetitive at times. I just tried to slightly refrace things in case they were not totaly clear the first time. My explanation is not very precise and I have tried to avoid jargon as much as possible. Also I realize this explanation is not the one you find in the books, but it represents a intuitive interpretation of what the math means. This way of looking at things is helping me. I hope it can also help you.
  6. Jul 18, 2004 #5
    QM is built on postulates that are taken for granted as always being true.
    For example one postulate states that for every observable in classical mechanics there is a linear QM operator. The operator is found from the classical mechanical expression for the observable written in terms of cartesian coordinates and conjugate momenta by replacing each coordinate q by itself and the conjugate component by -i h/2pi d/dq (the d's should be partial deriv. symbols, but I can't write them.)
  7. Jul 20, 2004 #6


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    Yes, that's how one usually first gets into contact with QM. And although one should go through that process, and first learn to use the tools to calculate simple things and get a feel for it, afterwards, it is a good idea to come back to the basics and see where they come from. One can simplify a lot the postulates of QM, and derive what you write above as their consequences. The way to do so is the following:

    choose what is a set of maximally compatible observables. If you have a classical equivalent, these are simply the variables defining the configuration space, but you might consider non-classical systems too.
    So you DEFINE yourself that variables: A, B, C and D are the things you'll be able to measure together. You put this in by hand. For a point particle, you say that its position X Y and Z are the maximal set of compatible observables (if the particle is spinless).

    define what are the possible combinations of values that these maximally compatible observables can take on. For our point particle, it is simple: all 3-tuples (x,y,z) are possible outcomes, if we consider a particle in free euclidean space. Now associate with each of these possible combinations, one single dimension of a hilbert space. So each ket |x,y,z> is a basis vector of the Hilbert space of states of the point particle.
    In our other example of A, B, C, and D, to each allowed combination (a,b,c,d) corresponds a basis ket |a,b,c,d>. Mind you, it is again BY HAND that the 'allowed values' are put into the theory.

    the observable operators corresponding to our variables A,B,C and D are defined to be diagonal in the basis introduced in step2, and have as eigenvalues the corresponding allowed value of the set of values that corresponds to it. So, the eigenvalue of B for the ket |a1,b2,c1,d4> is b2.
    Because we have defined these linear operators over a basis, they are defined in the entire hilbert space. Also, because they are diagonal and have real values in an orthonormal basis, they correspond to hermitean operators.
    For our point particle, X |x,y,z> = x |x,y,z>

    Relate, in one way or another, OTHER observables to these "defining" observables. For example, when there is a classical equivalent, the conjugate momenta correspond to the infinitesimal translation operators of the original defining observables.
    So we have, for our point particle, Px, which is the conjugate momentum to X, and it corresponds to the infinitesimal translation operator which maps the state |x,y,z> onto |x+dx, y,z>.
    It is a bit long but rather straightforward that this then implies the commutation relations [X, Px] = i hbar.
    If we don't have a classical equivalent, introducing other observables is less obvious, but can be accomplished by considering symmetries of the setup. For example, for a spin-1/2 system, if we took Sz as the single defining observable with the two allowed values -1/2 and 1/2 (so having a 2-dim hilbert space), the spin observable Sx and Sy can be found by considering rotations of 90 degrees around the x,y and z axes.

    In the case of a classical system, the commutation relation [X,Px] = ihbar AUTOMATICALLY implies that the representation of the Px operator is hbar/i d/dx.

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