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Confusion over quantum mechanics operators

  1. Dec 24, 2014 #1
    are operators solely used to find the expectation value of something?
    What does it mean to use the momentum operator over wavefunction? What does it give? I am guessing it doesnt give momentum since momentum can never be a function of space.
    How to calculate kinetic energy, given the wavefunction?
    My professor once told me "the difference between classical mechanics and quantum mechanics is that classical variables are operators in QM" i just wish to know what does that mean?

    I am abit confused, and i referred many books, but i didnt find anything that explains these things simply, its still first time to do QM for me.

    Thank you
     
  2. jcsd
  3. Dec 24, 2014 #2

    mfb

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    No, they are much more powerful tools. At some point, when you want to predict measurements, expectation values (or better, the distributions) can be interesting.
    Sure it can, in the same way it can be a function of time.
    Particles in quantum mechanics (in general) do not have a unique value as kinetic energy. You can ask "given this type of measurement, what is the distribution of kinetic energies I will measure?", but the answer depends on the measurement.
     
  4. Dec 24, 2014 #3
    How can momentum be a function of space! The heisenberg uncertainty doesnt allow it, right?

    But given a wavefunction, can we calculate the kinetic energy? Would that be using the kinetic energy operator?
     
  5. Dec 24, 2014 #4
    If you measure the possible values of momentum you immediately know what values of kinetic energy you can have.
     
  6. Dec 24, 2014 #5
    How do you measure the possible values of momentum?
     
  7. Dec 24, 2014 #6

    mfb

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    Why not?
    It gives a lower limit on the product of their widths, that is a different thing.
    You can calculate the expectation value, or its distribution. As I said, "the kinetic energy of this particle" in general is not a useful concept in quantum mechanics.
     
  8. Dec 24, 2014 #7
    Wait... So momentum can be a function of x? But if it is a function of x then the uncertainty in x is zero, which means that the uncertainty in momentum is huuge, right?

    Also, i am intrigued how do you calculate the kinetic energy in wave mechanics (schrödinger's equation is the limit of my knowledge)
     
  9. Dec 24, 2014 #8

    mfb

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    No. The uncertainty is an uncertainty of the particle, not an uncertainty of your coordinate value.
    Apply the Hamilton operator on your wave function. Or the part that resembles the kinetic energy if you are just interested in that.
     
  10. Dec 24, 2014 #9

    bhobba

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    To answer that you need to see an axiomatic treatment such as found in Ballentine which I know you have looked at, but, sigh, and this is the only major criticism I have of that beautiful book, it is not for the beginner student (unless they have a very strong math background), as you found.

    Operators have two purposes both of which are directly related to the two axioms in Ballentine. The first is their eigenvalues are the possible outcomes of an observation. The second is the Born Rule which is if a system is in state u and you observe it with observable O the expected outcome is <u|O|u>.

    Thanks
    Bill
     
    Last edited: Dec 25, 2014
  11. Dec 25, 2014 #10
    I am still confused over the uncertainty principle :/ please tell me more on how momentum can be a function of x? Does it give like the expectation value for each value of x? And gow is the uncertainty principle related to all this, if momentum can be a function of x.

    So to find the kinetic energy or energy we use their operators... Resonates with what my professor told me. But, why do we use operators? Isnt it somehow related to the uncertainty principle?

    According to my book (stewart) the uncertainty principle is related to the wave equation (wave packets having a larger array of wavelengths) so i seem to always get the uncertainty principle to try to undertstand everything.

    I am having an idea now... The wave equation gives the shape of the wave without measurement (right?) so the momentum found is really the momentum of the wave not of the actual particle (am i getting close to it?)

    Why is quantum mechanics so tough to understand for me? I understand my book really well, but its toooo simple to answer any of these questions...
     
  12. Dec 25, 2014 #11

    bhobba

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    I will let MFB answer the uncertainty thing.

    However the interesting thing about QM is axiomatically the most fundamental things are the operators.

    Hopefully you know a smattering of linear algebra to make sense of it - in particular you need the spectral theorem.

    A resolution of the identity are disjoint projection operators Ei that sum to 1 ie ΣEi = 1. The outcomes of observations are described by a resolution of the identity - that's it - that's all. Now what you do is associate a number yi with each outcome and you get the following hermitian operator O = Σyi Ei. This is the operator that describes the observation. Note the eigenvalues are the outcomes of the observation are its eigenvalues. Also by the spectral theorem given any Hermitian operator it can be broken down into a resolution of the identity and its associated eigenvalues.

    Now the really neat thing. We have this really beautiful theorem, called Gleason's theorem, that shows from the above the only probability you can associate with the Ei is via the Born rule. Proving that is a known very difficult problem but simply for completeness here is the proof, which at your level is likely to be gibberish:
    http://kof.physto.se/cond_mat_page/theses/helena-master.pdf [Broken]

    These two axioms (which from Gleason is basically, without delving too deeply into what basically means here, is just one) is all that's required for QM as detailed in Ballentine.

    All this is really neat, but unfortunately, as you are now glimpsing, QM requires some fairly deep math to get to grips with the kind of fundamental questions you are asking; which is why you will not find them answered in beginner texts. Its terrible for thinking students like you - but its the way it is.

    All I can suggest is persevere, take some advanced math and then study a more advanced text like Ballentine that will answer them.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  13. Dec 25, 2014 #12

    vanhees71

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    Well, before doing axiomatics, which is a very good thing after having some intuitive ideas about an subject, but doesn't help much to get the physics underlying the theory. That's particularly important in quantum theory, which is very abstract compared to all of classical physics (classical="not quantum" here, relativity is included in classical physics as long as it's not quantum :-)).

    There's also a problem with the didactics, because many books start with a kind of historical heuristics, which however confuses students more than it helps, because the socalled "old quantum theory" (1900-1925) is outdated and gives wrong pictures on nature even on a qualitative level.

    That's why I recommend as a first book

    J.J. Sakurai, Modern Quantum Mechanics (2nd edition), Addison Wesley

    It starts with a thorough analysis of the Stern-Gerlach experiment and spin observables. This is the most simple non-trivial example. So let me sketch it here.

    It's an experimental finding by Stern and Gerlach in one of the most important experiments ever conducted in the history of science. It was done in 1923, i.e., before "modern quantum theory" was discovered by Heisenberg, Born, and Jordan (1925), Schrödinger (1926) and in its final formulation by Dirac (1926). Here, I follow the modern version a la Dirac.

    The first observation is that particles like and electron have intrinsic angular momentum, called spin. It's very hard to imagine when trying to deal with it in a classical picture. Thus one shouldn't even try this! Phenomenologically the spin manifests itself as a magnetic moment of a particle like and electron or a neutron (the experiment was actually done with neutral silver atoms). This is familiar from classical electrodynamics: A current distribution looked from some distance looks in first approximation like a socalled dipole field. You can forget about the concrete current distribution and substitute it by a "point dipole", i.e., a point singularity in the current distribution at a certain place and the corresponding magnetic field it creates. Also such an "elementary magnetic dipole" is affected by an inhomogeneous magnetic field, i.e., it feels a force like a magnet (which is, looked at some distance to a good approximation such a dipole) in an inhomogeneous magnetic field.

    Now, when Stern and Gerlach run a beam of silver atoms, prepared by heating up silver in and oven and let out a beam through a small opening, through an inhomogeneous magnetic field and then detecting them on photo plates they found a surprising result that could not be explained with classical physics: The beam of silver atoms did not show some continuous spread as to be expected from a classical set of dipoles which are randomly oriented due to the thermal motion of the silver atoms inside the oven, but they found a split in two clearly separated lines. There was no way to predict in which of the two branches of the beam an individual silver atom might occur, but it were two discrete spots on the screen!

    The modern explanation is as follows: The magnetic moment of an electron is quantized, i.e., letting it run through a magnetic field with a large (quasi homogeneous) magnetic field in ##z## direction plus some inhomogeneous field in another direction there are two possible values for the spin and, proportional to it, its magnetic moment. As angular momenta the spin component is measured in units of the modified Planck constant, and for an electron the possible values of its spin-z component are ##s_z=\pm \hbar/2##. The direction of the large homogeneous part of the magnetic field determines the spin to point in z-direction (at least on average, we'll come back to this) and the inhomogeneous piece leads to the corresponding force on the electron. This explains, why there is not a continuous line on the screen as predicted by classical mechanics but only two spots.

    Now, using another Stern-Gerlach experiment of the same type (i.e., one that measures the z-component of the spin) on one of the partial beams, no more splitting of the beam is found, but it always comes out the spin-z value of the chosen partial beam.

    If however, the 2nd Stern-Gerlach experiment is set up to meausure a component of the spin in perpendicular direction, say in x direction, again you find two spots, and it's impossible to predict for an individual silver atom which value of ##\sigma_x =\pm \hbar/2## you'll measure. There's only a probability of ##1/2## to find ##+\hbar/2## and a probability of ##1/2## to find ##-\hbar/2##.

    This is pretty analogous with the well-known phenomenon of polarization of the electromagnetic field, which is described by the classical electromagnetic waves predicted by Maxwell. After some work on the foundations, the quantum physicists came to the following set of rules (the axioms bhobba is talking about), which I formulate for spin here.

    (1) The spin state of an electron is described by vectors in a two-dimensional complex Hilbert space. That's a two-dimensional vector space with complex numbers as scalars, which has a scalar product.

    (2) The spin components are represented by self-adjoint operators ##\hat{s}_j##, ##j \in \{x,y,z \}##, which obey the commutation relations of the Lie algebra of the rotation group, i.e., they represent infinitesimal rotations in three-dimensional Euclidean space. From this one can derive the commutation relations
    ##[\hat{s}_j,\hat{s}_k] = \mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l.##

    (3) The possible outcome of a measurement of the j-component of the spin are the eigenvalues of the corresponding operator ##\hat{\sigma}_j##. One can find the eigenvalues and the structure of the Hilbert space by analyzing the commutation relations carefully. It turns out that you can specify the values of ##\vec{s}^2## and one spin component, usually one takes the z-component, simultaneously, because the corresponding operators commute:
    ##[\hat{\vec{s}}^2,\hat{s}_z]=0.##
    Then it comes out that the eigenvalues of ##\hat{\vec{s}}^2## are ##\hbar^2 s(s+1)## with ##s \in \{0,1/2,1,3/2,\ldots \}##. For a given eigenvalue of ##s## (which determines the spin of the particle) the z-component of the spin can take the values ##s_z=\hbar \sigma_z## with ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##, i.e., for each ##s## there are ##(2s+1)## eigenvalues of ##\hat{\sigma}_z##. From Stern and Gerlach's experiment thus an electron must have spin ##s=1/2##, so that there are exactly two possible values for ##s_z \in \{-1/2,+1/2 \}##. The eigen vectors are called ##|s_z \rangle##. These vectors are assumed to be normalized, and they are orthogonal to each other, because according to a theorem from linear algebra (or Hilbert space theory, which here is the same because we deal with a finite-dimensional Hilbert space) the eigenvectors of self-adjoint operators with different eigenvalues have perpendicular eigenvectors.

    (4) Filtering out the partial beam corresponding to the value ##s_z=+1/2##, i.e., ##\sigma_z=\hbar/2##, leads to silver atoms whose valence electron is in the state represented by ##|s_z=+1/2 \rangle##.

    (5) The physical meaning of these state vectors is the following: When measuring any observable (e.g., the spin-x component) on a beam of particles prepared in the state represented by ##|s_z=+1/2 \rangle##, then the probability to measure a certain value of ##\sigma_x = \pm \hbar/2## is given by
    $$P_{s_z=1/2}(s_x)=|\langle s_x|s_z=1/2 \rangle|^2.$$
    As the calculation for this case shows, one finds indeed the observed values:
    $$P_{z_z=1/2}(s_x=+1/2)=1/2, \quad P_{s_z=-1/2}(s_x=-1/2)=1/2.$$
    This implies that for an electron with definite ##s_z## component the ##s_x## component's value is totally unknown and vize versa. That's because, there are no common eigenvectors of ##\hat{\sigma}_z## and ##\hat{\sigma}_x##, because these operators don't commute: ##[\hat{\sigma}_z,\hat{\sigma}_x]=\mathrm{i} \hbar \hat{\sigma}_y \neq 0##.
     
  14. Dec 25, 2014 #13

    mfb

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    If you do not like the book you have now, you can try another one.
    Sorry, I cannot replace a book.
     
  15. Dec 25, 2014 #14
    The answer to your question lies in the fundamental postulates of quantum mechanics. The first of which states that for every physical observable there exists a linear hermitian operator. If there does not exist a linear hermitian operator for some physical observable then this observable cannot be observed within the framework of quantum mechanics. In other words there would exist an observable which you cannot calculate expectation values and QM is either inconsistent or incomplete. The converse is however not true, there exists linear hermitian operators which do not necessarily correspond to a physical observable.

    An operator of some space is a mathematical transformation in which when applied to an element of that space it maps it to another element of that space. Linear Hermitian Operators in Hilbert Space take infinite dimensional complex vectors and map them to other infinite dimensional complex vectors. If this mapping returns the original vector by a multiplicative constant we have an eigenvalue equation in which the constant may or may not be a physical observable.
     
    Last edited: Dec 25, 2014
  16. Dec 26, 2014 #15
    So the eigen value itself is the momentum for the momentum operator? And is this so for all functions, i mean shouldnt the function be an eigen function first (of the operator)?

    I only understood the last sentence to be honest, i am seriously just starting QM :/
     
  17. Dec 26, 2014 #16
    Thank you for the book recommendation. I am just a beginner, so i will read what you wrote later when i know what these symbols even mean :D
     
  18. Dec 26, 2014 #17
    Yes, the eigenvalue of the momentum operator is the momentum and 1/sqrt(2)*e^(ikx) is the eigenfunction of the momentum operator, i.e. it is a solution to the differential equation you get from the momentum operator.
     
  19. Dec 26, 2014 #18
    Ok as far as i understand, the eigen value must be a constant, so momentum is a constant for the momentum eigen function, what does that mean? Is it the expectation value of momentum or is it the actual value?

    Also this seems to work only for exponential functions (right?) and psi can be a function of say sinx as in the particle in a box, so we get no eigen value for sinx when we apply the momentum operator since sinx is not an eigen function of the momentum operator. Does this indicate that momentum is not measurable for a particle in a box? Or what do we do here to calculate momentum?
     
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