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What is quantum angular momentum

  1. Jul 24, 2014 #1
    Definition/Summary

    Angular momentum carries over from classical mechanics into quantum mechanics, but quantum-mechanical angular momentum is restricted to discrete values. There are two main types: orbital or kinematic angular momentum, and spin or intrinsic angular momentum, carried by a field's internal geometry. Spin appears in the classical limit in the circular polarization of electromagnetic waves, which carries angular momentum.

    For orbital angular momentum, one can find wavefunctions by turning the angular-momentum operators into differential operators in the angular coordinates, and then finding the solutions of the appropriate differential equations. One finds spherical harmonics, functions that also appear in multipole expansions of fields.

    Both orbital angular momentum and spin can be described with an elegant ladder-operator algebraic formulation, the Lie algebra U(1) in the 2D case and the Lie algebra SO(3) ~ SU(2) in the 3D case. This formulation also applies to sums of angular momenta, meaning that its results apply to mixture angular momenta as well as to "pure" ones.

    Angular-momentum operators are closely related to rotation generators, and that is why a field's internal geometry can carry angular momentum.

    From the ladder-operator formulation, one finds that total angular momentum j can be a half-odd number as well as an integer. Orbital angular momentum can only be an integer, while spin can be any value. The relationship with rotation has an interesting consequence. Rotating by 360 degrees multiplies the wavefunction by (-1)^(2j), meaning that it reverses sign if j is half-odd, but stays the same if j is an integer.

    Angular momenta can be combined, with j1 and j2 yielding total angular momenta ranging in integer steps between |j2-j1| and (j1+j2). The eigenstates are sums of products of the original angular-momentum eigenstates and Clebsch-Gordan coefficients. One can find combined spin-orbit eigenstates in this way.

    Equations

    Angular momentum: [itex]{\mathbf L} = {\mathbf x} \times {\mathbf p} ,\ L_i = \epsilon_{ijk} x_j p_k[/itex]

    Its commutators: [itex][L_i,L_i] = i \hbar \epsilon_{ijk}L_k[/itex]
    and [itex][L_i,L^2] = 0[/itex]

    Theoretical work often uses the convention hbar = 1, and that convention will be used here. These commutators are also true of spin and combined angular momentum.

    2D angular momentum has eigenvalues m, and 2D orbital angular momentum restricts m to integer values. Its orbital wavefunctions are [itex]e^{im\phi}[/itex].

    3D angular momentum J has eigenvalues of its square j(j+1), where j is a multiple of 1/2, and eigenvalues of some selected component m, where m goes in integer steps between -j and j. 3D orbital angular momentum restricts j to integer values, and its wavefunctions are spherical harmonics: [itex]Y_{jm}(\theta,\phi)[/itex]

    Extended explanation

    Here is the ladder-operator derivation of the eigenstates of angular momentum J.

    The square, J2 = Jx2 + Jy2 + Jz2, commutes with all its components, and thus has its own eigenvalue. But since angular-momentum components do not commute with each other, the angular-momentum states can be eigenstates of the angular momentum component in only one direction. This is conventionally taken to be the z-direction. Thus,
    J2|C,m>= C|C,m>
    Jz|C,m> = m|C,m>

    We can construct raising and lowering operators
    J+ = Jx + i*Jy
    J- = Jx - i*Jy

    and they have commutators
    [Jz,J+] = J+
    [Jz,J-] = -J-
    [J+,J-] = 2Jz

    The square is
    J2 = J+J- + Jz(Jz - 1) = J-J+ + Jz(Jz + 1)

    The raising and lowering operators relate the eigenstates with
    J+|C,m> ~ |C,m+1>
    J-|C,m> ~ |C,m-1>

    Let the minimum and maximum eigenstates be defined by
    J+|C,mmax> = 0
    J-|C,mmin> = 0

    Plugging into the equation for J2, we get
    C = mmax(mmax + 1) = mmin(mmin - 1)

    This yields mmin = - mmax, and for convenience, we can rename mmax j, the total angular momentum. Thus
    C = j(j+1)
    and using j instead of C in the eigenstate designations,
    [itex]J_z |j,m> = m |j,m>[/itex]
    [itex]J_+ |j,m> = \sqrt{(j-m)(j+m+1)} |j,m+1>[/itex]
    [itex]J_- |j,m> = \sqrt{(j+m)(j-m+1)} |j,m-1>[/itex]

    Since m changes in integer steps, we find that the difference of its extremes, j - (-j) = 2j, must be an integer. Thus, j is either an integer or half-odd.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
  3. Jan 18, 2016 #2
    In mathematics, square of 2 is 2*2=4, but why in quantum J*J= J*(J+1), that means square of 2 is 2*2 = 2*3 = 6? thanks.
     
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