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I Angular momentum ladder operator derivation

  1. Apr 6, 2016 #1
    In the Griffiths text book for Quantum Mechanics, It just gives the ladder operator to be

    L±≡Lx±iLy

    With reference to it being similar to QHO ladder operator. The book shows how that ladder operator is obtained, but it doesn't show how angular momentum operator is derived.

    Ive searched the internet for past hour and I just can't get what I'm looking for... can anyone help?
     
  2. jcsd
  3. Apr 6, 2016 #2

    stevendaryl

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    I'm not sure what you're asking. Angular momentum is defined in classical mechanics by [itex]\vec{L} = \vec{r} \times \vec{p}[/itex], where [itex]\vec{r}[/itex] is the position vector and [itex]\vec{p}[/itex] is momentum, and [itex]\times[/itex] is the vector cross-product. In quantum mechanics, it's the same, except that [itex]\vec{p}[/itex] is replaced by the operator [itex]-i \hbar \vec{\nabla}[/itex]. So

    [itex]L_x = -i \hbar (y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y})[/itex]
    [itex]L_y = -i \hbar (z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z})[/itex]
    [itex]L_z = -i \hbar (x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x})[/itex]
     
  4. Apr 6, 2016 #3
    I would like to know where this comes from
    L±≡Lx±iLy
     
  5. Apr 6, 2016 #4

    stevendaryl

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    I'm not sure what you mean. Do you mean: How did anyone ever think of forming that combination? You can show that

    [itex]L_z L_{\pm} = L_{\pm} (L_z \pm \hbar)[/itex]

    That means that if [itex]|l, m\rangle[/itex] is a state satisfying [itex]L_z |l, m\rangle = m\hbar |l, m\rangle[/itex], then [itex]L_{\pm} |l, m\rangle[/itex] will satisfy:
    [itex]L_z L_{\pm} |l, m\rangle = (m \pm 1) \hbar L_{\pm}|l, m\rangle[/itex]. So [itex]L_{\pm} |l, m\rangle \propto |l, m \pm 1 \rangle[/itex]. So [itex]L_+[/itex] is a raising operator for [itex]L_z[/itex] and [itex]L_-[/itex] is a lowering operator.

    I don't know what else you want to know about it.
     
  6. Apr 6, 2016 #5
    Is there any logical formulation how it is derived? Or did someone just guess it
     
  7. Apr 6, 2016 #6

    ShayanJ

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    Its about playing around with the equations long enough for an idea to pop out!
     
  8. Apr 6, 2016 #7

    stevendaryl

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    I don't think there is a hard-and-fast distinction between (1) deriving and (2) guessing, and then checking if your guess works. I don't know of a straightforward way to derive it that doesn't involve fooling around with commutation relations to see if you can get something interesting.
     
  9. Apr 6, 2016 #8

    stevendaryl

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    I'm not a theoretical physicist, but in my opinion, progress in theoretical physics requires a certain amount of playing around with equations.
     
  10. Apr 6, 2016 #9
    Hmm okay. Not the answer I wanted but its a trend in QM!
     
  11. Apr 6, 2016 #10

    strangerep

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    Although these things were (probably) originally found by playing around, there's also a more systematic approach which (iirc) goes under the name of "triangular decomposition of Lie algebras". The idea is that one first finds the Cartan subalgebra of mutually commuting basis elements, and then one forms certain linear combinations of the remaining basis elements such that the commutation relations between the latter set and the Cartan set satisfy certain conditions. (Sorry, I don't have my books with me right now, so can't be more specific.)

    The "trend" you should perceive in QM is rather that the quantum structure of a given class of systems can be found systematically by analysis of its associated dynamical algebra (and also its spectrum-generating algebra).
     
    Last edited: Apr 7, 2016
  12. Apr 8, 2016 #11
    Ladder operators come from Dirac's factoring trick for the Harmonic Oscillator. They're also called "raising and lowering", "destruction and creation", etc, and applied to angular momentum (as your question) and particle creation / annhilation in QFT. All from this one Dirac-trick! By the way he used a similar trick to come up with his relativistic equation for fermions; today they don't teach it (it's too understandable), instead get there using spinors.

    Since I don't know how familiar you are with the QHO let's go through it first.

    ---- QHO RAISING / LOWERING OPERATORS TRICK ----

    The Hamiltonian for the QHO can be written

    H = 1/2m (p^2 + (mwq)^2) where w = omega, the frequency

    I'm going to cheat, just ignore everything but p and q. Obviously m and w are vital to get the right answers but have nothing to do with the trick.

    H =~ p^2 + q^2

    Now he wanted to factor this, and figured if only it had a minus sign he could. So here's the first part of the trick, introduce i^2 = -1

    H =~ p^2 - (iq)^2 = (p-iq)(p+iq) = a a'

    a =~ p-iq is the lowering operator, a' =~ (p+iq) the raising operator.

    Actually hbar comes in also but it's just another constant and I'll ignore it, like m and w. Also there's a "+ 1/2 hbar w" factor which is the ground state energy. With "cheating" the ground state will be 0 instead. None of it affects the trick.

    Now we merely have to show that a and a' really do raise and lower.

    First compute [a, a'], where square brackets denote the commutator of a and a'. This turns out to be exactly 1. Recall that [q, p] = i hbar. Continuing to cheat just ignore the hbar. Then,

    [a a'] = (p-iq)(p+iq) - (p+iq)(p-iq) = i[p, q] - i[q, p] = i*(-i) - i*i = 2 =~ 1

    (The 2 becomes 1 when the left-out factors are included.)

    Ok, now suppose e is an eigenvalue of H, that is, the energy of an eigenstate. Then we want to show that

    H a|e> =~ (e-1) a|e>

    The equation means: apply a to e to get a new ket. Apply H to it, and find out that it's an eigenket (eigenstate, eigenvector) with eigenvalue e-1.

    That will justify calling a the "lowering" operator; it lowers the energy state down a level. (Of course the next level is actually 1/2 hbar below, not ~ 1). The a' operator does the opposite, adds 1, following the same logic.

    Now we finally get to the "hard" part:

    H a =~ (a a') a = a * (a' a) =~ a * (H - [a a']) = a (H - 1)

    (The third step uses [a a'] = a a' - a' a =~ H - a' a). So

    H a|e> =~ a (H - 1) |e> =~ a (e-1) |e> = (e-1) a|e>

    The second step uses H|e> = e|e>, because e is the eigenvector of |e>.

    That's all there is to it! So a does, in fact, lower an eigenstate down to the next one, and a' raises.

    To complete this you must (trivially) put back in the m, w, and hbar's. There's one more step which is not so trivial, showing that the lowering operator stops at the ground state and doesn't keep lowering into negative energies. Dirac's handling of it is interesting, but it has nothing to do with this factoring trick.

    ---- GENERAL CONSIDERATIONS FOR APPLYING THE TRICK ----

    The trick for the QHO depends on just two things:

    - you need two non-commuting operators to play the part of the conjugates p and q, with standard commutator relation [p, q] = -i hbar
    - you need an expression that plays the part of the QHO Hamiltonian, that looks like p^2 + q^2

    Any time you have this situation you can produce raising and lowering operators by the factoring trick.

    ---- ANGULAR MOMENTUM ----

    It comes from the definition of L^2, which can be writtten:

    L^2 - Lz^2 = Lx^2 + Ly^2

    On the right-hand-side Lx and Ly look like p and q in the QHO. So trick-factor them, and call the raising and lowering operators L± :

    L± ≡ Lx±iLy, so
    L+ * L_ = Lx^2 + Ly^2

    There are a few points involved with the left-hand side. Since L+ and L- commute with L^2 they don't affect it, so the only thing they can be related to (on the left side) is Lz^2. The square corresponds to the usual squares appearing in Hamiltonian energy terms ... so it's not too hard to believe L± do in fact work to raise and lower Lz eigenvalues. The details are important, but this should answer your question: how did Dirac come up with the L± operators in the first place?
     
    Last edited: Apr 8, 2016
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