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The physical derivation of annihilation operator?

  1. Oct 29, 2015 #1
    From P. Meystre's book elements of quantum optics (Many labels of equations are wrong:H) Page 83, the annihilation operator and creation operator, which are helpful to discuss harmonic oscillator, are defined as

    a=\frac{1}{\sqrt{2\hbar\Omega}}(\Omega q+ip),\\
    a^\dagger=\frac{1}{\sqrt{2\hbar\Omega}}(\Omega q-ip).

    This definition can't satisfy anyone's physical curiosity. How we deduce these operators more naturally?:nb)
  2. jcsd
  3. Oct 29, 2015 #2


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  4. Oct 29, 2015 #3
  5. Oct 29, 2015 #4


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    I don't understand what you mean. The document a linked to shows how one can posit which form operators ##\hat{a}## and ##\hat{a}^+## should have, and why they are ladder operators.

    That's how I interpreted your original question
    with en emphasis on deduce.

    If this is not what you are looking for, please give a more detailed question.
  6. Oct 29, 2015 #5
    Thank you. I want to deduce the ladder operators based on physical reasons, not a mathematical skill and definition like this:

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  7. Oct 29, 2015 #6


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    I'm not aware of a more "physical" way of finding out what these operators are.

    You know, it is often like that in physics. Clever people play around with the maths, and figure out things that were unexpected. I don't know if some ever sat down with the intention of figuring out "What is the operator that will take any eigenfunction of the harmonic oscillator and give back the next higher eigenfunction?", compared to someone just realizing, "Hey, look at what this operator I found by playing around with the algebra does!"
  8. Oct 29, 2015 #7
    Yeah, many historical events prove your viewpoint, such as Dirac equation, positron. But more important thing is that we should understand it more deeply and find more simpliy deduction, rather than such mathematical skills.
  9. Oct 29, 2015 #8
    There is an "explanation" given in the fantastic textbook on Quantum Mechanics by Griffiths. The Schrodinger equation is a 2nd order differential equation. For a harmonic oscillator, the hamiltonian H is quadratic in position and momentum. If we can write the hamiltonian as a product of two first order terms, then the original 2nd order equation can be written as two first order equations, which are easier to solve. So the attempt is to find first order factors of H. When you try that, (ref: Griffiths), you find that because of the commutation relation between x and p, there is a leftover term, and the best you can do is, H = (aa + 1/2)ħω, where a and a are the usual creation and annihilation operators. Once they are defined thus, the algebra follows.
  10. Oct 29, 2015 #9


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    It is the same approach that was in the link I posted.
  11. Oct 29, 2015 #10
    Absolutely. Thought I would put it into words and add the reference of Griffiths.
  12. Oct 30, 2015 #11


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    I think the introduction of these operators is very natural given the large symmetry of the harmonic-oscillator Hamilton, which goes beyond the obvious space-time symmetries. It has a socalled dynamical symmetry. The homogeneous ##N##-dimensional oscillator has a ##\mathrm{SU}(N)## symmetry.

    For ##N=1## this can be seen as follows. You start with the Hamiltonian (e.g., from the heuristics of "canonical quantization" of the classical harmonic oscillator)
    $$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{m \omega^2}{2} \hat{x}^2.$$
    Now you define
    $$\sqrt{m \omega^2} \hat{x}^2=\hat{\xi}, \quad \frac{1}{\sqrt{2m}}=\hat{\Pi},$$
    and then the Hamiltonian reads
    $$\hat{H}=\frac{1}{2} (\hat{\Pi}^2+\hat{\xi}^2).$$
    You immideately see that this is symmetric under ##\mathrm{O}(2)## rotations
    $$\begin{pmatrix} \hat{\xi}' \\ \hat{\Pi}' \end{pmatrix} = \begin{pmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{pmatrix} \begin{pmatrix} \hat{\xi} \\ \hat{\Pi} \end{pmatrix}.$$
    Now it is much simpler to deal with the combination
    $$\hat{a}=\sqrt{\frac{1}{2 \hbar \omega}}(\hat{\xi} +\mathrm{i} \hat{\Pi}),$$
    because then the above ##\mathrm{O}(2)## transformation can be written as ##\mathrm{U}(1)## transformation
    $$\hat{a}'=\exp(-\mathrm{i} \varphi) \hat{a}.$$
    The Hamiltonian must thus be a function of ##\hat{a}^{\dagger} \hat{a}## and indeed you find, using the Heisenberg commutators for ##\hat{x}## and ##\hat{p}##
    $$\hat{H}=\hbar \omega \left (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \right ).$$
    You can solve the complete eigenvalue problem for ##\hat{H}## by analyzing the algebra in terms of ##\hat{a}## and ##\hat{a}^{\dagger}##, which generate the harmonic-oscillator energy eigenstates in terms of a Fock space, because ##\hat{a}## turns out to be a ladder operator, a socalled quasiparticle single-mode annihilation operator. The quasiparticles in this case can be called the most simple example for phonons.

    Indeed you can quantize a solid crystal in the most simple model as a set of harmonic oscillators, describing the crystal-lattice vibrations has harmonic oscillations around the equilibrium positions of the atoms/molecules making up the solid. That's then equivalent to a free gas of bosonic quasiparticles, which are called phonons, because the lattice vibrations are nothing else than sound waves.
  13. Oct 30, 2015 #12


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    Just to give a handwavy version of what vanhees71 said: it comes about naturally if one considers the "Fourier" domain, that solutions to the free wave equation can be considered superpositions of sinusoids. These are the sound waves that vanhees71 refers to.
  14. Oct 30, 2015 #13

    A. Neumaier

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  15. Nov 1, 2015 #14


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    Here's another perspective on the harmonic oscillator.

    Do you know about the Heisenberg picture? In the usual Schrodinger picture, [itex]p[/itex] and [itex]x[/itex] are time-independent operators, and the wave function [itex]\psi[/itex] is a function of time. In the Heisenberg picture, it's the other way around: [itex]p[/itex] and [itex]x[/itex] are time-dependent. One nice thing about the Heisenberg picture is that the Heisenberg equations of motion look almost like the classical ones, but with some quantum "correction terms". These correction terms turn out to vanish for the harmonic oscillator.

    So let's work in the Heisenberg picture. Since classically we know that the harmonic oscillator results in periodic motion with angular frequency [itex]\omega = \sqrt{\frac{k}{m}}[/itex], we can guess that quantum-mechanically, [itex]p[/itex] and [itex]x[/itex] are similarly periodic:

    [itex]x = A e^{-i \omega t} + A^{\dagger} e^{+i \omega t}[/itex]

    That's a perfectly general form for a Hermitian operator with angular frequency [itex]\omega[/itex].

    [itex]p = m \dot{x} = -i m \omega (A e^{-i \omega t} - A^\dagger e^{+i \omega t})[/itex]

    The commutation relation for [itex]p[/itex] and [itex]x[/itex] gives us:

    [itex]-i m \omega ([A, A^\dagger] - [A^\dagger, A]) = -i \hbar[/itex]

    or [itex][A, A^\dagger] = \frac{\hbar}{2m \omega}[/itex]

    In terms of [itex]A[/itex] and [itex]A^\dagger[/itex],

    [itex]E = \frac{m \omega^2}{2} x^2 + \frac{1}{2m} p^2 = m \omega^2(A A^\dagger + A^\dagger A)= m \omega^2 (2 A^\dagger A + [A, A^\dagger]) = 2 m \omega^2 A^\dagger A + \frac{\hbar \omega}{2}[/itex]

    The usual raising and lowering operators are:

    [itex]a = \sqrt{\frac{2m \omega}{\hbar}} A[/itex]
    [itex]a^\dagger = \sqrt{\frac{2m \omega}{\hbar}} A^\dagger[/itex]
    Last edited: Nov 1, 2015
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