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Quantum harmonic oscillator, creation & annihilation operators?

  1. Nov 7, 2013 #1
    For a set of energy eigenstates [itex]|n\rangle[/itex] then we have the energy eigenvalue equation [itex]\hat{H}|n\rangle = E_{n}|n\rangle[/itex].

    We also have a commutator equation [itex] [\hat{H}, \hat{a^\dagger}] = \hbar\omega\hat{a}^{\dagger}[/itex]

    From this we have [itex]\hat{a}^{\dagger}\hat{H}|n\rangle = (\hat{H}\hat{a}^{\dagger} - \hbar\omega\hat{a}^{\dagger})|n\rangle = E_{n}\hat{a}^{\dagger}|n\rangle[/itex]

    and thus [itex] \hat{H}\hat{a}^{\dagger}|n\rangle = (E_{n} + \hbar\omega)\hat{a}^{\dagger} |n\rangle [/itex]. So, comparing this to [itex]\hat{H}|n\rangle = E_{n}|n\rangle[/itex], we can see that the state [itex]\hat{a}^{\dagger}|n\rangle[/itex] has more energy by an amount [itex]\hbar\omega[/itex]. Fair enough.


    My question is to do with why this operator [itex]a^{\dagger}[/itex] is interpreted as a "creation" operator. If I'm not mistaken, it's interpreted as increasing the number of particles in a given state by 1, ie. creating a particle. Why is it not interpreted as just one particle existing before and after, but that particle now has an extra [itex]\hbar\omega[/itex] of energy, and just oscillates more violently in a higher excited state?
     
  2. jcsd
  3. Nov 8, 2013 #2
    When doing a simple oscillator in a one-particle Hilbert space, it does indeed make more sense to just think of one particle that is being given more and more energy. That's the way this problem is normally presented at first. However, there's some benefit to thinking about it more abstractly, because of the way in which this same set of equations pops up again in more advanced contexts.

    Specifically, in Quantum Field Theory, you have an entire field of operators, each of which acts like its own separate harmonic oscillator. Once you do the math, you discover that the system can be described by an entire set of creation/annihilation operators, which are called [itex]a_k[/itex] and [itex]a^\dagger_k[/itex], one for each value of momentum [itex]k[/itex]. Now instead of just hitting the vacuum state with the same creation operator over and over again, you can hit it with various combinations of them--a typical eigenstate of a QFT Hamiltonian will be something like [itex]a^\dagger_{k1}a^\dagger_{k2}a^\dagger_{k2}a^\dagger_{k3}|0\rangle[/itex].

    Like the single-particle case, each creation operator takes you from one eigenstate to another, with a higher value of energy. But now, the eigenstates are degenerate--there are many different eigenstates which have the same value of energy, but with various different values of momentum. In order to keep them straight, it's convenient to label them by the various momenta of the creation operators that made up each state (this structure is called a Fock space). So in this context, it starts making more sense to think about each creation operator as adding a single particle with momentum [itex]k[/itex] into the state--by thinking about it that way, it allows you to easily label all of the various eigenstates of the Hamiltonian.
     
    Last edited: Nov 8, 2013
  4. Nov 8, 2013 #3
    Right, thanks. I am working through some introductory QFT stuff now actually and I am vaguely familiar with what you've said - where the field operator [itex]\hat{\phi}(x)[/itex] is expressed as an integral over d^3k of a function containing [itex]\hat{a}(k), \hat{a}^{\dagger}(k)[/itex]. However I'm too busy picking my way through all the maths to have learned any new physics, stuff hasn't quite sunk in yet.
     
  5. Nov 8, 2013 #4

    naima

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    Divide this by [itex]\hbar\omega[/itex].
    As |n> is the eigenvector of N = [itex]{a}^{\dagger}a [/itex]. Where N is the occupation number operator
    You see that n becomes n+1
     
  6. Nov 8, 2013 #5
    thanks for response but I will have to return to this tomorrow as it is 3:45 AM where I am and I must give in and go to bed...
     
  7. Nov 8, 2013 #6
    The key takeaway is this: just like how in single-particle QM you can use the creation/annihilation operators to factor the Hamiltonian: [itex]\hat{H} = \hbar\omega \hat{a}^\dagger \hat{a}[/itex], you can do the same thing in QFT, but using the whole set of C/A operators: [itex]\hat{H} = \int d^3 k\: \omega_k \hat{a}^\dagger_k \hat{a}_k[/itex], where [itex]\omega_k^2 - k^2 = m^2[/itex]. Or, as naima said, you can also write it as [itex]\int d^3 k\: \omega_k \hat{N}_k[/itex], where [itex]\hat{N}_k = \hat{a}^\dagger_k \hat{a}_k[/itex] is the number operator for particles of momentum [itex]k[/itex].

    In this form, you can interpret the Hamiltonian in a very simple way: it's essentially "probing" the state for how many particles of each momentum there are, and adding in the appropriate amount of energy ([itex]\omega_k[/itex]) for each one that it finds. You can construct other operators this way as well, like the momentum operator [itex]\vec{\hat{K}} = \int d^3 k\: \vec{k} \hat{N}_k[/itex], and the charge operator [itex]\hat{Q} = \int d^3 k\: q \hat{N}_k[/itex]. In this context, the occupation number idea starts making a lot of sense--in fact, you can just take that to be the definition of a particle: it's a discrete excitation of the field which is an eigenstate of the number operator. And since all of these other operators can be written in terms of the number operator, they commute with it, so these excitations are eigenstates of all of them as well. This means that this "particle" will have a well-defined energy, momentum, charge, etc. just like we would expect.

    In that sense, you can also think of the single-variable harmonic oscillator as having a single kind of "particle" with energy [itex]\omega[/itex], of which any number can be added to the system using the creation operator. It's not a very useful idea when dealing with the equation in single-particle QM, but it helps lay the groundwork for understanding the more complicated variant that comes up in QFT.
     
    Last edited: Nov 8, 2013
  8. Nov 8, 2013 #7
    As others have suggested, the language of "creating and annihilating particles" is borrowed from QFT. If we are dealing with a single particle in a harmonic potential in nonrelativistic QM, then obviously we are not actually creating (or annihilating) particles. As you said, we are just exciting our single particle to a higher energy state. But in QFT there arise operators with the same commutation relations that should be interpreted as creating and annihilating particles, so that the ground state ##| 0 \rangle## is the empty vacuum and ##| n \rangle## is a state with ##n## particles.
     
  9. Nov 8, 2013 #8
    In first quantization the states are members of the Hilbert space and this operator is called a rising operator because it rises the energy of the oscillator by one unit [itex]\hbar\omega[/itex].

    But in second quantization the states are members of the Fock space and this operator is called a creator operator because it creates a brand new particle with the energy [itex]\hbar\omega[/itex].
     
  10. Nov 15, 2013 #9
    Very good question Doofy!
    But as far as I understood from your explanation and question, you are taking just one oscillator in your mind, so by solving the eigenvalue problem, you found different possible states that Oscillator could take based on the number of times that (a†) operator operated on state |n⟩. so far everything is right but then you asked why do we assign a particle to each state?
    You need to be careful , just like what Chopin said, when we talk about quantization, we are talking about several oscillators and not just only one, it means that each state is related to one mode of cavity, in the other words we assign an oscillator to each mode.
     
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