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Creation and anihillation operators

  1. Nov 23, 2013 #1
    The fundamental idea of these operators is that we can use them to add particles to our system to a specific eigenstate. Now my book has examples of these operators of which the harmonic oscillator ladder operators are used. But thinking about it, this example does not make sense for me.
    The idea of the ladder operators for the SHM was that when used on eigenstate number n it took us to eigenstate n+1. How is this the same as adding a new particle to the system? E.g. if we let a_+ denote the creation operator in second quantization and l1> the first excited state my books idea is that:
    a_+l1> = l2>
    But this is not true? a_+ acts on a vector specifying the number of particles in each eigenstate and by using it on l0,1,0,0,0,0...> we get the vector l0,1,1,0,0,0,0,0,0...>, that is a 2-particle state - one in first singleparticle eigenstate and one in the second.
    What am I missing?
     
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  3. Nov 23, 2013 #2

    hilbert2

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    In the case of the simple harmonic oscillator, the raising and lowering operators can't be interpreted as changing the number of particles. To handle systems with variable particle number, we need relativistic quantum field theory. When a free field (Klein-Gordon, EM, etc.) is quantized, each of the infinite number of Fourier components (normal modes) of the field behaves as a harmonic oscillator. Each mode has its own raising and lowering operators. In that case one can interpret the excitation level of a mode as the number of particles in that mode. Of course, if the field is fermionic, each mode can be occupied by at most one particle.
     
  4. Nov 23, 2013 #3
    Okay so the example is just showing the idea of an operator raising the number of something.
    Actually I have another question for you. The term diagonalized for an operator means that it is represented in a basis of eigenstates. But my book says this is equivalent to it being written in the form:
    Ʃka_k(dagger) a_k
    I would assume that in this case a_k(dagger),a_k refer to the creation an anihillation operators for the eigenstate k because else I don't see how to make sense of it. Do you agree?
     
  5. Nov 23, 2013 #4

    hilbert2

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    ^ Yes, I think the ##a## and ##a^{\dagger}## refer to creation and annihilation operators. If ##a_{k}## and ##a_{k}^{\dagger}## are the creation and annihilation operators for a field mode labelled ##k##, then the operator ##N=a_{k}^{\dagger}a_{k}## is the particle number operator for that mode.

    Usually, saying that an operator ##A## is diagonalized means that we write it in the form ##A=\sum_{n=0}^{\infty}a_{n}|n><n|##, where the ##|n>## are the eigenstates of the operator ##A## and the ##a_{n}## are the corresponding eigenvalues.
     
  6. Nov 23, 2013 #5
    Yes exactly. So what is the relation between the outer product of two eigenstates and the creation and anihillation operator for that state. Is it simply the same?
     
  7. Nov 23, 2013 #6

    hilbert2

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    ^ They are not the same. An outer product of two normalized vectors is a projection operator and its only eigenvalues are 0 and 1. The number operator ##a^{\dagger}a## has all natural numbers as its eigenvalues. What's the name of the book where you found that definition of diagonalization?
     
  8. Nov 23, 2013 #7
    It appeared in some exercises, for instance the attached. But transformation to k-space you find a hamiltonian which doesn't have crossterms for the operators c_kc_k(dagger) product.
     

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  9. Nov 23, 2013 #8
    I think it is more apparant in this exercise what I mean. Here the operators in the diagonal matrix are not even lowering/raising operators. On what ground can you call the H given diagonal.
    Also I have some other questions regarding the attached exercise:
    1) Why does it distinguish between h and H?
    2) A matrix is unitary if U*=U^(-1) so the diagonalization condition for H in the eigenbasis becomes H=UHU^(-1) (or something like that right?) - how on earth does this relate to the transformation formula the exercise gives?
     

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  10. Nov 24, 2013 #9

    hilbert2

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    I'm a bit confused by those exercises, even though I had condensed matter physics last spring and the tight-binding model was discussed there. In the physics of crystals, the "k-space" probably refers to writing the Hamiltonian in terms of the normal modes of crystal vibrations. The normal modes behave similarly to those of quantum fields, and their excitations are called phonons (quasiparticles).

    Maybe someone else can help you more with this problem.
     
  11. Nov 24, 2013 #10
    On the ground that the operator can be written [itex]H=\alpha_i^\dagger H_{ij} \alpha_j[/itex]. If [itex]H_{ij}[/itex] is diagonal, so that
    [tex]H=\sum_i \alpha_i^\dagger H_{ii} \alpha_i = \sum_i \epsilon_i \alpha_i^\dagger \alpha_i [/tex],
    it makes sense to call the operator diagonal. There is also the argument that a diagonal operator leaves the system in whatever eigenstate it is in (quite unlike a non-diagonal perturbation) and, as the operator [itex]\alpha_i^\dagger \alpha_i[/itex] preserves the particle number, the system remains in whatever number state it was in before.

    I guess this is just to make it explicit that there is a change of basis between [itex]h=a_i^\dagger h_{ij} a_j[/itex] and [itex]H=\alpha_i^\dagger H_{ij} \alpha_j[/itex]. However, both should still represent the same abstract Hamiltonian operator.

    Usually, we write down the Hamiltonian as some operator in terms of some energy basis and that's it. Then it is enough to diagonalize the matrix [itex]H_{ij}[/itex] in the manner you described. But if the Hamiltonian is written in terms of other operators, one must first consider how these change with a change of basis.
     
  12. Nov 25, 2013 #11
    I don't understand a lot of this, so let me try and start from where I understand it and we can work it out together? I have attached what would be my starting point and tried to work from there:
    Question, what are the matrix elements of H in the a,a_dagger basis - or whatever the basis is. I guess a main problem is that I don't see what basis the hamiltonian with h_ij is written in and I don't see what the matrix elements are. I mean if it said:
    H=h_ij lei><ejl, then obviously the matrix elements of H were just h_ij and the basis was lej> but this is not the case here.
     

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  13. Nov 26, 2013 #12
    We can do that. What book are you using? This is described quite well in Altland & Simon's book Condensed Matter Field Theory, for example, if you need a second resource.

    Let's start by looking what these things actually means. [itex]a^\dagger[/itex] adds a particle in some state, and [itex]a[/itex] removes it, right. So, for the entire system, we can write
    [tex]|n_1, n_2, \dots \rangle \equiv |n_1\rangle \otimes |n_2\rangle \otimes \dots [/tex]
    to label that we have [itex]n_1[/itex] particles of the first kind, [itex]n_2[/itex] of the second and so on. In terms of the creation operatpors, this means that we started from the vacuum and created these particles, i.e.
    [tex] |n_1, n_2, \dots \rangle = \prod_i \frac{1}{\left( n_i! \right)^{1/2}}\left( a_i^\dagger \right)^{n_i} |0\rangle[/tex]

    By ordering the particles like this, we have picked a one-particle basis, haven't we? Maybe it's based on position (i.e. [itex]n_i[/itex] particles at position i), maybe on momentum, or something else. What happens if we change that basis?

    Yeah, it's not quite that simple here. H is certainly written in the number representation, so it will always act on a state like the one above and thus depend on the basis of that number representation. But let us wait with this for a moment, since it gets much simpler once one knows how the creation and annihilation operators change under a change of state.
     
  14. Nov 27, 2013 #13
    I think I saw somewhere:

    a_k = Ʃ_r <rlk>lr>

    . but to prove it what do I do? Isn't it simply to inser identity in the slater determinant basis and then use the basis change equation?

    Edit: Indeed, in my book I found the attached equation which explains it - though I think some of the indices are wrong but never mind that.

    But still the eigenvector overlap could be quite complicated - I don't see where this leads us?
     

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    Last edited: Nov 27, 2013
  15. Nov 27, 2013 #14

    DrDu

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    That's not quite true. In (non-relativistic) solid state physics, you also interprete the exited states of the harmonic oscillators describing lattice vibrations as particles, namely phonons. The number of phonons is also not conserved but can fluctuate.
     
  16. Nov 27, 2013 #15
    That formula seems to have an operator on one side, and a state on the other side. In general,
    [tex]|k \rangle = \sum_r0 |r\rangle \langle r| k\rangle[/tex]
    which indeed follows directly from the resolution of identity. Taking [itex]|k\rangle[/itex] and [itex]|r\rangle[/itex] to be our one-particle basis states, i.e.
    [itex]|k\rangle = a_k^\dagger |0\rangle[/itex] and [itex]|r\rangle = a_r^\dagger |0\rangle[/itex]. One then immediately finds the transformation law
    [tex]a_k^\dagger = \sum_r \langle r|k\rangle a_k^\dagger[/tex]
    and the corresponding law for the annihilation operator is found by taking the Hermitian conjugate of this equation.

    The overlap could certainly be quite complicated, but if we want to be able to handle arbitrary bases we'll just have to accept that. Let us turn now to what is usually (in many-body theory) called single-particle or one body operators. I don't quite like the name, but I see where it comes from. They basically have the form
    [tex]\hat{O}_1 = \sum_{n=1}^N \hat{o}_n[/tex]
    where [itex]\hat{o}_n[/itex] is some operator acting only on the n:th particle. [itex]\hat{O}_1[/itex] could be the kinetic energy of the system, total spin operator, etc. Well, basically anything that doesn't involve interaction between two different particles. Hence its name.

    The reason I wanted to go through the transformation laws before, is that it is easiest to describe [itex]\hat{O}_1[/itex] in its diagonal basis and then transform to the general case. So, assuming [itex]\hat{O}_1[/itex] is diagonal in the [itex]|\lambda \rangle[/itex] basis, can you see what form it will get? It may help to first define the number operator [itex]\hat{n}_\lambda = a_\lambda^\dagger a_\lambda[/itex].
     
  17. Nov 27, 2013 #16
    Well in first quantization each one-particle hamiltonian is in diagonal form:
    ƩEilei><eil (1)
    The basis of λ is the set of many particle slaterdeterminants containing the ei, which just correspond to vectors of the form
    l1 1 0 .. >
    l0 1 0 1 ..>
    So I think next step is to put up and operator equation. Actually I will return tomorrow with a more qualified answer so don't bother answering this one.
     
    Last edited: Nov 27, 2013
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