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Again Question on Particle Birth

  1. Feb 24, 2006 #1
    Dear PF,

    There are some Baisic questions in QM/QFT that always creat some discomfort when try to think over. Could you pls have a look on my question which I put in word attachment, since formulas do no appear when I write them.

    Thank you a lot

    Attached Files:

  2. jcsd
  3. Feb 24, 2006 #2
    Ok, I'll give it a shot. I'll use real, scalar Klein-Gordon field for simplicity. Expanded as a fourier integral over three-momenta

    [tex]\phi (x)=\int \frac{d^3p}{(2\pi )^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left( a_{\mathbf{p}}e^{-ip\cdot x}+a_{\mathbf{p}}^{\dagger}e^{ip\cdot x}\right) [/itex]

    a and its hermitian conjugate are anihilation and creation operators, respectively.

    Say we want an out-going external boson line, then we need to create a final state Klein-Gordon particle defined at x. This final state particle of momentum p and energy [itex]E_{\mathbf{p}}[/itex] is, following the conventions for relativistic normalisation

    [tex]\langle \mathbf{p}|=\langle 0|a_{\mathbf{p}}\sqrt{2E_{\mathbf{p}}}[/tex]

    The state vector [itex]|\mathbf{p}\rangle [/itex] is, as you say, a vector of Hilbert space. It is an excited state of the vacuum state with momentum [itex]\mathbf{p}[/itex]. The physical momentum [itex]\mathbf{p}[/itex] is extracted as an eigenvalue by operating on [itex]|\mathbf{p}\rangle [/itex] with the momentum operator [itex]\mathbf{P}|\mathbf{p}\rangle =\mathbf{p}|\mathbf{p}\rangle[/itex].

    To create this state at x we operate on it with a creation operator from the quantised field [itex]\phi [/itex], as in QFT we treat matter as an excitation of a field. Your equation is missing a creation opertor from the negative frequency part of the field, and then the following step.

    [tex]\langle \mathbf{p}|\phi ^-(x) =\langle 0|a_{\mathbf{p}}\sqrt{2E_{\mathbf{p}}}\int \frac{d^3p}{(2\pi )^3}\frac{1}{\sqrt{2E_{\mathbf{q}}}}a_{\mathbf{q}}^{\dagger}e^{iq\cdot x}[/tex]

    [itex]\phi^-[/itex] being the negative frequency part of the field. Commuting [itex]a_{\mathbf{p}}[/itex] and [itex]a^{\dagger}_{\mathbf{q}}[/itex], we get a delta function and a term that will anihilate with the vacuum state. Explicitly

    [tex][a_{\mathbf{p}}, a^{\dagger}_{\mathbf{q}}]=\delta^{(3)}(\mathbf{p}-\mathbf{q})[/tex]
    [tex]\langle \mathbf{p}|\phi ^-(x) =\langle 0|\sqrt{2E_{\mathbf{p}}}\int \frac{d^3p}{(2\pi )^3}\frac{1}{\sqrt{2E_{\mathbf{q}}}}\left( \delta^{(3)}(\mathbf{p}-\mathbf{q})+a_{\mathbf{q}}^{\dagger}a_{\mathbf{p}}\right) e^{iq\cdot x}[/tex]
    [tex]=\langle 0|e^{ip\cdot x}[/tex]

    Which gives us a final state Klein-gordon particle of well defined four-momentum p and with the exponential describing its position distribution.

    Maybe your missunderstanding is coming from not being familiar with the creation/anihilation operators. A good way of getting accustomed to them is to go through the quantisation of the harmonic oscillator using such operators, or quantising the Klein-Gordon field using methods analogous to the quantisation of a harmonic oscillator (second quantisation).
    Last edited: Feb 24, 2006
  4. Feb 24, 2006 #3


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    Science Advisor

    Quantum fields are simply operators, really no different than a Hamiltonian. The whole idea of creation and destruction operators goes way back to Heisenberg's matrix mechanics, particularly applied to the oscillator. To see that the standard "a" operators do what is claimed for them requires a modest amount of work -- finding the eigenstates of the number operator, among other things.

    All of this is very basic to QM, and, fortunately, is discussed in many, many books and Googles. Landau and Lifschitz, Cohen-Tannoudji, Zee, Gross, Weinberg, ...... all write about your issue. So, read.

    Reilly Atkinson
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