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I Understanding the scalar field quantization

  1. Mar 22, 2017 #1
    I am getting started with QFT and I'm having a hard time to understand the quantization procedure for the simples field: the scalar, massless and real Klein-Gordon field.

    The approach I'm currently studying is that by Matthew Schwartz. In his QFT book he first solves the classical KG equation with Fourier transform and shows that the general solution is:
    [tex]\phi(t,x)=\int \dfrac{d^3 \mathbf{p}}{(2\pi)^3}(a_{\mathbf{p}}e^{-ix^\mu p_\mu}+a_{\mathbf{p}}^\ast e^{i x^\mu p_\mu}).[/tex]
    Here there is nothing fancy, it is the usual way to solve a DE by using Fourier transform and applying the additional conditions (in this case, that the field is real). In this setting, [itex]a_{\mathbf{p}}[/itex] are complex numbers.

    After this, he simply states the following
    A little after that he says:

    Now I must confess I'm quite confused with this. It is completely unclear to me what do we actually assume that we have and what we are actually deriving and how.

    In Quantum Mechanics, in particular in the case of the SHO, it is pretty clear what we assume that we have and what do we derive. There we assume we have one state space of kets [itex]\mathcal{E}[/itex] on which there is one position observable [itex]X[/itex] together with its basis [itex]|x\rangle[/itex]. We also assume that there exists one momentum observable [itex]P[/itex] which acts as the generator of spatial translations, which in turn is equivalent to satisfy [itex][X,P]=i[/itex]. In that sense we know how [itex]P[/itex] acts on the position representation and we can relate both representations. All of this is assumed to be there, and this is not so hard to grasp.

    Quantization then only means to pick the classical Hamiltonian [itex]H = p^2/2m + m\omega^2 x^2/2[/itex] and replace the classical dynamical variables by the operators, assumed to exist. This leads to the quantum Hamiltonian observable. We then try to factor the Hamitonian and we derive in terms of [itex]X,P[/itex] which we have assumed to exist, the ladder operators. Finally we prove several properties of these operators that allows us to find the spectrum of the Hamiltonian and the eigenstates.

    In QFT everything is blurry. I mean, we have to find one function that gives operators to events in spacetime. This is not the same as assuming one operator [itex]X[/itex] exists. We must find the dependency [itex]\phi(x)[/itex] so that the field obeys a differential equation. It is not as in QM, where we suppose there exists the position and momentum operator satisfying the canonical commutation relations. Here we must find a specific functional form obeying a DE.

    Considering all this there are some points to mention:
    1. It is not clear how does one find out what is [itex]\phi(x)[/itex] in terms of the ladder operators. I mean, there is a huge jump from the classical solution to the quantum one. I don't see how this is obvious, actually it seems pretty not obvious.
    2. One usually writes down [itex]\phi(x)[/itex] in terms of ladder operators. But neither [itex]\phi(x)[/itex] neither those ladder operators have been defined. In truth this is in sharp contrast to the SHO: there we clearly have [itex]X,P[/itex] and their representations as starting point from which we derive the rest. Here it seems we have no starting point.
    3. The space is also not clear. In QM as I said we assume there is one Hilbert space spanned by some representation physically relevant based on the observable's algebra. Here it is not even clear what Hilbert space spanned by what basis do we have.
    As I said this is being all confusing, and I wanted to make this extremely clear and settled as in QM, since this seems to be extremely important to really understanding QFT. So, considering all the points I mentioned, how can we deal with each of them and how can we actually understand the quantization of the scalar massless KG field?
     
  2. jcsd
  3. Mar 22, 2017 #2
    The crucial point is this:
    This is a free field, so each mode is an independent harmonic oscillator. For each value of p there is a harmonic oscillator equation. We quantize it. That means that the space of states of the theory (called Fock space) is the tensor product of an uncountable infinity of harmonic oscillator Hilbert spaces, one for each p.

    As far as what the creation and annihilation operators are in terms of field operators, well, you have the field operators. You can use the expression you have to find phi-dot in terms of phi. Phi-dot is sometimes given the letter pi and called the field momentum density. You can then find an expression for a and a-dagger in terms of phi and phi-dot.

    If this is all unclear to you, you can look for Sidney Coleman's lecture notes. They're easy to find, and they jump right into canonical quantization, with a suitable amount of detail. The expression I mentioned for a and a-dagger in terms of the field is in page 29 of the typeset notes.
     
  4. Mar 23, 2017 #3

    vanhees71

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    The most simple heuristic way to understand field quantization is "canonical quantization", i.e., you start with the Lagrangian for the fields. In your case of a free neutral scalar field (neutral Klein-Gordon field), the Lagrangian reads (I use the mainly-minus convention of the Minkowski metric, ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)##, as in Schwartz's book:
    $$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2, \quad \phi \in \mathbb{R}.$$
    Then you calculate the canonical field momenta,
    $$\Phi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}.$$
    This leads to the canonical equal-time commutation relations for the field operators (in the Heisenberg picture!)
    $$[\hat{\phi}(t,\vec{x}),\hat{\phi}(t,\vec{y})]=0 \\ \quad [\hat{\Pi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=[\dot{\hat{\phi}}(t,\vec{x}),\dot{\hat{\phi}}(t,\vec{y})]=0, \\
    [\hat{\phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
    Also the field operator must fulfill the equation of motion, which turns out to be the Klein-Gordon equation as for free fields
    $$(\Box+m^2) \hat{\phi}=0.$$
    Now you plug in the mode decomposition as given in #1 with the ##a(\vec{p})## now becoming operators ##\hat{a}(\vec{p})##. Then using the commutator relations for the field operators you indeed get the commutator relations for the annihilation and creation operators,
    $$[\hat{a}(\vec{p}),\hat{a}(\vec{q})]=0, \quad [\hat{a}(\vec{p}),\hat{a}(\vec{q})]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{q}).$$
     
  5. Mar 23, 2017 #4
    So as I understand what you mean is: the meanining of quantizing the field is to find operator valued fields such that (i) the commutation relations are obeyed and (ii) the operator valued field obeys the equation of motion of the classical field. Is that right?

    Because of that I still have two doubts:
    1. Where the mode decomposition for the quantum field comes from? I mean, sincerely it seems like it comes out of thin air. Both Schwartz and Peskin just present it without much explanation of where it came from. Actually I do understand where the classical analog comes from: we solve the KG equation with Fourier transform and impose the field is real-valued. How does one find out that decomposition for the quantum case is something I really don't understand. What is the right way to derive that decomposition for the quantum field?
    2. This thing of writing the field in terms of these ladder operators seems circular at first to me. The reason: we don't know yet what is the quantum field [itex]\phi(x)[/itex]. In QM we do know what is the position operator [itex]X[/itex] and the momentum operator [itex]P[/itex]. Then we derive the ladder operators in terms of these. So it is pretty clear what we assume and what we derive in terms of what we have. In QFT, we don't have [itex]\phi(x),\pi(x)[/itex] in the first place. Then we write them in terms of other operators which we also don't know yet. Actually we don't know what is the action of the operator [itex]a(\mathbf{p})[/itex], not even the space where it acts! I don't know, I think its because I'm new to this but this seems blurry.
    Actually the best I've been able to grasp is that: we want to field [itex]\phi(x),\pi(x)[/itex] obeying the commutation relations and the differential equation. We don't know them yet but somehow we know that to obey the differential equation is the same as having that decomposition. Finally we try to impose the commutation relations and we end up discovering that having [itex]\phi(x),\pi(x)[/itex] obeying the differential equation and the commutation relations is equivalent to having them written in terms of [itex]a(\mathbf{p})[/itex] and [itex]a^\dagger(\mathbf{p})[/itex] which in turn obey these other commutation relations. Is this the idea? We exchange the problem of finding [itex]\phi(x),\pi(x)[/itex] for the problem of finding [itex]a(\mathbf{p}),a^\dagger(\mathbf{p})[/itex]?

    If that's true, frankly I'm still confused for I still don't get where the decomposition came from, nor where the ladder operators came from. They, together with the Hilbert space where they act, still seem undefined to me.
     
  6. Mar 23, 2017 #5
    I explained that to you in my post.
     
  7. Mar 24, 2017 #6

    vanhees71

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    The decomposition comes from solving the Klein-Gordon equation for free particles in terms of a Fourier decomposition, which in fact should read
    $$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3 \sqrt{2 \omega_{\vec{p}}}} [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x)+\hat{a}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x)]_{p^0=+\omega_{\vec{p}}}.$$
    From the equal-time commutation relations of the fields and canonical field momenta you get the commutation relations for the ##\hat{a}(\vec{p})## and ##\hat{a}^{\dagger}(\vec{p})## by using the inverse Fourier transformation and then evaluate these commutators. This should be in Schwartz's book (I don't have it here right now to check).
     
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