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I don't get what is it we're trying to do in QFT. I'm currently in the beginning of the course and a clear picture of what we're trying to achiever hasn't been painted yet to me.

From what I've been able to gather is, for a spin 0 field, we wish to have an operator-density field which satisfies the Klein Gordon equation and then another operator-density field which satisfies the momentum-position like commutation relation with this field. Now after this, we construct a Hamiltonian-density operator field and integrate it over space to get the Hamiltonian operator from the scalar field.

Now is this Hamiltonian operator supposed to be applied in the Schrodinger's equation in QM? What is the vector space this Hamiltonian operator is going to act upon? When/How are the particle creation-annihilation process going to come into the picture?

Can someone please provide me with a picture/roadmap of the things we are trying to do in QFT. Like in QM, we replaced knowledge of particle with a wave function/quantum state and then had an evolution operator for this state.

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  • $\begingroup$ A free fields fourier expansion in the spatial variables will give a harmonic oscillator-like equation, which allows the introduction of creation and annihilation operators. The field operator is expanded in terms of them. These have the interpretation of particle creation and annihilation operators. The state space consists of a direct sum of $N$-particle state spaces (for all $N$s). The Schrodinger equation is not used usually, instead, the only relevant state is the vacuum. Every other state is given by creation/annihilation operators. $\endgroup$ – Bence Racskó May 25 '17 at 13:11
  • $\begingroup$ What I said works for free fields only, hence the perturbative approach to interactions. I don't know much QFT either, so in case somebody corrects me, listen to them, but I think this comment of mine is correct. $\endgroup$ – Bence Racskó May 25 '17 at 13:12
  • $\begingroup$ @Isomorphic I believe this starting was with you tube lectures, isn't it? $\endgroup$ – Wrichik Basu May 25 '17 at 13:14
  • $\begingroup$ @Isomorphic for a road map to QFT, you need a book like that of Quantum Field Theory Demystified by David McMohan. Buy it. It's a good one. $\endgroup$ – Wrichik Basu May 25 '17 at 13:16
  • $\begingroup$ In a nutshell, you are combining QM with special relativity. This motivation is for sure described in many opening chapters of QFT books $\endgroup$ – innisfree May 25 '17 at 13:24
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The quick and dirty version is that you model all the particles of a given type as excitations of a series of quantum harmonic oscillators:

$$ H = \int\frac{\mathrm{d}^{3}\vec{p}}{(2\pi)^3} E_{\vec{p}} \left(a_{\vec{p}}^{\dagger}a_{\vec{p}} + \frac{1}{2}\right) $$

so a particle of momentum $\vec{p}$ would be the $|1\rangle$ state of the harmonic oscillator of momentum $\vec{p}$. Note $E_{\vec{p}}^2 - \vec{p}^2 = m^2$ in natural units and $E_{\vec{p}}$ is an angular frequency by the de Broglie relation. To simplify this you define a thing called a 'field operator' that allows you work in position instead of momentum space:

$$ \phi = \int\frac{\mathrm{d}^{3}\vec{p}}{(2\pi)^3}\frac{1}{2E_{\vec{p}}}\left(a_{\vec{p}} \mathrm{e}^{-ipx} + a^{\dagger}_{\vec{p}} \mathrm{e}^{ipx}\right) $$

where $p$ and $x$ without arrows indicates four-vectors and four-position. If you plug this in and chug through the algebra you get the standard field theoretic Hamiltonian for a free (scalar) field:

$$ H = \int \mathrm{d}^{3}\vec{x}\left(\left(\frac{\partial\phi}{\partial t}\right)^2 + \left(\nabla\phi\right)^2 - m^2\phi^2\right) $$

The Hilbert space for this Hamiltonian is just what you'd expect from a set of harmonic oscillators:

$$ \mathcal{H} = \bigotimes_{\vec{p}}\mathcal{H}_{\vec{p}} $$

where $\mathcal{H}_{\vec{p}}$ is the Hilbert space for a single harmonic oscillator, and in the expression for the Hamiltonian we've really suppressed an uncountable series of $\otimes \mathbb{I} \otimes$ before and after each ladder operator. Sometimes people call this a Fock space, but it isn't really a Fock space. It has similar properties, but its construction is very different [1].

For dynamics, you use the Heisenberg picture, and in particular you use the Heisenberg equation (not the Schrodinger equation):

$$ \frac{\mathrm{d}\phi}{\mathrm{d}t} = i\left[H,\phi\right] \\ \frac{\mathrm{d}\pi}{\mathrm{d}t} = i\left[H,\pi\right] $$

where $\pi = \frac{\partial\phi}{\partial t}$ is the momentum conjugate of the field defined in the usual way from the Lagrangian. Again, ploughing through the algebra you will find that the field obeys the Klein-Gordon equation:

$$ \left(\Box + m^2\right)\phi = 0 $$

Naturally this is a rather bizarre statement to make about the universe. Why are all particles excitation of a harmonic oscillator? Is it just an approximation, like so many things in physics which are modeled by harmonic oscillators, or is there something more fundamental going on?

Obviously, the answer is that there is something more fundamental. To see it, you have to look at the differential geometric structure of the spacetime manifold, and in particular the different representations of its isotropy group (the Lorentz group). In doing this, you see that the position space picture is the natural starting point and it amazingly turns into harmonic oscillators when you do a Fourier transform. Essentially this is the true mathematical formalism of canonical quantisation.

I am happy to go into the technical details of this construction if you want (it explains vector fields and spinor fields as well, which the above approach does not), but it's mostly of mathematical and philosophical interest rather than anything practical with calculations. (It's also useful if you want to look at unification and stuff I suppose.)


[1]: In particular, $\mathcal{H}$ comes already equipped with the idea of indistinguishability built-in, because if you're going to call the state $|2\rangle$ of a harmonic oscillator a 2-particle state (where both have the same momentum), there's already no concept of 'which particle is 1 and which is 2'.

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  • $\begingroup$ Is there an experiment we've in mind when beginning to quantise a Klein Gordon field? $\endgroup$ – Isomorphic May 25 '17 at 15:54
  • $\begingroup$ @Isomorphic Particle accelerator experiments. What you want to do is create a 'free field' state $|i\rangle$ of some particles, and then evolve it under the unitary evolution operator of the full interacting Hamiltonian, $U$, and then find out the probability it has turned into some other free field state $|f\rangle$, given by $P = |\langle f|U|i \rangle|^2$. You can expand $U$ out as a Taylor series in powers of the coupling constant and then the $n$th term in the Taylor series corresponds to a sum over all Feynman diagrams with $n$ vertices. $\endgroup$ – gautampk May 25 '17 at 16:16
  • $\begingroup$ @Isomorphic I'd also point out that 'quantising a field' is just a mathematical trick we use as part of the process for guessing the Lagrangian. The classical field doesn't mean anything physically, only the quantum field does. $\endgroup$ – gautampk May 25 '17 at 16:31
  • $\begingroup$ Is it ultimately about computing the S-Matrix? $\endgroup$ – Isomorphic May 27 '17 at 19:26
  • $\begingroup$ @Isomorphic Yes. I wasn't sure how much QFT you had done. As I said, you take your unitary time evolution operator, $U(t',t)$, and evolve some initial state and find its overlap with a final state. However, to ensure that these initial and final states are eigenstates of the free field operator, you must let $t \to -\infty$ and $t' \to \infty$. The object $\langle f|U(\infty,-\infty)|i \rangle$ is then called an 'S-matrix element' and we'd usually write $S = U(\infty,-\infty)$. $\endgroup$ – gautampk May 27 '17 at 21:50
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To answer your specific questions:

Now is this Hamiltonian operator supposed to be applied in the Schrodinger's equation in QM?

Yes. This operator describes the evolution of the quantum state in exactly the same way you are used to. Namely, the state at any particular time is a vector in a Hilbert space, say $\lvert\text{state}\rangle$, and the state some time $t$ later is $e^{-iHt}\lvert\text{state}\rangle$. This raises the question...

What is the vector space this Hamiltonian operator is going to act upon?

In general, the Hilbert space of a QFT is the complex span of the space of field configurations. For example, for a real scalar field, the field configurations are all the functions from space $\mathbb{R}^d$ (not spacetime, just space) to $\mathbb{R}$. Symbolically the set of field configurations $B$ is

$$B=\{\phi\,|\,\phi:\mathbb{R}^d\rightarrow\mathbb{R}\}.$$

Now take $B$ to be the formal basis for a vector space $\mathcal{H}$. This is the Hilbert space of QFT. So if $\phi_1$ and $\phi_2$ are two different functions from $\mathbb{R}^d$ to $\mathbb{R}$, the Hilbert space will include states like $\vert\phi_1\rangle$, $\vert\phi_2\rangle$, and $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$. (Note that it is not the case that $\alpha\vert\phi\rangle=\vert\alpha\phi\rangle$, and it is not the case that $\vert\phi_s\rangle+\vert\phi_2\rangle=\vert\phi_1+\phi_2\rangle$. The linear combinations like $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$ are formal. Also note that we take the different elements of $B$ to be formally orthogonal. So if $\phi_1\neq\phi_2$ we have $\langle\phi_2\vert\phi_1\rangle=0$.) This Hilbert space $\mathcal{H}$ is indeed the Hilbert space that the Hamiltonian operator acts on. So, for example, at some time the state of the universe might be $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$. Then the state of the universe some time $t$ later will be $$ e^{-iHt}(\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle). $$ Just like you're used to.

(One might consider having a Hilbert space of this form as a definition of what a QFT is. It's in the name after all: a quantum field theory is just a quantum theory where the states are superpositions of field configurations, rather than, say, superpositions of particle configurations. All the other objects/properties that commonly get talked about in a QFT course, like Lagrangians, Lorentz symmetry, etc. are all just extras. There are indeed proper QFTs without Lagrangian formulations, or without Lorentz symmetry, and so on.)

When/How are the particle creation-annihilation process going to come into the picture?

We now have a Hilbert space $\mathcal{H}$, and we have a basis for it, $B$. As with any vector space there are many choices of basis for $\mathcal{H}$. The basis $B$ turns out to not be the only (or even most) useful basis. Remember that in one-particle QM, alongside the position basis $\{\vert x\rangle\}_{x\in\mathbb{R}}$, a common basis for the Hilbert space is the harmonic oscillator eigenstate basis: $\{\vert0\rangle,a^\dagger\vert0\rangle,a^\dagger a^\dagger\vert0\rangle,\ldots\}$. In QFT one often talks about the "Fock space" basis, which is analogous to the harmonic oscillator eigenstate basis you're familiar with from one-particle QM.

The elements of $B$ have the physical interpretation of field configurations. The elements of the Fock basis, on the other hand, have the physical interpretation of particles. These two bases for $\mathcal{H}$ are, of course, related by something like a unitary transformation. So states from the Fock basis like $a^\dagger_p a^\dagger_q\vert 0\rangle$ can be written as a "sum" of field configuration states like $\vert \phi\rangle$. And field configuration states like $\vert \phi_1\rangle$ or $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$ can be written as "sums" of Fock basis states. In practice, the way to go back and forth between these two bases is via the relation $$ \hat{\phi}(x)=\int\!\frac{\mathrm{d}^dp}{(2\pi)^d}\,\frac{1}{\sqrt{2\omega_p}}(a_pe^{-ip\cdot x}+a^\dagger_pe^{ip\cdot x}), $$ where $\hat{\phi}(x)$ are the field operators, the operators of which the elements of $B$ are eigenstates. (e.g. the operator $\hat{\phi}(x)$ acting on $\vert\phi_1\rangle\in B$ gives $ \hat{\phi}(x)\vert\phi_1\rangle=\phi_1(x)\vert\phi_1\rangle. $)

Realize that the above is all just a rough sketch. But it's the sketch you should be holding in your head when learning about QFT. Now for some editorializing. Many textbooks and courses do a bad job of explaining these fundamentals. In fact, QFT pedagogy is rife with such bad concepts as "second quantization" and false statements like "QFT is QM made compatible with special relativity," "The Klein-Gordon and Dirac equations are relativistic versions of the Schrodinger equation," "In QFT we use the Heisenberg equation, not the Schrodinger equation," "We replace the wavefunction from QM with the field operator," "There are no wavefunctions in QFT," and a million others.

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  • $\begingroup$ Thanks, that's a great answer. It makes intuitive sense too and is exactly what I needed. $\endgroup$ – Isomorphic May 27 '17 at 19:34
  • $\begingroup$ Can you recommend a resource for learning QFT the right way? $\endgroup$ – Isomorphic May 27 '17 at 19:36
  • $\begingroup$ Also, what do you mean by a rough sketch? $\endgroup$ – Isomorphic May 27 '17 at 19:46
  • $\begingroup$ Your last para basically summarises the root of my motivation to even ask this question $\endgroup$ – Isomorphic May 27 '17 at 23:49
  • $\begingroup$ If I remember correctly, Quantum Field Theory in a Nutshell by Zee has a good discussion about what QFT is. "Rough sketch" is because the category of QFTs includes things more general than the picture above. One common example is that, in a gauge theory, the Basis for the Hilbert space is smaller than what I described above. Via gauge symmetry we identify some of the different field configurations of $A$. $\endgroup$ – Jennex May 28 '17 at 19:44

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