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From the answer to this question Computing $\langle0|T[Q(t_2)Q(t_1)]|0\rangle$, I have discovered that there is two perspectives to QFT. I am doing a course which is unfortunately a summary of QFT and does not provide a clear reason as to why we consider the "second quantisation" and path integrals.

Could someone please explain why sometimes we consider the "second quantisation" and sometimes we consider path integrals. Moreover I cannot see the point of the different path integrals I have which use the Lagrangian or the Hamiltonian.

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    $\begingroup$ What do you mean, "why?"? Both are valid approaches (at least at the physicist's level of rigour), just like Lagrangian and Newtonian and Hamiltonian mechanics in classical mechanics are different, valid approaches. $\endgroup$
    – ACuriousMind
    Mar 11 '15 at 18:05
  • $\begingroup$ Well are the advantages of each, as far as I can see using path integrals is unecessarily compliated $\endgroup$
    – Trajan
    Mar 11 '15 at 18:06
  • $\begingroup$ Possible link: physics.stackexchange.com/q/169602 $\endgroup$
    – Trajan
    Mar 11 '15 at 20:43
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    $\begingroup$ By far, the most important advantage of the path integral approach is that it allows to understand non-perturbative dynamics of the system, something that is completely absent in canonical quantization approach (barring only very few systems) $\endgroup$
    – Prahar
    Mar 22 '15 at 5:26
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Second quantization is named only to differentiate it from first quantization, but essentially the process is the same in each case: take a physical quantity and upgrade it to an operator.

In the case of first quantization, this refers to taking position and momentum and upgrading them to operators: $$ [\hat{x},\hat{p}]=i $$

The main reason you do this, is because you want to be able to understand how operating on the specific state of a system gives rise to dynamics. This is done by defining creation and annihilation operators $a$ and $a^\dagger$. In non-relativistic quantum mechanics, these operate on a Hilbert space $\mathcal{H}$: the space of states of our system.

In QFT, the second quantization procedure upgrades fields to operators, giving rise to commutators of the type $$ [\phi(\vec{x}),\partial_t\phi(\vec{y})] = i\delta^3(\vec{x}-\vec{y}) $$ Where the fields are themselves made up of creation and annihilation operators. These now act on Fock space, which is essentially a space of Hilbert spaces.

The key reason that these quantization procedures are done is to be able to calculate physical quantities of interest, usually by defining a multi-particle state in terms of creation/annihilation operators acting on the vacuum. For example, once you have defined a quantum operator (through second quantization) a multi-particle state is defined as: $$ a_1^\dagger(p_1)a_2^\dagger(p_2)\cdots a_n^\dagger(p_n)|0> $$

Once you have defined a multi-particle state, you can calculate quantities of interest like the Hamiltonian, the total momentum, the stress-energy tensor, correlation functions etc, which can all be defined in terms of quantum operators.

The canonical way of doing things does have problems though. One example is with the commutator of fields I wrote down: $[\phi(\vec{x}),\partial_t\phi(\vec{y})] = i\delta^3(\vec{x}-\vec{y})$. This object is not necessarily Lorentz invariant, since it only depends on space, not spacetime.

The path integral approach skips all of this: Everything about the path integral is classical, and it is also manifestly relativistic and Lorentz invariant. Using the path integral allows you to directly calculate quantities of interest from the classical action of a system. This can often be very hard for certain systems, which is why canonical quantization is used. For example, systems with non-quadratic momenta in their actions are hard to calculate with path integrals. The path integral gets its "quantumness" from the fact that each path you sum over is "weighted" by a complex phase.

The use of both canonical and path integrals is dependent on the situation in question. Both are equivalent, but which you use is a matter of taste and/or convenience.

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