Why Path Integral?¶
If you have any previous touch with Quantum Field Theory, you may know that there is a much more popular way to quantise the classical fields — canonical quantisation. However, in this introduction, we will not mention a word about it. If you are curious about the reason, this section will tell you why. However, if you are just a new learner, you can well skip this, it won’t affect.
Take the scalar field for example. The canonical quantisation claims the following “quantisation condition”
This is convenient since similar commutation relation has been studied thoroughly in Quantum Mechanics. However, it is this very first assumption that ruins the whole field theory.
To see this, let’s first review what the Quantum Mechanics claims. In Quantum Mechanics, we have the following commutation relation
There is nothing wrong here if both \(x_i\) and \(p_i\) are operators, and the product refers to operator composition. As an example, explicitly write an representation as \(p_i = -\mathrm i\partial_i\) will satisfy the above relation.
However, although many physicists claim that the commutation relation of fields is nothing but a continuous version of the point particle one, it is still mathematically impossible for such a “continuous” commutation relation to hold.
This can be seen through the following argument: field \(\varphi\) and \(\pi\) are both scalar field; according to the definition of scalar field product and subtraction, \(\varphi\pi\) and \(\pi\varphi\) will both be scalar field, so will their commutator \([\varphi, \pi]\). Thus, the left hand side of the commutation relation will be a scalar field (or more specifically, operator valued scalar field).
However, the right hand side is not a scalar field. Instead, according to the most moderate point of view, \(\delta(\vec x - \vec x')\) should be at most a generalized function. Therefore, the two sides of the equation can never be equal.