Editing Path integral formulation (section)

=== Canonical commutation relations ===

The formulation of the path integral does not make it clear at first sight that the quantities {{mvar|x}} and {{mvar|p}} do not commute. In the path integral, these are just integration variables and they have no obvious ordering. Feynman discovered that the non-commutativity is still present.<ref>{{harvnb|Feynman|1948}}</ref>

To see this, consider the simplest path integral, the brownian walk. This is not yet quantum mechanics, so in the path-integral the action is not multiplied by {{mvar|i}}:
: <math>S=  \int \left( \frac{dx}{dt} \right)^2\, dt</math>

The quantity {{mvar|''x''(''t'')}} is fluctuating, and the derivative is defined as the limit of a discrete difference.
: <math>\frac{dx}{dt} = \frac{x(t+\varepsilon) - x(t)} \varepsilon </math>

The distance that a random walk moves is proportional to {{math|{{sqrt|''t''}}}}, so that:
: <math>x(t+\varepsilon) - x(t) \approx \sqrt{\varepsilon}</math>
This shows that the random walk is not differentiable, since the ratio that defines the derivative diverges with probability one.

The quantity {{mvar|xẋ}} is ambiguous, with two possible meanings:
: <math>[1] = x \frac{dx}{dt}  = x(t) \frac{x(t+\varepsilon) - x(t) }{\varepsilon } </math>
: <math>[2] = x \frac{dx}{dt} = x(t+\varepsilon) \frac{x(t+\varepsilon) - x(t) }{\varepsilon} </math>

In elementary calculus, the two are only different by an amount that goes to 0 as {{mvar|ε}} goes to 0. But in this case, the difference between the two is not 0:
: <math>[2] - [1] =  \frac{\big( x(t + \varepsilon) - x(t)\big )^2}{\varepsilon} \approx \frac \varepsilon \varepsilon</math>

Let
: <math>f(t) = \frac{\big(x(t+\varepsilon)- x(t)\big)^2 }{\varepsilon}</math>

Then {{math|''f''(''t'')}} is a rapidly fluctuating statistical quantity, whose average value is 1, i.e. a normalized "Gaussian process". The fluctuations of such a quantity can be described by a statistical Lagrangian
: <math>\mathcal L =  (f(t)-1)^2 \,,</math>
and the equations of motion for {{mvar|f}} derived from extremizing the action {{mvar|S}} corresponding to {{mathcal|L}} just set it equal to 1. In physics, such a quantity is "equal to 1 as an operator identity". In mathematics, it "weakly converges to 1". In either case, it is 1 in any expectation value, or when averaged over any interval, or for all practical purpose.

Defining the time order to ''be'' the operator order:
: <math>[x, \dot x] = x \frac{dx}{dt} - \frac{dx}{dt} x = 1</math>

This is called the [[Itō lemma]] in [[stochastic calculus]], and the (euclideanized) canonical commutation relations in physics.

For a general statistical action, a similar argument shows that
: <math>\left[x , \frac{\partial S }{ \partial \dot x} \right] = 1</math>
and in quantum mechanics, the extra imaginary unit in the action converts this to the canonical commutation relation,
: <math>[x,p ] = i</math>