Editing Path integral formulation (section)

=== Equations of motion ===

Since the states obey the Schrödinger equation, the path integral must reproduce the Heisenberg equations of motion for the averages of {{mvar|x}} and  {{mvar|ẋ}} variables, but it is instructive to see this directly. The direct approach shows that the expectation values calculated from the path integral reproduce the usual ones of quantum mechanics.

Start by considering the path integral with some fixed initial state
: <math>\int \psi_0(x) \int_{x(0)=x} e^{iS(x,\dot{x})}\, Dx\,</math>

Now {{mvar|''x''(''t'')}} at each separate time is a separate integration variable. So it is legitimate to change variables in the integral by shifting: {{math|''x''(''t'') {{=}} ''u''(''t'') + ''ε''(''t'')}} where {{math|''ε''(''t'')}} is a different shift at each time but {{math|''ε''(0) {{=}} ''ε''(''T'') {{=}} 0}}, since the endpoints are not integrated:
: <math>\int \psi_0(x) \int_{u(0)=x} e^{iS(u+\varepsilon,\dot{u}+\dot{\varepsilon})}\, Du\,</math>

The change in the integral from the shift is, to first infinitesimal order in {{mvar|ε}}:
: <math>\int \psi_0(x) \int_{u(0)=x} \left( \int \frac{\partial S }{ \partial u } \varepsilon + \frac{ \partial S }{ \partial \dot{u} } \dot{\varepsilon}\, dt \right) e^{iS} \,Du\,</math>
which, integrating by parts in {{mvar|t}}, gives:
: <math>\int \psi_0(x) \int_{u(0)=x} -\left( \int \left(\frac{d}{dt} \frac{\partial S}{\partial \dot{u}} - \frac{\partial S}{\partial u}\right)\varepsilon(t)\, dt \right) e^{iS}\, Du\,</math>

But this was just a shift of integration variables, which doesn't change the value of the integral for any choice of {{mvar|''ε''(''t'')}}. The conclusion is that this first order variation is zero for an arbitrary initial state and at any arbitrary point in time:
: <math>\left\langle \psi_0\left| \frac{\delta S}{\delta x}(t) \right|\psi_0 \right\rangle = 0</math>
this is the Heisenberg equation of motion.

If the action contains terms that multiply {{mvar|ẋ}} and {{mvar|x}}, at the same moment in time, the manipulations above are only heuristic, because the multiplication rules for these quantities is just as noncommuting in the path integral as it is in the operator formalism.