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In QFT the principle of stationary action states that we choose fields that will make the action stationary but what if the action has many stationary points (for a fixed choice of boundary conditions)? What's the significance of these other solutions?

Qmechanic
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Quark
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2 Answers2

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There can be more than one stationary classical solution to an action principle with pertinent boundary conditions. E.g. because of instantons or gauge symmetry.

Example: A rotating rigid body with moment of inertia $I$. The action is $$\tag{1} S~=~\int_{t_i}^{t_f}\! dt ~L , \qquad L ~:=~\frac{I}{2}\dot{\theta}^2, $$ with Dirichlet boundary conditions (BC) $$\tag{2} \theta(t_i)-\theta_i~\in~2\pi\mathbb{Z} \qquad\text{and}\qquad\theta(t_f)~-\theta_f\in~2\pi\mathbb{Z} . $$ It is possible to satisfy the EOM $\ddot{\theta}\approx 0$ and BC (2) in infinitely many ways corresponding to different number of windings.

Instanton solutions are an important feature of a quantum field theory. On the other hand, gauge ambiguities are typically removed by gauge-fixing.

Qmechanic
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As an example, consider the real scalar field with action,

$$S=\int \mathrm{d}^4 x \, \left( \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi^2\right)$$

The principle of stationary action insists $\delta S = 0$, and the fields which allow this are those which satisfy the associated Euler-Lagrange equations, which are simply,

$$(\square+m^2)\phi=0$$

As the OP noted, there are many solutions to the equations of motion; this is not an issue. For the purposes of canonical quantization, we usually expand the field as a plane wave, and promote the Fourier coefficients to operators, etc. A theory may have other solutions, and in many cases it is interesting to study these, c.f. solitons.

JamalS
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