Editing Instanton (section)

=== Path integral interpretation via instantons ===
Alternatively, the use of [[path integral formulation|path integrals]] allows an ''instanton'' interpretation and the same result can be obtained with this approach. In path integral formulation, the transition amplitude can be expressed as

:<math>K(a,b;t)=\langle x=a|e^{-\frac{i\mathbb{H}t}{\hbar}}|x=b\rangle =\int d[x(t)]e^{\frac{iS[x(t)]}{\hbar}}.</math>

Following the process of [[Wick rotation]] (analytic continuation) to Euclidean spacetime (<math>it\rightarrow \tau</math>), one gets

:<math>K_E(a,b;\tau)=\langle x=a|e^{-\frac{\mathbb{H}\tau}{\hbar}}|x=b\rangle =\int d[x(\tau)]e^{-\frac{S_E[x(\tau)]}{\hbar}},</math>

with the Euclidean action

:<math>S_E=\int_{\tau_a}^{\tau_b}\left(\frac{1}{2}m\left(\frac{dx}{d\tau}\right)^2+V(x)\right) d\tau.</math>

The potential energy changes sign <math> V(x) \rightarrow - V(x) </math> under the Wick rotation and the minima transform into maxima, thereby <math> V(x) </math> exhibits two "hills" of maximal energy.

Let us now consider the local minimum of the Euclidean action <math>S_E</math> with the double-well potential <math>V(x)={1\over 4}(x^2-1)^2</math>, and we set <math>m=1</math> just for simplicity of computation. Since we want to know how the two classically lowest energy states <math>x=\pm1</math> are connected, let us set <math>a=-1</math> and <math>b=1</math>. 
For <math>a=-1</math> and <math> b=1</math>, we can rewrite the Euclidean action as

:<math> S_E=\int_{\tau_a}^{\tau_b}d \tau {1\over 2}\left({d x\over d \tau}-\sqrt{2V(x)}\right)^2 + \sqrt{2}\int_{\tau_a}^{\tau_b}d \tau{d x\over d \tau}\sqrt{V(x)} </math>
:<math> \quad =\int_{\tau_a}^{\tau_b}d \tau {1\over 2}\left({d x\over d \tau}-\sqrt{2V(x)}\right)^2 + \int_{-1}^{1}d x {1\over \sqrt{2}}(1-x^2). </math>
:<math> \quad \ge {2\sqrt{2}\over 3}. </math>

The above inequality is saturated by the solution of <math> {d x\over d \tau}=\sqrt{2V(x)}</math> with the condition <math>x(\tau_a)=-1</math> and <math>x(\tau_b)=1</math>. Such solutions exist, and the solution takes the simple form when <math>\tau_a=-\infty</math> and <math>\tau_b=\infty</math>. The explicit formula for the instanton solution is given by

:<math> x(\tau)=\tanh\left({1\over \sqrt{2}}(\tau-\tau_0)\right).  </math>

Here <math>\tau_0</math> is an arbitrary constant. Since this solution jumps from one classical vacuum <math>x=-1</math> to another classical vacuum <math>x=1</math> instantaneously around <math>\tau=\tau_0</math>, it is called an instanton.