Editing Path integral formulation (section)

=== Simple harmonic oscillator ===
{{see also|Propagator#Basic examples: propagator of free particle and harmonic oscillator| Mehler kernel}}
The Lagrangian for the simple harmonic oscillator is<ref>{{cite web |last1=Hilke |first1=M. |title=Path Integral |work=221A Lecture Notes |url=http://hitoshi.berkeley.edu/221A/pathintegral.pdf}}</ref>
: <math>\mathcal{L} = \tfrac12 m \dot{x}^2 - \tfrac12 m \omega^2 x^2.</math>

Write its trajectory {{math|''x''(''t'')}} as the classical trajectory plus some perturbation, {{math|''x''(''t'') {{=}} ''x''<sub>c</sub>(''t'') + ''δx''(''t'')}} and the action as {{math|''S'' {{=}} ''S''<sub>c</sub> + ''δS''}}. The classical trajectory can be written as

: <math>x_\text{c}(t) = x_i \frac{\sin\omega(t_f - t)}{\sin\omega(t_f - t_i)} + x_f \frac{\sin\omega(t - t_i)}{\sin\omega(t_f - t_i)}.</math>

This trajectory yields the classical action
: <math>
\begin{align}
S_\text{c} & = \int_{t_i}^{t_f} \mathcal{L} \,dt = \int_{t_i}^{t_f} \left(\tfrac12 m\dot{x}^2 - \tfrac12 m\omega^2 x^2 \right) \,dt \\[6pt]
& = \frac 1 2 m\omega \left( \frac{(x_i^2 + x_f^2) \cos\omega(t_f - t_i) - 2 x_i x_f}{\sin\omega(t_f - t_i)} \right)~.
\end{align}
</math>

Next, expand the deviation from the classical path as a Fourier series, and calculate the contribution to the action {{mvar|δS}}, which gives
: <math>S = S_\text{c} + \sum_{n = 1}^\infty \tfrac12 a_n^2 \frac{m}{2} \left( \frac{(n \pi)^2}{t_f - t_i} - \omega^2(t_f - t_i) \right).</math>

This means that the propagator is
: <math>
\begin{align}
K(x_f, t_f; x_i, t_i) & = Q e^\frac{i S_\text{c}}{\hbar} \prod_{j=1}^\infty \frac{j \pi}{\sqrt{2}} \int da_j \exp{\left( \frac{i}{2\hbar}a_j^2 \frac{m}{2} \left( \frac{(j \pi)^2}{t_f - t_i} - \omega^2(t_f - t_i) \right) \right)} \\[6pt]
& = e^\frac{i S_\text{c}}{\hbar} Q \prod_{j=1}^\infty \left( 1 - \left( \frac{\omega(t_f - t_i)}{j \pi} \right)^2 \right)^{-\frac12}
\end{align}
</math>
for some normalization
: <math> Q = \sqrt{\frac{m}{2\pi i \hbar (t_f - t_i)}}~. </math>

Using the infinite-product representation of the [[sinc function]],
: <math>\prod_{j=1}^\infty \left( 1 - \frac{x^2}{j^2} \right) = \frac{\sin\pi x}{\pi x}, </math>
the propagator can be written as
: <math> K(x_f, t_f; x_i, t_i) = Q e^\frac{i S_\text{c}}{\hbar} \sqrt{ \frac{\omega(t_f - t_i)}{\sin\omega(t_f - t_i)} } = e^\frac{i S_c}{\hbar} \sqrt{ \frac{m\omega}{2\pi i \hbar \sin\omega(t_f - t_i)}}.</math>

Let {{math|''T'' {{=}} ''t<sub>f</sub>'' − ''t<sub>i</sub>''}}.  One may write this propagator in terms of energy eigenstates as
: <math>
\begin{align}
K(x_f, t_f; x_i, t_i) & = \left( \frac{m \omega}{2 \pi i \hbar \sin\omega T } \right)^\frac12 \exp{ \left( \frac{i}{\hbar} \tfrac12 m \omega  \frac{ (x_i^2 + x_f^2) \cos \omega T - 2 x_i x_f }{ \sin \omega T } \right) } \\[6pt]
& = \sum_{n = 0}^\infty \exp{ \left( - \frac{i E_n T}{\hbar} \right) } \psi_n(x_f) \psi_n(x_i)^{*}~.
\end{align}
</math>

Using the identities {{math|''i'' sin ''ωT'' {{=}} {{sfrac|1|2}}''e''<sup>''iωT''</sup> (1 − ''e''<sup>−2''iωT''</sup>)}} and {{math|cos ''ωT'' {{=}} {{sfrac|1|2}}''e''<sup>''iωT''</sup> (1 + ''e''<sup>−2''iωT''</sup>)}}, this amounts to
: <math>K(x_f, t_f; x_i, t_i) = \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T} 2 \left( 1 - e^{-2 i \omega T} \right)^{-\frac12} \exp{ \left( - \frac{m \omega}{2 \hbar} \left( \left(x_i^2 + x_f^2\right) \frac{ 1 + e^{-2 i \omega T} }{ 1 - e^{- 2 i \omega T}} - \frac{4 x_i x_f e^{-i \omega T}}{1 - e^{ - 2 i \omega T} }\right) \right) }.</math>

One may absorb all terms after the first {{math|''e''<sup>−''iωT''/2</sup>}} into {{math|''R''(''T'')}}, thereby  obtaining
: <math> K(x_f, t_f; x_i, t_i) = \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T } 2 \cdot R(T).</math>

One may  finally expand {{math|''R''(''T'')}} in  powers of {{math|''e''<sup>−''iωT''</sup>}}: All  terms in this expansion get multiplied by the {{math|''e''<sup>−''iωT''/2</sup>}} factor in the front,  yielding terms  of the form
: <math>e^\frac{-i\omega T}{2} e^{-i n\omega T} = e^{-i \omega T \left( \frac12 + n\right) } \quad\text{for } n = 0, 1, 2, \ldots.</math>
Comparison  to the above eigenstate expansion  yields the  standard energy spectrum for the simple harmonic oscillator,
: <math>E_n = \left( n + \tfrac12 \right) \hbar \omega~.</math>