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In Sakurai the derivation of the propagator leads to the expression

$$u_n(x)\exp{\left(\frac{-iE_nt}{\hbar}\right)} = \left(\frac{1}{2^{n/2}\sqrt{n!}}\right) \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp{\left(\frac{-m\omega x^2}{2\hbar}\right)} H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)\exp{\left(-i\omega(n+\frac{1}{2})t\right)}$$

Which it says leads to the expression for the propagator by using the formula $$\frac{1}{\sqrt{1-\zeta^2}}\exp{\left(\frac{-\xi^2-\eta^2+2\xi\eta\zeta}{(1-\zeta^2)}\right)} = \exp{\left[-(\xi^2+\eta^2)\right]\sum_{n=0}\left(\frac{\xi^n}{2^nn!}\right)H_n(\xi)H_n(\eta)}.$$

I’m wondering, what exactly is the reasoning for using this formula? Or how exactly does the transformation from this formula to the propagator start? Am I setting the prefactor equal to $\frac{1}{\sqrt{1-\zeta^2}}$ and solving for $\zeta$? For the inner product $\langle x’’ |\psi|x’\rangle$ Am I calling $\xi=x’’$ and $\eta=x’$? Any help would be much appreciated!

EDIT: Formula has been fixed. Same questions as before remain

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  • $\begingroup$ Can you give the precise citation of this formula in the text you are attributing to? It does not look like the standard Mehler formula at all! $\endgroup$ – Cosmas Zachos Sep 22 '18 at 19:51
  • $\begingroup$ You've copied it wrong! It is (2.6.19) in Sakurai-Napolitano. The sum of Hermites is not in the exponent. The closing square bracket should precede the summation symbol. It is basically the celebrated completeness relation of Hermite polynomials, first invented by Mehler. There are dozens of ways to derive it, including the functional integral. But please correct your formula first, and specify exactly what troubles you. $\endgroup$ – Cosmas Zachos Sep 22 '18 at 20:04
  • $\begingroup$ Please see the corrected formula. I’m not sure how this summation formula is supposed to be applied to the transition amplitude to get the propagator. $\endgroup$ – Некто Sep 22 '18 at 20:56
  • $\begingroup$ Have you first written down the expression of the propagator in terms of the eigenstates of the Hamiltonian? $\endgroup$ – Cosmas Zachos Sep 22 '18 at 21:45
  • $\begingroup$ It is a bland plug-in: plug (2.6.17) of Sakurai-Napolitano into the definition (2.6.8) and utilize Mehler's (2.6.19), a given, to obtain (2.6.18). $\endgroup$ – Cosmas Zachos Sep 22 '18 at 22:04
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You can find in Sakurai this expression for the propagator $$K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\left\langle {x''\left| {a'} \right\rangle } \right.} \left\langle {a'\left| {x'} \right\rangle } \right.\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right]$$

Taking into account the QHO wavefunctions

$${a_n}(x) = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega {x^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x} \right),\qquad n = 0,1,2, \ldots . $$

and

$$\left\langle {x''\left| {a'} \right\rangle } \right. = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right), $$

We simply can derive the expression for $K({\bf{x''}},t;{\bf{x'}},t)$

$$\begin{array}{l}K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {{\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)}^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)} \times \\ \times \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right)\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right] = \\\frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]} = \\ = \frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{\hbar }}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{x{'^2} + x'{'^2} - 2x''x'{e^{ - i\omega (t - {t_0})}}}}{{1 - {e^{ - 2i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{{e^{\frac{{i\omega }}{2}(t - {t_0})}}\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{{2\hbar }}\left[ {\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{i\sin (\omega (t - {t_0}))}} - (x{'^2} + x'{'^2})} \right]} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{im\omega }}{{2\hbar }}\frac{{(x{'^2} + x'{'^2})\cos (\omega (t - {t_0})) - 2x''x'}}{{\sin (\omega (t - {t_0}))}}} \right)\\\end{array} $$

The last moment you can see the Hermitian polinomial functions is when we apply your formula

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