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How does one derive the fact that $$\psi(t,x) = (\tfrac{2 \pi \hbar t}{m})^{-d/2}\int_{\mathbb{R}^d} e^{im\tfrac{(x-y)^2}{2\hbar t}}\psi_0(y)dy$$ is a solution of the time-dependent free Schrödinger equation $$i\hbar \tfrac{\partial }{\partial t}\psi(t,x) = -\tfrac{\hbar^2}{2m} \Delta \psi(t,x), \\ \psi(0,x) = \psi_0(x)~?$$

It may follow from the operator solution of Schrödinger

$$\psi(t,x) = e^{\tfrac{i}{\hbar}H_ot}\psi_0(x)$$

i.e. start with

$$ e^{\tfrac{i}{\hbar}H_ot}\psi_0(x) $$

and insert a delta function or something and end up with

$$\psi(t,x) = e^{\tfrac{i}{\hbar}H_ot}\psi_0(x) = (\tfrac{2 \pi \hbar t}{m})^{-d/2}\int_{\mathbb{R}^d} e^{im\tfrac{(x-y)^2}{2\hbar t}}\psi_0(y)dy$$

as page 3 of S. Mazzucchi, Mathematical Feynman Path Integrals and Their Applications, seems to imply (readable on amazon's search feature).

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closed as off-topic by JamalS, tpg2114, ACuriousMind, Danu, Colin McFaul Nov 23 '14 at 23:34

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  • $\begingroup$ Please note that Schrödinger is written with ö, not o. If you can't type it, write oe instead. $\endgroup$ – Ruslan Nov 23 '14 at 19:32
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    $\begingroup$ Is your question about how to confirm the wavefunction satisfies the time dependent Schrodinger equation - or is it how from first principles this solution is derived?.. sorry - mixed up windows and put answer to different question in here- many apologies $\endgroup$ – tom Nov 23 '14 at 19:46
  • $\begingroup$ @Ruslan "Schrodinger" is a common spelling, whether originally correct or not, and is listed as one of the spellings on Erwin's wiki page. $\endgroup$ – Jold Nov 23 '14 at 19:51
  • $\begingroup$ How to derive it from first principles :) It seems like the key step in deriving the Path integral without any of the physicsey ket notation :) $\endgroup$ – bobby Nov 23 '14 at 20:04
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This question is about Guassian wave-packet propagation and the corresponding Green's function in ordinary quantum mechanics.

Assuming $\hbar=m=1$ for simplicity, consider the solution (with its initial condition) to the following Schrodinger equation:

$$i\partial_t G=-\partial^2_x G \\ G(t=0,x)=\delta(x) $$

Now assume the Fourier ansatz for $G$: $$G(t,x)=\int dk A(t,k)e^{ikx}$$ Putting it back to the equation above determines the form of $A$: $$i\partial_t A=k^2A$$ therefore $$A(t,k)=C(k)e^{-ik^2t}$$ which gives $$ G(t,x)=\int dk C(k)e^{-ik^2t+ikx} $$ Imposing the initial condition now yields: $$ G(t=0,x)=\int dk C(k)e^{ikx}=\delta(x) \implies C(k)=1/2\pi \\ G(t,x)=\int \frac{dk}{2\pi} e^{-ik^2t+ikx} $$ The last integral is the usual Gaussian integral, which can be evaluated by making a complete square out of the integrand: $$ -ik^2t+ikx = -it\left[k^2-\frac{kx}{t}\right] = -it\left[\left(k-\frac{x}{2t}\right)^2-\frac{x^2}{4t^2}\right] $$ therefore the integral will evaluate to $$ G(t,x) \propto \frac{1}{\sqrt{t}}\exp\left(\frac{ix^2}{4t}\right) $$ Now if you imagine $\psi_0$ is made up of weighted $\delta$ contributions, i.e.: $$ \psi_0(x)=\int dy \psi_0(y) \delta(y-x) $$ then you arrive at your original $\psi$ function in your question. In plain language, the $\psi$ function integrates the result of propagating the delta functions which make up the initial condition $\psi_0$. Mathematically: $$ \psi(t,x)=\int dy G(t,x-y)\psi_0(y) $$

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