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This question is closely releted to this question: How do we solve the laplace transform of the Heat Kernel?

Let $A>0$ and $$f(t) = \frac{A^2}{2\sqrt{\pi}t^\frac{3}{2}}e^{-\frac{A^2}{4t}}$$ Following the same computations as in the question I linked, one can prove that for $s \in \mathbb{R}$, $s\geq 0$ one has $$\mathcal{L[f](s) :=\int_0^{+\infty}e^{-st}f(t)\,dt}= e^{-A\sqrt{s}}$$ (see also https://projecteuclid.org/download/pdf_1/euclid.aoms/1177731708 at page 252 )

Is this formula true also for $s \in \mathbb{C}, \mathrm{Re(s)} \geq 0$, thinking the square root of $s$ as the principal branch of the root (i.e. the square root of $s$ with positive real part)?

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    $\begingroup$ Yes and it relates to Quantum Mechanics because one must use the complex conjugate to arrive at a sensible answer; I would check out Griffiths Quantum Mechanics to see this result for yourself. $\endgroup$ – JohnColtraneisJC Jun 12 '18 at 16:22
  • $\begingroup$ @JohnColtraneisJC If you can, could you please give me a more precise reference? $\endgroup$ – foo90 Jun 12 '18 at 17:13
  • $\begingroup$ amazon.com/Introduction-Quantum-Mechanics-David-Griffiths/dp/… $\endgroup$ – JohnColtraneisJC Jun 12 '18 at 18:07
  • $\begingroup$ A wave is a complex number of form ω+it, this is why when we have $exp(ω+it)$ we must take the conjugate to deal with the complex part of the exponential. $\endgroup$ – JohnColtraneisJC Jun 12 '18 at 18:09
  • $\begingroup$ @foo90 I got $\sqrt{A} e^{-\sqrt{A s} }$ by two methods, so you might want to double check the $A$ terms. $\endgroup$ – Benedict W. J. Irwin Jun 18 '18 at 9:47
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I hope this is helpful, I was originally only validating your quoted identity with my statement. I think the validity of the complex plane can be carried through the following procedure.

If we consider the function to be a function of $A$, we can take the Mellin transform with respect to $A$. This is essentially the Mellin transform of $A^2 f(A^2)$ which is then equivalent to the transform of $\frac{1}{2}f(\frac{2+s}{2})$ by transform rules.

The transform of an exponential is valid for the entire complex plane, because the gamma function can be extended to the whole plane except for non-positive integers.

We can then calculate that $$ \mathcal{M}_A[f](s) =\frac{2^{\frac{s}{2}+\frac{s+2}{2}-1} \left(\frac{1}{t}\right)^{\frac{1}{2} (-s-2)} \Gamma \left(\frac{s}{2}+1\right)}{\sqrt{\pi } t^{3/2}} \\ \mathcal{M}_A[f](s) =\frac{2^s \left(\frac{1}{t}\right)^{-s/2} \Gamma \left(\frac{s}{2}+1\right)}{\sqrt{\pi } \sqrt{t}} $$ it should then be true that $$ \mathcal{L}_t[t^k](q) = \frac{\Gamma(1+k)}{q^{1+k}} $$ for $q>0$ and $k>-1$, so $$ \mathcal{L}_t[\mathcal{M}_A[f](s)](q) =\frac{2^s q^{-\frac{s}{2}-\frac{1}{2}} \Gamma \left(\frac{s}{2}+1\right) \Gamma \left(\frac{s+1}{2}\right)}{\sqrt{\pi }} , \;\; s>-1 \\ \mathcal{L}_t[\mathcal{M}_A[f](s)](q) = q^{-\frac{s}{2}-\frac{1}{2}} \Gamma (s+1) $$ so now we need to take the inverse Mellin transform which is also well defined over then complex plane for this function, which gives $$ A e^{-A \sqrt{q}} $$ let me know if anything is unclear.

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