Can Discrete Parameters Be Used in Limit Calculations?

[friend]

this would mean the square roots have angles -π/4 (first term), π/4 (second term), respectively.

Thanks for the hint, zinq. So let me start over with your suggestion.
I want to find the limit
$$\mathop {\lim }\limits_{{t_1} \to 0} \,\,{\left( {\frac{m}{{2\pi \hbar i{t_1}}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x' - x)}^2}/2\hbar {t_1}}}\,\,\,\,\, + \,\,\,\,\,\mathop {\lim }\limits_{{t_2} \to 0} \,\,{\left( {\frac{m}{{2\pi \hbar ( - i){t_2}}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x - x')}^2}/2\hbar {t_2}}}$$
because it represents a particle in superposition with its antiparticle. If I can get this limit to equal zero, then it can represent virtual particle/antiparticle cancellations that have no physical effects on their own. To simplify a bit, I will make
$$A = {\left( {\frac{m}{{2\pi \hbar i}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$$
and
$$B = m{(x' - x)^2}/2\hbar$$
Then the above limit becomes
$$A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} + i\mathop {\lim }\limits_{{t_2} \to 0} \frac{{{e^{ - iB/{t_2}}}}}{{{t_2}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} \right)$$
And since $i = {e^{\frac{{i\pi }}{2}}}$, this becomes
$$A\left( {\mathop {\lim }\limits_{{t_1} \to 0} \frac{{{e^{iB/{t_1}}}}}{{{t_1}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}} + \mathop {\lim }\limits_{{t_2} \to 0} \frac{{{e^{ - iB/{t_2} + i\pi /2}}}}{{{t_2}^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} \right)$$
If we are talking about virtual particles that pop into existence at the same place at the same time and pop out of existence at the same place at the same time, we must have ${t_1} = {t_2} = t$, and the limit becomes
$$A\mathop {\lim }\limits_{t \to 0} \left( {\frac{{{e^{iB/t}}}}{{{t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} + \frac{{{e^{ - iB/t + i\pi /2}}}}{{{t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}}} \right) = A\mathop {\lim }\limits_{t \to 0} \frac{1}{{{t^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}}\left( {{e^{iB/t}} + {e^{ - iB/t + i\pi /2}}} \right)$$
Now, here's the new part that I just thought up, thanks to zinq. Let
$$\frac{{iB}}{t} = \frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n$$
with n any integer. Then $t \to 0$ means that $n \to \infty$. And we also have
$$t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}$$
Putting this into the limit gives
$$A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}\left( {{e^{\frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n}} + {e^{ - \frac{{i\pi }}{4} - \frac{{i\pi }}{2} - i2\pi n + \frac{{i\pi }}{2}}}} \right)$$
The first and last terms in the second exponent add to give $\frac{{i\pi }}{4}$ so that we have
$$A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}\left( {{e^{\frac{{i\pi }}{4} + \frac{{i\pi }}{2} + i2\pi n}} + {e^{\frac{{i\pi }}{4} - \frac{{i\pi }}{2} - i2\pi n}}} \right)$$
The factor ${e^{\frac{{i\pi }}{4}}}$ in both exponents can be pulled out to give
$$A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {{e^{\frac{{i\pi }}{2} + i2\pi n}} + {e^{ - \frac{{i\pi }}{2} - i2\pi n}}} \right)$$
And since it is true that ${e^{ix}} = {e^{ix \pm i2\pi n}}$, we have
$$A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {{e^{\frac{{i\pi }}{2}}} + {e^{ - \frac{{i\pi }}{2}}}} \right)$$
And with ${e^{\frac{{i\pi }}{2}}} = i$ and ${e^{ - \frac{{i\pi }}{2}}} = - i$, the limit reduces to
$$A\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}{B}} \right)^{\frac{1}{2}}}{e^{\frac{{i\pi }}{4}}}\left( {i - i} \right) = 0$$
for any n whatsoever.

I got convergence by specifying a path of ${t_1} = {t_2} = t$ and making t discrete, $t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}$. So the last equation is not equal to the first equation. But I still have to wonder if they are equal in the limit.

 

[Samy_A]

I got convergence by specifying a path of ${t_1} = {t_2} = t$ and making t discrete, $t = \frac{B}{{\frac{\pi }{4} + \frac{\pi }{2} + 2\pi n}}$. So the last equation is not equal to the first equation. But I still have to wonder if they are equal in the limit.

As both diverge, they can't be "equal in the limit".

 

[zinq]

Never mind, friend, I had overlooked that in your definition of A, you had included i in the denominator.

 

[friend]

It's possible my concerns about convergence of the limit may be rendered mute since it is identically 0 under certain circumstance. Maybe all I need do is specify under what conditions these virtual particle pairs can exist. I still don't know, however, if my condition for t is the only one that results in cancellation.

 

[friend]

Never mind, friend, I had overlooked that in your definition of A, you had included i in the denominator.

Still, it was your mention of iπ/4 that got me thinking. Thanks.