# Limit acting on exponential function

## Property 1

Suppose we want to take a limit like below.

\begin{align*} \lim_{t \to \infty} e^{- \iota t} & = \hspace{0.1cm} ? \\ & = \text{Apply the Euler formula}, e^{\iota x} := \cos(x) + \iota \sin(x) \\ & = \lim_{t \to \infty} \left(\cos(t) - \iota sin(t)\right) \\ & = \text{Since, cosine and sine doesnot vanishes as } t \to \infty. \\ & \hspace{0.45cm} \text{Because } t \text{ is a phase for these functions}.\\ & \hspace{0.45cm} \text{Convince yourself by plotting sine and cosine function :)}\\ \therefore \lim_{t \to \infty} e^{- \iota t} & \neq 0. \end{align*}

This gives you a number which is not indeterminate.

## Property 2

Suppose we want to take limits like below.

\begin{align*} \lim_{t \to \infty} \lim_{x \to 0} e^{- \iota xt} & = \hspace{0.1cm} ? \\ \end{align*}

For this, it depends upon which limit you take first.

Note: These type of questions you usually encounter while studying quantum mechanics or quantum field theory (QFT).

## Application

Let me give you a real problem taken from Peskin & Schroeder QFT book at page 86. Suppose you have this equation $$$$e^{-\iota H T} \ket{0} = e^{-\iota E_{0} T} \ket{\Omega}\braket{\Omega \vert 0} + \sum_{n \neq 0} e^{-\iota E_{n} T} \ket{n}\braket{n \vert 0},$$$$

and you need to prove this equation ($4.27$ in above mentioned book) $$$$\ket{\Omega} = \lim_{T \to \infty (1 - \iota \epsilon)} \left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{- \iota H T} \ket{0}$$$$

Please see the mentioned book for the symbols' meaning.

Proof: First multiply equation $(1)$ by $\left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1}$. i.e.

\begin{align*} \left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{-\iota H T} \ket{0} & = \left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} ( e^{-\iota E_{0} T} \ket{\Omega}\braket{\Omega \vert 0}\\ & \hspace{0.3cm} + \sum_{n \neq 0} e^{-\iota E_{n} T} \ket{n}\braket{n \vert 0} ) \end{align*}

Since $e^{-\iota E_{0} T} \text{and} \braket{\Omega \vert 0}$ are just numbers. So, $\left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} = \frac{e^{\iota E_{0} T}}{\braket{\Omega \vert 0}}$. Thus, we get \begin{align*} \left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{-\iota H T} \ket{0} & = \ket{\Omega} + \frac{1}{\braket{\Omega \vert 0}} \sum_{n \neq 0} e^{-\iota (E_{n} - E_{0}) T} \ket{n}\braket{n \vert 0}. \end{align*}

Note that $E_{n} > E_{0}$. Now, we take limit $T \to \infty (1 - \iota \epsilon)$ both side of above equation, and let’s observe especially this part

\begin{align*} \lim_{T \to \infty (1 - \iota \epsilon)} e^{-\iota (E_{n} - E_{0}) T} & = e^{-\iota (E_{n} - E_{0}) \infty (1 - \iota \epsilon)} \\ & = e^{-\iota (E_{n} - E_{0}) (\infty - \iota \epsilon \infty)} \\ & = e^{-\iota (E_{n} - E_{0})\infty} . e^{-(E_{n} - E_{0})\epsilon \infty} \end{align*}

Since $E_{n} - E_{0} > 0$ and $\epsilon \to 0$. We can notice in the second factor (i.e. $e^{-(E_{n} - E_{0})\epsilon \infty}$) look like property 2. Our rule here is that we don't use $\epsilon \to 0$ first than $T \to \infty$ because $\epsilon$ is our artificial construction in-order to get rid off ill-defined integral, but $T$ is a physical quantity. Thus, $e^{-(E_{n} - E_{0})\epsilon \infty} \to 0$.

Using the property 1, we can notice that

\begin{align*} \lim_{T \to \infty (1 - \iota \epsilon)} e^{-\iota (E_{n} - E_{0}) T} & = (\text{number})\times 0 \\ \therefore \lim_{T \to \infty (1 - \iota \epsilon)} e^{-\iota (E_{n} - E_{0}) T} & = 0. \end{align*}

Thus, we finally proved the equation ($4.27$).

Published on May 8, 2021

Last revised on Aug 6, 2021