# Limit acting on exponential function

October 2, 2022  704 words 4 mins read

In this post, I will show you two properties of limit acting on the exponential function. These properties are very useful when we are dealing with quantum mechanics or quantum field theory (QFT). I will also show you an example of using these properties.

### Property 1

Suppose we want to know $\displaystyle{\lim_{t \to \infty} e^{- \iota t} =}$ ? You may simply answer that it is zero. But, it is not true. Let me show you why?

We know that $\displaystyle{e^{\iota x} = \cos(x) + \iota \sin(x)}$ which is known as famous “Euler formula”. So, we can write $\displaystyle{e^{- \iota t} = \cos(-t) + \iota \sin(-t)}$ $= \cos(t) - \iota \sin(t)$. Now, we can take the limit of this expression as $\displaystyle{t \to \infty}$. Since, cosine and sine do not vanishes as $t \to \infty$ because they have range $[-1, 1]$, and they cannot go beyond this bound. Thus, this cannot be zero. This means it’s indeterminate form and you cannot able to determine the value of this limit expression. Let’s say this is property no. 1.

### Property 2

Suppose we also want to know $\displaystyle{\lim_{t \to \infty} \lim_{x \to 0} e^{- \iota xt} =}$ ? In this case, it depends upon which limit you take first. What I mean by this is that if you do $\displaystyle{\lim_{t \to \infty} \left(\lim_{x \to 0} e^{- \iota xt}\right)}$ then, the answer is 1. But, if you do $\displaystyle{\lim_{x \to 0} \left(\lim_{t \to \infty} e^{- \iota xt}\right)}$ then, the answer is indeterminate. Let’s say this is property no. 2.

### Application

Let me give you a real problem taken from Peskin & Schroeder QFT book at page 86. Suppose you have this equation $$e^{-\iota H T} \ket{0} = e^{-\iota E_{0} T} \ket{\Omega}\braket{\Omega \vert 0} + \sum_{n \neq 0} e^{-\iota E_{n} T} \ket{n}\braket{n \vert 0},$$ say equation $(1)$. You want to prove this equation ($4.27$ in the above mentioned book) $$\ket{\Omega} = \lim_{T \to \infty (1 - \iota \epsilon)} \left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{- \iota H T} \ket{0},$$ say equation $(2)$.

Please refer 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. $$\left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{-\iota H T} \ket{0} =$$ $$\left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} ( e^{-\iota E_{0} T} \ket{\Omega}\braket{\Omega \vert 0} +$$ $$\hspace{6em} \sum_{n \neq 0} e^{-\iota E_{n} T} \ket{n}\braket{n \vert 0} )$$

Since $e^{-\iota E_{0} T}$ 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 $$\left( e^{-\iota E_{0} T} \braket{\Omega \vert 0} \right)^{-1} e^{-\iota H T} \ket{0} = \ket{\Omega} +$$ $$\frac{1}{\braket{\Omega \vert 0}} \sum_{n \neq 0} e^{-\iota (E_{n} - E_{0}) T} \ket{n}\braket{n \vert 0}.$$

Note that $\displaystyle{E_{n} > E_{0} \implies (E_{n} - E_{0}) > 0}$. Now, we take limit $T \to \infty (1 - \iota \epsilon)$ both side of above equation. Note that $\epsilon$ is our artificial construction in-order to get rid off possible ill-defined integral, but $T$ is a physical quantity. 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*}

We can notice the property 1 in the first factor (i.e. $\displaystyle{e^{-\iota (E_{n} - E_{0})\infty}}$). Since, we do not know exactly its value so, we assume it is some number.

For second factor (i.e. $\displaystyle{ e^{-(E_{n} - E_{0})\epsilon \infty} }$), we first evaluate $\infty$. This is similar to the property 2. Thus, $e^{-(E_n - E_0)\epsilon \infty} \to 0$.

Thus, we get

\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*}

This way we finally derive the equation ($4.27$).

Final words, when you see this type of scenario, you need to remember these above properties.

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