- About Fourier Transforms

1. Complex Inner Product Space $L^2[a, b]$ of Square Integrable Functions Represented by Fourier Series on a Finite Interval #

• Definition of inner product: $$( f | g ) = \frac{1}{b-a} \int_a^b dx , f^*(x) g(x)$$

• Normalization of basis: Kronecker delta function, i.e., $( e_m | e_n ) = \delta_{mn}$

Here, $e_n(x) = e^{i \frac{2 \pi}{(b-a)/n}x} = e^{i \frac{2n \pi}{b-a}x}$, that is, $e_n(x) = e^{i \frac{2n \pi}{b-a}x}$

2. Complex Inner Product Space $L^2(\mathbb{R})$ of Square Integrable Functions Represented by Fourier Transform on $\mathbb{R}$ #

Transform format: position $x \leftrightarrow$ momentum $p$ (or wave number $k$) or time $t \leftrightarrow$ angular frequency $\omega$ (or frequency $f$)

3. Using Dirac Notation #

In $( v | w )$, the vectors are $v, w$, and in Dirac notation $\langle a | b \rangle$, the vectors are $\langle a |, |b \rangle$, where $a, b$ typically denote eigenvalues, but are sometimes used as names when converted to functions. Here, we will use them only as eigenvalues, and when converted to functions, they will be represented as $\psi_a$, $\phi_b$, etc.

4. Functions and Dirac Notation #

$\psi_\alpha (x) = \langle x | \alpha \rangle$, $\hat{x} | x \rangle = x | x \rangle$

5. Representing Functions in $L^2(\mathbb{R})$ as Infinite Linear Combinations of Basis Vectors Using Dirac Notation #

• Basis selection: We can use either $|x\rangle$ or $|p\rangle$, and if we represent $|\alpha\rangle$ using $|p\rangle$, then $$|\alpha\rangle = \int_{-\infty}^\infty dp , \phi_\alpha(p) |p\rangle$$ and taking the inner product of $|x\rangle$ and $|\alpha\rangle$ gives $$\psi_\alpha(x) = \langle x | \alpha \rangle = \int_{-\infty}^\infty dp , \phi_\alpha(p) \langle x | p \rangle$$

• Definition of inner product: $$\langle \alpha | \beta \rangle = \int_{-\infty}^{\infty} dx , \psi_{\alpha}^*(x) \phi_{\beta}(x)$$

• Normalization of basis: Dirac delta function, i.e., $\langle x | x’ \rangle = \delta(x-x’)$, $\langle p | p’ \rangle = \delta(p-p’)$

• $\langle x | p \rangle$: $$\langle x | p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} e^{i \frac{p}{\hbar} x} = \frac{1}{\sqrt{2 \pi \hbar}} e^{\frac{i}{\hbar}px},$$

$$\langle p | x \rangle = \langle x | p \rangle^*$$

To help remember the function of $\langle x | p \rangle$, first remember the following: $$\hat{p} |p\rangle = p |p\rangle, \quad \langle x | \hat{p} |p\rangle = \hat{p}_x \langle x | p \rangle = p \langle x | p \rangle,$$

$$\hat{p}_x = -i \hbar \frac{\partial}{\partial x}$$

Therefore, for the exponential function to yield the value $p$ when differentiated with respect to $x$, it should be $e^{i \frac{px}{\hbar}}$, and it’s good to remember that the normalization constant is $\frac{1}{\sqrt{2 \pi \hbar}}$.

When confused about whether it’s $i$ or $-i$, remember the momentum operator as a differential operator, and you’ll know it’s $i$.

Thus, for $\langle p | x \rangle$, you can know it’s $-i$.

• $\langle p | x \rangle$: Also, $|\alpha\rangle$ can be represented as an infinite linear combination of the basis $|x\rangle$. $$|\alpha\rangle = \int_{-\infty}^\infty dx , \psi_\alpha(x) |x\rangle,$$

Applying the same method: $$\phi_\alpha(p) = \langle p | \alpha \rangle = \int_{-\infty}^\infty dx , \psi_\alpha(x) \langle p | x \rangle,$$

$$\langle p | x \rangle = \frac{1}{\sqrt{2 \pi \hbar}} e^{-i \frac{px}{\hbar}}$$