# Plancherel's theorem

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David
J. (2005), Introduction to Quantum
Mechanics, 2nd Edition; Pearson Education – Chapter 2, Post 20.

While analyzing the free particle, we saw that we could construct a normalizable combination of stationary states by writing

$\displaystyle \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}e^{-i\hbar k^{2}t/2m}dk \ \ \ \ \ (1)$

We can find the function ${\phi(k)}$ by specifying the initial wave function:

 $\displaystyle \Psi(x,0)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{ikx}dk \ \ \ \ \ (2)$

This relation can be inverted by using Plancherel’s theorem, which states

 $\displaystyle \phi(k)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)e^{-ikx}dx \ \ \ \ \ (3)$

Here we run through a plausibility argument which is a sort of physicist’s proof of Plancherel’s theorem. We start with Dirichlet’s theorem which says that any (physically realistic, anyway) function can be written as a Fourier series. We can show that this is equivalent to a series in complex exponentials. That is

 $\displaystyle f(x)$ $\displaystyle =$ $\displaystyle \sum_{n=-\infty}^{\infty}c_{n}e^{in\pi x/a}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=-\infty}^{\infty}c_{n}\left[\cos\frac{n\pi x}{a}+i\sin\frac{n\pi x}{a}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}+\sum_{n=1}^{\infty}\left(c_{n}+c_{-n}\right)\cos\frac{n\pi x}{a}+i\sum_{n=1}^{\infty}\left(c_{n}-c_{-n}\right)\sin\frac{n\pi x}{a} \ \ \ \ \ (6)$

We’ve used the facts that cosine is even and sine is odd. This is equivalent to a Fourier series:

$\displaystyle f(x)=\sum_{n=0}^{\infty}\left[a_{n}\sin\frac{n\pi x}{a}+b_{n}\cos\frac{n\pi x}{a}\right] \ \ \ \ \ (7)$

where the coefficients are related by

 $\displaystyle b_{0}$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (8)$ $\displaystyle b_{n}$ $\displaystyle =$ $\displaystyle c_{n}+c_{-n}\ \ \ \ \ (9)$ $\displaystyle a_{n}$ $\displaystyle =$ $\displaystyle i\left(c_{n}-c_{-n}\right) \ \ \ \ \ (10)$

Inverting the relations we get, for ${n>0}$

 $\displaystyle c_{n}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(b_{n}-ia_{n}\right)\ \ \ \ \ (11)$ $\displaystyle c_{-n}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(b_{n}+ia_{n}\right) \ \ \ \ \ (12)$

We can get the coefficients in terms of ${f(x)}$ by integration:

$\displaystyle \frac{1}{2a}\int_{-a}^{a}f(x)e^{-in\pi x/a}dx=\frac{1}{2a}\sum_{m=-\infty}^{\infty}c_{m}\int_{-a}^{a}e^{i\left(m-n\right)\pi x/a}dx \ \ \ \ \ (13)$

The integral is zero if ${m\ne n}$ and ${2a}$ if ${m=n}$, so the right hand side comes out to just ${c_{n}}$ and we get

$\displaystyle c_{n}=\frac{1}{2a}\int_{-a}^{a}f(x)e^{-in\pi x/a}dx \ \ \ \ \ (14)$

Now we can make the substitutions

 $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{n\pi}{a}\ \ \ \ \ (15)$ $\displaystyle F(k)$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{2}{\pi}}ac_{n} \ \ \ \ \ (16)$

If ${\Delta k}$ is the increment in ${k}$ from one ${n}$ to the next, then ${\Delta k=\pi/a}$. We can then write the original series as

 $\displaystyle f(x)$ $\displaystyle =$ $\displaystyle \sum_{n=-\infty}^{\infty}\sqrt{\frac{\pi}{2}}\frac{1}{a}F(k)e^{ikx}\left(\frac{a}{\pi}\Delta k\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}F(k)e^{ikx}\Delta k \ \ \ \ \ (18)$

The formula for ${c_{n}}$ now becomes

 $\displaystyle \sqrt{\frac{\pi}{2}}\frac{1}{a}F(k)$ $\displaystyle =$ $\displaystyle \frac{1}{2a}\int_{-a}^{a}f(x)e^{-ikx}dx\ \ \ \ \ (19)$ $\displaystyle F(k)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-a}^{a}f(x)e^{-ikx}dx \ \ \ \ \ (20)$

Now we can take the limit as ${a\rightarrow\infty}$. In this case, ${\Delta k\rightarrow dk}$ (that is, it becomes a differential) and the sum becomes an integral, so we get

$\displaystyle f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(k)e^{ikx}dk \ \ \ \ \ (21)$

In the second formula, the limits on the integral become infinite, and we get the other half of Plancherel’s theorem:

$\displaystyle F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx \ \ \ \ \ (22)$

## 9 thoughts on “Plancherel's theorem”

1. Ziruo Zhang

Hi! I was just wondering if there should be a third case apart from equations (11) and (12) which specifies that c0=b0? Otherwise how should we deal with the case where n=0? Thank you very much! Your answers to these problems have been really helpful!

1. gwrowe Post author

I have already specified that ${c_{0}=b_{0}}$ in equation 8. Equation 14 works for ${n=0}$ as well, as it just says that ${c_{0}}$ is the average of ${f\left(x\right)}$ over the interval ${\left[-a,a\right]}$.