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. Pingback: Free particle: Gaussian wave packet « Physics tutorials

  2. Pingback: Free particle – travelling wave packet « Physics tutorials

  3. Pingback: Delta function – Fourier transform « Physics tutorials

  4. Pingback: Momentum: eigenvalues and normalization « Physics tutorials

  5. Pingback: Energy-time uncertainty principle: Gaussian free particle « Physics tutorials

  6. Pingback: Energy-time uncertainty principle: Gaussian free particle « Physics tutorials

  7. Pingback: Fourier transform of superposition of plane waves | Physics pages

  8. 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!

    Reply
    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]}.

      Reply

Leave a Reply

Your email address will not be published. Required fields are marked *