Sunday, March 17, 2013

Proof of Plancherel's Theorem: QM (Griffiths)

Problem 2.20 from chapter two of David J. Griffiths' "Introduction to Quantum Mechanics: 2nd Edition" takes the student through a step-by-step proof of Plancherel's Theorem:

f(x)=12π−−√∫∞−∞F(k)eikxdk⇔F(k)=12π−−√∫∞−∞f(x)e−ikxdx

where F(k) is the Fourier transform of f(x), and f(x) is the inverse Fourier transform of F(k).

Step 1
First, Griffiths asks us to show that a function f(x) on the interval [-a, a] expanded as a Fourier series

f(x)=∑n=0∞[ansin(nπx/a)+bncos(nπx/a)][1]

    

          

can be written equivalently as

f(x)=∑n=−∞∞cneinπx/a[2]

and to write cn in terms of an and bn. 

If we look at n = 0, [1] --> f(x)=b0

Now, use Euler's identity eiϕ=cos(ϕ)+isin(ϕ) to rewrite [1] as

f(x)=b0+∑n=1∞an2i(einπx/a−e−inπx/a)+∑n=1∞bn2(einπx/a+e−inπx/a)

And, rearranging this a little...

f(x)=b0+∑n=1∞{an2i+bn2}einπx/a+∑n=1∞{−an2i+bn2}e−inπx/a[3]

Evaluating [3] with n = 1 gives

f(x)=12(−ia1+b1)eiπx/a+12(ia1+b1)e−iπx/a[4]

Now, let's evaluate [2] with both n = -1 and n = 1

n = 1:   

f(x)=c1eiπx/a

n = -1: 

f(x)=c−1e−iπx/a

These two terms look like the two terms in [4] if c1 = 12(−ia1+b1) and  c−1 = 12(ia1+b1) (Remember, [3] sums on n from 1 to infinity, but [2] sums on n from negative infinity to positive infinity, so including the -n terms and the +n terms in [2] will give you the same two terms you get when evaluating [3] with only the +n terms.)

So, in general, f(x)=∑∞n=0[ansin(nπx/a)+bncos(nπx/a)] can be written equivalently as

f(x)=∑n=−∞∞cneinπx/a

with cn = 12(−ian+bn), c−n = 12(ian+bn), and c0 = b0 

QED

Step 2
Next, we are asked to show that

cn=12a∫+a−af(x)e−inπx/adx[5]

   

by using what Griffiths refers to as 'Fourier's Trick', which exploits the orthonormality of wavefunctions of different excited states.

The 'trick' is to multiply both sides of [2] by ψ∗m and integrate. Of course, in this case ψn=einπx/a.

f(x)=∑n=−∞∞cneinπx/a

∫+a−aψm(x)∗f(x)dx=∑n=−∞∞cn∫+a−aψ∗m(x)ψn(x)dx=∑n=−∞∞2acnδmn=2acm

The orthonormality of ψ∗m and ψm knocks out all the terms except for the n = m one, thus the appearance of the Kronecker delta.

So now we have

∫+a−aψ∗m(x)f(x)dx=2acm

Swapping indices (m for n) gives

∫+a−aψ∗n(x)f(x)dx=2acn

Finally, solving for cn and remembering that ψn=einπx/a gives us

12a∫+a−af(x)e−inπx/adx=cn

QED

Step 3
Next, we are told to eliminate n and cn and use the new variables k=(nπ/a) and F(k)=2π√acn and show that [2] becomes

f(x)=12π−−√∑n=−∞∞F(k)eikxΔk

and [5] (Step 2) becomes

F(k)=12π−−√∫+a−af(x)e−ikxdx

F(k)=2π√acn solved for cn gives cn=1aF(k)π2√. Also, we know that Δk=Δnπa Δn=1 so Δkπ=1a

Substituting these results into [2] gives

f(x)=Δkππ2−−√∑n=−∞∞F(k)eikx=12π−−√∑n=−∞∞F(k)eikxΔkQED

That takes care of f(x); Now for F(k)...

cn=12a∫+a−af(x)e−inπx/adx[5]

But cn=1aF(k)π2√, so [5] becomes

1aF(k)π2−−√=12a∫+a−af(x)e−inπx/adx

A little algebra and an appropriate substitution gives

F(k)=12π−−√∫+a−af(x)e−ikxdxQED

Final Step
To complete the proof of Plancherel's Theorem, we are asked to take the limit a →∞ for the two results from Step 3

1st, f(x)...

limΔk→0f(x)=limΔk→012π−−√∑n=−∞∞F(k)eikxΔk=12π−−√∫∞−∞F(k)eikxdk

Now, F(k)...

lima→∞F(k)=lima→∞12π−−√∫+a−af(x)e−ikxdx=12π−−√∫∞−∞f(x)e−ikxdx

So, finally we have Plancherel's Theorem as our result

f(x)=12π−−√∫∞−∞F(k)eikxdk⇔F(k)=12π−−√∫∞−∞f(x)e−ikxdx

QED