### Wave packets and the Fourier transform

Fourier transform

#### Problem:

Find the Fourier transform of the δ function δ(x - x0), and then use the inverse Fourier transform to show that
δ(x - x0) = (2π)-1-∞ exp(ik(x - x0)) dk.

Solution:

• Concepts:
The Fourier transform
• Reasoning:
Let us denote the Fourier transform of δ(x - x0) by δxo(p).
• Details of the calculation:
δxo(p) = (2πħ)-∞ exp(-ipx/ħ) δ(x-  x0)dx = (2πħ) exp(-ipx0/ħ)
by the definition of the Fourier transform.
In particular δxo(0) = (2πħ).

The inverse Fourier transform then yields
δ(x - x0) = (2πħ)-∞ exp(ipx/ħ) δxo(p) dp = (2πħ)-1-∞ exp(ipx/ħ) exp(-ipx0/ħ) dp
= (2πħ)-1-∞ exp(ip(x - x0)/ħ) dp = (2π)-1-∞ exp(ik(x - x0)) dk.
This is an equivalent definition of the Dirac δ function.

#### Problem:

The wave function of a particle at t = 0 is
Ψ(x) = 1/L½,  |x| < L/2,  Ψ(x) = 0 otherwise.
At t = 0, what possible values of the momentum of the particle can be found, and with what probability?

Solution:

• Concepts:
The postulates of quantum mechanics
• Reasoning:
P is a Hermitian operator.  The possible results of a measurement are the eigenvalues of the P, -∞ < p < ∞.
When a physical quantity described by the operator P is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue p is given by dP(p) = |<p|Ψ>|2dp = |Ψ(p)|2dp, where |p> is the eigenvector corresponding to the eigenvalue p; we assume p is a non-degenerate continuous eigenvalue of P.
• Details of the calculation:
Ψ(p)  = (2πħ)∫Ψ(x,0) exp(-ipx/ħ)dx,
Ψ(p)  = (2πħL)-L/2+L/2 exp(-ipx/ħ)dx
= (2πħL)(iħ/p) [exp(-ipL/(2ħ)) - exp(-pL/(2ħ))]
= (2πħL)(2ħ/p) sin(pL/(2ħ)).
The probability of finding the particle with momentum between p and p + dp  is |Ψ(p)|2dp.
|Ψ(p)|2  = (2ħ/(πLp2) )sin2(pL/(2ħ)).

#### Problem:

Consider a free particle which is described at t = 0 by the normalized Gaussian wave function ψ(x,0) = Nexp(ik0x)exp(-ax2).
(a)  Normalize the wave function.
(b)  Find the probability density |ψ(x,0)|2 of the particle.
(c)  Find its Fourier transform Φ(k,0) of the wave function and the probability density
|Φ(k,0)|2 in k-space.

Solution:

• Concepts:
The Fourier transform
• Reasoning:
We are asked to find the Fourier transform of a wave packet.
• Details of the calculation:
(a)  N2-∞+∞ exp(-2ax2) dx = N2 √(π/2a)   = 1
ψ(x,0) = (2a/π)¼exp(ik0x)exp(-ax2).
(b)  |ψ(x,0)|2 = (2a/π)½ exp(-2ax2).
(c)  ψ(x,0) = (2π)-∞+∞ Φ(k,0) exp(ikx) dk.
Φ(k,0) = (2π)-∞+∞ ψ(x,0) exp(-ikx) dx
= (a/(2π3))¼-∞+∞ exp(-ax2) exp(-i(k-k0)x) dx.
ax2 + i(k-k0)x = ax2 + i(k-k0)x - (k-k0)2/(4a) + (k-k0)2/(4a)
= (a½x + i(k-k0)/(2a½)2 + (k-k0)2/(4a).  (completing the square)
Φ(k,0) = (a/(2π3))¼exp(-(k-k0)2/(4a))∫-∞+∞ exp(-(ax2 + i(k-k0)x  - (k-k0)2/(4a))dx
= (a/(2π3))¼exp(-(k-k0)2/(4a))∫-∞+∞ exp(-(a½x + i(k-k0)/(2a½)2) dx
= (1/(a2π3))¼exp(-(k-k0)2/(4a))∫-∞+∞ exp(-x'2) dx'
= (a2π)exp(-(k-k0)2/(4a)).
|Φ(k,0)|2 = (a2π)exp(-(k-k0)2/(2a)) is the probability density in k-space.

Evolution of a wave packet

#### Problem:

In one dimension, at t = 0 the normalized wave function of a free particle of mass m in k-space is
Φ(k,0) = Nexp(-k2/(2b2)).
(a)  Find the normalization constant N.  Find the expectation value <p> = ħ<k>.
(b)  Find the FWHM in of |Φ(k,0)|2 in k-space.
(c)  Find the corresponding wave packet Ψ(x,0) in coordinate space.  Find <x>.
(d)  Find the FWHM in of |Ψ(x,0)|2 in coordinate space.
(e)  Find the FWHM in of |Ψ(x,t)|2 an some later time t.  Does it change with time?
(f)  Find the FWHM in of |Φ(k,t)|.  Does it change with time?

Hint:  ∫-∞+∞exp(-a2(x + c)2)dx = √π/a

Solution:

• Concepts:
Postulates of QM, the Fourier transform
• Reasoning:
We investigate the properties of a Gaussian wave packet representing a free particle.
• Details of the calculation:
(a)  N2-∞+∞exp(-k2/b2)dk = 1.  ∫-∞+∞exp(-k2/b2)dk = b√π
N = 1/(b√π) ½.  <p> = <k> = 0.

(b)  exp(-k2/b2) = ½,   k = b√(ln2),  FWHM = 2√(ln2)b.

(c)  Ψ(x',0) = [1/(2π)½][1/(b√π)½] ∫-∞+∞exp(-k2/(2b2))exp(ikx') dk.
-∞+∞exp(-k2/(2b2))exp(ikx') dk = ∫-∞+∞exp(-k2/(2b2))cos(kx') dk
= 2 ∫0+∞exp(-k2/(2b2))cos(kx') dk = 2b(π/2)1/2 exp(-x'2b2/2).
We can also evaluate the integral by grouping the k-dependent terms into a perfect square.
-k2/(2b2) + ikx' = [-1/(2b2)][k - ix'b2]2 - x'2b2/2.
Now we can evaluate the integral using  ∫-∞+∞exp(-a(x+b)2)dx = √(π/a).
Ψ(x',0) = (b2/π)¼exp(-x'2b2/2).  <x'> = 0.
x' = x - x0 depends on the choice of the origin of the coordinate system.

(d)  exp(-(x - x0)2b2) = ½.   FWHM = 2√(ln2) (1/b).

(e)  Choose x0 = 0.
The wave function of a free particle is a linear superposition of plane waves.
Ψ(x,t) = [1/(2π)½][1/(b√π)½] ∫-∞+∞exp(-k2/(2b2))exp(i(kx - ωt)) dk.
For a plane wave E = ħω = ħ2k2/(2m),  ω = ħk2/(2m).
We can group the k-dependent terms into a perfect square
-k2/(2b2) + i(kx - ħk2t/(2m)) = -k2[1/(2b2) + iħt/(2m)] + ikx
= -A(k2 - ikx/A + (ix/2A)2 - (ix/2A)2]
= -A[k - ix/2A]2 - x2/(4A),
where A = 1/(2b2) + iħt/(2m), |A|2 = 1/(4b4) + ħ2t2/(4m2).                                                                                  =
-∞+∞exp(-A[k - ix/2A]2 - x2/(4A)) dk = exp(-x2/(4A)∫-∞+∞exp(-A[k - ix/2A]2) dk
= exp(-x2/(4A)√(π/A), using ∫-∞+∞exp(-a(x+b)2)dx = √(π/a).
Ψ(x,t) = [1/(2π)½][1/(b√π)½] √(π/A) exp(-x2/(4A)).
Ψ(x,t) = (1/(b2π))¼[1/(1/b2 + iħt/m)]½ exp(-x2/(4A)).

|Ψ(x,t)|2 = (1/(b2π))½[(1/(1/b4 + (ħt/m)2) ½ exp(-(b2x2/(1 + (b2ħt/m)2)).
= b/(π(1 + (b2ħt/m)2) exp(-(b2x2/(1 + (b2ħt/m)2)).
To find the FWHM we use
exp(-(b2x2/(1 + (b2ħt/m)2)) = ½.   b2x2 = ln(2)(1 + (b2ħt/m)2),
FWHM = 2√(ln2)(1/b2 + (bħt/m)2)½ = 2√(ln2) (1/b)(1 + (b2ħt/m)2).
The FWHM of |Ψ(x,t)|2 increases with time.

(b)  iħ∂Φ(k,t)/∂t = [p2/(2m)]Φ(k,t) = ħω Φ(k,t).
Φ(k,t) = exp(-iωt)) Φ(k,0).  FWHM of |Φ(k,t)| = 2√(ln2)b.  It does not change with time.

#### Problem:

In one dimension, at t = 0 the normalized wave function of a free particle in momentum space is
Φ(p,0) = Nexp(-(p - p0)2/(2b2ħ2))exp(-ipx0/ħ).
(a)  Find the normalization constant N.  Find the expectation value <p>.
(b)  Find the FWHM in of |Φ(p,0)|2 in momentum space.
(c)  Find the corresponding wave packet Ψ(x,0) in coordinate space.  Find the expectation value <x>.
(d)  Find the FWHM in of |Ψ(x,0)|2 in coordinate space.

Solution:

• Concepts:
Postulates of QM, the Fourier transform
• Reasoning:
We investigate the properties of a Gaussian wave packet representing a free particle.
• Details of the calculation:
(a)  N2-∞+∞exp(-(p - p0)2/(b2ħ2))dp = 1.  ∫-∞+∞exp(-(p - p0)2/(b2ħ2))dp = bħ√π.
N = 1/(bħ√π) ½.  <p> = p0.
(b)  exp(-(p - p0)2/(b2ħ2)) = ½,   (p - p0) = aħ√(ln2),  FWHM = 2√(ln2)bħ.
(c)  Ψ(x,0) = [1/(2πħ)½][1/(bħ√π)½] ∫-∞+∞exp(-(p - p0)2/(2b2ħ2))exp(ip(x-x0)/ħ) dp.
Let us group the p-dependent terms into a perfect square.
-(p - p0)2/(2b2ħ2) + ip(x-x0)/ħ
= [-1/(2b2ħ2)][p - p0 - i(x-x0)b2ħ]2 - (x-x0)2b2/2 + ip0(x-x0)/ħ.
Then, using ∫-∞+∞exp(-a2(x+b)2)dx = √π/a, we have
Ψ(x,0) = (b2/π)¼exp(-(x-x0)2b2/2) exp(ip0(x-x0)/ħ.  <x> = x0.
(d)  exp(-(x-x0)2b2) = ½.   FWHM = 2√(ln2) (1/b).

#### Problem:

The wave packet for a quantum mechanical particle of mass m in one dimension is described by

Ψ(x,t) = [1/(2π)½] lim(R-->∞) ∫-R+R dk Φ(k) exp(i(kx - ω(k)t)),

where  Φ(k) = N exp(-(k - k0)2/(4(Δk)2)) is a "strongly peaked" distribution around k = k0 with ΔkΔx = ½ at t = 0.
(a)  Show that  ψ(x, 0) =  [Δx(2π)] exp(ik0x)exp(-x2/(4Δx2)) (evaluate the integral),
and find the probability density |ψ(x, 0)|2 of the particle.  Use ΔkΔx = ½  to eliminate Δk.  Show your work!
(b)  For a free particle E = ħω = ħ2k2/(2m).  Show that for a free particle
|ψ(x, t)|2 = exp[-½ (x - ħk0t/m)2/Δx(t)2]/[ (2π)½ Δx(t)],
with Δx(t) = (Δx2 + ħ2t2/(4m2Δx2)½.
(c)  Determine the group velocity of the wave packet.
(d)  Evaluate the time it takes for the wave-packet to double in spatial extent, specifically if the particle is an electron and Δx ~10 nm, at t = 0.

Useful integrals:
-∞+∞exp(-αx2)dx = √(π/α)
-∞+∞exp(-αx2 + iβx))dx = (π/α)½exp(-β2/(4α))
Note:  You can complete some parts of the problem by using the given results without evaluating the integrals.

Solution:

• Concepts:
Time evolution of a free wave packet
• Reasoning:
We are asked to evaluate integrals to better visualize the time evolution of a free wavepacket.
• Details of the calculation:
(a)  Normalize Φ(k):
N2-∞+∞exp(-(k - k0)2/(2(Δk)2))dk = 1.    ∫-∞+∞exp(-(k - k0)2/(2(Δk)2))dk = Δk√(2π).
N = 1/(Δk√(2π))½.
ψ(x, 0) = [exp(ik0x)N/√(2π) ]∫-∞+∞exp(-(k - k0)2/(4(Δk)2) + ix(k - k0))dk.
(k - k0)2/(4(Δk)2) - ix(k - k0)) - x2(Δk)2 + x2(Δk)2
= ((k - k0)/(2 Δk) - ixΔk)2 + x2(Δk)2.  (completing the square)
ψ(x, 0) = [N/√(2π)]exp(ik0x)exp(-x2(Δk)2)∫-∞+∞exp(-k'2)k'.
ψ(x, 0) = [(Δk)½2¼¼]exp(ik0x)exp(-x2(Δk)2) = [Δx(2π)] exp(ik0x)exp(-x2/(4Δx2)).
|ψ(x, 0)|2 = [1/(Δx(2π)½] exp(-x2/(2Δx2)).
This is a Gaussian with σ = Δx centered at x = 0.

(b)  ψ(x, t) = (2π)-∞+∞ Φ(k, t) exp(ikx) dk
Φ(k, t) = N exp(-(k - k0)2/(4(Δk)2) - iħk2t/(2m ))
N = 1/(Δk√(2π))½.
-(k - k0)2/(4(Δk)2) + ixk  - iħk2t/(2m)
= -(k - k0)2/(4(Δk)2) + ix(k - k0) - iħ(k - k0)2t/(2m) + iħk02t/(2m) - iħkk0t/m + ik0x
= -(k - k0)2[1/(4(Δk)2) + iħt/(2m)] + i(k - k0)[x - ħk0t/m] + iħk02t/(2m) - iħkk02t/m + ik0x
= -(k - k0)2[Δx2 + iħt/(2m)] + i(k - k0)[x - ħk0t/m] - iħk02t/(2m)  + ik0x
ψ(x, t) =  [N/√(2π)]exp(ik0x - iħk02t/(2m)) ∫-∞+∞exp(-αk'2 + iβk'))dk'
= [N/√2] exp(ik0x - iħk02t/(2m))[(Δx2 + iħt/(2m))]exp[-¼(x - ħk0t/m)2/(Δx2 + iħt/(2m))].
ψ(x, t) = Δx½ [(Δx2 + iħt/(2m))](2π) ]*
exp(ik0x - iħk02t/(2m)) exp[-¼(x - ħk0t/m)2/(Δx2 + iħt/(2m))].
|ψ(x, t)|2 = exp[-½(x - ħk0t/m)2/Δx(t)2]/[ (2π)½Δx(t)], with Δx(t) = (Δx2 + ħ2t2/(4m2Δx2)½.
(c) |ψ(x, t)|2 is a Gaussian with  σ = Δx(t) centered at x = ħk0t/m.  The center of the wave packet moves with speed ħk0t/m.  This is the group velocity vg = dω/dk of the wave packet.

(d)  Δx(t) = (Δx2 + ħ2t2/(4m2Δx2)½ = 2Δx
Δx2 + ħ2t2/(4m2Δx2) = 4Δx2.  t2 = 12 Δx4 m2/ ħ2.
t2 = 12*10-32*(9.1*10-31)2/(1.05*10-34)2 s2 =  ~10-23 s2,  t ~ 3*10-12 s.
We get the same order of magnitude result by just using the uncertainty principle.
(Δv ~ ħΔk/m = ħ/(2mΔx),  t ~ 2Δx/Δv = 4mΔx2/ħ).