Solutions to Homework Assignment 2Physics 501

1. For the state specified by: ψ(x) =(

12πσ2

)1/4e−(x−a)2/4σ2

x

show that 〈x〉 = a 〈x2〉 = σ2x + a2

and using the momentum operator p = −ih ddx

show that 〈p〉 = 0 〈p2〉 = h2

4σ2x

〈x〉 =

∫ψ∗xψ dx =

(1

2πσ2x

)1/2 ∫e−(x−a)2/2σ2

xx dx

=

(1

2πσ2x

)1/2 ∫e−y

2/2σ2x(y + a) dy

=

(1

2πσ2x

)1/2 [∫e−y

2/2σ2xy dy +

∫e−y

2/2σ2xa dy

]

=

(1

2πσ2x

)1/2 ∫e−y

2/2σ2xa dy = a

〈x2〉 =

∫ψ∗x2ψ dx =

(1

2πσ2x

)1/2 ∫e−(x−a)2/2σ2

xx2 dx

=

(1

2πσ2x

)1/2 ∫e−y

2/2σ2x(y + a)2 dy

=

(1

2πσ2x

)1/2 [∫e−y

2/2σ2xy2 dy +

∫e−y

2/2σ2x2ya dy +

∫e−y

2/2σ2xa2 dy

]

=

(1

2πσ2x

)1/2 [∫e−y

2/2σ2xy2 dy +

∫e−y

2/2σ2xa2 dy

]= σ2

x + a2

〈p〉 =

∫ψ∗pψ dx = −ih

∫ψ∗dψ

dxdx

= −ih(

1

2πσ2x

)1/2 ∫e−(x−a)2/4σ2

xd

dx

(e−(x−a)2/4σ2

x

)dx

= −ih(

1

2πσ2x

)1/2 ∫ (−2(x− a)

4σ2x

)e−(x−a)2/2σ2

xdx

=

(ih

2

)(1

2πσ2x

)1/2 [∫e−(x−a)2/2σ2

xx dx −∫e−(x−a)2/2σ2

xa dx

]

=

(ih

2

)(a− a) = 0

〈p2〉 =

∫ψ∗p 2ψ dx = −h2

∫ψ∗d2ψ

dx2dx

= −h2

(1

2πσ2x

)1/2 ∫e−(x−a)2/4σ2

xd2

dx2

(e−(x−a)2/4σ2

x

)dx

= −h2

(1

2πσ2x

)1/2 ∫e−(x−a)2/4σ2

xd

dx

[(−x− a

2σ2x

)e−(x−a)2/4σ2

x

]dx

= −h2

(1

2πσ2x

)1/2 ∫e−(x−a)2/4σ2

x

[− 1

2σ2x

+(x− a)2

4σ2x

]e−(x−a)2/4σ2

xdx

= −h2

(1

2πσ2x

)1/2 ∫ [− 1

2σ2x

+(x− a)2

4σ4x

]e−(x−a)2/2σ2

xdx

= −h2

[− 1

2σ2x

+1

4σ2x

]=

h2

4σ2x

2. Show that the free-particle one-dimensionalSchrodinger equation is invariantwith respect to Galilean transformations. do this by showing that, when thetransformation

x′ = x − vt t′ = t

is applied, the transformed wave function

ψ′(x′, t′) = f(x, t)ψ(x, t)

satisfies the Schrodinger equation with respect to the primed variables, wheref involves only x, t, h, m and v. Find the form of f and show that the planewave solution

ψ(x, t) = Aei(px−Et)/h transforms as expected.

x′ = x − vt ⇒ x′ = x − v ⇒ p′ = p − mv

p′x′ − E ′t′ = p′x′ − p′2

2mt′ = (p − mv)(x − vt) − (p − mv)2

2mt

= px − pvt − mvx + mv2t − p2

2mt + pvt − mv2

2t

= px − mvx − p2

2mt +

mv2

2t = px − Et − mvx +

mv2

2t

=

∫a(p)ei(px−Et)/h e

−ihmvx−mv

2

2ti/hdp

= ψ(x, t)e−ihmvx−mv

2

2ti/h

= f(x, t)ψ(x, t)

f(x, t) = e−ihmvx−mv

2

2ti/h

Aei(p′x′−E′t′)/h = Aei(px−Et−mvx+mv2

2t)/h

= Aei(px−Et)/he−ihmvx−mv

2

2ti/h

= Aei(px−Et)/hf(x, t)

3. Show that for a three dimensional wave packetd

dt〈x2〉 =

1

m(〈xpx〉 + 〈pxx〉)

d

dt〈x2〉 =

d

dt

∫ψ∗x2ψ dx =

∫ (∂ψ∗

∂tx2ψ + ψ∗x2 ∂ψ

∂t

)dx

= − ih

2m

∫ (∂2ψ∗

∂x2x2ψ − ψ∗x2 ∂

2ψ

∂x2

)dx

= − ih

2m

[d

dx

∫ (∂ψ∗

∂xx2ψ − ψ∗x2 ∂ψ

∂x

)dx −

∫ (∂ψ∗

∂x

∂

∂x

(x2ψ

)− ∂

∂x

(ψ∗x2

) ∂ψ∂x

)dx

]

=ih

2m

∫ (∂ψ∗

∂x

∂

∂x

(x2ψ

)− ∂

∂x

(ψ∗x2

) ∂ψ∂x

)dx

=ih

2m

∫ (∂ψ∗

∂x2xψ +

∂ψ∗

∂xx2∂ψ

∂x− ∂ψ∗

∂xx2 ∂ψ

∂x− ψ∗2x

∂ψ

∂x

)dx

=ih

2m

∫ (∂ψ∗

∂x2xψ − ψ∗2x

∂ψ

∂x

)dx

=ih

m

[d

dx

∫(ψ∗2xψ − ψ∗2xψ) dx −

∫ (ψ∗

∂

∂x(xψ) − ∂

∂x(ψ∗x)ψ

)dx

]

= −ihm

∫ (ψ∗

∂

∂x(xψ) − ∂

∂x(ψ∗x)ψ

)dx = −ih

m

∫ (ψ∗

∂

∂x(xψ) + ψ∗x

∂ψ

∂x

)dx

=1

m

∫(ψ∗ px xψ + ψ∗x px ψ) dx =

1

m(〈xpx〉 + 〈pxx〉)

4. The expression ψ(x, t) =

∫ [∑E

u∗E(x′)uE(x)e−iE(t−t′)/h

]ψ(x′, t′) dx′

enables one to calculate ψ at time t in terms of ψ at time t′. If the expressionin brackets above is called iG(x, t : x′, t′), then

ψ(x, t) = i

∫G(x, t : x′, t′)ψ(x′, t′) dx′

a) Show that for a free particle in one dimension

G0(x, t; x′, t′) = −i[

m

2πih(t− t′)

]1/2

eim(x−x′)2/2h(t−t′)

G0(x, t; x′, t′) = −i∞∫

−∞

u∗p(x′)up(x) e−ip

2(t−t′)/2mh dp =1

2iπh

∞∫−∞

eip(x−x′)/h e−ip

2(t−t′)/2mh dp

=1

2iπh

∞∫−∞

eβp e−αp2

dp =1

2iπheβ

2/4α(πα

)1/2

β =i(x− x′)

hα =

i(t− t′)2mh

⇒ G0(x, t; x′, t′) =

(1

2iπh

)(π2mh

i(t′ − t)

)1/2

eim(x′−x)2/2(t−t′)h

= −i[

m

2πih(t− t′)

]1/2

eim(x−x′)2/2h(t−t′)

b) Assume that ψ has the form of the normalized wave packet

ψ(x′, 0) =

(1

2πσ2x

)1/4

e−x′2/4σ2

x

at t′ = 0; use the above result to find ψ(x, t) and |ψ(x, t)|2 at another timet. This G0 is called the free-particle Green’s function in one dimension.

ψ(x, t) =( m

2πiht

)1/2∞∫

−∞

eim(x−x′)2/2ht

(1

2πσ2x

)1/4

e−x′2/4σ2

x dx′

I’m going to simplify the algebra by setting m = h = σx = 1. At the end, we’llput them back in by dimensional analysis.

ψ(x, t) =

(1

2πit

)1/2∞∫

−∞

ei(x−x′)2/2t

(1

2π

)1/4

e−x′2/4 dx′

=

(1

2π

)3/4(1

it

)1/2∞∫

−∞

ei(x−x′)2/2te−x

′2/4 dx′

=

(1

2π

)3/4(1

it

)1/2∞∫

−∞

ei(x2−2xx′+x′

2)/2te−x

′2/4 dx′

=

(1

2π

)3/4(1

it

)1/2

eix2/2t

∞∫−∞

e−ixx′/te−x

′2 (1/4−i/2t) dx′

=

(1

2π

)3/4(1

it

)1/2

eix2/2t

∞∫−∞

eβx′e−αx

′2

dx′ =

(1

2π

)3/4(1

it

)1/2

eix2/2teβ

2/4α(πα

)1/2

β =−ixt

α =t− i2

4t

β2

4α= − x2

t(t− 2i)

ψ(x, t) =

(1

2π

)1/4(2

2 + it

)1/2

e−x2[ 1t(t−2i)

− i2t ]

|ψ(x, t)|2 =

(1

2π

)1/2 ∣∣∣∣ 2

2 + it

∣∣∣∣ e2Re[−x2/t(t−2i)] =

(1

2π

)1/2 ∣∣∣∣ 2

2 + it

∣∣∣∣ e−2x2/(4+t2)

Now put h, m and σ2x back in by dimensional analysis.

ψ(x, t) =

(1

2πσ2x

)1/4(2σ2

x

2σ2x + iht/m

)1/2

e−x2m/ht

„1

(ht/mσ2x−2i)

− i2

«

|ψ(x, t)|2 =

(1

2πσ2x

)1/2 ∣∣∣∣ 2σ2x

2σ2x + iht/m

∣∣∣∣ e−x2/(2σ2x+h2t2/2m2σ2

x)

5. Show that the free particle propagator

G0(x, t; x′, t′) = −i[

m

2πih(t− t′)

]1/2

eim(x−x′)2/2h(t−t′)

satisfies the free particle Schrodinger equation(ih∂

∂t+

h2

2m

∂2

∂x2

)G0(x, t; x′, t′) = 0

We can dispense with the constants out in front so

G0(x, t; x′, t′) ∝ (t− t′)−1/2eim(x−x′)2/2h(t−t′)

ih∂

∂t

[(t− t′)−1/2eim(x−x′)2/2h(t−t′)

]= −ih

2(t−t′)−3/2

[1 +

im(x− x′)2

h(t− t′)

]eim(x−x′)2/2h(t−t′)

h2

2m

∂2

∂x2

[(t− t′)−1/2eim(x−x′)2/2h(t−t′)

]=

h2

2m(t−t′)−1/2 ∂

∂x

[im(x− x′)h(t− t′)

eim(x−x′)2/2h(t−t′)]

=ih

2(t− t′)−3/2 ∂

∂x

[(x− x′)eim(x−x′)2/2h(t−t′)

]

=ih

2(t− t′)−3/2

[1 +

im(x− x′)2

h(t− t′)

]eim(x−x′)2/2h(t−t′)

ih∂

∂t

[(t− t′)−1/2eim(x−x′)2/2h(t−t′)

]+h2

2m

∂2

∂x2

[(t− t′)−1/2eim(x−x′)2/2h(t−t′)

]= 0

6. We showed in class, and you showed in the problem above, that a free-particleGaussian wave packet will spread out as a function of time, σx(t) increases withtime. What about σp?

a) First calculate σp(t) directly for ψ(x′, 0) =

(1

2πσ2x

)1/4

e−x′2/4σ2

x .

b) Then, be clever and make an argument involving essentially no calculationto arrive at the same answer.

c) Now try to resolve this paradox. We initially have a state for which both theposition and the momentum probability density distributions are Gaussian.We found in class that for a Gaussian wave function we have the exactrelation σxσp = h/2. Now after time t the probability density distributionsin both x and p are still Gaussian but σx(t) > σx(0) while σp(t) = σp(0) sothat now we have σxσp > h/2.

ψ(x, t) = Ne−x2/(4σ2

x+2iht2/m)

φ(p, t) = N

∞∫−∞

e−ipx/he−x2/(4σ2

x+2iht2/m) dx

use

∞∫−∞

eβxe−αx2

dx = eβ2/4α

√π

αwith α =

1

(4σ2x + 2iht2/m)

β = −ip/h

φ(p, t) = N

(π

4σ2x + 2iht2/m

)1/2

e−p2(σ2

x+iht2/2m)/h2

|φ(p, t)|2 = |N |2(

4σ4x

h2

)e−p

22σ2x/h

2

|φ(p, t)|2 is time independent.

We could have anticipated this results since, for a free particle, momentumeigenstates are also energy eigenstates and are therefore stationary states (timeindependent except for a time-dependent complex phase.) We can also see this

by expanding ψ(x, t) in momentum eigenstates and taking the Fourier trans-form.

ψ(x, t) =1√2π

∞∫−∞

a(p′)e−ip′x/he−ip

2t/2mh dp′

φ(p, t) =1

2πh1/2

∞∫−∞

a(p′)eipx/he−ip′x/he−ip

2t/2mh dp′ dx = h1/2a(p)e−ip2t/2mh

|φ(p, t)|2 = h2|a(p)|2 independent of t.

As for the paradox, the condition that σxσp = h/2 is true only if the coefficientof the the the −x2 term in the exponent is real, ψ(x) ∝ e−αx

2with α real. In

the above case, α is a complex number. It still leads to a Gaussian probabilitydensity for both x and p, i.e., |ψ(x)|2 and |ψ(p)|2 are both Gaussian, but itavoids the condition that σxσp = h/2.

7. Determine the probability current density for the wave function

ψ(x) = Aeipx/h + Be−ipx/h

and show that it is the same as the current density of two beams of classicalparticles with densities |A|2 and |B|2 moving in opposite directions. Is this trueif p is replaced by p′ in the second term?

~j =h

2mi

(ψ∗~∇ψ − ψ~∇ψ∗

)Sine this is effectively a one-dimensional problem use

j =h

2mi

(ψ∗

∂

∂xψ − ψ ∂

∂xψ∗)

=h

2mi

[(A∗e−ipx/h +B∗eipx/h

)(iph

)(Aeipx/h −Be−ipx/h

)−(Aeipx/h +Be−ipx/h

)(−iph

)(A∗e−ipx/h −B∗eipx/h

)]

=p

2m

(A∗A+B∗Ae2ipx/h − A∗Be−2ipx/h −B∗B + AA∗ − AB∗e2ipx/h +BA∗e−2ipx/h −BB∗

)=

p

m

(|A|2 − |B|2

)This corresponds to two classical beams one of density |A|2 moving to the rightwith speed p/m and the other of density |B|2 moving to the left with speedp/m. if p is replaced by p′ in the second term, then the cross terms will notcancel out as they did above and there will be a non-classical interference term.

j =h

2mi

(ψ∗

∂

∂xψ − ψ ∂

∂xψ∗)

=1

2m

[(A∗e−ipx/h +B∗eip

′x/h)(

pAeipx/h − p′Be−ip′x/h)

−(Aeipx/h +Be−ip

′x/h)(−pA∗e−ipx/h + p′B∗eip

′x/h)]

=1

2m

(pA∗A+ pB∗Aei(p+p

′)x/h − p′A∗Be−i(p+p′)x/h − p′B∗B

+pAA∗ − p′AB∗ei(p+p′)x/h + pBA∗e−i(p+p′)x/h − p′BB∗

)

=1

m

[p|A|2 − p′|B|2 + (p− p′)

(AB∗ei(p+p

′)x/h + A∗Be−i(p+p′)x/h

)]