Advanced kinetics Exercise 2 May 2, 2014

1 Free Particle

1.1 For the free particle, show that the Time-Dependent Schrodinger Equation (TDSE)

i~∂

∂tψ(r, t) = Hψ(r, t) (1)

becomes the equation below

β∂

∂tψ(r, t) = ∇2ψ(r, t). (2)

Where β is a complex constant.

Hint: The representation of the momentum operator in position space is known.

1.2 If the Hamiltonian is time independent, the solutions of the TDSE can be separated as:

ψ(r, t) = ψ(r)ψ(t). (3)

Show that the 1-D solutions for Equation (1) are

ψk(x, t) = [Ak exp(−ikx) +Bk exp(ikx)] exp(−iωkt) (4)

and express ωk explicitly as a function of k.

1.3 Explain why the solutions in task 1.2 cannot represent a free-particle wavefunction bythemselves. Hint: Evaluate 〈ψk | ψk〉.

1.4 Assume that the free particle is inside a box with infinitely spaced walls. Show that themost general solution of the wave equation can be written as

ψ(x, t) =

∫ ∞−∞

a(k) exp(ikx) exp(−iωkt)dk. (5)

From Equation (5), it is clear that any spatially Fourier transformable function is alegitimate free particle wavefunction. In the next exercise we will construct a spatiallyconfined wave function and evaluate its time evolution.

2 Gaussian wave packet

Now let’s assume the 1-D electron wave function takes the form

ψ(x, t = 0) = (2πa2)−14 exp(− x2

4a2), a > 0. (6)

2.1 Show that the wave function represents an object with a probability to be within x =[−∞,+∞] of 1 and calculate the position of the wave packet.Hint: The position is given by the expectation value of the x operator.

2.2 By calculating the Fourier transform of Equation (6) calculate a(k) from Equation (5).

1

Advanced kinetics Exercise 2 May 2, 2014

2.3 Write a short script using the programming language of your choice to calculate ψ(x, t).Plot | ψ(x, t) |2 for various time steps and describe your observation.

Hint: Use a = 1 A, this is the typical extent for an electron created, for example,by interaction of an intense laser field with an atom. In order to describe the wavepacket, an infinite set of ks would have to be considered, which is however infeasible.Use k = [− 1

4a ,14a ]. As you will see, this will yield a rather coarse reconstruction but the

overall tendency is correct. Use t = [0, 50] fs.

2.4 Now consider

ψ(x, t = 0) = (2πa2)−14 exp(− x2

2a2) exp(ik0x). (7)

Use the Fourier-shift theorem to calculate the a(k) from your results in 2.2. Do a timepropagation as in 2.3 and explain the difference in the outcome. (k0 = 1

10a)

3 Different pictures in quantum mechanics

As shown in the lecture (chapter 2) the time evolution operator has the property of temporallytranslating a wavefunction ψs(x, t0) to ψs(x, t), which can be expressed as

ψs(x, t) = U(t, t0)ψs(x, t0). (8)

3.1 Give the explicit form of the time-evolution operator for the free particle. Compare yourresult to the result of task 1.4.Hint: Use the closure relation for the time evolution operator U =

∫U |k〉〈k|dk.

3.2 In the lecture different pictures for understanding quantum mechanics have been dis-cussed. In the Schrodinger picture, the eigenstates of a system are containing the timeevolution:

ψs(x, t) = U(t, t0)ψs(x, t0), (9)

while in the Heisenberg picture this time evolution is associated with the individualoperators

HH = U∗(t, t0)HsU(t, t0). (10)

Show that the observables of a system do not depend on the picture used.Hint: In quantum mechanics every observable is the expectation value of an operator.

3.3 There is a set of important cases when the Hamiltonian can be split into a time dependentinteraction Hamiltonian (Hint(t)) and a time independent Hamiltonian (H0)

H(t) = H0 + Hint(t). (11)

We define now the interaction picture by choosing the following wave function

| ψI(t)〉 = U∗0 (t, t0) | ψs(t)〉, (12)

where U0 = exp(−i H0(t−t0)~ ). By inserting Equation (12) into the Schrodinger equation

show that

i~∂

∂t| ψI(t)〉 = HI

int | ψI(t)〉. (13)

Where HIint corresponds to the Hamiltonian Hint in the interaction picture.

2

Advanced kinetics Exercise 2 May 2, 2014

3.4 From Equation (13) it is possible to obtain

UI(t, t0) = 1− i

~

∫ t

t0

HIint(t

′)U I(t′, t0)dt′. (14)

Show that equation (14) is equivalent to

U(t, t0) = U0(t, t0)−i

~

∫ t

t0

U0(t, t′)Hint(t

′)U0(t′, t0)dt

′ (15)

and explain in simple terms why this is useful.

This exercise was created by Yoann Pertot and Aaron von Conta. For questions regardingthe task, please contact yoann.pertot@phys.chem.ethz.ch.

3