Problem Set6-Sol.pdf · PHYSICS 621 - Fall Semester 2013 - ODU Graduate Quantum Mechanics - Problem

Transcript

PHYSICS 621 - Fall Semester 2013 - ODU

Graduate Quantum Mechanics - Problem Set 6 - Solution

Problem 1) The normalized wave function ψ(x, t) satisfies the time-dependent Schrödinger equation for a free particle of mass m, moving along the x-axis (in 1 dimension). Consider a second wave function of the form φ(x, t) = exp(i(ax − bt))ψ(x − vt, t) . Show that φ(x, t) obeys the same time-dependent Schrödinger

equation provided the constants a, b and v are related by b = a2

2m, v = a

Calculate the expectation value of position <X>, momentum <P> and energy <H> for a particle in the state φ(x, t) in terms of those for a particle in the state ψ(x, t) . Show that the uncertainty in momentum is the same in both states. What physical interpretation can be given to the transformation from the state ψ(x, t) to the state φ(x, t)?

Ans.:

We need to show that

i ddtφ(x, t) = P2

2mφ(x, t) = −

2m∂2

∂x2φ(x, t) .

Plugging in the form given, we find for the l.h.s.

i ddtφ(x, t) = i d

dtexp(i(ax − bt))ψ(x − vt, t) = i (−ib)exp(i(ax − bt))ψ + exp(i(ax − bt)) ∂ψ

∂x(−v)+ ∂ψ

∂t#

)

and for the r.h.s.

−2

2m∂2

∂x2φ(x, t) = −

2m∂∂xexp(i(ax − bt)) iaψ(x − vt, t)+ ∂ψ

∂x#

'(= −

2mexp(i(ax − bt)) −a2ψ + 2ia ∂ψ

∂x+∂2ψ∂x2

Dividing by the exponential on both sides and subtracting the Schrödinger equation for the original ψ:

bψ − iv ∂ψ∂x

= −2

2m−a2ψ + 2ia ∂ψ

∂x#

Both sides are equal with the substitutions given; φ(x, t) satisfies the Schrödinger Equation.

We next calculate the requested expectation values:

Xφ= exp(−i(ax − bt))ψ *(x − vt, t)∫ xexp(i(ax − bt))ψ(x − vt, t)dx =

∫ x ψ(x − vt, t) 2 dx = ∫ [y+ vt]ψ(y, t) 2 dy = Xψ+ vt

PHYSICS 621 - Fall Semester 2013 - ODU

Pφ= exp(−i(ax − bt))ψ *(x − vt, t)∫

i∂∂xexp(i(ax − bt))ψ(x − vt, t)dx =

a∫ ψ(x − vt, t) 2 +ψ *(x − vt, t) i∂∂xψ(x − vt, t)dx = a+ ψ *(y, t)

i∂∂yψ(y, t)∫ dy = P

ψ+mv

P2φ= exp(−i(ax − bt))ψ *(x − vt, t)∫ −2( ) ∂

∂x2exp(i(ax − bt))ψ(x − vt, t)dx =

−2 ψ *(x − vt, t)∫ −a2ψ(x − vt, t)+ 2ia ∂ψ∂x(x − vt, t)+ ∂

2ψ∂x2

(x − vt, t)$

()= 2a2 + 2a P

ψ+ P2

From this it follows that Hφ=12m

P2φ=12mv2 + v P

ψ+ H

ψ and

Δpφ = P2φ− P

2= m2v2 + 2mv P

ψ+ P2

ψ− P

2+ 2mv P

ψ+m2v2( ) = P2

ψ− P

2= Δpψ

which is unchanged. The physical interpretation is as follows: φ(x, t) describes the same state as ψ(x, t) , except from a coordinate system that is moving towards the left with velocity v. In that coordinate system, the system seems to be moving to the right with additional velocity v and therefore additional momentum mv. The total kinetic energy increases accordingly.

Problem 2)

A particle is in the ground state of a box of length L with infinitely high walls at –L/2 and +L/2 (see Shankar pp. 157 and our first HW problem set). Suddenly, the box expands (symmetrically) to length 2L, leaving the wave function momentarily undisturbed. Calculate the probability that measuring the energy of the system afterwards yields as result the ground state energy of the new box. (Hint: the

answer is 83π!

%&2

Ans.:

First we write down the fully normalized eigenstate with the lowest energy (=ground state) for the box of length L:

ψL (x) =2Lsin π x

L+π2

Correspondingly, the new ground state of the double as large box must be

ψ2L (x) =1Lsin π x

2L+π2

If we expand the initial wave function in terms of coefficients for the basis of new eigenstates, the coefficient with respect to the new ground state will be

PHYSICS 621 - Fall Semester 2013 - ODU

a = ψ2L ψL = ψ*2L−L

L∫ (x)ψL (x)dx =

sin π x2L

+π2

−L/2

L/2∫ sin π x

L+π2

'(dx

=2L

cos π x2L#

−L/2

L/2∫ cos π x

'(dx =

cos π x2L#

−L/2

L/2∫ 1− 2sin2 π x

2L#

)

-.dx

where the change in limits is due to the fact that ψL=0 outside of ±L/2.

Using y = sin(πx/2L) we can integrate to find

a = 2L2Lπ

1− 2y2"# $%dy−sin(π /4)

sin(π /4)∫ =

8π

y− 23 y

3"# $%−1/ 21/ 2

=8π

2 1223

#'$

%(=83π

The probability to find the particle in this new ground state is the square of a, q.e.d.

Problem 3)

Consider a particle (confined in one dimension along the x-axis) within a potential given by the delta-function at the origin: V (x) = −aV0δ(x) . (You could consider this the extreme limit of a particle in a box – the box has infinite depth and infinitesimally small width). Surprisingly, there is a bound state solution to the stationary (time-independent) Schrödinger equation, and your task is to find it (and the corresponding energy eigenvalue). You can find hints in Shankar on page 163. See also the discussion on page 156 – you can not assume that ψ’(x) is continuous everywhere in this special case. In fact, the second derivative is (obviously) not even finite everywhere; however, integrating it over a very small interval centered around zero gives a finite answer which allows you to solve the problem.

Ans.:

Since the potential is zero at large distance from the origin, a bound state must have E < 0. We know already what the solutions look like for any point other than zero: ψ(x) = Aeκx , x < 0ψ(x) = Ae−κx , x > 0

where κ = 2m E / . All that remains to do is to check what happens at x=0. First of all, we know that

the wave function should be continuous, so ψ(0) = A . However, we realize that ψ’(x) cannot be continuous at x=0. In fact, ψ '(−ε) = Aκ ;ψ '(+ε) = −Aκ ⇒ψ '(+ε)−ψ '(−ε) = −2Aκ This must be compared to the Schrödinger equation, which we can formally write

−2

2m∂2

∂x2ψ(x) = −

2m∂∂xψ '(x) = E −V (x)( )ψ(x) = Eψ(x)+ aV0δ(x)ψ(x)

If we integrate both sides over the small interval -ε…ε, we get

−2

2m∂∂xψ '(x)

−ε

∫ dx = − 2

2mψ '(+ε)−ψ '(−ε)( ) =

2Aκm

= Eψ(x)−ε