Solutions5.pdf

7/27/2019 Solutions5.pdf

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Quantum Mechanics I

Solutions 5.

HS 2012Prof. Ch. Anastasiou

Exercise 1. Uncertainty Relation

Compute

(Sx)2

S2

x Sx2

,where the expectation value is taken for the |Sz, + state. Using your result check the generalizeduncertainty relation

(A)2(B)2 14|[A, B]|2

with A Sx, B Sy.

Solution. The solution to this exercise is attached on page 5.

Exercise 2. Momentum-Space Wave Function

Prove the following:

i) p|x | = i p

(p) ii) |x | =

dp(p

)i

p(p

) ,

where (p) = p| and (p) = p| are momentum-space wave functions.

Solution. The solution to this exercise is attached on page 7.

Exercise 3. Gaussian Wave Packets

Let us introduce the state | defined by

x| = (x) = A exp(x x0)

2

2d2

.

(a) Knowing that the wave function extends over all space (i.e. (x) L2(R)), find thenormalization constant A such that ||| = 1.

Solution.

| = +

dx|(x)|2

= |A|2 +

dx e

(xx0)2

d2

= |A|2

d+ dy e

y2

= |A|2

d (S.1)where we made the change of variables y = (x x0)/d. Setting | = 1 and taking A real we getA = 1

d

.

(b) Compute the expectation values x, p, x2 and p2 over the state | and describe theirdependence on d. Show that | is a minimum for the Heisenberg uncertainty relation, i.e.show that it satisfies

(x)2 (p)2 = 2

4.

(c) What is the interpretation of

x

,

p

,

(x)2

and

(p)2

if you consider

|(x)

|2 as a

probability density ?

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Solution. We can already note that when (x x0) d, |(x)|2 0, so that the particle is localizedaround x x0.

x =+

dx(x)x(x) =

1

d

+

dxxe(xx0)

2

d2 =1

d

d

+

dy(dy + x0)ey2

=1

x0 +

dyey2

= x0

(S.2)

where we made again the change of variables y = (x x0)/d, dx = ddy and x = x0 + dy, and the integral+ dy ye

y2 = + dy yey2 vanishes for parity.p = i

+

dx(x)x(x) = i

d

+

dxe (xx0)

2

d2 (2) (x x0)2d2

=i

d3

x x0 = 0(S.3)

where we used x(x) = (x)(2) (xx0)2d2 and x = x0.Using +

dyy( ye

y2 ) = 0 (S.4)

as can be seen calculating the primitive at the extrema, we get:+

dyy( yey2 ) =

+

dy

ey2 2y2ey2 = 0 (S.5)

so that +

dyy2ey2

=1

2

+

dyey2

=

2(S.6)

we can calculate:

x2 = 1d

+

dxx2e (xx0)

2

d2 =1

+

dy(x20 + d2y2 + 2x0dy)e

y2 = x20 +d2

2(S.7)

with x = x0 + dy. In the same way:

p2

=

2

d +

dxe

(xx0)2

2d2 2

xe (xx0)

2

2d2 =2

d +

dx 4

(x

x0)

2

4d4e (xx0)

2

d2 =

=2

d5

+

dxe (xx0)

2

d2

x2 + x20 2x0x

=2

d4

x20 +d2

2+ x20 2x0x0) =

2

2d2

(S.8)

where we used previous results and+

dx(x)2x(x) =

+

dxx(x)x(x) (S.9)

We then see that the larger d, the larger the position uncertainty, while the smaller d, the larger themomentum uncertainty, but the product is constant and independent of d, and reads:

(x)2(p)2 =2

4(S.10)

(d) Consider now the wave-function

x| = (x) = A exp(x x0)

2

2d2+ ikx

.

Describe how (x) changes when d is varied.

(e) Compute again the expectation values x, p, x2, p2 and (x)2 (p)2 , now overthe state |. Are there any differences with respect to point (b)?

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Solution. when d (x x0), d , (x) eikx and approaches a momentum eigenstate.when d (x x0), d 0, (x) (x x0) and approaches a position eigenstate.x and x2 remain equal as the additional term is just a phase factor. While

p = i+

dx(x)x(x) = i

d

+

dxe (xx0)

2

d2

(2) (x x0)

2d2+ ik

= k (S.11)

where we used x(x) = (x)

(2)(xx0)

2d2 + ik] and previous result. while

p2 = 2

d

+

dxe (xx0)

2

2d2ikx

2xe (xx0)

2

2d2+ikx

=2

d

+

dxx e (xx0)

2

2d2ikx

xe (xx0)

2

2d2+ikx

=2

d

+

dxe (xx0)

2

d2

(2) (x x0)2d2

+ ik = 2

2d2+ 2k2

(S.12)

So that: (x)2(p)2 doesnt change.

(f) Calculate the momentum representation of the state |, i.e.,p| = (p)

What is the relation between (x) and (p), i.e., through which mathematical operationare they linked? What can we say about that? Describe how (p) varies when d changes.Compare with previous results.

Solution.

(p) =

+

dx(x)ei p

x (S.13)

the relation is the Fourier Transform. We have to complete the square in the exponent to get:

(x

x0)

2d2 i(k p

)x =

1

2d2

(x x0)2

+ i(

p

k)2d2

(x x0 + x0) ==

1

2d2

(x x0)2 + 2id2(p

k)(x x0)

+ i(

p

k)x0

=1

2d2

[(x x0 + id2(p

k)]2 + d4(p

k)2

+ i(

p

k)x0

=1

2d2

(x x0 + id2(p

k)

2+

d2

2(p

k)2 + i(p

k)x0

(S.14)

where we initially collected a minus sign. The x integration is straight again with the usual change ofvariables (since the other part id2( p

k) in the square doesnt affect the variable change) and we are left

with the remaining part of the exponent:

(p) =

2

d

exp

d

2

2(p

k)2 i(p

k)x0

(S.15)

and the normalization is different as in the starting integral we have A and not A2 as usual. so the FT

of a gaussian centered at x0 in real space is a gaussian in momentum space centered at k. so when d is

large the gaussian in momentum space is small, and has contributions only when (p k) 0, therefore is

sharp in momentum space; when when d is small the gaussian in momentum space is very extended and

momentum is not well defined, as we concluded also from (x)2, (p)2.

(g) We want now to understand how such wave packets spread in space when time flows. Letus consider the wave packet (x) with x0 = 0. Its time evolution is governed by theSchrodinger equation with appropriate boundary conditions

2

2m

2

x2(t, x) = i

t (t, x) with (0, x) = (x)|x0=0 = A expx2

2d2

.

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For positive time t > 0 show that (t, x) can be written as a gaussian wave packet with atime dependent width d(t) given by

d(t) = d

1 +

t

md2

2

.

Hint. Make the following Ansatz

(t, x) =

dk C(k)exp

ikx i k

2

2mt

.

First show that (t, x) satisfies the Schrodinger equation and then solve the boundarycondition for C(k). You will also need (and prove if you want) the following identity

dx eikxex2

=

ek2

4 , () > 0 .

Also do not care about global phases in your final result.

Solution. Plugging our Ansatz in the Schrodinger equation gives

2

2m(ik)2 = i

i k

2

2m

,

what shows that is a solution. The boundary condition then reads

(0, x) =

dk C(k)eikx = A exp

x

2

2d2

which can be directly solved by a simple Fourier transformation. We obtain

C(k) =A

2dx eikxe x22d2 = Ad2 e

k2d2

2 ,

using the hint. For positive time t > 0, the full solution can then be written as

(t, x) =

dk C(k)exp

ikx i k

2

2mt

=

Ad2

dk e

k2d2

2 eikxik2

2mt

=Ad

2

dk ek

2(i t2m+d2

2 )eikx = Ad

i

t

m+ d2

1exp

x

2

2(i tm + d2)

. (S.16)

We have to take care of the non vanishing imaginary part for the factor in the exponential

(it

m+ d2)1 =

d2 i tm2

t2

m2+ d4

=1

d2 + 2 t2

m2d2

id4mt +

tm

.

If we do not care about the imaginary parts (that do not contribute to the envelope of the wave functionbut only to a global phase) we can rewrite the square root using

x =

|x|ei arg(x)/2 as

it

m+ d2

1=

ei...d2

1 +

tmd2

2 =ei...

d

d(t)

where d(t) as in the exercise. Plugging A we finally obtain

(t, x) =1

14

d(t)

exp

x

2

2d(t)2

exp {i . . .} .

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Solution to Exercise 1.

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