Solutions2.pdf

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Quantum Mechanics I

Solutions 2.

HS 2012Prof. Ch. Anastasiou

Exercise 1. Classical Treatment of the Stern-Gerlach experiment

We consider here the Stern-Gerlach experiment in a classical setup, described by the figure

above.

Consider a particle in a magnetic field B and no electric field. Here we are only interested inthe force induced by the magnetic moments of the particles and will not consider effects due tothe Lorentz force. This force is given by

F = ( B) ,where is the magnetic moment of the particle.

(a) Can you build a magnetic field along the z direction with non vanishing gradient ? i.e.

B = (0, 0, Bz) and zBz

= 0 .

(b) Now assume that the magnetic moments of the incoming particles are directed along thez axis, positively and negatively. What do you observe in this case for Bz = z? Computethe separation between the two beams on the screen.

(c) We consider a magnetic field B with zBz = 0. What do you expect to observe forrandomly oriented magnetic moments ?

Solution. The solution to this exercise is attached on page 8.

Exercise 2. Stern Gerlach Experiment with Spin 12 Particles1

In this exercise we will explore the Stern-Gerlach experiment with spin 12 particles and extendit to have more complex arrangements. We consider the setup shown in the figure below.

For a beam of electrons, one-half will follow the upper path, one half the lower (left). If we putin a blocker in the path of the spin down particles (center), the beam will only emit spin upparticles. In this case the setup acts as a spin-filter. Let us draw a blue box to abstract such afilter and indicate the direction of the selected spin by a red arrow (right).

What fraction of incident particles emerge in the following cases ?

1Images sourced from Physics Virtual Bookshelf.

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(a) (b) (c) (d)

(e) (f) (g)

Solution. (a) 1/2, (b) 1/2, (c) 1/2, (d) 1, (e) 0, (f) 1/4, (g) 1/8

Exercise 3. Analogy with Polarization of Light

Let us compare the outcome of the Stern-Gerlach experiment with polarized light going throughtwo different sets of polarizing filters. Consider a monochromatic light wave propagating in thez-direction going through three different polarizing filters (see figure 1), where

the x-filter selects only beams polarized in the x-direction, the x-filter selects only beams polarized in the direction, which makes an angle of 45

with the x-direction in the x-y plane,

the y-filter selects only beams polarized in the y-direction.

x y

x x y

Figure 1: Light beams going through a sequence of polarizing filter.

(a) Describe in words what happens step by step in each case.

Solution. First setup: After the first filter the light beam is completely polarized in the x-direction,

therefore nothing nothing comes out after the y-filter, which is orthogonal polarized in respect to the x-

filter.

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Second setup: The second filter selects light polarized in the x-direction, which is a linear combination of

x and y. Since the light beams enters the x-filter polarized in the x-direction, the outcome is nonzero.

Now it has also an polarization component in the y-direction, that is why this time the beam comes out

the last y-filter as y-polarized light.

(b) To which cases of exercise 2 do these setups correspond?

Solution. They correspond to the cases e) and f) of Exercise 2, respectively. This can be seen by

identifying Sx atoms with x-, y-polarized light and Sz atoms with x-, y-polarized light.

(c) A linearly polarized light with polarization in the x-direction has a electric field oscillatingin the x-direction

E = E0x cos(kz wt). (1)Write down the electric field of the beams after each filter, using x and y as base vectors.

Solution.

After x-filter: E1 = E0x cos(kz wt), Ater x-filter: E2 = E02 x cos(kz t) = E02

x cos(kz t) + y cos(kz t)

,

Ater y-filter: E3 = E02 y cos(kz wt).

We now turn back to the Stern-Gerlach experiment. As above we want to try to represent allthe different spin-states as a linear combination of two base objects, i.e. by some kind of vectorin a new kind of two-dimensional vector space. Without loss of generality we choose |Sz, +and

|Sz,

to be our base vectors, where

|Si,

are the spin-up/spin-down states lying in the

i-direction.

(d) Write |Sx, + and |Sx, as linear combination of the |Sz, states.

Solution. In analogy to

x =1

2x +

12

y, (S.1)

y = 12

x +1

2y (S.2)

we may conjecture

12

|Sz ; + + 12

|Sz; , (S.3)

12

|Sz ; + + 12

|Sz; . (S.4)

Strictly speaking, this formalism is based on guessing; Later we will see how these expressionsindeed come out using the general formalism of quantum mechanics.

We want now to represent the spin-states |Sy, + and |Sy, in our |Sz, basis. But it seemsthat we already used up the available possibilities in writing

|Sx,

. Again, we get inspiration

from polarized light. Consider this time a circularly polarized light which is nothing more than a

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linear combination of an x-polarized light and a y-polarized light with a relative phase-differenceof /2,

E = E0

1

2x cos(kz wt) + 1

2y cos

kz wt +

2

. (2)

(e) Show that (2) is identical to

E = E0 Re

1

2xei(kzwt) +

i2

yei(kzwt)

. (3)

Solution. We substitute exp (i(kz t)) = cos (kz t)+i sin(kz t) in (3), use sin (x + 2 ) = cos (x),take the real part of the expression and we end up with (2).

(f) What is the crucial point in the expression (3)?

Solution. The second term in (3) has a purely imaginary coefficient, i.e. we need a two-dimensional

complex vector space to describe the spin states.

(g) Write |Sy, + and |Sy, as linear combination of the |Sz, states.

Solution. Analogously to (d) we conjecture

|Sy ; = 12

|Sz ; + i2

|Sz ; . (S.5)

Exercise 4. Hilbert and Lp spaces

Given any real number p [1, ) and a subset X of Rn, let us define the following sets ofcomplex valued functions

Lp(X, ) =

f : X C so thatX

|f|pd < , (4)where is a measure on X.

(a) Prove that L1(X, ) is a vector space.

Solution. We have to show that if f belongs to L1(X, ), then also f belongs to L1(X, ) with C,and that if f and g belong to L

1

(X, ), then f + g belongs to L1

(X, ) as well. Using the definition ofL1(X, ), we write

X

|f|d = ||X

|f|d < when || < X

|f + g|d X

|f| + |g|d < (S.6)

so that f and f + g belong to L1(X, ) and L1(X, ) is a vector space.

(b) Making use of the inequality

(s + t)p 2p1(sp + tp) s, t [0, +) (5)

show that also Lp(X, ) is a vector space for any p > 1.

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Solution. Here the scalar multiplication works as in part a), for the linear combination we need to usethe inequality given, to get:

X

|f + g|pd X

|f| + |g|pd 2p1X

|f|p + |g|pd < (S.7)note that the inequality given is only valid for positive numbers.

For f Lp(X, ) we can introduce the so called p-norm defined by

fp =

X

|f|pd 1p

. (6)

It can be shown to be a norm, i.e. it has the following properties

i) kfp = |k| fp, k C, (positive scalability)ii) f + gp fp + gp, (triangular inequality)

iii) fp = 0 f = 0 almost everywhere. (separate points)

It can further be shown that Lp(X, ) equipped with the p-norm given by equation (6) is com-plete, i.e. it is a Banach space. When the norm in a Banach space can be induced by an innerproduct, the space can be made a Hilbert space.

(c) Let H be a Hilbert space. Show that the parallelogram law is satisfied, i.e.f + g2 + f g2 = 2(f2 + g2), f, g H, (7)

where the norm is induced by the inner product, i.e. f =

f, f.

Solution. We use the definition of norm induced by the scalar product and the linearity property of it,to get:

||f + g||2 + ||f g||2 = f + g, f + g + f g, f g = f, f + f, g + g, f + g, g++ f, f f, g g, f + g, g = 2||f||2 + 2||g||2

(S.8)

(d) Prove that Lp(X, ) can be a Hilbert space only if p = 2.

Hint. With an educated choice of f and g, show that (7) is not valid for p = 2.

Solution. An educated choice ;-) can be to split X into two subsets X1, X2, so that X = X1 X2 andX1 X2 = and with

f(x) =1

p

(X1)

when x X1

f(x) = 0 when x / X1g(x) =

1p

(X2)

when x X2

g(x) = 0 when x / X2

(S.9)

where (X1) is the measure of the set X1. so that:X

|f|pd =X1

|f|pd =X1

1

(X1)d = 1

X |g|

p

d =X2 |g|

p

d =X2

1

(X2) d = 1

(S.10)

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and all integrals taken over the intersection (involving the product of the two functions) are zero as one ofthe two functions is zero there. then we compute the left hand side of the parallelogram law:

||f + g||2 + ||f g||2 =

X

|f + g|pd 2p

+

X

|f g|pd 2p

=

=

X

|f|pd +X

|g|pd 2p

+

X

|f|pd +X

|g|pd 2p

=

= 22p + 2

2p = 2(1+

2p)

(S.11)

while the rhs gives:2(||f||2 + ||g||2) = 2(1 + 1) = 22 (S.12)

comparing the two we see that they are different unless p = 2. As in a vector space in which the norm

is induced by the scalar product the parallelogram law is true (as we proved in c)) then this excludes all

Lp(X, ) spaces with p = 2 from candidates of being Hilbert.

(e) Show that

f, g =X

f g d (8)

is an inner product in L2

(X, ) and deduce the norm induced by it. What do we conclude?

Solution. The properties of the internal product follow from the linearity of the integral. Therefore, wehave an internal product which induces a norm, and we now that space to be Banach with respect to thenorm; so it is Hilbert.some maybe important remarks: the property of the Lp norm, that if the norm off is zero then f is zero, isvalid for f = 0 almost everywhere, as the points in which f is non zero have zero measure in integrationtheory, so they dont count.the parallelogram law is also valid in the converse: if a norm satisfies it, through the polarization identitya scalar product can be defined from which the norm induced satisfies the parallelogram law. Polarizationidentity in complex spaces:

f, g = 14

||f + g||2 ||f g||2 + i||f + ig||2 i||f ig||2 (S.13)

Let us now look at two examples of L2 Hilbert spaces.

(f) Let X = R and dX =1

2

ex2

dx. Furthermore, let us define the three polynomials

H1(x) =

2 , H2(x) = 2x2 1 , F(x) = x2 . (9)

(i) Show that Hi, Hj = ij for i, j {1, 2}.(ii) Calculate F, F and Ci := F, Hi for i {1, 2}.

Hint. dx e

x2

= .Solution.

(i) From dx e

x2 =

, we find (by rescaling the integration variable x) that

dx eax2

=

a. (S.14)

We can then use the well-known derivative-trick to find the result for all even powers of x timesthe measure:

dx x2n eax

2

= (1)n dn

dan

dx eax2

a=1

(S.15)

This yields

/2 for n = 1 and 3

/4 for n = 2, which is enough to prove the orthonormality of

the functions H1 and H2.

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(ii) The squared norm of F is found via the same formulae and is found to be F, F = 3/8. For thecoefficients C1 and C2, we find:

C1 =1

8, C2 =

1

2, C21 + C22 = 3

8= F, F (S.16)

(g) Let X = [1, 1] R and dX = dx, the Lebesgue measure. Furthermore, let us definethe three polynomials

L1(x) =

3

2x , L2(x) =

7

8(5x3 3x) , G(x) = (x 1)3 . (10)

(i) Show that Li, Lj = ij for i, j {1, 2}.(ii) Calculate G, G and Di := G, Li for i {1, 2}.

Solution.

(i) Orthonormality of L1 and L2 can only be shown numerically using a multidimensional adaptive

Monte-Carlo integrator. It seems to be true to five digits, at least thats what I found after runningon the cluster for a day (code will follow...).

(ii) For the squared norm ofG, we find G, G = 02 dy y6 = 1287 . The coefficients D1 and D2 are

D1 =6

6

5, D2 =

2

14

7 D21 + D22 = 304

35

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