LN-27S Stern-Gerlach experiment

1

Chapter 27 Supplement Stern-Gerlach experiment

Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton

(Date: November 19, 2010) In 1921, the Stern-Gerlach experiment demonstrated the quantization of angular momentum . _____________________________________________________________________ Otto Stern (17 February 1888 – 17 August 1969) was a German physicist and Nobel laureate in physics. As an experimental physicist Stern contributed to the discovery of spin quantization in the Stern-Gerlach experiment with Walther Gerlach in 1922, demonstration of the wave nature of atoms and molecules; measurement of atomic magnetic moments; discovery of the proton's magnetic moment; and development of the molecular ray method which is utilized for the technique of molecular beam epitaxy. He was awarded the 1943 Nobel Prize in Physics, the first to be awarded since 1939.

http://en.wikipedia.org/wiki/Otto_Stern ________________________________________________________________________ Walt(h)er Gerlach (1 August 1889 - 10 August 1979) was a German physicist who co-discovered spin quantization in a magnetic field, the Stern-Gerlach effect. In 1920, he became a teaching assistant and lecturer at the Johann Wolfgang Goethe University of Frankfurt am Main. The next year, he took a position as extraordinarius professor at Frankfurt. It was in November 1921 that he and Otto Stern discovered space quantization in a magnetic field, known as the Stern-Gerlach effect.

2

______________________________________________________________________ 27S.1 Stern-Gerlach (SG) experiment

We consider the Stern-Gerlach experiment, which provides a direct evidence of the quantization of magnetic moment and angular momentum. One way of measuring the angular momentum is by means of a Stern-Gerlach experiment. Suppose that we want to measure the angular momentum of the electrons in a given type of atom. A beam of these atoms is prepared by evaporation from the solid, and passing the evaporated atoms through a set of collimating slits. This beam then enters a region in which there is an inhomogeneous magnetic field that is normal to the direction of motion of atoms. The apparatus is shown schematically in Fig. The angular magnetic moment is related to the orbital angular momentum as

Lμmc

eL 2

,

where e>0. In an inhomogeneous magnetic field, we have an interaction energy called the Zeeman energy,

Bμ LV . The atoms experience a force given by

3

)()( BμBμF LLV .

We consider the case when the magnetic field zzB eB is applied along the z axis. Then we have the force along the z axis,

z

BL

z

BL

mc

e

z

BF zz

Bzzz

Lzz

2

where B (= )2/ mce is the Bohr magneton. Thus, each atom experiences a force which is proportional to the z component of the orbital angular momentum. The beam is collected some distance from the magnet at a point that is far enough away so that atoms of different Lz is separated. By measuring the deflection one can calculate Lz.

mlmmlLz ,,ˆ

where m = -l, -l+1,..., l.

The experiment can also be used to reveal the existence of electron spin. For example, if we send a beam of hydrogen atoms in their ground state, the beam split into two parts. Note that the spin magnetic moment is related to the spin angular momentum as

Sμ BS 2

where s is the spin magnetic moment and S (= σ2

) is the spin angular momentum, and

2

ˆ zS ,

2ˆ

zS .

4

http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment Fig. Stern-Gerlach (SG) apparatus. A beam of particles with magnetic moment enters

the inhomogeneous magnetic field. Classically, the beam is expected to fan out and a produce a continuous trace. In fact, the atomic beam is split into two beams, indicating that the magnetic moments of the atoms are quantized to two orientation in space.

_________________________________________________________________ 27S.2 Stern-Gerlach for S = 1/2 with the magnetic field along the z axis

Spin angular momentum is related to the Paili matrices as

xxS 2

ˆ , yyS

2ˆ zzS

2ˆ

The eigenkets of zS are given by

0

1,

1

0

The Pauli matrices are defined as

01

10ˆ x ,

0

0ˆ

i

iy ,

10

01ˆ z

The commutation relations

5

zyx i ˆ2]ˆ,ˆ[ xzy i ˆ2]ˆ,ˆ[ , yxz i ˆ2]ˆ,ˆ[

SGz stands for an apparatus with the inhomogeneous magnetic field along the z direction.

We assume that z

Bz

>0. The atom with µz > 0 (Sz < 0) experiences a downward force,

while the atom with µz < 0 (Sz > 0) experiences a upward force, where the force Fz along the z axis is given by

z

BSF zz

Bz

2 ,

where

Sμ B2

where B is the Bohr magneton, and the magnetic moment is antiparallel to the spin angular momentum S.

The beam is then expected to get slit according to the values of µ (or Sz). In other words, the SG apparatus measures the z-component of µ, or equivalently, the z-componenet of S.

2

ˆ2

ˆ zzS

2

ˆ2

ˆ zzS

where

10

01ˆ z ,

0

1,

1

0

We have a closure relation.

SG-z

+>z

->z

6

1

22

)(ˆˆ zz SS

27S.4 Stern-Gerlach for S = 1/2 with the magnetic field along the x axis

Here we discuss the expression for x

xxx

01

10ˆ x

)(2

1

2

12

1

x

)(2

1

2

12

1

x

((Mathematica))

SG-x

+>x

->x

7

Clear"Global`";

SuperStar : expr_ : expr . Complexa_, b_ Complexa, b;

x 0, 1, 1, 0;

eq1 Eigensystemx1, 1, 1, 1, 1, 1

2 Normalizeeq12, 1 1

2,

1

2


2,

1

2

UT 1, 2 1

2,

1

2, 1

2,

1

2

Unitary operator U

U TransposeUT 1

2,

1

2, 1

2,

1

2

U MatrixForm

12

12

12

12

Hermite conjugate of U

UH UT

1

2,

1

2, 1

2,

1

2

UH MatrixForm

12

12

12

12

UH.U

1, 0, 0, 1

U.UH

1 0 0 1

8

((Summary)) Eigenvalues: ±1

Ux

ˆ

Eigenkets:

0

1ˆ

2

12

1

Ux

,

1

0ˆ

2

12

1

Ux

The Unitary operator:

2

1

2

12

1

2

1

U

((Another method))

2

1

2

1

01

10

c

c

c

c

or

0

0

1

1

2

1

c

c

det 1

1 2 1 0

= ±1.

For = 1,

0

0

11

11

2

1

c

c

9

2

12

1

x

Similarly, For = -1,

1 1

1 1

c1

c2

0

0

2

12

1

x

2

1

2

12

1

2

1

U

______________________________________________________________________ 27S.5 Stern-Gerlach for S = 1/2 with the magnetic field along the y axis

Expression for

y

yyy

0

0ˆ

i

iy

SG-y

+> y

-> y

10

)(2

1

2

2

1

iiy ,

)(2

1

2

12

1

i

y

((Mathematica))

11

Clear"Global`";

SuperStar : expr_ : expr . Complexa_, b_ Complexa, b;

y 0, , , 0;

eq1 Eigensystemy1, 1, , 1, , 1


2,

2


2,

2

UT 1, 2 1

2,

2, 1

2,

2

Unitary operator U

U TransposeUT 1

2,

1

2,

2,

2

U MatrixForm

12

12

2

2

Hermite conjugate of U

UH UT

1

2,

2, 1

2,

2

UH MatrixForm

12

212

2

UH.U

1, 0, 0, 1

U.UH

12

((Summary)) The eigenkets:

0

1ˆ

2

2

1

Uiy ,

1

0ˆ

2

12

1

Uy

,

The Unitary operator:

22

2

1

2

1

ˆii

U

((Another method)) Expression for

y

yyy

0

0ˆ

i

iy

We assume that

2

1

C

Cx

We solve the eigenvalue problem

yyy

where is an eigenvalue.

13

22

11

0

0

C

C

C

C

i

i

or

02

1

C

C

i

i

i

iM

01det 2

i

iM

or

= ±1 For = 1

01

1

2

1

C

C

i

i

or

21 iCC

The normalization condition: 12

2

2

1 CC . We choose 2

121 iCC . Then we have

)(2

1

2

2

1

iiy

Similarly, for = -1

01

1

2

1

C

C

i

i

or

14

21 iCC

The normalization condition: 12

2

2

1 CC . We choose 2

121 iCC . Then we have

)(2

1

2

2

1

i

ii

y

Unitary operator U

22

2

1

2

1

ˆii

U

22

122

1

ˆi

i

U

((Note))

10

01ˆˆˆ UU y

10

01

10

01

10

01ˆˆˆˆˆˆˆˆˆ 2 UUUUUU yyy

10

01

10

01

10

01ˆˆˆ 3UU y

In general

n

nn

y UU)1(0

01ˆˆˆ

15

2/

2/

0

0ˆ)ˆ2

exp(ˆ

i

i

ye

eU

iU

From this we can calculate the matrix of y .

2cos

2sin

2sin

2cos

ˆ0

0ˆ)ˆ2

exp(2/

2/

Ue

eU

ii

i

y

________________________________________________________________________ 27S.6 Stern-Gerlach for S = 1/2 with the magnetic field along the n direction

n (sin cos,sin sin,cos)

cossin

sincosˆˆ

i

i

ne

en

SG-n

+>n

->n

16

((Eigenvalue problem))

Here we use the rotation operator for j = 1/2.

nnn nnn

Unitary operator U is given by the rotation operator for j = 1/2.

)2

cos()2

sin(

)2

sin()2

cos(),(ˆ

22

22

)2/1(

ii

ii

ee

eeDU

The eigenkets

n and

n are obtained as

)2

sin(

)2

cos(ˆ

2

2

i

i

n

e

eU ,

and

x

y

z

n

q

f

17

)2

cos(

)2

sin(ˆ

2

2

i

i

n

e

eU .

We note that

UUnˆˆ UU n

ˆˆˆ

nnn

UUnˆˆ UU n

ˆˆˆ

10

01ˆˆˆ UU n

((Another method)) The above formula can be derived without Mathematica.

(n)

Since (

n)2

1 ,

(n)2 (

n) 2 (eigenvalue problem)

We get 2 = 1, or = ±1. Thus

(n) n n

(n) n n

n

U

U11 U12

U21 U22

1

0

U11

U21

n

U

U11 U12

U21 U22

0

1

U12

U22

where

18

U11* U21

*

U12* U22

*

U11 U12

U21 U22

1 0

0 1

or Derivation of the eigenket n

cos sinei

sinei cos

U11

U21

U11

U21

or

cosU11 sineiU21 U11

sineiU11 cosU21 U21 or

111121 2tan

)cos1(

sinUeU

eU i

i

Since

U11

2 U21

21 (normalization),

we get

2cos22

11

U .

When we choose

2cos2

11

i

eU

we have

2sin2

21

i

eU

Derivation of the eigenket n

19

cos sinei

sinei cos

U12

U22

U12

U22

cosU12 sine iU22 U12

sineiU12 cosU22 U22

U22 (1 cos )ei

sinU12 cot(

2

)eiU12

Since

U12

2 U22

2 1 (normalization),

we get

U22

2 cos2 (

2

).

When we choose

2cos2

22

i

eU

we have

2sin2

12

i

eU

27S.7 Derivation of the eigenkets in each SG configuration from the above formula (i) SGx experiment:

= /2 and = 0.

2

12

1

x,

2

12

1

x

The eigenket of

x thus obtained is different from the conventional eigenket

20

2

12

1

x

except for the phase factor exp(i). (ii) SGy experiment:

= /2 and = /2.

2

2

1

2

12

1

4/

4/

4/

ie

e

ei

i

i

y

,

2

2

1

2

12

1

4/

4/

4/

ie

e

ei

i

i

y

The eigenket of

y is different from the conventional

y except for the phase factor

4/ie . The eigenket of y

is different from the conventional y

except for the phase

factor (- 4/ie ). 27S.8 SG Thinking experiment _______________________________________________________________________ 1. Experiment

______________________________________________________________________ 2. Experiment

______________________________________________________________________ 3. Experiment

SG-z

+>z

->z

SG-z

+>z

SG-z

+>z

->z

SG-x

+>x

->x

21

Analysis of experiment-3

2

12

1

x,

2

12

1

x

2

122

1 xx

P

2

122

2 xx

P

2

12

3 x

P

2

12

4 x

P

0

2

12

1

10

01

2

1

2

1

2

zS

or

0)2

(2 43 PPSz

42

12

1

10

01

2

1

2

1

4

222

zS

or

SG-z+>z

->z

SG-x+>x

->x

SG-z+>z

->z

22

444

2

4

2

3

22

PPSz

4

2222 zzz SSS

________________________________________________________________________ 4. Experiment

_______________________________________________________________________ 5. Experiment

_______________________________________________________________________ 6. Experiment

Analysis of experiment-6

2sin

2cos

n,

2

12

1

x,

2

12

1

x

SG-x

+>x

->x

SG-y

+> y

-> y

SG-n

f=0

+>n

->n

SG-z

+>z

->z

SG-n

f=0

+>n

->n

SG-x

+>x

->x

23

)]42

[cos(2

1)

2sin

2(cos

2

1

2sin

2cos

2

1

2

1

nx

)]42

[cos(2

1)

2sin

2(cos

2

1

2sin

2cos

2

1

2

1

nx

)sin1(4

1)

2sin

2(cos

4

1 22

nxxP ,

),sin1(4

1)

2sin

2(cos

4

1 22

nxxP

sin22

cos2

sin

2sin

2cos

01

10

2sin

2cos

2ˆ

nxnx SS

or

sin2

)2

(2

xxx PPS

cos2

)2

sin2

(cos2

2sin

2cos

10

01

2sin

2cos

2ˆ

22

nznz SS

________________________________________________________________________ 27S.9 Stern-Gerlach experiment with J = 1

24

2

cos1

2

sin

2

cos12

sincos

2

sin2

cos1

2

sin

2

cos1

)(ˆ

i

y

e

R

Using the Mathematica, one can get the matrix representation

cos2

sin0

2

sin0

2

sin

02

sincos

ˆˆˆˆˆ

ii

iiii

ii

zzyyxxn

e

ee

e

JJJJJ nnnn

The above result can be obtained by solving the eigenvalue problem:

nnJ 11)ˆ( n , 00)ˆ(

nJ n ,

nnJ 11)ˆ( n

Use the Mathematica to obtain the eigenkets and the eigenvalues: Eigensystem[Jn]

0

0

1

1 ,

0

1

0

0 ,

1

0

0

1

For = π/2 and = 0 (corresponding to the x axis)

12

10

2

11

2

1

2/1

2/1

2/1

1)2

(ˆ1

yx

R ,

12

11

2

1

2/1

0

2/1

0)2

(ˆ0

yx

R ,

25

12

10

2

11

2

1

2/1

2/1

2/1

1)2

(ˆ1

yx

R

More generally for the unit vector n in the x-z plane,

12

cos10

2

sin1

2

cos11

n

,

12

sin0cos1

2

sin0

n

,

12

cos10

2

sin1

2

cos11

n

,

or, inversely

2

cos1

2

sin

2

cos12

sincos

2

sin2

cos1

2

sin

2

cos1

2

cos1

2

sin

2

cos12

sincos

2

sin2

cos1

2

sin

2

cos11

When = /2, this matrix is expressed by

2

1

2

1

2

12

10

2

12

1

2

1

2

1

or

xxx1

2

10

2

11

2

11

xx1

2

11

2

10

26

xxx1

2

10

2

11

2

11

more generally

nnn1

2

cos10

2

sin1

2

cos11

nnn1

2

sin0cos1

2

sin0

nnn1

2

cos10

2

sin1

2

cos11

For = π/2 and = π/2 (corresponding to the y axis)

2/1

2/

2/1

2/

2/1

2/

1ˆ ii

i

i

R ,

2/1

0

2/1

2/

0

2/

0ˆ i

i

i

R ,

2/1

2/

2/1

2/

2/1

2/

1ˆ ii

i

i

R .

Conventionally we use

2/1

2/

2/1

1 iy

,

2/1

0

2/1

0y

,

27

2/1

2/

2/1

1 iy

or inversely,

yyy1

2

10

2

11

2

11

yy

ii1

21

20

yyy1

2

10

2

11

2

11

______________________________________________________________________ Calculation of the rotation matrix with J = 1 without the use of Mathematica Taylor expansion:

..)ˆ(!4

1)ˆ(

!3

1)ˆ(

!2

1)ˆ(

!1

11)ˆexp( 432 yyyyy J

iJ

iJ

iJ

iJ

i

where

i

JJJ y 2

ˆˆˆ

Note that

01ˆ J , 120ˆ J , 021ˆ J

021ˆ J , 120ˆ J , 01ˆ J

02

1ˆ iJ y , )11(

20ˆ

iJ y , 0

21ˆ i

J y

28

02

0

20

2

02

0

ˆ

i

ii

i

J y ,

101

020

101ˆ 22

yJ

ˆ J y

3 2 ˆ J y, ˆ J y

4 ˆ J y3 ˆ J y

2 ˆ J yˆ J y

2 ˆ J y2 ,

yyyyyyy JJJJJJJ ˆˆˆˆˆˆˆ 4322245

Therefore

2

cos1

2

sin

2

cos12

sincos

2

sin2

cos1

2

sin

2

cos1

)1(cosˆ

)sin(ˆ

1

..])(!4

1)(

!2

1[

ˆ

..])(!5

1)(

!3

1)[(

ˆ1)ˆexp(

2

2

422

2

53

yy

y

yy

Ji

J

iiJ

iiJ

Ji

We also get

exp(i

ˆ J z )

ei 0 0

0 1 0

0 0 ei

which is a diagonal matrix. 27S.10 Examples of SG experiments ______________________________________________________________________ 1. A spin-1 particle exists an SGy device in a state with yS . The beam then

enters an SGz device. What is the probability that the measurement of Sz yields the value 0, + , and - ?

29

12

10

21

2

1

2/1

2/

2/1

1

ii

y

The probability for finding the state 1 is

4

11111

22

1 yy

P


2

11001

22

1 yy

P


4

11111

22

1 yyP

________________________________________________________________________ 2. ((Townsend 3.16)) A spin-1 particle exists an SGz device in a state with

Sz . The beam then enters an SGx device. What is the probability that the measurement of Sx yields the value 0, + , and - ?

SG-y

+1> y

0> y

-1> y

SG-z

+1>z

0>z

-1>z

30

12

11

2

10

x

The probability for finding the state

x1 is

2

10110

22

1 xx

P


x0 is

0000022

1 xx

P


x1 is

2

10110

22

1 xxP

_______________________________________________________________________ 3. ((Shankhar)) A beam of spin 1 particles, moving along the y axis, is incident

on two collinear SG apparatuses, the first with B along the z axis and the second with B along the z' axis, which lies in the x-z plane at an angle relative to the z axis. Both apparatuses transmit only the uppermost beams. What fraction leaving the first will pass the second?

The intial state after passing the first SGz, is 11

z

SG-z

+1>z

0>z

-1>z

SG-x

+1>x

0>x

-1>x

31

We note that

12

cos10

2

sin1

2

cos11

n

,

12

sin0cos1

2

sin0

n

,

12

cos10

2

sin1

2

cos11

n

,


n1 is

222

1 )cos1(4

11111

nnP .


n0 is

222

0 sin2

10110

nnP


n1 is

222

1 )cos1(4

11111 nn

P

The total probability is

1101 PPP

________________________________________________________________________ 4. ((Townsend 3.20)) A beam of spin-1 particle is sent through a series of three

Stern-Gerlach measuring devices. The first SGz device transmits particles

with Sz = and filters out particles with Sz = 0 and Sz = -. The second

SG-z

+1>z

0>z

-1>z

SG-n

f=0

+1>n

0>n

-1>n

32

device, an SGn device, transmits particles with Sn = and filters out particles

with Sn = 0 and Sn = -, where the axis n makes an angle (0≤≤π/2) in the x-z plane with respect to the z axis. A last SGz device transmits particles with

Sz = 0 and filters out particles with Sz = and Sz = -.

(a) What fraction of the particles transmitted by the first SGz device will survive the third measurement?

(b) How must the angle of the SGn device be oriented so as to maximize the number of particles that are transmitted by the final SGz device? What fraction of the particles survive the third measurements for this value of ?

(c) What fractions of the particles with Sx = , Sx =0, and Sx = -, respectively, survive after the fourth device, SGx which device

transmits particles with Sx = , Sx =0, and Sx = -?

We note that

12

cos10

2

sin1

2

cos11

n

,

12

sin0cos1

2

sin0

n

,

12

cos10

2

sin1

2

cos11

n

,

SG-z

+1>z

0>z

-1>z

SG-nf=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

SG-z+1>z0>z-1>z

SG-nf=0+1>n0>n-1>n

SG-z+1>z0>z-1>z

SG-x+1>x0>x-1>x

33

12

10

2

11

2

11

x,

12

11

2

10

x,

12

10

2

11

2

11

x

(a) The fraction of the particles transmitted by the first SGz device will survive the third measurement is

8

)cos1(sin100111

22222

0

nnn

P

(b)

)1cos2(2

sin2

cos2 50 d

dP.

P0 takes a maximum (= 0.210938) at = 60 (c)

20 40 60 80q Degrees

0.05

0.10

0.15

0.20

P0

34

The fractions of the particles with Sx = ,

210 0

2

0

PP

x

The fractions of the particles with Sx =0,

0002

0 x

P

The fractions of the particles with Sx = -,

210 0

2

0

PP

x

________________________________________________________________________ 5.

A spin-1 particle is in the state

)13021(14

1

3

2

1

14

1

i

i

(a) What are the probabilities that a measurement of Sz will yield the values , 0, or -

for this state? (b) What is <Sz>?

(c) What is the probability that a measurement of Sx will yield the values , 0, or - for this state?

(d) What is <Sx> for this state?

12

10

2

11

2

1

2/1

2/1

2/1

1

x,

12

11

2

1

2/1

0

2/1

0

x,

35

12

10

2

11

2

1

2.1

2/1

2/1

1

x

(a) and (b)

14

11)1(

2 P

7

2

14

40)0(

2 P

14

91)1(

2 P

57143.07

4

14

8

14

9

14

)1()0(0)1(

PPPSz

(c) and (d)

28

2291)1(

2 xx

P

14

50)0(

2 xx

P

SG-z

+1>z

0>z

-1>z

SG-x

+1>x

0>x

-1>x

36

28

2291)1(

2 xx

P

20231.07

2

)]1()1([

)1()0(0)1(

xx

xxxx

PP

PPPS

27S.11 Feynman's thinking SG experiment Richard Phillips Feynman (pronounced /ˈfaɪnmən/, May 11, 1918 – February 15, 1988) was an American physicist known for his work in the path integral formulation of quantum mechanics, the theory of quantum electrodynamics and the physics of the superfluidity of supercooled liquid helium, as well as in particle physics (he proposed the parton model). For his contributions to the development of quantum electrodynamics, Feynman, jointly with Julian Schwinger and Sin-Itiro Tomonaga, received the Nobel Prize in Physics in 1965. He developed a widely used pictorial representation scheme for the mathematical expressions governing the behavior of subatomic particles, which later became known as Feynman diagrams. During his lifetime, Feynman became one of the best-known scientists in the world. http://en.wikipedia.org/wiki/Richard_Feynman

http://www.atomicarchive.com/Bios/FeynmanPhoto.shtml 1. Experiment-1

37

2. Experiment-2

Two Stern-Gerlach type filters in series; the second is tilted at the angle from the z axis in the x-z plane. 3. Experiment-3

4. Experiment 4

SG-z

+1>z

0>z

-1>z

SG-n

f=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

SG-n

f=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

SG-nf=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

SG-z

+1>z

0>z

-1>z

SG-nf=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

38

The probability that an atom that comes out of SGz (the first) will also go through both SGn and SGz (the second) is

22

4 0001zzP nn

5. Experiment-5

The probability that an atom that comes out of SGz (the first) will also go through both SGn and SGz (the second) is

22

5 0000zzP nn

Then we have

22

2

2

2

5

4 tan2

1

00

01

00

01

n

n

n

n

z

z

P

P

This ratio does not depend on which state is selected by the first SGz. 6. Experiment-6

_______________________________________________________________________ REFERENCES 1. R.P. Feynman, R.,B. Leighton, and M. Sands, The Feynman Lectures in Physics,

6th edition (Addison Wesley, Reading Massachusetts, 1977).

SG-z

+1>z

0>z

-1>z

SG-nf=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

SG-z

+1>z

0>z

-1>z

SG-nf=0

+1>n

0>n

-1>n

SG-z

+1>z

0>z

-1>z

39

2. J.J. Sakurai, Modern Quantum Mechanics, Revised Edition (Addison-Wesley, Reading Massachusetts, 1994).

3. J.S. Townsend, A Modern Approach to Quantum Mechanics (McGraw-Hill, Inc., New York, 1992).

4. B. Friedrich and D. Herschbach, Phys. Today December 2003, p.53. "Stern and Gerlach: How a bad cigar helped reorient atomic physics."

LN-27S Stern-Gerlach experiment

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