Today’s Outline - October 26, 2017csrri.iit.edu/~segre/phys405/17F/lecture_17.pdf · Today’s Outline - October 26, 2017 ... Reading Assignment: Chapter 4.4 Homework Assignment

Today’s Outline - October 26, 2017

• The Stern-Gerlach experiment

• Total spin of hydrogen

• Adding angular momentum

• More chapter 4 problems

Reading Assignment: Chapter 4.4

Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017


Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4

C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17

















































































Angular momentum in a non-uniform B-field

An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque

~F = ∇(~µ · ~B)

If a beam of heavy neutral atoms traveling in the y direction passesthrough a field

~B = −αxx + (B0 + αz)z

it will experience a force

~F = γα(−Sx x + Sz z)

since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz

Suppose we use a beam of atoms with spin 12 . What do we expect to see

in an experiment?




~F = ∇(~µ · ~B)


~B = −αxx + (B0 + αz)z





in an experiment?




~F = ∇(~µ · ~B)


~B = −αxx + (B0 + αz)z





in an experiment?




~F = ∇(~µ · ~B)


~B = −αxx + (B0 + αz)z





in an experiment?




~F = ∇(~µ · ~B)


~B = −αxx + (B0 + αz)z





in an experiment?




~F = ∇(~µ · ~B)


~B = −αxx + (B0 + αz)z





in an experiment?


The Stern-Gerlach experiment (SGE)

In 1922, Stern and Gerlach performed an experiment designed to test theBohr-Sommerfeld theory of the atom which predicted that the electronorbits had a property called “space quantization” in a magnetic field

In the classical regime we would ex-pect that the relative orientation of thespin with respect to the magnetic fieldwould determine where the atom is de-tected spatially.

Because of the Bohr quantized circularorbits, the direction of the orbital mo-tion with respect to the magnetic field,would cause the atoms to be deflectedin opposite directions

N

S






N

S






N

S


The SGE result

When the experiment was performed with neutral silver atoms two distinctspots on the screen were seen.

Gerlach sent the above picture on a postcard to announce this result.


The SGE result


Gerlach sent the above picture on a postcard to announce this result.


The SGE result


Gerlach sent the above picture on a postcard to announce this result.C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 4 / 17

Reaction to the SGE

Arnold Sommerfeld – Through clever experimental arrangement Stern andGerlach not only demostrated ad oculus the space quantization of atomsin a magnetic field, but they also proved the quantum origin of electricityand its connection with atomic structure

Albert Einstein – The most interesting achievement at this point is theexperiment of Stern and Gerlach. The alignment of the atoms withoutcollisions via radiative exchange is not comprehensible based on currenttheoretical methods.

James Franck – More important is whether this proves the existence ofspace quantization. Please add a few words of explanation to your puzzle,such as what’s really going on.

Niels Bohr – I would be very grateful if you or Stern could let me know, ina few lines, whether you interpret your experimental results in this waythat the atoms are oriented only parallel or opposed, but not normal to thefield, as one could provide theoretical reasons for the latter assertion.


Reaction to the SGE






Reaction to the SGE






Reaction to the SGE






Interesting SGE facts

Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.

The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1

2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum

The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0

The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1

2spin wavefunction.





2?

As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum



2spin wavefunction.








2spin wavefunction.






The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925.

Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0


2spin wavefunction.








2spin wavefunction.








2spin wavefunction.


Space quantization

Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is

start with a beam of mixed spin12 with a wave function

H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2

χ(t) = aχ+e−iE+t/~ + bχ−e

−iE−t/~

χ(T ) =(ae iγTB0/2χ+

)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2

There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field


Space quantization

Consider the frame of the par-ticles in the beam.

At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is


H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization

Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T .

Ignor-ing the x-component, the timedependent Hamiltonian is


H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T )

= |a|2αγT~2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T ) = |a|2αγT~

2− |b|2αγT~

2



Space quantization



H(t) =

0, t < 0

−γ(B0 + αz)Sz , 0 ≤ t ≤ T

0, t > T

χ(0) = aχ+ + bχ−

E± = ∓γ(B0 + αz)~2


−iE−t/~


)e i(αγT/2)z +

(be−iγTB0/2χ−

)e−i(αγT/2)z

〈pz〉 = χ(T )∗~i

∂

∂zχ(T ) = |a|2αγT~

2− |b|2αγT~

2



Total spin of the hydrogen atom

Suppose we have a hydrogen atom in its ground state (L = 0).

Theelectron, and the proton, each have spin S = 1

2. What is the total angular

momentum of the atom?

Each particle can have spin up or down, for 4 combinations.

↑↑ ↑↓ ↓↑ ↓↓

The total angular momentum is just the sum

~S ≡ ~S (1) + ~S (2)

each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum

Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2



Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1

2.

What is the total angularmomentum of the atom?


↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑

↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓

↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑

↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2

= (S(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2)

= ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2


Triplet state

We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1.

Let’s check this

S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state

We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this

S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2))

= (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑

=~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)

=~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)

=~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑

=~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)

= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)

= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑

=~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)

=~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)

=~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ =

~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑

+ ~2 12(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑

+ 2(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

= (S (1))2 + (S (2))2 + 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~2 12(32

)↑↑+ ~2 12

(32

)↑↑+ 2

(~24 ↓↓ −

~24 ↓↓+ ~2

4 ↑↑)

= 2~2↑↑


Triplet state (cont.)

Use the lowering operator to generate the state with total s = 1 and m = 0

S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)

Apply the lowering operator again to get s = 1, m = −1

S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓

The triplet state is thus

|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑

= (S(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓)

= ~(↓↑+ ↑↓) =√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)

=√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]

=1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓


Singlet state

Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be

|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑)

= (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑)

= 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

= S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑)

= 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑)

= 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state


|0 0〉 = 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

S (1)S (1) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = 1√2

(14~

2↓↑ − 14~

2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = 1√2

(−i2 14~

2↓↑+ i2 14~2↑↓)

= −14~

2 1√2

(↑↓ − ↓↑)

S(1)z S

(2)z

1√2

(↑↓ − ↓↑) = 1√2

(−1

4~2↑↓+ 1

4~2↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[

34~

2 + 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2)

]1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[

34~

2 + 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2)

]1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[

34~

2 + 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2)

]1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[34~

2

+ 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2)

]1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[34~

2 + 34~

2

+ 2(−1

4~2 − 1

4~2 − 1

4~2)

]1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[34~

2 + 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2) ]

1√2

(↑↓ − ↓↑)

= 0


Singlet state

S (1)S (1) 1√2

(↑↓ − ↓↑) = S (2)S (2) 1√2

(↑↓ − ↓↑) = 34~

2 1√2

(↑↓ − ↓↑)

S(1)x S

(2)x

1√2

(↑↓ − ↓↑) = S(1)y S

(2)y

1√2

(↑↓ − ↓↑) = S(1)z S

(2)z

1√2

(↑↓ − ↓↑)

= −14~

2 1√2

(↑↓ − ↓↑)

S2 1√2

(↑↓ − ↓↑) = (S (1))2 1√2

(↑↓ − ↓↑) + (S (2))2 1√2

(↑↓ − ↓↑)

+ 2(S(1)x S

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z ) 1√

2(↑↓ − ↓↑)

=[34~

2 + 34~

2 + 2(−1

4~2 − 1

4~2 − 1

4~2) ]

1√2

(↑↓ − ↓↑)

= 0


Orthonormality

The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.

〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0

What about the singlet state?

〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0

We have recovered all four orthonormal states.


Orthonormality


〈1 1|1 −1〉 ≡ 0

〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0

〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉

= (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)†

1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑)

= 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉

= (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)†

1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑)

= 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉

= 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉 = 1√2

(↑↓+ ↓↑)†

1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉 = 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑)

= 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉 = 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1)

= 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉 = 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Orthonormality


〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0


〈1 1|0 0〉 = (↑↑)† 1√2

(↑↓ − ↓↑) = 0

〈1 −1|0 0〉 = (↓↓)† 1√2

(↑↓ − ↓↑) = 0

〈1 0|0 0〉 = 1√2

(↑↓+ ↓↑)† 1√2

(↑↓ − ↓↑) = 12(1− 1) = 0



Eigenvalues of Sz

We can also check that these are eigenfunctions of the Sz operator

Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑

= ~(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑

= +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓

= ~(−1

2 −12

)↓↓ = −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓

= −1~↓↓


Eigenvalues of Sz


Sz |0 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓ − ↓↑)

= 1√2

(S(1)z ↑↓+ S

(2)z ↑↓ − S

(1)z ↓↑ − S

(2)z ↓↑)

= 1√2~(12↑↓ −

12↑↓+ 1

2↓↑ −12↓↑)

= 0

Sz |1 1〉 = (S(1)z + S

(2)z )↑↑ = ~

(12 + 1

2

)↑↑ = +1~↑↑

Sz |1 0〉 = (S(1)z + S

(2)z ) 1√

2(↑↓+ ↓↑)

= 1√2~(12↑↓ −

12↑↓ −

12↓↑+ 1

2↓↑)

= 0

Sz |1−1〉 = (S(1)z + S

(2)z )↓↓ = ~

(−1

2 −12

)↓↓ = −1~↓↓


“Good” quantum numbers

|0 0〉 = 1√2

(↑↓ − ↓↑)

|1 1〉 = ↑↑, |1 0〉 = 1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓

Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,

S(1)z and S

(2)z

Thus the valid quantum numbers for these states are s, m, s, s – a generalresult



|0 0〉 = 1√2

(↑↓ − ↓↑)

|1 1〉 = ↑↑, |1 0〉 = 1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓

Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2

but not of their z-components,

S(1)z and S

(2)z




|0 0〉 = 1√2

(↑↓ − ↓↑)

|1 1〉 = ↑↑, |1 0〉 = 1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓


S(1)z and S

(2)z




|0 0〉 = 1√2

(↑↓ − ↓↑)

|1 1〉 = ↑↑, |1 0〉 = 1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓


S(1)z and S

(2)z



Adding angular momentum

Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources

For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps

s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|

the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states

|s m〉 =∑

m1+m2=m

C s1s2m1m2m|s1 m1〉|s2 m2〉

The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m





s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|


|s m〉 =∑

m1+m2=m







s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|


|s m〉 =∑

m1+m2=m







s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|


|s m〉 =∑

m1+m2=m







s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|


|s m〉 =∑

m1+m2=m







s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|


|s m〉 =∑

m1+m2=m




Clebsch-Gordan coefficients

consider the 2× 1 table of Clebsch-Gordan coefficients

|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 >

=√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 >

+√

35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 >

+√

15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 >

=√

35 |3 0 > −

√25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 >

−√

25 |1 0 >




|3 0 > =√

15 |2 1 > |1−1 > +

√35 |2 0 > |1 0 > +

√15 |2−1 > |1 1 >

|2 0 > |1 0 > =√

35 |3 0 > −

√25 |1 0 >


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Today’s Outline - October 26, 2017csrri.iit.edu/~segre/phys405/17F/lecture_17.pdf · Today’s Outline - October 26, 2017 ... Reading Assignment: Chapter 4.4 Homework Assignment