Spin One-Half and Pauli Spin Matrices | Physics Optional Notes for UPSC PDF Download

 Page 1


SPIN ONE-HALF AND THE PAULI SPIN MATRICES
In our introduction to spin, we saw that spin is an intrinsic, ?xed quantity
for each elementary particle. It doesn’t really have a classical analogue, as
it is not the result of any motion of the particle. Despite that, spin really is
a form of angular momentum, as it combines with the orbital angular mo-
mentumL to form a total angular momentum for a given particle. It seems
that elementary particles (at least those with non-zero spin) just have an in-
strinsic angular momentum that is ?xed, depending on the type of particle.
The lack of analogy with classical angular momentum extends to the fact
that spin does not depend on spatial coordinates. As such, the eigenfunc-
tions of the spin operators are not functions of position, so we don’t need
to worry about Hilbert space, or their behaviour at in?nity or any of those
things that make the other quantum operators so tricky (and interesting, it
has to be said) to deal with.
We’ve seen that the eigenvalues of the square of the total spin S
2
are
¯ h
2
s(s+ 1), wheres is any non-negative integer or half-integer. Ifs= 0, no
spin is present and there isn’t anything more to say. The simplest non-trivial
case is therefores=
1
2
. In this case, the eigenvalues ofs
z
are
1
2
, so there
are only 2 possible states. Since the spin of a particle is ?xed, a particle
withs=
1
2
can existonly in a linear combination of these 2 states, no matter
how much you poke it or excite it by passing electric ?elds through it or do
anything else to it.
How does this affect the Schrödinger equation? After all, it involves
derivatives with respect to both time and position, so if we’re dealing with a
system that makes no use of position, does that mean that the¯ h
2
Ñ
2
Y=2m
is zero? If that were the case then the time derivative would also be zero, and
it would seem to indicate that any particle’s spin state would never change.
This isn’t the case, and the reason is that we need to step back a bit in our
interpretation of the Schrödinger equation. The term¯ h
2
Ñ
2
Y=2m is based
on our taking the energy of the system as p
2
=2m and then translating the
momentum into its quantum operator form. In the case of spin, we need to
write the energy (or more precisely, the Hamiltonian) in terms of the spin,
so the spatial derivative is replaced by a term involving the spin variables,
1
Page 2


SPIN ONE-HALF AND THE PAULI SPIN MATRICES
In our introduction to spin, we saw that spin is an intrinsic, ?xed quantity
for each elementary particle. It doesn’t really have a classical analogue, as
it is not the result of any motion of the particle. Despite that, spin really is
a form of angular momentum, as it combines with the orbital angular mo-
mentumL to form a total angular momentum for a given particle. It seems
that elementary particles (at least those with non-zero spin) just have an in-
strinsic angular momentum that is ?xed, depending on the type of particle.
The lack of analogy with classical angular momentum extends to the fact
that spin does not depend on spatial coordinates. As such, the eigenfunc-
tions of the spin operators are not functions of position, so we don’t need
to worry about Hilbert space, or their behaviour at in?nity or any of those
things that make the other quantum operators so tricky (and interesting, it
has to be said) to deal with.
We’ve seen that the eigenvalues of the square of the total spin S
2
are
¯ h
2
s(s+ 1), wheres is any non-negative integer or half-integer. Ifs= 0, no
spin is present and there isn’t anything more to say. The simplest non-trivial
case is therefores=
1
2
. In this case, the eigenvalues ofs
z
are
1
2
, so there
are only 2 possible states. Since the spin of a particle is ?xed, a particle
withs=
1
2
can existonly in a linear combination of these 2 states, no matter
how much you poke it or excite it by passing electric ?elds through it or do
anything else to it.
How does this affect the Schrödinger equation? After all, it involves
derivatives with respect to both time and position, so if we’re dealing with a
system that makes no use of position, does that mean that the¯ h
2
Ñ
2
Y=2m
is zero? If that were the case then the time derivative would also be zero, and
it would seem to indicate that any particle’s spin state would never change.
This isn’t the case, and the reason is that we need to step back a bit in our
interpretation of the Schrödinger equation. The term¯ h
2
Ñ
2
Y=2m is based
on our taking the energy of the system as p
2
=2m and then translating the
momentum into its quantum operator form. In the case of spin, we need to
write the energy (or more precisely, the Hamiltonian) in terms of the spin,
so the spatial derivative is replaced by a term involving the spin variables,
1
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 2
while the time derivative remains as it is. We’ll see some examples of this
in due course; for now, just be assured that the Schrödinger equation does
still work in systems involving spin.
Returning to spin
1
2
, since we are allowed only 2 states, we can use a 2-
dimensional vector to represent the state of a particle. These vectors form a
vector space, and we can take the basis of this space to be the two stationary
states, commonly called spin up and spin down. We represent these two
stationary states as

+
=

1
0

; 

=

0
1

(1)
We could, of course, have chosen any two independent vectors as the
basis, but these two make life simpler.
With this de?nition, a general spin state of a spin
1
2
particle is a linear
combination of these two states:
=a
+
+b

=

a
b

(2)
Since the state of a particle is now a 2-dimensional vector, any opera-
tor that operates on such a vector must be a matrix. Since we want the
spin operators to have eigenvalues, these matrixes must be 2 2. We can
work out the S
2
matrix since we know that its eigenvalues must both be
¯ h
2
s(s+ 1)= 3¯ h
2
=4 from above. By specifying the basis above, we have
effectively speci?ed the eigenvectors ofS
2
, so we get
S
2

+
=

s
11
s
12
s
21
s
22

1
0

=
3¯ h
2
4

1
0

(3)
from which we get
s
11
=
3¯ h
2
4
(4)
s
21
= 0 (5)
with the other two elements undetermined, so far.
Applying the same logic to the other eigenvector gives us
s
12
= 0 (6)
s
22
=
3¯ h
2
4
(7)
Thus
Page 3


SPIN ONE-HALF AND THE PAULI SPIN MATRICES
In our introduction to spin, we saw that spin is an intrinsic, ?xed quantity
for each elementary particle. It doesn’t really have a classical analogue, as
it is not the result of any motion of the particle. Despite that, spin really is
a form of angular momentum, as it combines with the orbital angular mo-
mentumL to form a total angular momentum for a given particle. It seems
that elementary particles (at least those with non-zero spin) just have an in-
strinsic angular momentum that is ?xed, depending on the type of particle.
The lack of analogy with classical angular momentum extends to the fact
that spin does not depend on spatial coordinates. As such, the eigenfunc-
tions of the spin operators are not functions of position, so we don’t need
to worry about Hilbert space, or their behaviour at in?nity or any of those
things that make the other quantum operators so tricky (and interesting, it
has to be said) to deal with.
We’ve seen that the eigenvalues of the square of the total spin S
2
are
¯ h
2
s(s+ 1), wheres is any non-negative integer or half-integer. Ifs= 0, no
spin is present and there isn’t anything more to say. The simplest non-trivial
case is therefores=
1
2
. In this case, the eigenvalues ofs
z
are
1
2
, so there
are only 2 possible states. Since the spin of a particle is ?xed, a particle
withs=
1
2
can existonly in a linear combination of these 2 states, no matter
how much you poke it or excite it by passing electric ?elds through it or do
anything else to it.
How does this affect the Schrödinger equation? After all, it involves
derivatives with respect to both time and position, so if we’re dealing with a
system that makes no use of position, does that mean that the¯ h
2
Ñ
2
Y=2m
is zero? If that were the case then the time derivative would also be zero, and
it would seem to indicate that any particle’s spin state would never change.
This isn’t the case, and the reason is that we need to step back a bit in our
interpretation of the Schrödinger equation. The term¯ h
2
Ñ
2
Y=2m is based
on our taking the energy of the system as p
2
=2m and then translating the
momentum into its quantum operator form. In the case of spin, we need to
write the energy (or more precisely, the Hamiltonian) in terms of the spin,
so the spatial derivative is replaced by a term involving the spin variables,
1
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 2
while the time derivative remains as it is. We’ll see some examples of this
in due course; for now, just be assured that the Schrödinger equation does
still work in systems involving spin.
Returning to spin
1
2
, since we are allowed only 2 states, we can use a 2-
dimensional vector to represent the state of a particle. These vectors form a
vector space, and we can take the basis of this space to be the two stationary
states, commonly called spin up and spin down. We represent these two
stationary states as

+
=

1
0

; 

=

0
1

(1)
We could, of course, have chosen any two independent vectors as the
basis, but these two make life simpler.
With this de?nition, a general spin state of a spin
1
2
particle is a linear
combination of these two states:
=a
+
+b

=

a
b

(2)
Since the state of a particle is now a 2-dimensional vector, any opera-
tor that operates on such a vector must be a matrix. Since we want the
spin operators to have eigenvalues, these matrixes must be 2 2. We can
work out the S
2
matrix since we know that its eigenvalues must both be
¯ h
2
s(s+ 1)= 3¯ h
2
=4 from above. By specifying the basis above, we have
effectively speci?ed the eigenvectors ofS
2
, so we get
S
2

+
=

s
11
s
12
s
21
s
22

1
0

=
3¯ h
2
4

1
0

(3)
from which we get
s
11
=
3¯ h
2
4
(4)
s
21
= 0 (5)
with the other two elements undetermined, so far.
Applying the same logic to the other eigenvector gives us
s
12
= 0 (6)
s
22
=
3¯ h
2
4
(7)
Thus
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 3
S
2
=
3¯ h
2
4

1 0
0 1

(8)
Incidentally, since the eigenvalues ofS
2
are degenerate (both the same),
then any linear combination of the two eigenvectors that we’ve used above
is also an eigenvector, which is just another way of saying that we could
have chosen any two independent vectors to represent the two stationary
states.
We can do a similar calculation forS
z
, since we know that
S
z

+
=
¯ h
2

+
(9)
S
z


=
¯ h
2


(10)
This gives
S
z
=
¯ h
2

1 0
0 1

(11)
How aboutS
x
andS
y
? Since the raising and lowering operators that we
derived for orbital angular momentum depended only on the commutators,
we can write similar de?nitions for spin. In particular
S
+
=S
x
+iS
y
(12)
S

=S
x
iS
y
(13)
The action of these operators on a stationary state is also the same as for
L:
S

jsm
s
i= ¯ h
p
s(s+ 1)m(m 1)jsm
s
 1i (14)
We can invert the relations above and get
S
x
=
1
2
(S
+
+S

) (15)
S
y
=
1
2i
(S
+
S

) (16)
Then we can apply these operators to the stationary states to get
Page 4


SPIN ONE-HALF AND THE PAULI SPIN MATRICES
In our introduction to spin, we saw that spin is an intrinsic, ?xed quantity
for each elementary particle. It doesn’t really have a classical analogue, as
it is not the result of any motion of the particle. Despite that, spin really is
a form of angular momentum, as it combines with the orbital angular mo-
mentumL to form a total angular momentum for a given particle. It seems
that elementary particles (at least those with non-zero spin) just have an in-
strinsic angular momentum that is ?xed, depending on the type of particle.
The lack of analogy with classical angular momentum extends to the fact
that spin does not depend on spatial coordinates. As such, the eigenfunc-
tions of the spin operators are not functions of position, so we don’t need
to worry about Hilbert space, or their behaviour at in?nity or any of those
things that make the other quantum operators so tricky (and interesting, it
has to be said) to deal with.
We’ve seen that the eigenvalues of the square of the total spin S
2
are
¯ h
2
s(s+ 1), wheres is any non-negative integer or half-integer. Ifs= 0, no
spin is present and there isn’t anything more to say. The simplest non-trivial
case is therefores=
1
2
. In this case, the eigenvalues ofs
z
are
1
2
, so there
are only 2 possible states. Since the spin of a particle is ?xed, a particle
withs=
1
2
can existonly in a linear combination of these 2 states, no matter
how much you poke it or excite it by passing electric ?elds through it or do
anything else to it.
How does this affect the Schrödinger equation? After all, it involves
derivatives with respect to both time and position, so if we’re dealing with a
system that makes no use of position, does that mean that the¯ h
2
Ñ
2
Y=2m
is zero? If that were the case then the time derivative would also be zero, and
it would seem to indicate that any particle’s spin state would never change.
This isn’t the case, and the reason is that we need to step back a bit in our
interpretation of the Schrödinger equation. The term¯ h
2
Ñ
2
Y=2m is based
on our taking the energy of the system as p
2
=2m and then translating the
momentum into its quantum operator form. In the case of spin, we need to
write the energy (or more precisely, the Hamiltonian) in terms of the spin,
so the spatial derivative is replaced by a term involving the spin variables,
1
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 2
while the time derivative remains as it is. We’ll see some examples of this
in due course; for now, just be assured that the Schrödinger equation does
still work in systems involving spin.
Returning to spin
1
2
, since we are allowed only 2 states, we can use a 2-
dimensional vector to represent the state of a particle. These vectors form a
vector space, and we can take the basis of this space to be the two stationary
states, commonly called spin up and spin down. We represent these two
stationary states as

+
=

1
0

; 

=

0
1

(1)
We could, of course, have chosen any two independent vectors as the
basis, but these two make life simpler.
With this de?nition, a general spin state of a spin
1
2
particle is a linear
combination of these two states:
=a
+
+b

=

a
b

(2)
Since the state of a particle is now a 2-dimensional vector, any opera-
tor that operates on such a vector must be a matrix. Since we want the
spin operators to have eigenvalues, these matrixes must be 2 2. We can
work out the S
2
matrix since we know that its eigenvalues must both be
¯ h
2
s(s+ 1)= 3¯ h
2
=4 from above. By specifying the basis above, we have
effectively speci?ed the eigenvectors ofS
2
, so we get
S
2

+
=

s
11
s
12
s
21
s
22

1
0

=
3¯ h
2
4

1
0

(3)
from which we get
s
11
=
3¯ h
2
4
(4)
s
21
= 0 (5)
with the other two elements undetermined, so far.
Applying the same logic to the other eigenvector gives us
s
12
= 0 (6)
s
22
=
3¯ h
2
4
(7)
Thus
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 3
S
2
=
3¯ h
2
4

1 0
0 1

(8)
Incidentally, since the eigenvalues ofS
2
are degenerate (both the same),
then any linear combination of the two eigenvectors that we’ve used above
is also an eigenvector, which is just another way of saying that we could
have chosen any two independent vectors to represent the two stationary
states.
We can do a similar calculation forS
z
, since we know that
S
z

+
=
¯ h
2

+
(9)
S
z


=
¯ h
2


(10)
This gives
S
z
=
¯ h
2

1 0
0 1

(11)
How aboutS
x
andS
y
? Since the raising and lowering operators that we
derived for orbital angular momentum depended only on the commutators,
we can write similar de?nitions for spin. In particular
S
+
=S
x
+iS
y
(12)
S

=S
x
iS
y
(13)
The action of these operators on a stationary state is also the same as for
L:
S

jsm
s
i= ¯ h
p
s(s+ 1)m(m 1)jsm
s
 1i (14)
We can invert the relations above and get
S
x
=
1
2
(S
+
+S

) (15)
S
y
=
1
2i
(S
+
S

) (16)
Then we can apply these operators to the stationary states to get
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 4
S
x




1
2
1
2

=
1
2
(S
+
+S

)

1
0

=
¯ h
2

0
1

(17)
S
x




1
2

1
2

=
1
2
(S
+
+S

)

0
1

=
¯ h
2

1
0

(18)
From this we get
S
x
=
¯ h
2

0 1
1 0

(19)
A similar calculation gives usS
y
:
S
y
=
¯ h
2

0 i
i 0

(20)
It is usual to extract the factor of ¯ h=2 from these three matrixes and de?ne
S=
¯ h
2
 with

x
=

0 1
1 0

; 
y
=

0 i
i 0

; 
z
=

1 0
0 1

(21)
These are called the Pauli spin matrices.
The commutation relations can be veri?ed by direct calculation, so we
give only one as an example.
[S
x
;S
y
]=
¯ h
2
4

i0
0i



i0
0i

(22)
=
¯ h
2
2
i

10
0 1

(23)
=i¯ hS
z
(24)
By direct calculation, we can show that

2
x
=
2
y
=
2
z
=

1 0
0 1

(25)
Also, by direct calculation we see that

j

k
=i
l
=
k

j
(26)
ifjkl is a forward permutation ofx;y;z (that is,jkl is one ofxyz,yzx or
zxy). These results can be combined into the formula

j

k
=
jk
+i
å
l

jkl

l
(27)
Page 5


SPIN ONE-HALF AND THE PAULI SPIN MATRICES
In our introduction to spin, we saw that spin is an intrinsic, ?xed quantity
for each elementary particle. It doesn’t really have a classical analogue, as
it is not the result of any motion of the particle. Despite that, spin really is
a form of angular momentum, as it combines with the orbital angular mo-
mentumL to form a total angular momentum for a given particle. It seems
that elementary particles (at least those with non-zero spin) just have an in-
strinsic angular momentum that is ?xed, depending on the type of particle.
The lack of analogy with classical angular momentum extends to the fact
that spin does not depend on spatial coordinates. As such, the eigenfunc-
tions of the spin operators are not functions of position, so we don’t need
to worry about Hilbert space, or their behaviour at in?nity or any of those
things that make the other quantum operators so tricky (and interesting, it
has to be said) to deal with.
We’ve seen that the eigenvalues of the square of the total spin S
2
are
¯ h
2
s(s+ 1), wheres is any non-negative integer or half-integer. Ifs= 0, no
spin is present and there isn’t anything more to say. The simplest non-trivial
case is therefores=
1
2
. In this case, the eigenvalues ofs
z
are
1
2
, so there
are only 2 possible states. Since the spin of a particle is ?xed, a particle
withs=
1
2
can existonly in a linear combination of these 2 states, no matter
how much you poke it or excite it by passing electric ?elds through it or do
anything else to it.
How does this affect the Schrödinger equation? After all, it involves
derivatives with respect to both time and position, so if we’re dealing with a
system that makes no use of position, does that mean that the¯ h
2
Ñ
2
Y=2m
is zero? If that were the case then the time derivative would also be zero, and
it would seem to indicate that any particle’s spin state would never change.
This isn’t the case, and the reason is that we need to step back a bit in our
interpretation of the Schrödinger equation. The term¯ h
2
Ñ
2
Y=2m is based
on our taking the energy of the system as p
2
=2m and then translating the
momentum into its quantum operator form. In the case of spin, we need to
write the energy (or more precisely, the Hamiltonian) in terms of the spin,
so the spatial derivative is replaced by a term involving the spin variables,
1
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 2
while the time derivative remains as it is. We’ll see some examples of this
in due course; for now, just be assured that the Schrödinger equation does
still work in systems involving spin.
Returning to spin
1
2
, since we are allowed only 2 states, we can use a 2-
dimensional vector to represent the state of a particle. These vectors form a
vector space, and we can take the basis of this space to be the two stationary
states, commonly called spin up and spin down. We represent these two
stationary states as

+
=

1
0

; 

=

0
1

(1)
We could, of course, have chosen any two independent vectors as the
basis, but these two make life simpler.
With this de?nition, a general spin state of a spin
1
2
particle is a linear
combination of these two states:
=a
+
+b

=

a
b

(2)
Since the state of a particle is now a 2-dimensional vector, any opera-
tor that operates on such a vector must be a matrix. Since we want the
spin operators to have eigenvalues, these matrixes must be 2 2. We can
work out the S
2
matrix since we know that its eigenvalues must both be
¯ h
2
s(s+ 1)= 3¯ h
2
=4 from above. By specifying the basis above, we have
effectively speci?ed the eigenvectors ofS
2
, so we get
S
2

+
=

s
11
s
12
s
21
s
22

1
0

=
3¯ h
2
4

1
0

(3)
from which we get
s
11
=
3¯ h
2
4
(4)
s
21
= 0 (5)
with the other two elements undetermined, so far.
Applying the same logic to the other eigenvector gives us
s
12
= 0 (6)
s
22
=
3¯ h
2
4
(7)
Thus
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 3
S
2
=
3¯ h
2
4

1 0
0 1

(8)
Incidentally, since the eigenvalues ofS
2
are degenerate (both the same),
then any linear combination of the two eigenvectors that we’ve used above
is also an eigenvector, which is just another way of saying that we could
have chosen any two independent vectors to represent the two stationary
states.
We can do a similar calculation forS
z
, since we know that
S
z

+
=
¯ h
2

+
(9)
S
z


=
¯ h
2


(10)
This gives
S
z
=
¯ h
2

1 0
0 1

(11)
How aboutS
x
andS
y
? Since the raising and lowering operators that we
derived for orbital angular momentum depended only on the commutators,
we can write similar de?nitions for spin. In particular
S
+
=S
x
+iS
y
(12)
S

=S
x
iS
y
(13)
The action of these operators on a stationary state is also the same as for
L:
S

jsm
s
i= ¯ h
p
s(s+ 1)m(m 1)jsm
s
 1i (14)
We can invert the relations above and get
S
x
=
1
2
(S
+
+S

) (15)
S
y
=
1
2i
(S
+
S

) (16)
Then we can apply these operators to the stationary states to get
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 4
S
x




1
2
1
2

=
1
2
(S
+
+S

)

1
0

=
¯ h
2

0
1

(17)
S
x




1
2

1
2

=
1
2
(S
+
+S

)

0
1

=
¯ h
2

1
0

(18)
From this we get
S
x
=
¯ h
2

0 1
1 0

(19)
A similar calculation gives usS
y
:
S
y
=
¯ h
2

0 i
i 0

(20)
It is usual to extract the factor of ¯ h=2 from these three matrixes and de?ne
S=
¯ h
2
 with

x
=

0 1
1 0

; 
y
=

0 i
i 0

; 
z
=

1 0
0 1

(21)
These are called the Pauli spin matrices.
The commutation relations can be veri?ed by direct calculation, so we
give only one as an example.
[S
x
;S
y
]=
¯ h
2
4

i0
0i



i0
0i

(22)
=
¯ h
2
2
i

10
0 1

(23)
=i¯ hS
z
(24)
By direct calculation, we can show that

2
x
=
2
y
=
2
z
=

1 0
0 1

(25)
Also, by direct calculation we see that

j

k
=i
l
=
k

j
(26)
ifjkl is a forward permutation ofx;y;z (that is,jkl is one ofxyz,yzx or
zxy). These results can be combined into the formula

j

k
=
jk
+i
å
l

jkl

l
(27)
SPIN ONE-HALF AND THE PAULI SPIN MATRICES 5
where
jkl
is the Levi-Civita symbol, which is +1 ifjkl is a forward permu-
tation ofxyz,1 ifjkl is a non-forward permutation ofxyz and 0 if any
two ofjkl are equal.
Ifj6=k only one term in the sum is non-zero, while ifj=k all terms in
the sum are zero.