Introduction to Quantum Mechanics
Notes from various sources (listed below)
June 29, 2009
1
Angular Momentum
Any Quantum system has to be identified with some property that is conserved in
that system. This is true for classical systems as well. In many cases we define a
quantum system by its total Hamiltonian (as its total energy remains conserved).
We can also consider another observable, the total Angular Momentum. It is
represented by J. Every quantum system has a total angular momentum associated with it (just like how every quantum system has a position and momentum
associated with it). Also just like X and P, J is also an vector observable (unlike
the Hamiltonian, which is a scalar). In the cartesian system of coordinates
−
→
J = Jx î + Jy ĵ + Jz k̂
(1)
Going back to our classical concepts, we se that the angular momentum is defined
as:
−
→ −
→
L =→
r ×−
p
(2)
The same definition still carries on to the quantum case. The difference is that all
those quantities those are variables in the classical case (eq. 2) are now operators.
The quantum case becomes:
−
→ −
→
J =→
x ×−
p
(3)
From expanding the above equation (eq. 3) we can extract the relation between
the total angular momentum and the known observable. Let us now solve the cross
product by considering the components or the physical quantities. To solve the
cross product we may consider the determinant form of the cross product.
!
!
! î
!
ĵ
k̂
!
!
−
→ −
→ !
A × B = ! Ax Ay Az !!
! Bx By Bz !
1
−
→
Solving for J , we get:
−
→
J = (ypz − zpy )î + (zpx − xpz )ĵ + (xpy − ypx )k̂
−
→
On equating the components of J , we get:
Jx = (ypz − zpy )
Jy = (zpx − xpz )
Jz = (xpy − ypx )
(4)
(5)
(6)
The above three equations (eq. 4, eq. 5 and eq. 6) represent the various components of the angular momentum operator. From this we can extract more about
the operators. Firstly we should see if they commute. Let us start by finding the
commutator of Jx and Jy operators. For this we need to consider equations (eq.
4, eq. 5 and eq. 6).
[Jx , Jy ] = [(ypz − zpy ), (zpx − xpz )]
⇒ [ypz , zpx ] − [ypz , xpz ] − [zpy , zpx ] + [zpy , xpz ]
To solve the above commutations we need to use the expansions:
[A, BC] = [A, B]C + B[A, C]
[AB, C] = A[B, C] + [A, C]B
From the above two expressions,we can make a new commutation expansion for
[AB,CD] which shall be useful in evaluating the terms of the above expression:
[AB, CD] = [AB, C]D + C[AB, D]
⇒ (A[B, C] + [A, C]B)D + C(A[B, D] + [A, D]B)
⇒ A[B, C]D + [A, C]BD + CA[B, D] + C[A, D]B
Therefore, we write:
[AB, CD] = A[B, C]D + [A, C]BD + CA[B, D] + C[A, D]B
(7)
Using the above equation (eq. 7) we can see how each term becomes. In each
term we can make the right substitutions for A, B, C and D. Then we get the
results:
Table 1: examining the terms of the commutator
Terms [B,C] [A,C] [B,D] [A,D]
Result
term 1
−i!
0
0
0
A[B,C]D ≡ −i!ypx
term 2
0
0
0
0
0
term 3
0
0
0
0
0
term 4
0
0
0
i!
C[A,D]B ≡ i!xpy
2
Therefore, the result is: i!(xpy − ypx ). If we refer back to (eq. 6), we can see
that this is the expression for Jz , which is the z-component of the total angular
momentum (eq. 1). Therefore, we have the relation [Jx , Jy ] = i!Jz . Similarly,
we can work out the other relations too as [Jy , Jz ] = −i!Jx and [Jx , Jz ] = −i!Jy .
The relations can be summarized as:
[Jp , Jq ] = i!!pqr Jr
(8)
The equation above (eq. 8) gives the summary of all the commutation relations
−
→
discussed above. Let us now go back and see why we introduced J . We said
that we would like to associate a total angular momentum to a quantum system
in order to characterize it. Now we have seen some way of identifying a system. A
quantum system may also exist in various phases or forms. These various forms
−
→
or phases of a quantum system are called the states of a quantum system. J
does not say anything about the states in which the quantum system may exist.
To identify the states, we need some property that is different for each of the
−
→
individual components 1 and that can be compared with J . For this, we can
take some component of the total angular momentum. Take for example the zcomponent (this is just a convention) Jz . The number of values of Jz will tell us
how many states the system exists in, because each state will have a contribution
−
→
to Jz . So, on comparing J and Jz for a system, we can see how many states the
system exists in. There is another minor problem before we proceed. The two
−
→
properties J and Jz can be compared. When we say that two properties can be
compared, we mean that there exists a common basis for the two operators. In
−
→
other words, the two operators commute. Comparing the two properties J and
Jz is slightly weird because the former is a vector and the latter is a scalar. So, let
−
→
us consider J2 instead of the vector J . This would make the comparison logically
right. Now we have:
J2 = J2x + J2y + J2z
(9)
From the above expression itself, it is clear that [J2 , Jk ] = 0, ∀k ∈ {x, y, z}. Since
J2 commutes with Jz , these two operators can be diagonalized simultaneously.
This means they have simultaneous eigen kets. Let us consider the equation:
J2 |α, β( = α|α, β(
Jz |α, β( = β|α, β(
(10)
(11)
Here |α, β( denotes the simultaneous eigen ket of J2 and Jz , with different eigen
values. We know J2 is a property of a system that exists in various states, each
state corresponding to a value of Jz . If we expect that a particular system must
exist in some given number of states, then we must expect those many values of Jz ,
1 If
it is not different, the states shall be indistinguishable.
3
(corresponding to each state), for a single value of J2 (corresponding the system
as a whole). So, in mathematical terminology, if we expect a system to exist in
n different quantum states, then we must expect n different eigen values for Jz
(corresponding to each state) for a single eigen value of J2 (corresponding to the
system). Going back to the above eigenvalue equations (eq. 10 and eq. 11), we
can now claim that: for every α, there shall be a number of values of β. The
number of such β’s for a given α shall tell us the number of states that the system
exists in. So, till now we have argued with physical reasons. We can try to verify
this mathematically too. From (eq. 11) we have:
J2z |α, β( = βJz |α, β( ⇒ β 2 |α, β(
(12)
Therefore, on subtracting the equations (eq. 12) from (eq. 10), we get:
(J2 − J2z )|α, β( = (α − β 2 )|α, β(
⇒ )α, β|(J2 − J2z )|α, β( = )α, β|(α − β 2 )|α, β(
On the LHS, from equation (eq. 1) we can write (J2 − J2z ) as (J2x + J2y ). The RHS,
shall become (α−β 2 ) if we assume that the simultaneous eigenkets are normalized.
Hence, we have:
)α, β|(J2x + J2y )|α, β( = (α − β 2 )
The LHS of the above equation is the expectation value of the a positive operator.
The LHS is therefore a positive definite quantity. Hence, we can write that:
0 ≤ (α − β 2 )
⇒ β2 ≤ α
(13)
Therefore, from the above equation (eq. 13), we can see that the value of β 2 is
bounded by α. This means there are a finite number of states for a finite value
of α. In other words, there are two distinct bounds for the value of β. Let the
extreme values of β be denoted as βmin and βmax . But to examine all the states
(values of β between βmin and βmax ) we need to construct some operator that can
help us to traverse through each of the states from βmin up to βmax and vice-versa.
A higher value of β corresponds to a higher value of angular momentum in the
z-direction. We call call this a higher state.
Let us now define two operators that take us to a higher and a lower state respectively, from a given state. These are the raising and lowering operators
represented as J+ and J− . These operators are defined as:
J+ = Jx + iJy
J− = Jx − iJy
4
(14)
(15)
The action of these operators can be defined as:
J+ |α, β( = c+ |α, (β + !)(
J− |α, β( = c− |α, (β − !)(
and more importantly:
J+ |α, βmax ( = 0
J− |α, βmin ( = 0
(16)
(17)
(18)
(19)
We can just pause for a moment and explore the mathematics of the raising and
lowering operators. Let us look at some commutation relations:
[J+ , J− ] = [(Jx + iJy ), (Jx − iJy )]
⇒ RHS : [Jx , Jx ] + i([Jy , Jx ] − [Jx , Jy ]) + [Jy , Jy ]
using equation (eq. 8) ⇒ 0 + i(!Jz + !Jz ) + 0
⇒ [J+ , J− ] = 2i!Jz
(20)
Similarly, we can see:
[Jz , J+ ] = [(Jz , (Jx + iJy )]
RHS : [Jz , Jx ] + i[Jz , Jy ]
⇒ i!Jy + !Jx
⇒ [J+ , Jz ] = i!J+
similarly, we also have: [J− , Jz ] = i!J−
(21)
(22)
There is another important relations that we should work out as we would be using
it shortly. Let us simplify the products J+ J− and J− J+ :
J+ J− = (Jx + iJy )(Jx − iJy )
RHS :
J2x + i(Jy Jx − Jx Jy ) + J2y
with equation (eq. 9), we can replace J2x + J2y with J2 − J2z
⇒ J2 − J2z + i(Jy Jx − Jx Jy )
⇒ J2 − J2z + i[Jy , Jx ]
with equation (eq. 8) we have: J+ J− = J2 − J2z + !Jz
Similarly, J− J+ = J2 − J2z − !Jz
(23)
(24)
Hence with this many commutators, we can proceed to see what the ladder operators (the raising and the lowering operators) do to the states of a system and
how to find the number of states in which a system exists. Let us start by the
assumption made in the equation (eq. 18):
J+ |α, βmax ( = 0
5
since any operator acting on 0 would give 0, we can have:
J− J+ |α, βmax ( = 0
from equation (eq. 24): (J2 − J2z − !Jz )|α, βmax ( = 0
2
on expanding: (α − βmax
− !βmax )|α, βmax ( = 0
We know that |α, βmax ( cannot be a null ket since it represents a state of some
system. So expression preceding the ket must be equal to 0. On equating the eigen
value to 0, we have:
2
α − βmax
− !βmax = 0
⇒ α = βmax (βmax + !)
(25)
Similarly, on considering equation (eq. 23), we have:
α = βmin (βmin + !)
(26)
On comparing the above two equations (eq. 25 and eq. 26), we see that:
βmax = −βmin
(27)
We know that |α, βmin ( is the lowest possible state and |α, βmax ( is the highest
possible state. The raising operator can be applied repeatedly to the lowest state
to get it to the highest one. Each time we apply the raising operator, we get to a
higher state. So if we apply the raising operator n times (where n is the number
of states) to |α, βmin (, we reach |α, βmax (. Each time we act a state with J+ , the
value of β increases by !. Therefore:
βmax = βmin + n! n → number of states
n!
⇒ from equation (eq. 27): βmax =
2
To eliminate the factor of !, Let us define: j =
n
⇒j=
2
βmax
!
(28)
Since n, the number of states is an integer, j is either an integer or a half integer.
n!
From equation (eq. 25), we can get the form of α. On substituting βmax =
≡
2
βmax = !j in equation (eq. 25), we have:
α = !2 j(j + 1)
(29)
On referring equation (eq. 13), we have all the possible values of β. We know that
β is related to α by a power half law. β ≤ |α|. Since α has a factor of !2 , the
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expression of β must have a ! factor. The rest is just a constant. Hence, we may
write:
β = m!
(30)
where m is some number, depending on the value of j.On rearranging the above
β
βmax
equation, we get
= m. Even from the equation (eq. 28), we have
= j.
!
!
We discussed earlier that for every βmax , β takes all values from −βmax (which is
βmin ) to βmax . So, for every value of βmax , β takes (2βmax + 1) values. Hence, the
same relation can be said about j and m also since they are equivalent to βmax
and β up to some constant. Hence, for every value of j, we have (2j + 1) values
for m and m takes only integer values.
−j ≤ m ≤ j
(31)
Now, the number of values for m shall determine the number of states in which
the system (given by the value of j) shall exist. Now the simultaneous eigenstates
of Jz and J2 can now be labelled as |j, m(. Therefore, we have:
J2 |j, m( = j(j + 1)!2 |j, m(
Jz |j, m( = m!|j, m(
(32)
(33)
Where j tells us about the total angular momentum of the system and the number
of values of m give the number of states in which the system exists.
For any general quantum system, the total angular moment of the system is given
by:
J=L+S
(34)
where: J is the total angular momentum, L is the orbital angular momentum,
S is the spin angular momentum. If the wave function associated with the
quantum system is a scalar function, the J = S. If the wave function is a vector
function, then we use the relation given in equation (eq. 34).
7
2
Composing two quantum systems - addition of angular
momenta
3
References
Table 2: Reference Books used:
Book
Introduction To Quantum Mechanics
Author(s)
David J Griffiths
Modern Quantum Mechanics
Jun John Sakhurai
Fundamentals of Quantum Mechanics
R.Shankar
Quantum Computation and Quantum Information
Neilson and Chuang
Quantum Computing
Mikio Nakahara and Ohmi
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