Chapter 11

Angular Momentum–Revisited

11.1  INTRODUCTION

In Chapter 9, we studied orbital angular momentum. There, we defined orbital angular momentum operators and components We found out their commutation relations and found simultaneous eigenfunctions and eigenvalues of Let us rewrite some of these relations.

Here we have written the eigenstates Y_{l, m}(θ, ϕ) in the ket notation as In this chapter, we shall define raising and lowering operators. Using properties of these operators, we shall find out, in a very simple way, the eigenvalues and eigenstates of angular momentum operators. We shall also express angular momentum operators in matrix form.

11.2   RAISING AND LOWERING OPERATORS (THE LADDER OPERATORS)

In Chapter 9, we found eigenvalues and eigenstates of operators and by solving the corresponding Schrodinger equations (in the form of second/first order differential equations). In this chapter, we use an alternative method to arrive at same results of eigenvalues and eigenstates of these operators. This method is simpler and straight forward. The method uses the so-called raising and lowering operators (or the ladder operators) which we shall define below.

There are two classes of angular momenta: the orbital angular momentum (denoted by L) and the spin angular momentum (denoted by S). We start denoting the angular momentum, in general, by J. So J may represent L or S or the combination L + S. We define the commutation relations amongst the components of (similar to those in components of ), as

The raising and lowering operators and are defined as

Using Eqs (9.18b) and (9.18c), we may write these operators as

Now, one may easily see that

Similarly,

Also, we have

Other relations that and satisfy are

From Eqs (11.12), we may write

and

11.3  EIGENVALUES AND EIGENSTATES OF ORBITAL ANGULAR MOMENTUM OPERATORS: SECOND CONSTRUCTION OF SPHERICAL HARMONICS

Let us now turn to finding eigenvalues and eigenstates of operators and We start with eigenvalue equation of operator J_z,

where we have to find eigenvalues mħ and eigenstates φ_m(ϕ). We now consider the operation

This equation implies that is an (unnormalized) eigenstate of corresponding to eigenvalue (m + 1)ħ . That simply means

where C₊(m) is some constant (to be determined later) and for the time being we replace it by unity. So in the place of Eq. (11.17), we write

Now applying J₊ on φ_{m +1}, we get

Similarly, we get

But we shall take constant C_–(m) to be unity, so

and

Here we observe that operation of J₊ on an eigenstate φ_m of generates another eigenstate φ_m+1 of . The successive operations of generate eigenstates φ_m+2, φ_m+3 ... and so on. Therefore, we have found a scheme of generating a sequence of eigenstates of operator starting from the eigenstate φ_m: the successive eigenstates have values of m in the sequence differing by unity,

 

The corresponding eigenvalues of operator in these eigenstates being respectively

 

 

As the operators and commute, these two have common eigenstates. Let φ_m be the common eigenstate with eigenvalue K² ħ² of operator . So

where we have to determine values of K². Now, we know that operators and commute [(Eq. (11.11)], so

or

Equation (11.23) dictates that is an eigenstate of corresponding to eigenvalue K² ħ². Therefore, all eigenstates of found in Eq. (11.15) are the eigenstates of corresponding to the same eigenvalue K² ħ². Let us firstly see how many such eigenstates are there:

From Eqs (11.15) and (11.22), the expectation values of and in state φ_m are

But we also have

Therefore,

As and it follows from Eq. (11.25) that

 

or

 

It may be noted that for a given value of K (> 0), Eq. (11.27) dictates that the possible values of m in Eq. (11.21) must lie in between +K and –K. Therefore, if the maximum possible value of m, for a given value Kħ of the magnitude of angular momentum vector, is m₊, then we should have the condition

Similarly, if m_– is the minimum value of m for the angular momentum magnitude Kħ, we have

Now let the L.H.S. and R.H.S. operators of Eq. (11.12b) operate on state φ_m+, we get

or

or

 

Again with operators of Eq. (11.12a) operating on state φ_m–, we get

or

or

From Eqs (11.29) it follows that

 

 

Apart from the trivial solution m₊ = m_–= 0, this equation is satisfied if

 

 

Hence, for a given magnitude K² ħ² of J², the allowed values of m are like the following sequence (see Figure 11.1).

 

 

where m₊ (the maximum value of m) and m_– (the minimum value of m) are given by Eqs (11.29a) and (11.29b), respectively. Thus all possible values of m form a symmetric sequence about m = 0. The successive values of m differ by unity. Let us call maximum value of m as j.

 

 

Therefore, m runs from –j to +j in unit steps. Clearly, one of the two possibilities is there:

1. If m = 0 is present in the sequence of m values, then

    

   j = an integer         (11.34a)

    

2. If m = 0 is not present in the sequence of m values, then

   

Figure 11.1 For a given eigenvalue of , the allowed values of m are shown. Successive values of m differ by unity

In fact, it can be easily seen that if we take any other choice of j [other than those of Eqs (11.34a) and (11.34b)], then all possible values of m shall not satisfy the conditions which the sequence of m should satisfy, which we state again: The possible values of m should form a symmetric sequence about m = 0 and the successive values of m should differ by unity.

Let us take some particular examples.

•  If j is an integer say 3, the corresponding m values are 3, 2, 1, 0, –1, –2, –3.
• If j is an odd multiple of say the corresponding m values are
• If we take j different than those of category (i) and (ii) above, say the corresponding m values should be

It is clearly seen that the choice of j in category (iii) does not give m values that satisfy the above mentioned criterion for the sequence of m values.

In fact, in either case of category (i) and (ii), for a given value of j, the allowed values of m run from –j to +j (see Figure 11.2), so inserting j = m₊ = – m_– into Eq. (11.29), we get the eigenvalues of J² as

 

 

and eigenvalues of J_z as

 

 

where j is an integer or half an odd integer. The eigenvalue equations of and are for general angular momentum which could be orbital angular momentum spin angular momentum or their sum In the following work, we consider representing orbital angular momentum , where corresponding quantum numbers l and m₁ (we start writing now l and m₁ for j and m_j) have integral values. In a later section, we shall consider representing and then quantum numbers j and m_j shall have half an odd integer values.

For orbital angular momentum Eqs (11.35) become

and

 

Figure 11.2 Allowed values of quantum number m for a given value of orbital quantum number j. j may be either an integer or an odd multiple of one-half

As per Eqs (11.15) and (11.22), the state φ_ml is simultaneous eigenstate of and . These two operators have eigenvalues l (l + 1) ħ² and m₁ ħ, respectively, therefore, the eigenstate φ_ml should really be assigned two quantum numbers l and m_l and so we rewrite these eigenstates as φ_l,_ml.

At this stage, we can borrow a simple result from Chapter 9. There we found the eigenstates of operator [(Eq. (9.22)].

Now the simultaneous eigenstate φ_l,_ml of and may be written as product of two functions

 

We shall simply write m in place of m_l in what follows. Let us start now with Eq. (11.28a), which we rewrite as

(as maximum value of m is l)

or

Using the expression of [Eq. (11.9)], the above equation gives

or

or

Solution of this equation is

 

 

The multiplicative normalization constant will be obtained later. Eq. (11.38) gives

 

 

Let us start calling the functions φ_l,m(θ, ϕ)(when normalized) as Y_l,m(θ, ϕ).

So, Eq. (11.42) is written as

 

 

where A_l is the normalization constant. From Eq. (11.43) one may easily find expressions of Y_0,0(θ, ϕ), Y_1,1(θ, ϕ), Y_2,2(θ, ϕ), Y_3,3(θ, ϕ), ... which we write below

 

 

 

 

 

Here constants A₀, A₁, A₂, ... are determined by normalizing the eigenstate. One notices that these expressions are same as the corresponding expressions of Y_{1, 1} (θ, ϕ) in Table 9.1 of Chapter 9.

Let us now apply the lowering operator on both sides of Eq. (11.43). We have

or

 

where B_l is the normalization constant. From Eq. (11.45), we may write expressions of Y_1,0(θ, ϕ), Y_2,1(θ, ϕ), Y_3,2(θ, ϕ) and so on, which are

 

 

 

 

which are same as those obtained in Chapter 9 and shown in Table 9.1.

Proceeding in a similar way one may easily find expressions of other spherical harmonics like

Here A, B, ... D are normalization constants.

11.4  THE CONSTANTS C₊ AND C_

Let us recollect that constants C₊ (m) and C_–(m) were defined through the operation of raising and lowering operators and on eigenstate φ_m [Eqs (11.17) and (11.19)]. As discussed in the previous Section, the state φ_m is simultaneous eigenstate of and and, therefore, (state φ_m) should be assigned two quantum numbers j and m and should be rewritten as φ_j,m. Furthermore, this eigenstate φ_j,m can be expressed as the product of two functions Θ_j,m(θ) and Φ_m(ϕ) as

 

 

We now denote eigenstate φ_j,m(θ, ϕ) as eigenket and write various operator equations as

Equations (11.17) and (11.19) showing operations of and are rewritten as

Let us now determine coefficients C₊ and C_–. Taking adjoint of both sides of Eq. (11.50a), we get

(since is adjoint of )

Now multiplying Eqs (11.50c) and (11.50a), we have

Putting value of from Eq. (11.12b) gives

Thus,

Similarly, we may get

It is clear from above expressions that

 

 

11.5  MATRIX REPRESENTATION OF ANGULAR MOMENTUM OPERATOR CORRESPONDING TO j = 1

We shall now proceed to represent the angular momentum operators and components in matrix form in a basis in which operators and are diagonal. That simply means the simultaneous eigenkets of and are chosen as basis vectors. Let us take the case corresponding to angular momentum quantum number j = 1. For j = 1, the allowed values of m are 1, 0 and –1. So, there are three basis states, which we denote by and and write as

Using Eqs (11.49a) and (11.49b) we have

which in the notations of Eq. (11.53) are written as

Similarly,

In the basis set the (m, n)^th matrix element of an operator  is defined as

Therefore,

We find that, in fact, all off-diagonal terms are zero. So matrix form of may be written as

Similarly, we get

Therefore, the matrix form of is

Now using Eqs (11.50a) and (11.51a), we get

So matrix form of is

Similarly, one gets for

Now

Therefore, matrix forms of and are

It is clear from the matrix form of [Eq. (11.56a)] that it has the following three eigenstates and eigenvalues:

And same are the eigenstates of the matrix form of (with eigenvalues 2ħ² for each state). Therefore, we may write eigenkets in the (column) matrix form as

It may be easily checked that column matrices (or generally called column vectors) of Eq. (11.61) represent the correct eigenstates. For example, L.H.S. of Eq. (11.57b) gives

 

Also we may check, these column vectors are orthogonal to each other. For example

11.6  MATRIX REPRESENTATION OF ANGULAR MOMENTUM OPERATOR CORRESPONDING TO

It is to be noted that the results obtained in this chapter and Chapter 9 are in connection with the orbital angular momentum operator Though in defining the raising and lowering operators, we had used the general angular momentum operator which may represent orbital angular momentum operator or spin angular momentum operator or their combination, that is, Yet in Eq. (11.9), while writing the expressions for raising and lowering operators and we had used only x- and y- components of orbital angular momentum operator. Therefore, the results for the spectrum of magnetic quantum number m for a given quantum number j (i.e., m values lying symmetric about m = 0, ranging from –j to +j, differing by unity) should be applicable to the orbital angular momentum only. As we know the case of j = an integer corresponds to the orbital angular momentum. However, value of j = half an odd integer is also possible, as it allows m values lying symmetric about m = 0, ranging from –j to +j, differing by unity. Therefore, the values of which are not there in case of orbital angular momentum operator, may be there corresponding to some other angular momentum operator. Let us consider the case of and find out the matrix form of the corresponding angular momentum operator. We shall, in fact, see that the form of the resulting matrix operators is those of familiar Pauli spin matrices.

So we proceed to represent operator and component operators (in matrix form) corresponding to angular momentum quantum number in a basis in which and are diagonal. Now corresponding to , the magnetic quantum number and There are two basis states denoted by and where

Using Eqs (11.49a) and (11.49b), we have

which may be rewritten as

Also, we have

We may easily find the matrix elements of

So matrix form of is

Similarly, using Eqs (11.65) we may find out various matrix elements of giving the resulting matrix of form

Now using Eq. (11,50a), we have

where is calculated using Eq. (11.51a)

So, Eq. (11.68) gives

And it can be easily seen that

The matrix form of is

Similarly, is

From these equations, we find and as

 

The angular momentum component operators given by matrices of Eqs (11.71a), (11.71b), and (11.66), respectively, may be expressed as

where

are known as the Pauli spin matrices.

The 2 × 2 matrix representing has following two eigenstates and eigenvalues:

Therefore, the eigenkets and may be written in the column matrix form, generally termed as spinors:

Any linear combination of column matrices will be a column matrix and, therefore, shall also be termed as spinor. Now, we have seen the angular momentum operator for quantum number has two eigenstates which represent two states of electron spin. Hence, we start writing and in place of and and write quantum numbers s and m_s in place of j and m. Therefore, Eq. (11.72) is rewritten for spin angular momentum operator

EXERCISES

 

Exercise 11.1

Find the following expectation values in state

• 
• 
• 
• 
• 
• 
• 
• 

  where represents the anti-commutator

Exercise 11.2

Find out matrix form of following angular momentum operators, corresponding to quantum number in a basis in which and are diagonal.

• 
• 
• 
• 
• 
• 
• 

Exercise 11.3

Consider a system of angular momentum with quantum number l = 1. Its angular momentum state is represented by state vector.

Find the probability that a measurement of L_z gives value 0.

SOLUTIONS

Solution 11.1

• First method: From Eq. (11.12a)
  

  so

  
  = l(l+1)ħ² – m²ħ² + ħ²m
  = ħ²[(l + m)(l – m + 1)]        ( 11.77)

  Second method:

  
  = ħ²[(l + m)(l – m+ 1)]        (Refer to Eq. 11.77)
•  
  
  = ħ²[(l – m) (l + m + 1)]        (11.78)
•  
  
•  
  
•  
  
  
  
•  
  
  
• From (a) and (b)
  
  = 2mħ²        (11.81)
•  
  
  = 2ħ² [l(l + 1) – m²]         (11.82)

Solution 11.2

are the simultaneous eigenkets of operators and For angular momentum quantum number allowed values of m are So, there are four basis states, which we shall denote by and written explicitly as

• Using Eqs (11.50a) and (11.51a), we have
  

  From these relations, one may find out all matrix elements

  

  and may get matrix form of as

  
• Using Eqs (11.50b) and (11.51b), we have
  

  so

  
•  
  
  
•  
  
  
•  
  
•  
  
• We know
  
  

Solution 11.3

We know three basis state vectors of operator are and corresponding to eigenvalues of and –ħ, respectively. The state vector ϕ may be written as

So the probability for to have eigenvalue 0 is

REFERENCES

1. Griffiths, D.J. 1995. Introduction to Quantum Mechanics. New York, NY: Prentice Hall.
2. Tinkham, M. 1964. Group Theory and Quantum Mechanics. New York, NY: McGraw-Hill.
3. Sakurai, J.J. 1985. Modern Quantum Mechanics. Massachusetts: Benjamin/Cummings.
4. Liboff, R.L. 1992. Introductory Quantum Mechanics. Massachusetts: Addison-Wesley.
5. Merzbacher, E. 1999. Quantum Mechanics. 3rd edn., New York, NY: John Wiley & Sons.