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 Yl, 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.
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
Let us now turn to finding eigenvalues and eigenstates of operators and We start with eigenvalue equation of operator Jz,
where we have to find eigenvalues mħ and eigenstates φm(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 φm +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 φm of generates another eigenstate φm+1 of . The successive operations of generate eigenstates φm+2, φm+3 ... and so on. Therefore, we have found a scheme of generating a sequence of eigenstates of operator starting from the eigenstate φm: 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 φm 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 φm 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 φm+, we get
or
or
Again with operators of Eq. (11.12a) operating on state φm–, 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:
If m = 0 is present in the sequence of m values, then
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.
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 Jz 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 mj) have integral values. In a later section, we shall consider representing and then quantum numbers j and mj 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 φml is simultaneous eigenstate of and . These two operators have eigenvalues l (l + 1) ħ2 and m1 ħ, respectively, therefore, the eigenstate φml should really be assigned two quantum numbers l and ml and so we rewrite these eigenstates as φl,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 φl,ml of and may be written as product of two functions
We shall simply write m in place of ml 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
Solution of this equation is
The multiplicative normalization constant will be obtained later. Eq. (11.38) gives
Let us start calling the functions φl,m(θ, ϕ)(when normalized) as Yl,m(θ, ϕ).
So, Eq. (11.42) is written as
where Al is the normalization constant. From Eq. (11.43) one may easily find expressions of Y0,0(θ, ϕ), Y1,1(θ, ϕ), Y2,2(θ, ϕ), Y3,3(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 Y1, 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 Bl is the normalization constant. From Eq. (11.45), we may write expressions of Y1,0(θ, ϕ), Y2,1(θ, ϕ), Y3,2(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.
Let us recollect that constants C+ (m) and C–(m) were defined through the operation of raising and lowering operators and on eigenstate φm [Eqs (11.17) and (11.19)]. As discussed in the previous Section, the state φm is simultaneous eigenstate of and and, therefore, (state φm) should be assigned two quantum numbers j and m and should be rewritten as φj,m. Furthermore, this eigenstate φj,m can be expressed as the product of two functions Θj,m(θ) and Φm(ϕ) as
We now denote eigenstate φj,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
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
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
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
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 ms in place of j and m. Therefore, Eq. (11.72) is rewritten for spin angular momentum operator
Exercise 11.1
Find the following expectation values in state
where represents the anti-commutator
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 Lz gives value 0.
Solution 11.1
so
Second method:
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
From these relations, one may find out all matrix elements
and may get matrix form of as
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
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