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Quantum Mechanics

Quantum Mechanics. Chapter 3 . Angular Momentum and Superposition of States. §3.1 Eigenfunctions of angular momentum components. Rotation of Axes When we make a single measurement of an angular momentum component, that component is L z , by definition. The measurement defines the z axis.

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Quantum Mechanics

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  1. Quantum Mechanics • Chapter 3. Angular Momentum and Superposition of States

  2. §3.1 Eigenfunctions of angular momentum components • Rotation of Axes • When we make a single measurement of an angular momentum component, that component is Lz, by definition. The measurement defines the z axis. • If we now measure the component along another axis, we must give a different label to the second axis. If we call this axis the x axis, then the results of the measurement are dictated by the eigenfunctions of the Lx operator.

  3. Since we already know the eigenfunctions of the Lz operator, there is a simple way to find the eigenfunctions of the Lx operator: We write each eigenfunction of Lz in rectangular coordinates (instead of spherical coordinates), and then rotate the coordinate system. If we rotate the axes by 90° FIGURE 7.1 A rotation of 90°abour the y axis causes the x axis to replace the z axis.

  4. about the y axis, then the direction that was labeled z is now labeled x, and the direction previously labeled x becomes -z. (See Figure 7.1). • The effect of this rotation can be seen clearly if we write the eigenfunctions in rectangular coordinates rather than the spherical coordinates of Chapter 2. Thus we have, for l = 1,

  5. After rotation of the axes, we can consider the apparatus to be unchanged; only the coordinate labels are changed. Figure 7.1 shows that x → -z, y → y, and z → x. The wave functions relative to the apparatus are unchanged, but they are described in terms of the new labels as eigenfunctions of Lx. We identify them by using the subscript l, mx , as follows: • [Exercise] Verify that are indeed eigenfunctions of Lx and L2 with the indicated eigenvalues.

  6. Superposition of eigenfunctions of angular momentum • The functions of Eqs.(7.2) can now be used to solve the problems posed above, to determine the probabilities of the various possible results when measuring Lx after previously measuring Lz. • Suppose that we have found that l=1 and m=1,so that the wave function is . • To find the probabilities of the various results for Lx, we consider the function to be a superposition of the eigenfunctions of Lx : (7.3)

  7. where the coefficients a,b and c are amplitudes that determine the probabilities of obtaining the results +ћ, 0 and -ћ ,respectively, when Lx is measured. (Notice that the probability is the absolute square of the amplitude!) • Substitution of the expressions from Eqs.(7.1) and (7.2) into Eq.(7.3) gives (7.4) or (7.5) • Eq.(7.5) must hold for all values of x, y and z, so we equate the coefficients of x, y and z in turn,obtaining

  8. (7.6) with the result that • How can we verify this result? • First, we see that the sum of the probabilities is 1, as it should be. • Second, we can determine the value of <Lx2> that follows from the result. We deduce from Eqs.(7.8) that Lx2 = 0 half of the time, and half of the time, so that Because there is no difference between the x axis and the y axis in this situation

  9. (after we have just measured Lz), we can also deduce the This is consistent with the initial condition that and ; we require that . • Example Problem 3.1 Deduce the probabilities of the various results of measuring Lz for a system in which l = 2 and given that the probability of finding Lz = 0 is 3/8. • Solution. From the reasoning above, we know that the possible results are 0, From symmetry we know that the probability that I is equal to the probability that ; the same is true for

  10. where α is the probability that Lz will equal either and β is the probability that Lz will equal either • Thus we can write α+4β=1. Given that the total probability of a nonzero result α+β=1-3/8=5/8, we can solve for α and β to obtain α= 1/2 and β = l/8.

  11. §3.2 The Superposition Postulate • The expansion of wave function ψ • The expansion of the spherical harmonics can be applied to many other functions in quantum mechanics, to obtain probabilities of experimental results. The procedure is of such general use that we have expressed it as the sixth fundamental postulate to accompany postulates 2 and 3 of §1.2. • POSTULATE 6 Any acceptable wave function Ψ can be expanded in a series of eigenfunctions of the operator corresponding to any dynamical variable.

  12. As a postulate, this statement can not be proved directly,but it underlines all of quantum mechanics. • Postulate 6 has a useful corollary that concerns results of measurements. • Corollary to postulate 6: The probability of finding a particular value of a dynamical variable as a result of a measurement on a system with normalized wave function ψ is equal to the sum of the squares of the coefficients of the corresponding normalized eigenfunctions in the expansion of the function ψ. • According to the corollary, if the angular dependence of the wave function is any function f(θ,φ), then one can always write it as a sum of the

  13. functions Yl,m (θ,φ) with constant coefficients cl,m (7.9) • Then, for example, is the probability of finding when these two variables are both measured. When L2 alone is measured,the probability that l=3 is the sum of all seven possible values of . • The identification of this sum with the probabilities is required if we are to obtain expectation values . For example, the expectation value of L2 is given by

  14. Substitution of the expansion (7.9) into the integral yields (7.11) •After performing the indicated operation with the L2 operator, we obtain •To evaluate the integral, we use the fact that the spherical harmonics are orthogonal and normalized; that is •Thus every term of the infinite series vanishes except the terms for which l'=l and m' = m. This reduces Eq.(7.12) to

  15. But, by definition, where P(l) is the probability that L2 will be equal to l(l + 1) when it is measured. Therefore we find that in agreement with the corollary. •Calculation of the Coefficients • To find any particular coefficient cl’m’ in the expansion of any given function f(θ,φ), we make use of the fact that the Yl,m are orthogonal and normalized. •We multiply both sides of Eq.(7.9) by Y*l’,m’, and thenintegrate both sides of the equation

  16. over the surface of a sphere, using the area element sinθdθdφ. Because of the orthogonality of the Yl.m [Eq.(7.13)], all terms except one drop out of the series, and we obtain • Example. If f(θ,φ) = Nsin θ, then • Where is the normalizing factor for Y1,1, the result is c1,0=0. The same procedure shows that all of the other coefficients are zero except c1,1 and c1,-1, which must be equal in magnitude, because f(θ,φ) is independent of φ.

  17. If we use the correct normalizing factor N, then we will find that the normalizing factor N is given by the same condition used for normalizing the spherical harmonics: • Example Problem 7.2 • If f(θ,φ) = N(3 + cosθ), where is a normalizing factor, find the probability that l=1. What values of l are possible? • Solution. For l = 1. Eq.(7.16) shows that c1,1 = c1,-1 = 0 and

  18. Similarly, • Thus the probability that l = 1 is (3/5)2, or 9/25, and the probability that l = 0 is (4/5)2, or 16/25. The sum of these probabilities is l; therefore no other values of l are possible. You can verify this statement for l = 2 by referring to the spherical harmonics shown in Eqs.(6.49). All of them are orthogonal to the given function f(θ,φ). • Commuting Operators • In quantum mechanics there are many cases in which measurement of one dynamical variable disturbs the value of another variable, making it

  19. impossible to know the values of both variables simultaneously. There is a general rule for identifying such pairs of variables: • If it is possible to measure a dynamical variable A without losing the knowledge of another dynamical variable B, then the operators Oa and Ob representing the two variables must commute. • This means that OaObψ= ObOaψ for any function ψ, with the understanding that the operator on the right is applied and then the other operator is applied to the result. • The justification for this rule is as follows: • . If the value of a variable A is to be unchanged

  20. when variable A is measured, it must be possible to expand any eigenfunction of A in a series of eigenfunctions of both operators Oa and Ob. • . Since any function whatever may be expanded in a series of eigenfunctions of operator A, and each of these may be expanded in a series of eigenfunctions of both variables, we see that any function may be expanded in a series of eigenfunctions of both operators Oa and Ob. • . Let ψij be an eigenfunction of both Oa and Ob, with eigenvalues ai and bj, respectively. Then we may write Oaψij = aiψij and Obψij = bjψij

  21. Applying both operators gives OaObψij = Oabjψij = aibjψij and ObOaψij = Obaiψij = aibjψij • Therefore OaObψij = ObOaψij and these operators commute when applied to any of the functions ψij. But this means that they must commute when applied to any wave function, because any wave function can be expanded in a series of the functions ψij, according to Postulate 6. • Thus the commutation of operators Oa and Ob is a necessary condition if variables A and B are to be simultaneously measurable.

  22. For example, we have assumed that it is possible to measure E and L2 simultaneously; this requires that the operator Ĥ (the energy operator) and the operator must commute. • This is possible if the potential energy V is a function of r only, because in that case the Schroedinger equation [Eq.(6.32a)] may be written in the form where operates on r only. • Clearly commutes with Ĥ; the operator L must commute with each term in Ĥ , because L2 does not operate on the variable r. • On the other hand, the operators Lx and Lz do not

  23. commute. You may verify from Eqs.(6.28) that • Two additional commutation relations result from permuting the subscripts to obtain and

  24. §3.3 Matrix Methods • We can express the methods of the previous section in a more manageable form by introducing the methods of matrix algebra. • We can represent a quantum mechanical operator by a matrix that acts upon the quantum states, which are in turn represented as vectors in an abstract space. • This space (called Hilbert space after mathematician David Hilbert) is constructed by representing each normalized eigenfunction by a unit vector.

  25. These unit vectors are orthogonal, as the eigenfunctions themselves are. Then any other function can be represented by a linear combination of these vectors, just as we have written the function as a superposition of the eigenfunctions. • Each dynamical system (e.g., the hydrogen atom or a harmonic oscillator) has its own Hilbert space, which has as many dimensions as there are eigenfunctions of the operators for the dynamical variables of that system. • The angular momentum operators are uniquely suitable for illustrating how this works in practice.

  26. Consider the three eigenfunctions for l = 1. The corresponding subspace of Hilbert space is three dimensional, and we can write the three eigenfunctions as unit vectors just as we would write an ordinary column vector: • Then any function F for which l = 1 can be a three-dimensional vector:

  27. representing the combination aY1,1+bY1,0+cY1,-1. •The angular momentum operators can now be represented as matrices that operate on these vectors. •Knowing the results of applying these operators to the spherical harmonics, we can deduce that a correct representation is (7.19)

  28. Using this representation, we can determine the result of applying operators Lx, Ly, or Lz to any function that can be expanded in spherical harmonics. • For example, the operation LxY1,1 may be represented as • This means that which you may verify by applying the differential operator for Lx to the function Y1,1. • Example Problem 7.3 Using Eqs.(7.18) and (7.19), find a vector that is an eigenfunction of Lx, and express this function in terms of spherical harmonics.

  29. Solution. Let us consider the eigenvalue zero. In that case the result of the matrix operation must be • But the result of the indicated operation is the column vector • If this vector is zero, then each of its components must be zero, and thus we know that b = 0 and a + c = 0. or a = -c. The absolute values of a and c are given by normalization :

  30. and the eigenfunction is • This can obviously be verified by application of the differential operator for Lx. • The same procedure can be applied to the set of five eigenfunctions for l = 2, using vectors in five dimensions, operated on by 5-by-5 matrices. • For example, the matrix for Lx is

  31. with eigenvalues of • In each of these examples. the matrix operator is very convenient but is not necessary, because the eigenvalues have already been found by using the differential operators. • But in some situations (for example, in the case of spin angular momentum), there are no space coordinates and no differential operator. In those cases we have no alternative but to use an abstract operator and to do the computations by matrix algebra. • But, you may ask how do we know what the matrix is, when there is no differential operator to guide us?

  32. One way to proceed is to use the commutation relations. Eqs. (7.17). • Later we shall see how theserelations allow us to analyze spin, but here we illustrate the power of Eqs.(7.17) by using them to derive the eigenvalues of L2 in a completely independent way. • In doing this we shall introduce a powerful shorthand method, invented by P. A. M. Dirac, to write symbols for eigenstates, independently of the particular functions of space, time, or spin that might be involved.

  33. §Generalized Operators and Dirac Notation • Dirac Notation and Raising (lowering) Operator • Instead of dealing with the specific form of the wave function, Dirac focused on the state of the system, and developed a set of symbols for state vectors in Hilbert space. We illustrate with the angular momentum states. • The symbol for the state vector is representing a state whose angular coordinates are given by Yl,m. • Operating on this state with operators L2 and Lz gives the operator equations that we already know,

  34. now written •Can we derive the eigenvalue using only the eigenvalues of Lz, the definition of l as the maximum value of m, and Eqs.(7.17)? •We can do it by using a trick similar to the one used to find the harmonic oscillator eigenfunctions. We define a new operator L+ =(Lx + iLy), apply it to the state and then demonstrate that the resulting state is also an eigenstate of Lz, as follows: •Using the commutation relations. we now make the substitutions

  35. to obtain • Using we then have which can be written or • Thus we see that L+ is a raising operator; is another eigenstate of Lz, with eigenvalue • We could write the state vector as In a similar way we can generate the state vector by applying the lowering operator L- = Lx - iLy, and by repeated use of these operators we can generate a whole set of eigenstates of Lz with eigenvalues

  36. Eigenvalues of L2 • We can prove that each of these states is an eigenstate of L2 as well as Lz; we do this by applying the operator to the state and work it out with the aid of the commutation relations [Eqs.(7.21)]. •But if all of these states have the same value of L2, there must be an upper limit on Lz. By definition, that upper limit is and the corresponding state may be written . • This means that because there is no state . If we apply the operator (Lx - iLy), we obtain

  37. or • Using the commutation rule and the relation we have or the same eigenvalue obtained in Chapter 2, but this time it is obtained from the commutation relations without reference to the form of the eigenfunctions. •We shall encounter other uses of this method in later chapters.

  38. Expectation Values in Dirac Notation • Consider the problem of finding the expectation value of Lx2 for a particle in the state denoted by where l = 1 and m = 1. • Using the form of Eq.(7.10). we could write then perform the L2xop operation and integrate over all angles, eventually obtaining the result in this case. • But that is not necessary if we know the expansion of Yl,m in terms of eigenfunctions of Lx. In that case we simply use Eq.(7.11) and the orthogonality of the spherical harmonics to obtain the result without

  39. integration. • In the Dirac notation the equivalent definition of Lx2 is and we expand the bra and ket vectors in the same way that we expanded the spherical harmonics to obtain Eq.(7.11). • With l = 1 and m = 1, the expansions are found from Eqs.(7.3) and (7.7) to be and with

  40. We also know that • Combining (7.38) through (7.41) and using the orthogonality and normalization of the vectors, we obtain as we anticipated. • Hermitian Operators • We cannot measure imaginary quantities. This fact imposes a requirement on any operator O that can represent a dynamical variable. • It is required that the expectation value <O> be a real quantity, or that for any acceptable wave function ψ, the integral ∫ψ*Fψd(volume) must be real. In the notation of Dirac, <ψ∣O∣ψ>.

  41. must be real. • Notice two important features of this bracket symbol: • . The form of the bracket is always interpreted to mean that the bra is the conjugate of the ket, so no asterisk is needed. • . The bracket can be evaluated by performing the operation O oneither the bra or the ket. If the operation is performed on the bra, then in the integral form it is equivalent to ∫(Fψ)*ψd(volume). This requires that

  42. • An operator that satisfies this requirement is called a Hermitean operator. • Thus we have the requirement that Any operator that represents a dynamical variable must be a Hermitean operator

  43. The EndThank You for Your Attention!

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