# Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

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1 Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 + cotθ θ + 2 ) ' sin 2 θ φ 2 + * % 3 & 4 ) e iφ is an eigenfunction of the ˆ L 2 operator To show that the function Y θ,φ) is an eigenfunction of L ˆ 2, we apply L ˆ 2 to the function, ˆ L 2 Y θ,φ )! 2 2! 2 & 3 ) ' 4 *! 2 & 3 ) ' 4 * & θ 2 + cotθ θ + 2 )& 3 ) ' sin 2 θ φ 2 + *' 4 * & 2 ' θ 2-2 / θ cotθ θ + sin 2 θ φ 2 eiφ ) + cotθ e iφ ) + e iφ * eiφ ) + θ sin 2 θ φ 2 2 eiφ ) 2 Evaluating the derivatives separately gives eiφ θ ) e iφ ) θ e iφ cosθ 2 θ 2 eiφ ) e iφ 2 θ 2 e iφ ) 2 φ 2 eiφ ) 2 ) φ 2 eiφ i 2 e iφ e iφ

2 2 Continued Substituting, ˆ L 2 Y θ,φ & ) + ' 4 * )! 2 3 ˆ L 2 Y θ,φ! 2 & 3 ) ' 4 *! 2 & 3 ) ' 4 * & ) + ' 4 * )! / θ 2 - e iφ / eiφ ) + cotθ eiφ ) + θ + cotθ cosθ e iφ ) cot θ cosθ) + sin 2 θ ) / - + cos2 θ / 2 eiφ sin 2 θ φ 2 2 sin 2 eiφ ) θ 2 2 eiφ eiφ ) 2 The term in brackets can be simplified by use of a common denominator, + cos2 θ sin2 θ + cos2 θ sin2 θ + cos 2 θ sin2 θ + sin 2 θ 2sin2 θ 2 Note that the trigonometric identity cos 2 θ sin 2 θ was used in order to simplify the above expression Substituting, ˆ L 2 Y θ,φ & ) + ' 4 * )! 2 3! 2 & 3 ) ' 4 * 2! 2 & 3 ) ' 4 * L ˆ 2 Y θ,φ) 2! 2 Y θ,φ), + cos2 θ / - 2) e iφ e iφ So, we see that the function Y θ,φ) is an eigenfunction of the L ˆ 2 operator The eigenvalue is 2! 2 eiφ

3 3 2 Show that the spherical harmonic function Y 2 θ,φ) is normalized This function is given by Y 2 θ,φ) % 5 & 6 ) 3cos 2 θ ) To show that the function Y 2 θ,φ) is normalized, we must show that Y * 2 θ,φ)y 2 θ,φ) dθ dφ As shown in the integral above, the requisite integral is actually a double integral over the two angular coordinates, θ and φ In addition, the angular portion of the volume element in spherical polar coordinates, dθ dφ, must be included Substituting the expression for the spherical harmonic function Y 2 θ,φ) into the normalization integral yields Y * 2 θ,φ)y 2 θ,φ) dθ dφ ) & 5 ) 3cos 2 & 5 ) θ 3cos 2 θ dθ dφ ' 6 * ' 6 * ) 2 & 5 ) 3cos 2 θ dθ dφ ' 6 * ) Note that in this case, the function Y 2 θ,φ) does not include the imaginary number i, so the complex conjugate is the same as the original function In addition, the normalization constants have been pulled out and written in front of the integral The next step is to break the double integral into separate parts for θ and φ, Y * 2 θ,φ)y 2 θ,φ) dθ dφ ) 2 dθ dφ & 5 ) 3cos 2 θ ' 6 * The integral over φ may be evaluated immediately, Substituting, the normalization integral becomes 2 dφ 2 Y * 2 θ,φ)y 2 θ,φ) dθ dφ ) 2 dθ & 5 ) + 3cos 2 θ ' 8*

4 4 2 Continued Upon expansion of the factor 3cos 2 θ ) 2 9 cos 4 θ 6 cos 2 θ +, we obtain Y * 2 θ,φ)y 2 θ,φ) dθ dφ & 5 ), + 9 cos 4 θ dθ ' 8 * - 6 cos 2 θ dθ + dθ2 2 These integrals may be easily evaluated by using tables or by making the substitution u cosθ We then get cos 4 θ dθ 2 5, cos2 θ dθ 2 3, and dθ 2 Substituting these into the normalization expression yields Y * 2 θ,φ)y 2 θ,φ) dθ dφ & 5 ) ' 8 * / Therefore, the spherical harmonic function Y 2 θ,φ) is normalized

5 5 3 Show that the spherical harmonic functions Y θ,φ) and Y, θ,φ) are orthogonal These functions are given by Y θ,φ) Y, θ,φ) % 3 & 4 ) % 3 & 8 ) e iφ e iφ Two functions f x) and g x) are orthogonal if f * x) g x) dx functions given above, in order to prove orthogonality, we therefore must show that Y * θ,φ) Y, θ,φ) dθ dφ For the two spherical harmonic Note that the order of the functions does not matter, and it also does not matter which of the functions is given the complex conjugate Substituting the spherical harmonic functions into the orthogonality integral and pulling out the constant terms yields Y * θ,φ)y, θ,φ) dθ dφ ), 8 + ), 8 + e iφ ), 8 + sin 3 θ e 2iφ dθ dφ e iφ dθ dφ Next, the double integral may be separated into a product of integrals involving θ and φ, Y * θ,φ)y θ,φ) dθ dφ 2 & 3 ) e 2iφ dφ sin 3 θ dθ ' 8 * The integral involving the angle φ may be easily evaluated The indefinite integral has the solution e 2iφ dφ 2i e 2iφ i 2 e 2iφ The last form is obtained by multiplying by i /i and using the relation i 2 Evaluating this integral at the limits yields 2 e 2iφ dφ i 2 e i4 e ) i 2 e i4 )

6 6 3 Continued The exponential e i4 is evaluated using Euler s relation, e iα cosα + i sinα Thus, e i4 cos 4 i sin 4 Since cos 4 and sin 4, this becomes e i4 Therefore, the integral involving φ simplifies to 2 e 2iφ dφ i 2 e i4 ) i 2 The orthogonality integral for Y θ,φ) and Y, θ,φ) becomes ) Y * θ,φ)y, θ,φ ) dθ dφ ), sin 3 θ dθ 8 + Or, in other words, the functions are orthogonal regardless of the value of the integral involving θ, and therefore Y * θ,φ)y, θ,φ ) dθ dφ As a side note, the integral involving θ may be easily evaluated using tables This leads to the result sin 3 θ dθ 4 3

7 7 4 Determine the average value of cosθ for a system in the state described by the wavefunction Y θ,φ) % 3 & 4 ) e iφ The expectation value of cosθ for a system with wavefunction Y θ,φ) is defined as cosθ Y * θ,φ) cosθ Y θ,φ) dθ dφ Substituting the wavefunction and simplifying, cosθ cosθ Y * θ,φ) cosθ Y θ,φ) dθ dφ 2 ), e iφ 3 / cosθ 2 ), e iφ 3 dθ dφ / ), e iφ cosθ e iφ dθ dφ 4 + ), sin 3 θ cosθ dθ dφ 4 + Breaking the double integral into separate parts for θ and φ yields cosθ \$ 3 ' & ) sin 3 θ cosθ dθ dφ % 4 The integral over φ may be evaluated immediately, 2 dφ 2 The integral over θ may be easily evaluated by using tables or by making the substitution u The result is sin 3 θ cosθ dθ Substituting these into the expectation value expression yields cosθ

8 8 5 The spherical harmonics Y θ,φ) and Y θ,φ) are Y θ,φ) Y θ,φ) % & 4 ) % 3 & 4 ) cosθ Determine the result of measuring L 2, L z, and L x for systems described by each of these states Since the functions Y θ,φ) and Y θ,φ) are eigenfunctions of the operators L ˆ 2 and L ˆ z, measurement of L 2 and L z will yield the eigenvalues, with the result! 2 ll +) for L 2 and m l " for L z For L x, measurement will yield the expectation value, * L x Y lml θ,φ) L ˆ x Y lml θ,φ) dθ dφ To evaluate the expectation value we need to determine the outcome of the operation of L ˆ x on Y lml θ,φ ) For the function Y θ,φ) % & 4 ), we have ˆ L x Y θ,φ ) i! ' or L ˆ x Y θ,φ), & θ ) & ) cotθ cosφ + φ * ' 4 * since the derivative of a constant is zero Substituting this into the expectation value, we get L x for a system in the state Y θ,φ)

9 9 5 Continued For the function Y θ,φ) % 3 & 4 ) cosθ, we have ˆ L x Y θ,φ ) i! & ' θ ) & 3 ) cotθ cosφ + φ * ' 4 * cosθ Evaluating the derivatives, we have θ cosθ and φ cosθ Substituting, ˆ L x Y θ,φ ) i! & 3 ) ' 4 * sin 2 θ Substituting for the function Y * θ,φ) and the term L ˆ x Y θ,φ), the expectation value is L x Y * θ,φ L x Y θ,φ) dθ dφ ) ˆ, & 3 ) /, cosθ - ' 4 * i! & 3 ) / sin 2 θ dθ dφ - ' 4 * Pulling out the constants and dividing into separate integrals for θ and φ yields 2 \$ 3 ' L x i! & ) dφ sin 3 θ cosθ dθ % 4 The integral over the angle φ gives a factor of 2, while the integral over the angle θ may be evaluated using tables or by making the substitution u to give sin 3 θ cosθ dθ

10 5 Continued Therefore, the expectation value of L x for the function Y θ,φ) is L x To summarize, the results of measurement for the states Y θ,φ) and Y θ,φ) are shown in the table below State L 2! 2 ll +) L z m l " L x L x ) ) 2! 2 Y θ,φ Y θ,φ 6 Determine the possible angles that the quantum mechanical orbital angular momentum vector make with the z-axis for l 2! L can The angle θ that the orbital angular momentum vector makes with the z-axis is described by the relation cosθ L z L m l " " 2 ll +) m l ll +) For l 2, there are 5 possible values of m l, and therefore 5 possible angles Substituting l 2 we have, cosθ m l 6, with m l 2,,,, and 2 The equation may be solved for the angle θ by taking the inverse cosine The results are shown in the table below m l θ degrees)

11 7 Determine the result of measuring S 2 and S z for an electron with a wavefunction corresponding to the "spin up" state The wavefunction for the "spin up" state of an electron is denoted α The eigenvalue equations for the "spin up" state are ˆ S 2 α! 2 s s + ) α 3 4!2 α ˆ S z α m l! α 2!α Since the "spin up" wavefunction is an eigenfunction of the operators ˆ S 2 and ˆ S z, measurement of S 2 and S z will yield the eigenvalues, with the result 3 4!2 for S 2 and 2! for S z 8 Determine the possible angles that the quantum mechanical electron spin angular momentum vector! S can make with the z-axis The angle θ that the spin angular momentum vector makes with the z-axis is described by the relation cosθ S z S m s!! 2 ss +) For an electron, we have s 2 and m s ± 2 Substituting, the equation becomes cosθ ± 2!! ), or cosθ ± 3 The two angles that we get are θ 253! and θ 547!

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