Quantum Mechanics in 3-Dimensions

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1 Quantum Mechanics in 3-Dimensions Pavithran S Iyer, 2nd yr BSc Physics, Chennai Mathematical Institute pavithra@cmi.ac.in August 28 th, Schrodinger equation in Spherical Coordinates 1.1 Transforming to the spherical coordinates The Schrodinger equation in one dimensional case may be written as 2 2 ψ Vψ = Eψ 1 2m x2 Using the differential operators, this can be generalized to three dimensions. Here, we just need to replace the one dimensional differential operator which is also called the laplacian. Therefore, we have 2 x 2 with 2 2 2m 2 ψ Vψ = Eψ 2 With the normalization condition which is applied in the three dimensional space: ψ 2 dv = 1 3 Where dv is the volume element in the three dimensional space. This volume element depends on the coordinate system used to represent ψ. In three dimensional space we can try to work with the spherical coordinates, as it has some advantages in studying some properties of the system that deal with rotations and its spherical symmetry. So, in order to shift to spherical coordinates, we must first write the Schrodinger equation in spherical coordinates. Therefore, we should first express the action of operators in the Schrodinger equation in spherical coordinates. Taking the operator 2, we know that in cartesian coordinates, 2 1

2 2 x + 2 y + 2 z. Let us consider the coordinate transformations: r = x 2 + y 2 + z 2 4 x = r cos φ sin θ 5 y = r sin φ sin θ 6 z = r cos θ 7 Therefore, to find the corresponding operators for x, y and z, we need to differentiate the RHS of equations eq.??, eq.?? and eq.?? respectively. Therefore, we have: x = r x r + θ x θ + φ x φ Where, we can substitute the values of r from eq.??, θ from eq.?? and φ from dividing eq.?? by eq.??: θ x Since cos θ = z r φ x Since y x = tan φ Putting together, we have: r x = x r x cos θ = z θ r = sec 2 φ φ x x y x 8 θ x = 1 zx sin θ r 3 9 φ x = y cos2 φ x 2 10 x = x r r + 1 zx sin θ r 3 θ cos2 θ x 2 φ 11 On substituting the values of x, y and z from equations eq.??, eq.?? and eq.??, we have: x = cos φ sin θ r + cos θ cos φ θ sin φ r r sin θ φ 12 On doing similar calculations on y and z. By looking at equation eq.??, we can say that z does not have any φ dependence. Hence, we expect the coefficient of φ in z to be 0. cos θ sin φ y = sin θ sin φ r + θ + cos φ r r sin θ φ 13 z = cos θ r sin θ r θ 14 2

3 Therefore as expected, z expression does not have a φ term. From the above expressions, we can calculate x, 2 y 2 and z. 2 Then, on adding them, we get 2. 2 = 1 r sin θ 1 + r 2 r r r 2 sin θ θ θ r 2 sin 2 15 θ φ Therefore, we can now see the Schrodinger equation in spherical coordinates: [ 2 1 r 2 ψ + 1 sin θ ψ 1 2 ψ + 2m r 2 r r r 2 sin θ θ θ r 2 sin 2 + Vψ = Eψ 16 θ φ 2 2 Solving the Schrodinger equation The above equation eq.?? looks too difficult to be solved for the general case. We make certain assumptions here. Let us assume the solution for this equation can be decomposed into a radial part as well as a an angular part, which are independent of each other. That is: ψr, θ, φ = RrY θ, φ 17 Plugging this into the schrodinger equation eq.??, the partial differentials will yield: ψ r = Y θ, φr r ; ψ θ = RrY θ ; ψ φ = RrY φ Therefore, the schrodinger equation decomposes into two parts - one which is dependent only on r and independent of θ and the other only dependent on θ and independent of r. 2 2m [ Y r 2 r r 2 R r + R r 2 sin θ Y + θ θ 18 R 2 Y r 2 sin 2 + VRrY θ, φ = ERrY θ, φ θ φ 2 19 In order to make the above equation variable separable, we need to remove R from terms which are θ or φ dependent and remove Y form terms which are r dependent. To do this, we must divide the equation by RrY θ, φ. Also, since r 2 is a common denominator in the first term of the LHS, we can get rid of it by multiplying throughout with 2mr2. 2 [ 1 R r r 2 R r 2mr2 2 Vr E + 1 Y [ 1 sin θ sin θ Y Y θ θ sin 2 = 0 θ φ

4 Now we see that the first term of the LHS is independent of θ and only dependent on r. While the second is independent of r. Since these two terms are independent of each other, and add up to a constant, they individually must be equal to some constants. Let us set some constants to them. For now, let us write this arbitrary constant as ll +1, for some complex l 1. This does not lead to any loss of generality as we see that ll + 1 can take the value of any complex number. Therefore, we set the first term of the LHS to be ll + 1 and naturally the second term would be ll + 1. therefore, we have the equations: 1 r 2 R 2mr2 R r 1 sin θ θ r sin θ Y θ Vr E = ll Y sin 2 = ll θ φ2 The first equation can be called the radial equation and the second one - the angular equation. The angular equation is the famous Laplace s equation which we try to solve by using the variable separable method. 2.1 Angular equation We now employ the variable separable technique. So, we write the solution of the equation as a product to two functions, one which is independent of φ and the other independent of θ. Therefore, let: Y θ, φ = ΘθΦφ 23 On putting this in the equation eq.?? and as before, dividing the equation by ΘθΦφ, we get: [ 1 sin θ sin θ Θ + ll + 1 sin 2 θ Φ Θ θ θ Φ φ = As expected, we get two terms on the LHS, one which is independent of θ and the other independent of φ. Therefore, we again argue that both of them must individually equal some constant. This time let us take some arbitrary constant represented by m 2. Again there is no loss in generality as m 2 can take all values on a complex plane like any other complex number z. On equating the equations to m 2 1 Later on we see the constraint that l is real and ll + 1 is nothing but the eigenvalue of J z operator 4

5 and m 2 respectively, we get: [ 1 sin θ sin θ Θ + ll + 1 sin 2 θ = m 2 Θ θ θ Φ Φ φ = 2 m Solving the Φφ equation: The differential equation for Φφ eq.?? looks easier to solve. On solving this equation 2 we get that: Φφ = e imφ 27 We now put in some physical constraints to find out more about the value of m. Since Φφ is the φ dependence of the wave function, we have the requirement that Φφ must be a single valued function. Therefore, after a full 360 o rotation of φ, the functional value must remain unchanged. Φφ + 2nπ = Φφ 28 From the above equation, we have Φ0 = Φ2π. Therefore, from equation eq.??: Therefore, we see that m can take only integer values. e im2π = e 0 1 m = 0, 1, Solving the Θθ equation: The differential equation eq.?? for Θθ is not very easy to solve and here we take the solution for granted. The solution for this differential equation is given by: Θθ = AP m l cos θ 30 Where P m l is called the Associated Legendre Polynomial. It is defined as: m Pl m x = 1 x 2 m d 2 P l x 31 dx 2 To solve this, consider the operator equation: D 2 Φ = m 2 Φ yd 2 + m 2 = 0 This is an equation with solution x = ±im Φ = e imφ. We cover the negative solutions by allowing m to run negative. 5

6 where P l is the l th Legendre Polynomial of x. It is given by: P l x = 1 l d x 2 1 l 32 2 l l! dx Now that we have solved the equations for Θθ and Φφ, we know that angular part of the wave function, up to some normalization constant. Therefore we know the function Y θ, φ in equation eq.??. Y θ, φ = P m l cos θe imφ 33 We may now refer to this function as Yl m θ, φ for which the reasons are obvious. 2.2 Normalizing the angular wave function To normalize the wave function, consider the equation eq.??. We may replace the volume element dv by the volume element in spherical coordinates, given by: dv = r 2 sin θdrdθdφ r 2 RrdrYl m θ, φ = 1 34 We can try to normalize the radial and angular parts of the wave function separately. Since, we have not yet calculated the radial part, we assume that it is normalized. That is: r 2 Rr 2 dr = 1 35 We go ahead computing the normalization constant for the angular part. We use the convention that dω = dθdφ. sin θ Y m l θ, φ 2 dω 36 On putting the definition of Yl m θ, φ from equation eq.??, we get the normalization constants. The normalized angular wave function now takes the form: Yl m θ, φ = ɛ V 2l + 1l m! e imφ Pl m cosθ 37 4πl + m! Where ɛ is a constant that takes the value 1 m for all m 0 and the value 1 for all m < 0. These normalized angular wave functions are called Spherical Harmonics. 6

7 3 Radial Function The radial function refer to equation eq.?? is given by: 1 r 2 R 2mr2 Vr E = ll R r r 2 Notice that the angular part of the wavefunction is constant for all spherically symmetric potentials. It is the radial part alone that is affected. The above equation can be simplified by putting ur = rrr. It reduces to: 2 d 2 u 2m dr 2 [V r + 2 ll + 1 E 2m r 2 The normalization condition on the wavefunction is: r 0 u = 0 39 u 2 dr =

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