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h? [1 a 21 [r² dr () 1 1 sin (0) ý = Ev 4TEor + dr r²2 sin² (0) 00 r2 sin? ( (0) (14) This problem involves solving the differential equation (14) by converting the partial differential equation into three ordinary differential equations each consisting only one of the independent variables (r, 0 and ø). We will assume that (r, 0, ¢) is separable, that is we can write p in the form v (r, 0, ø) = F (r) G (8, 6), (15) where F depends only on r and G depends only on 0 and o. (a) With the help of equation (15) and (14) show that h? | G d 2µ r2 dr OG sin (0) · dF F dr r2 sin (0) d0 Q FG = EFG F p² sin² (0) ( d¢². (16) [4] (b) Using equation (16) show that 1 d 2µr? dr + E 4TEor dr ƏG 1 a ( sin (0) · 1 = 0 sin (0) d0 sin? (17) [4] Note that the part within the curly braces on the left depends only on r and the part on the right depends only on 0 and ø. Since r, 0 and o are completely independent of each other, both parts must differ by the same constant to be equal to zero. We will call this constant (j(j+1)), where j is a constant (if j is a constant so is j(j+1)). To satisfy equation (17), we must have: 2µr? + E) = j(j+1), 4περΓ 1 d (18) F dr dr and 1 a ( sin (0) 1 (19) G sin (0) Ə0 2 sin² (0) Ə6? = -j(j+1). (c) Show that equation (18) combined with equation (19) in a certain way gives us (17) (this means that equation (18) along with (19) is completely equivalent to equation(17)). [3]

h? [1 a 21 [r² dr () 1 1 sin (0) ý = Ev 4TEor + dr r²2 sin² (0) 00 r2 sin? ( (0) (14) This problem involves solving the differential equation (14) by converting the partial differential equation into three ordinary differential equations each consisting only one of the independent variables (r, 0 and ø). We will assume that (r, 0, ¢) is separable, that is we can write p in the form v (r, 0, ø) = F (r) G (8, 6), (15) where F depends only on r and G depends only on 0 and o. (a) With the help of equation (15) and (14) show that h? | G d 2µ r2 dr OG sin (0) · dF F dr r2 sin (0) d0 Q FG = EFG F p² sin² (0) ( d¢². (16) [4] (b) Using equation (16) show that 1 d 2µr? dr + E 4TEor dr ƏG 1 a ( sin (0) · 1 = 0 sin (0) d0 sin? (17) [4] Note that the part within the curly braces on the left depends only on r and the part on the right depends only on 0 and ø. Since r, 0 and o are completely independent of each other, both parts must differ by the same constant to be equal to zero. We will call this constant (j(j+1)), where j is a constant (if j is a constant so is j(j+1)). To satisfy equation (17), we must have: 2µr? + E) = j(j+1), 4περΓ 1 d (18) F dr dr and 1 a ( sin (0) 1 (19) G sin (0) Ə0 2 sin² (0) Ə6? = -j(j+1). (c) Show that equation (18) combined with equation (19) in a certain way gives us (17) (this means that equation (18) along with (19) is completely equivalent to equation(17)). [3]

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4. Charged Quantum Particles We have a pair of charged particles with differing masses, m1 and m2, differing nature of charges +qı and -q2, and we want to get some idea about their interaction. When we have a two-body problem, often the problem simplifies if we can turn it into a one-body problem, instead of having two masses, we use a concept known as equivalent mass, u defined as u = mim2 mi+m2 The two-body problem then becomes equivalent to a one-body problem with mass i. The following cartoon diagram shows the scenerio. + While solving differential equations, we often need the help of operators. In simple terms, the idea of Hamiltonian operator is that we sum the kinetic energy operator with the potential operator. The kinetic operator is momentum operator, p, squared divided by 2µ. From quantum mechanics p = -ihV Then p? becomes -hV². The potential energy due to two charged particles is given by the operator V law). Now, if we let q192 = Q, the Hamiltonian operator becomes: 192 (this results from Coulomb's 4TEor The Schrödinger's equation is Hp = E (where E is some constant); we will want to find the solution y from this equation. The Laplacian (V²) in spherical coordinates is (here r = radial distance, O = azimuthal angle polar angle and 1 + p2 sin (0) 0 1 sin (0) a0) + 72 sin² (0) p2 ar 20 Using the Laplacian from above, we can find expand the Schrödinger's equation (H acts on to produce the following, you can think of it as a sort of multiplication)
Transcribed Image Text:4. Charged Quantum Particles We have a pair of charged particles with differing masses, m1 and m2, differing nature of charges +qı and -q2, and we want to get some idea about their interaction. When we have a two-body problem, often the problem simplifies if we can turn it into a one-body problem, instead of having two masses, we use a concept known as equivalent mass, u defined as u = mim2 mi+m2 The two-body problem then becomes equivalent to a one-body problem with mass i. The following cartoon diagram shows the scenerio. + While solving differential equations, we often need the help of operators. In simple terms, the idea of Hamiltonian operator is that we sum the kinetic energy operator with the potential operator. The kinetic operator is momentum operator, p, squared divided by 2µ. From quantum mechanics p = -ihV Then p? becomes -hV². The potential energy due to two charged particles is given by the operator V law). Now, if we let q192 = Q, the Hamiltonian operator becomes: 192 (this results from Coulomb's 4TEor The Schrödinger's equation is Hp = E (where E is some constant); we will want to find the solution y from this equation. The Laplacian (V²) in spherical coordinates is (here r = radial distance, O = azimuthal angle polar angle and 1 + p2 sin (0) 0 1 sin (0) a0) + 72 sin² (0) p2 ar 20 Using the Laplacian from above, we can find expand the Schrödinger's equation (H acts on to produce the following, you can think of it as a sort of multiplication)
h? [1 a Q h = Ev 4Teor 1 1 sin (0) + r2 sin? ( + dr r2 sin? (0) 00 (14) This problem involves solving the differential equation (14) by converting the partial differential equation into three ordinary differential equations each consisting only one of the independent variables (r, 0 and ø). We will assume that (r, 0, 4) is separable, that is wve can write in the form v (r, 0, 6) = F (r) G (0, ¢), (15) where F depends only on r and G depends only on 0 and o. (a) With the help of equation (15) and (14) show that h? | G d dF dr F OG (sin (0) ) 2µ r2 dr r2 sin (0) 0 Q FG = EFG F r2 sin? (0) (a4? (16) 14] (b) Using equation (16) show that 1 d 2µr2 Q + E 4T€or dF' Fdr F dr dr OG a ( sin (0) 1 1 1 = 0 sin (0) d0 sin? (0) d02 (17) [4] Note that the part within the curly braces on the left depends only on r and the part on the right depends only on 0 and o. Since r, 0 and o are completely independent of each other, both parts must differ by the same constant to be equal to zero. We will call this constant (j(j+ 1)), where j is a constant (if j is a constant so is j(j+1)). To satisfy equation (17), we must have: 1 d 2jur? dF F dr + E) = j(j+ 1), (18) %3D dr and ƏG ( sin (0)) 1 1 (19) G | sin (0) d0 sin² (0) d6² =-j(j+1). (c) Show that equation (18) combined with equation (19) in a certain way gives us (17) (this means that equation (18) along with (19) is completely equivalent to equation(17)). [3]
Transcribed Image Text:h? [1 a Q h = Ev 4Teor 1 1 sin (0) + r2 sin? ( + dr r2 sin? (0) 00 (14) This problem involves solving the differential equation (14) by converting the partial differential equation into three ordinary differential equations each consisting only one of the independent variables (r, 0 and ø). We will assume that (r, 0, 4) is separable, that is wve can write in the form v (r, 0, 6) = F (r) G (0, ¢), (15) where F depends only on r and G depends only on 0 and o. (a) With the help of equation (15) and (14) show that h? | G d dF dr F OG (sin (0) ) 2µ r2 dr r2 sin (0) 0 Q FG = EFG F r2 sin? (0) (a4? (16) 14] (b) Using equation (16) show that 1 d 2µr2 Q + E 4T€or dF' Fdr F dr dr OG a ( sin (0) 1 1 1 = 0 sin (0) d0 sin? (0) d02 (17) [4] Note that the part within the curly braces on the left depends only on r and the part on the right depends only on 0 and o. Since r, 0 and o are completely independent of each other, both parts must differ by the same constant to be equal to zero. We will call this constant (j(j+ 1)), where j is a constant (if j is a constant so is j(j+1)). To satisfy equation (17), we must have: 1 d 2jur? dF F dr + E) = j(j+ 1), (18) %3D dr and ƏG ( sin (0)) 1 1 (19) G | sin (0) d0 sin² (0) d6² =-j(j+1). (c) Show that equation (18) combined with equation (19) in a certain way gives us (17) (this means that equation (18) along with (19) is completely equivalent to equation(17)). [3]
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