PHYS3060 Quantum Mechanics Solution to Problem Set 6

Transcription

1 PHYS3060 Quantum Mechanics Solution to Problem Set 6 Multi-Electron Atoms (a) Write down the time-independent Schrödinger equation for a lithium atom (Z=3; i.e. three electrons) and explain the purpose of every term in this equation (there should be 0). You may use the shorthand notation (e.g. H, V 2, see lecture notes). If you do, make sure you define the shorthand symbols that you use. In short hand notation, the time-independent Schrödinger equation for the lithium atom reads [Ĥ + Ĥ2 + Ĥ3 + V 2 + V 3 + V 23 ] ψ ( r, r 2, r 3 ) = Eψ ( r, r 2, r 3 ) () The three Ĥi operators are defined as follows Ĥ i = h2 2m 2 3e2 4πɛ 0 r i (2) in this, the first and second term represent the kinetic energy of the electron and the electron-nucleus Coulomb attraction, respectively. (Remember that the atomic number of lithium is Z=3). For the three electrons i=,2, and 3, there are a total of six terms (three kinetic energy terms and three electron-nucleus attraction terms). The electron-electron repulsion between pairs of electrons is represented by the three V ij terms, which are defined as follows: V ij = + e 2 4πɛ 0 r i r j (3) Note the + sign (this was a fairly common mistake!); electrons are negatively charged, and thus the Coulomb energy between two electrons must be positive. The last (0th) term on the right hand side of Eq. is the total energy term. (b) Assume that the three particle eigenfunction ψ(r,r 2,r 3 ) can be written as a product of three single electron eigenfunction, i.e. ψ(r, r 2, r 3 ) = ψ n (r )ψ n2 (r 2 )ψ n3 (r 3 ). (4) Assume further that the electron-electron repulsion experienced by electron i is simply a constant value U i dependent on the main quantum number (n i ) of the electron. Use this to simplify the Schrödinger equation. Use the method of separation of variables to separate this simplified equation into three Schroedinger equations; one for each electron. Be judicious in your choice of separation variables, such that E, E 2, and E 3 are the energies of the first, second, and third electron, respectively, and that E=E +E 2 +E 3. As per instructions, we simplify the Schrödinger equation and use single-index constants U i to represent the electron-electron repulsion. We write the three-electron eigenfunction ψ( r, r 2, r 3 ) as product of three singleelectron eigenfunctions ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ). With these changes, Eq. now reads [Ĥ + Ĥ2 + Ĥ3 + U + U 2 + U 3 ] ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) = Eψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) (5) In this form, all terms are one-index terms, that is, they depend on the position of one electron only, which allows us to apply the method of separation of variables. We will go through this here step-by-step. First evaluate the square bracket on the left hand side.

2 2 Ĥ ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + Ĥ2ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + Ĥ3ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + (6) U ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + U 2 ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + U 3 ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) = Eψ n ( r )ψ n2 ( r 2 )ψ n 3( r 3 ) It is important to recognize here that the Ĥi s are not just simple factors and that the ψ s on their right cannot simply be cancelled. Ĥ i is a differential operator (see definition in Eq. 2) that modify those ψ s that are associated with the coordinate r i ; it leaves the others unaffected. This means for the Ĥψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) term that we must leave ψ n ( r ) on the right hand side of Ĥ. However, we can pull over ψ n2 ( r 2 ) and ψ n3 ( r 3 ) to the left hand side of Ĥ (... and, will in due course, cancel these factors). Doing this for all terms, we get: ψ n2 ( r 2 )ψ n3 ( r 3 )Ĥψ n ( r ) + ψ n ( r )ψ n3 ( r 3 )Ĥ2ψ n2 ( r 2 ) + ψ n ( r )ψ n2 ( r 2 )Ĥ3ψ n3 ( r 3 ) + (7) U ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + U 2 ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) + U 3 ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) = Eψ n ( r )ψ n2 ( r 2 )ψ n 3( r 3 ) Division by ψ n ( r )ψ n2 ( r 2 )ψ n3 ( r 3 ) leads to ψ n ( r )Ĥψ n ( r ) + ψ n2 ( r 2 )Ĥ2ψ n2 ( r 2 ) + ψ n3 ( r 3 )Ĥ3ψ n3 ( r 3 ) + U + U 2 + U 3 = E (8) Reorder such that terms dependent on electron are on the left hand side and all other terms on the right hand side. ψ n ( r )Ĥψ n ( r ) + U = E U 2 U 3 ψ n2 ( r 2 )Ĥ2ψ n2 ( r 2 ) ψ n3 ( r 3 )Ĥ3ψ n3 ( r 3 ) (9) In this form, everything associated with electron is on one side, and everything associated with the other two electrons is on the other side. We can therefore separate the equation into two, introducing a separation variable that is conveniently called E n. The left and right hand side thus read respectively and ψ n ( r )Ĥψ n ( r ) + U = E n (0) E U 2 U 3 ψ n2 ( r 2 )Ĥ2ψ n2 ( r 2 ) ψ n3 ( r 3 )Ĥ3ψ n3 ( r 3 ) = E n () Multiplication by ψ n ( r ) turns the left-hand-side equation (Eq. 0) into the familiar form of a Schrödinger Equation for a single electron, with a single extra (potential energy) term associated with U. Ĥ ψ n ( r ) + U ψ n ( r ) = E n ψ n ( r ) (2) The right-hand-side equation (Eq. ) depends on the position of electrons 2 and 3. We reorder this equation so as to bring all terms associated with electron 2 to the left hand side, and all terms associated with electron 3 to the right hand side. This gives us U 2 + ψ n2 ( r 2 )Ĥ2ψ n2 ( r 2 ) = E E n U 3 ψ n3 ( r 3 )Ĥ3ψ n3 ( r 3 ) (3)

3 3 We can again separate this into two equations, using E n2 Eq. 3 we get as the separation variable. For the left hand side of U 2 + which we can rewrite into the same form as Eq. 2: ψ n2 ( r 2 )Ĥ2ψ n2 ( r 2 ) = E n2 (4) For the right hand side of Eq. 3 we obtain Ĥ 2 ψ n2 ( r 2 ) + U 2 ψ n2 ( r 2 ) = E n2 ψ n2 ( r 2 ) (5) which we can reorder into E E n U 3 ψ n3 ( r 3 )Ĥ3ψ n3 ( r 3 ) = E n2 (6) Ĥ 3 ψ n3 ( r 3 ) + U 3 ψ n3 ( r 3 ) = [E E n E n2 ]ψ n3 ( r 3 ) (7) This equation can be further simplified by introducing E n3 such that which gives us after substitution into Eq. 7) E = E n + E n2 + E n3 (8) Ĥ 3 ψ n3 ( r 3 ) + U 3 ψ n3 ( r 3 ) = E n3 ψ n3 ( r 3 ) (9) Together, Eqs. 2, 5, and 9 are the three separated one-electron equations for the lithium atom. These have the general form: Ĥ i ψ ni ( r i ) + U i ψ ni ( r i ) = E ni ψ ni ( r i ) i =, 2, 3 (20) (c) Consider the special case, where the e-e repulsion energy is approximated as U i =0. Show that in this case one can use the energy Bohr/Schrödinger energy expression for a single-electron atom E n = 3.6 ev Z2 n 2 to work out the energy levels of a lithium atom as a function of the main quantum numbers n,n 2 and n 3 of the three electrons. Work out the energy of the three lowest electronic states of lithium in this simplified model. Make sure your result takes Pauli s exclusion principle into account. With the approximation U i =0, the one-electron equations (Eq. 20) simplify to Ĥ i ψ ni ( r i ) = E ni ψ ni ( r i ) (2) This is the Schödinger equation for a hydrogen-like atom, and we know that the solutions to this equation are hydrogen-like eigenfunctions and the energy is a function of the main quantum number n i of the electron given by

4 4 E ni = 3.6eV Z2 (22) As introduced above (Eq. 8), the total energy E of the lithium atom is the sum of the three single electron energies E n, E n2, and E n3, thus ( E n,n 2,n 3 = = E n + E n2 + E n3 = 3.6eV Z 2 n 2 + n 2 + ) 2 n 2 3 (23) The question now asks for the energy of the three lowest electronic states. For this you have to find the three electron configurations (combinations of the three quantum numbers n, n 2, and n 3 ) that give the lowest energy according to Eq. 23. The combinations are further subject to the Pauli principle which dictates that any two electrons in an atom must differ in at least one of the four quantum numbers n i, l i, m l,i, m s,i. The upshot of this is that a level with quantum number n can accommodate a maximum of 2n 2 electrons (see the discussion of degeneracy in E&R 7.5); that is, we have a maximum of two electrons with n= (and a maximum 8 with n=2). This means the configuration where all three electrons are in the n= is forbidden. The three lowest energy configurations are: n = n 2 = n 3 = 2 E 2 = ev (24) n = n 2 = n 3 = 3 E 3 = ev (25) n = n 2 = n 3 = 4 E 4 = ev (26) Note that raising the one electron into successively higher levels is more favorable than raising two electrons from n= into n = 2, i.e. n = n 2 = 2 n 3 = 2 E 2 = 83.6 ev (27) (d) How does the energy expression change when U i adopts a finite value? Because U i was introduced as a constant; the energy equation becomes ( E n,n 2,n 3 = 3.6 ev Z 2 n 2 + n 2 + ) 2 n 2 + U + U 2 + U 3 (28) 3 This can be shown by rewriting the general single-electron solution Eq. 20 as follows and define F ni = [E ni U i ] which gives us Ĥ i ψ ni ( r i ) = [E ni U i ] ψ ni ( r i ) (29) Ĥ i ψ ni ( r i ) = F ni ψ ni ( r i ) (30) This too has the general form of a hydrogen-like Schrödinger equation, and the known energy solution for F n is F ni = 3.6eV Z2 (3)

5 5 Substituting back F ni = [E ni U i ] gives us E ni = 3.6eV Z2 And for all three electrons in combination, we get Eq. 28 above. + U i (32) (e) Briefly explain why the product of single electron eigenfunctions (Eq. 4) does not satisfy the particle indistinguishability principle. Propose a better, antisymmetric, form for the three-electron eigenfunction ψ(r,r 2,r 3 ). Explain why this form does not allow two electrons to have the same quantum numbers. As a simple product of one-electron functions, Eq. 4 does not satisfy the indistinguishability principle. The exchange of two particles leads to a change in the probability density (see discussion E&R chapter 9-2). An improved form uses a determinant ψ ( r, r 2, r 3 ) = 6 ψ n ( r ) ψ n ( r 2 ) ψ n ( r 3 ) ψ n2 ( r ) ψ n2 ( r 2 ) ψ n2 ( r 3 ) ψ n3 ( r ) ψ n3 ( r 2 ) ψ n3 ( r 3 ) (33) In the determinant form, the exchange of two electrons is equivalent to the exchange of two columns, which turns the determinant into its negative as required for an antisymmetric wavefunction. If two electrons have the same quantum number (e.g. n 2 =n 3 ), then two of the rows of the determinant would be identical. This in turn means the determinant evaluates to zero. Note that this argument includes the spin quantum number. (f) With the improved, antisymmetric form of the three-electron eigenfunction, would it still be possible to use the method of separation of variables to separate the three-electrons Schrödinger equation into three single-electron equations? Provide a brief illustrative response, no need to go through the full algebra. The answer is no, separation of variables is not possible. Try it out (say, for the simpler case of a antisymmetric two electron eigenfunction)!