Lectures 3-4: Quantum mechanics of one-electron atoms The Schrödinger equation o One-electron atom is simplest bound system in nature. o Schrödinger equation for one-electron atom. o Consists of electron moving in 3D Coulomb potential of nucleus : o Solving the Schrödinger equation. o Z =1 for hydrogen, Z =2 for helium, etc. o Wavefunctions and eigenvalues. o As mass of nucleus (M) >> mass of electron, electron moves moves relative to nucleus as if nucleus was fixed, and the mass (m) of were slightly reduced to µ. o Atomic orbitals. o Classically, the total energy of system is: " = "(x,y,z,t) ! PY3004 PY3004 The Schrödinger equation The Schrödinger equation o Using the Equivalence Principle, the classical dynamical quantities can be replaced with their o Since V(x,y,z) does not depend on time, "(x,y,z,t) =# ( x , y , z)e$iEt / h is a solution to the associated differential operators: Schrödinger equation. o The time-independent Schrödinger! equation can therefore be written: o Substituting, we obtain the operator equation: o As V = V(r), convenient to use spherical polar coordinates. o Assuming electron can be described by a wavefunction of form (1) can write where 2 1 # # 1 # # 1 # "2 = (r 2 )+ (sin$ )+ r 2 #r #r r 2 sin$ #$ #$ r 2 sin 2$ #% 2 or o Can now use separation of variables to split the partial differential o This is the Schrodinger equation for the system, where, is the equation! into a set of ordinary differential equations. Laplacian operator. PY3004 PY3004 Separation of the Schrödinger equation Separation of the Schrödinger equation o As the LHS of Eqn 3 does nor depend on r or ! and RHS does not depend on " their o Assuming that the eigenfunction is separable: common value cannot depend on any of these variables. (2) o Substituting (2) into the time-independent Schrodinger equation (1) and using the Laplacian: o Therefore set the LHS of Eqn 3 to a constant: (4) and o Carrying out the differentiations, o As LHS only depends on r and RHS on !, both sides must equal a constant, which we choose as l(l+1): o Note total derivatives now used, as R is a function of r alone, etc. (5) o Now multiply through by "2µ r2 sin2 # /R $% 2 and taking transpose, (6) h (3) o We have now separated the time-independent Schrödinger equation into three ! differential equations, which each only depend on one of # (4), $ (5) and R(6). PY3004 PY3004 Azimuthal solutions (#(")) Polar solutions (!(!)) o A particular solution of (4) is o By making a change of variables using z = rcos", Eqn. 5 is transformed into a differential equation called the associated Legendre equation: o As the einegfunctions must be single valued, i.e., #(0)=#(2%) => (7) o Solutions to Eqn. 7 are of the form or using Euler’s formula, 1= cosml 2" + isinml 2" o This is only satisfied if ml = 0, ±1, ±2, ... where F (cos") are associated Legendre polynomial functions. ! l|ml | o Therefore, acceptable solutions to (4) only exist when m can only have certain integer values, o The requirement that ! remains finite leads to integer values of l and certain l restrictions on m as follows: i.e. it is a quantum number. ! l l = 0, 1, 2, 3, ... ml = -l, -l+1, .., 0, .., l-1, l o ml is called the magnetic quantum number as plays role when atom in magnetic field. o l is the orbital or angular momentum quantum number. PY3004 PY3004 Polar solutions (!(!)) and spherical harmonics Spherical harmonic solutions o We can write the associated Legendre functions o The first few spherical harmonics are: 0 Y 0= 1 with the quantum number subscripts as: 0 Y 1= cos" Y ±1 = (1-cos2")1/2 e±i% ! = 1 1 00 Y 0 = 1-3cos2" ! = cos" ! = (1-cos2")1/2 2 10 1±1 ±1 2 1/2 ±i% 2 2 1/2 Y 2= (1-cos ") cos" e # !20 = 1-3cos " !2±1 = (1-cos ") cos" 2 # !2±2 = 1-cos " o Customary to multiply the $(") and !(!) functions to form the so-called spherical harmonic functions which can be written as: i.e., product of trigonometric and polynomial functions. PY3004 PY3004 Radial solutions (R( r )) Radial solutions (R( r )) o What is the ground state of hydrogen (Z=1)? Assuming that the ground state has n o In general, the radial wave equation has many solutions, one for each positive = 1, l = 0, the radial wave equation (Eqn. 6) can be written: integer of n. Can therefore write o Taking the derivative o More generally (Appendix N of Eisberg & Resnick) solutions are of form (7) "r / a0 o Try a solution of the form R = Ae , where A is a constant and a0 is a constant where a0 is the Bohr radius. Bound-state solutions are only acceptable if with the dimension of length. Sub into Eqn. 7: ! where n is the principle quantum number, defined by n = l +1, l +2, l +3, … o To satisfy this Eqn. for any r, both expressions in brackets must equal zero. Setting the second expression to zero => o En only depends on n: all l states for a given n are degenerate (i.e. have the same Same as Bohr’s results energy). o Setting first term to zero => eV PY3004 PY3004 Radial solutions (R( r )) Summary of separation of variables o Gnl(Zr/a0) are called associated Laguerre polynomials, which depend on n and l. o Express electron wavefunction as product of three functions: "(r,#,$) = R(r)%(#)&($) o Several resultant radial wavefunctions (Rnl( r )) for the hydrogen atom are given o As V ! V(t), attempt to solve time-independent Schrodinger equation. below ! o Separate into three ordinary differential equations for R ( r ), " ( # ) and " ( # ). o Eqn. 4 for " ( # ) only has acceptable solutions for certain value of m . ! ! l o Using these values for ml in Eqn. 5, !(!) only has acceptable values for certain values! of l. o With these values for l in Eqn. 6, R(r) only has acceptable solutions for certain values of En. o => three quantum numbers! PY3004 PY3004 Born interpretation of the wavefunction Born interpretation of the wavefunction o Principle of QM: the wavefunction contains all the dynamical information about the o In H-atom, the ground state orbital has the same sign everywhere => sign of orbital system it describes. must be all positive or all negative. o Born interpretation of the wavefunction: The probability (P(x,t)) of finding a o Other orbitals vary in sign. Where orbital changes sign, & = 0 (called a node) => particle at a position between x and x+dx is proportional to |&(x,t)|2dx: probability of finding electron is zero. o Consider first excited state of hydrogen: sign of wavefunction is insignificant (P = &2 = (-&)2). o P(x,t) is the probability density. o Immediately implies that sign of wavefunction has no direct physical significance. &(x,t) P(x,t) PY3004 PY3004 Born interpretation of the wavefunction Radial solutions (R( r )) o Next excited state of H-atom is asymmetric about origin. Wavefunction has o The radial probability function Pnl(r ), is the probability that the electron is found opposite sign on opposite sides of nucleus. between r and r + dr: o Some radial probability functions are given at right: o Some points to note: o The r2 factor makes the radial probability density o The square of the wavefunction is identical on vanish at the origin, even for l = 0 states. opposite sides, representing equal distribution of electron density on both side of nucleus. o For each state (given n and l), there are n - l - 1 nodes in the distribution. o The distribution for states with l = 0, have n maxima, which increase in amplitude with distance from origin. PY3004 PY3004 Radial solutions (R( r )) Radial probability density o Radial probability distributions for an electron in several of the low energy orbitals o The radial distribution function, P(r ) gives the of hydrogen. probability that the electron will be found in a shell of radius r. o The abscissa is the radius in units of a0. o For a 1s electron in hydrogen, P(r ) is a maximum when r = a0/Z. o For hydrogen, Z = 1 => rmax= a0 o Most likely to find the electron at the Bohr radius. PY3004 PY3004 Hydrogen eigenfunctions s orbitals -0.2 0 0.2 o The eigenfunctions for the state described by the quantum numbers (n, l, m ) are of o Named from “sharp” spectroscopic lines. l 0.2 the form: 0 o l = 0, ml = 0 -0.2 -0.2 o Depend on quantum numbers: 0 o &n,0,m = Rn,0 (r ) Y0,m (!, ") 0.2 n = 1, 2, 3, … o Angular solution: l = 0, 1, 2, …, n-1 m = -l, -l+1, …, 0, …, l-1, l l o Value of Y0,0 is constant over sphere. o Energy of state on dependent on n: o For n = 0, l = 0, ml = 0 => 1s orbital: 13.6Z 2 E n = " 2 n o The probability density is o Usually more than one state has same energy, i.e., are degenerate. ! PY3004 PY3004 p orbitals d orbitals o Named from “principal” spectroscopic lines. o Named from “diffuse” spectroscopic lines. 2 1 o l = 1, m = -1, 0, +1 (n must therefore be >1) l o l = 2, ml = -2, -1, 0, +1, +2 (n must 0 therefore be >2) -1 -1 -0.5 o &n,1,m = Rn1 (r ) Y1,m (!, ") -2 0 0.5 -1 -0.5 0 0.5 1 o &n,2,m = Rn1 (r ) Y2,m (!, ") 1 o Angular solution: o Angular solution: o A node passes through the nucleus and separates the two lobes of each orbital.