Quantum Chemistry and Spectroscopy

Size: px

Start display at page:

Download "Quantum Chemistry and Spectroscopy"

Luke Cunningham
5 years ago
Views:

1 Quantum Chemistry and Spectroscopy Text book Engel: Quantum Chemistry and Spectroscopy (3 ed., QCS) or Engel and Reid: Physical Chemistry (3 ed.) Chapter 1. Basic Quantum Mechanics (Engel: QCS chapters 1-6) The basic topics of quantum mechanics has been studied in Physics course PHYS-A140 Aineen rakenne (CHEM) and if needed the student should be familiarized the topics from any modern Physical Chemistry (or physics) textbook. (The Engels book is good also same topic can be found form Atkins Physical Chemistry textbook and from many others.) Why Quantum mechanics is important Quantum mechanics (QM) is the key theory for chemistry. QM describe the world at atomic and subatomic scale and chemistry is about atoms and bonds between them. Chemical bond is quantum mechanical in nature. All spectroscopy is based on QM. Many of the molecules and materials properties are based on QM. Do you need more reasons? In the following I list few of the main topics but I do not go to the details. Energy is quantized A good example is the atomic spectra. With any accurate measurements of molecules, every energy related quantity is quantized. This is the basis of spectroscopy.

2 Observed spectra of Na. Particle-wave dualism Every particle is both a particle and a wave. This is strange but so it is. The phenomena has been observed in many experiments. Electrons as waves One of the best example of the particle-wave dualism the wave nature of an electron. X-rays can be used to determine the structure of mater because light waves that reflect from the atoms will interfere and they form distinguished diffractions pattern. Similar effect has been used when an electron beam with uniform energy is hit to a surface. The electrons formed a diffraction pattern and thus they have to behave like waves. This phenomena is now used to determine surface structures of different materials. (EELS electron energy loss spectroscopy)

3 Another example of electrons wave-nature is seen in electron microscope, which in principle works like an optical microscope.

4 The particle-wave dualisms apply to all particles. The photons have particle character and electrons, protons etc. have wave character. The wavelength of the particle can be calculated from the de Broglien law l = h/p Where the p is the particles impulse, E kin =p /m. Last example of the particle-wave dualism is the double slit experiment where electrons will travel from two slits that are very close to each other. Like in the case of light the electrons will interfere and they from a diffraction pattern behind the slits. The electrons wavelength can be calculated from the de Broglie law. A very interesting version of this experiment is that it can be done with single electron. Thus only one electron will go through the double slit and still the diffraction pattern is formed. In some sense the electron will go through both of the slits! This is strange but so it goes. Also the particle nature of the electron is clearly seen since a electron will left only one spot to the detector.

Schrödinger equation As explained earlier everything in quantum mechanics can

(Well almost everything since the Schrödinger equation is a non-relativistic

The relativistic equation is the Dirac equation, which explains the spin and

5 Schrödinger equation As explained earlier everything in quantum mechanics can be derived from the Schrödinger equation. (Well almost everything since the Schrödinger equation is a non-relativistic equation. The relativistic equation is the Dirac equation, which explains the spin and certain phenomena seen in heavy atoms.) The Schrödinger equation is an eigenvalue equation and it defines wave functions fn(r) and energy E n. The time independent Schrödinger equation

6 - h m f ( r) + V ( r) f ( r) = n n E f ( r) n n = d dx + d dy + d dz The time dependent Schrödinger equation is (in this course we do not use much of the time dependent Schrödinger eq.) - h m f ( r, t) + V ( r, t) f ( r, t) = n n fn( r, t) t Here m is the mass of the particle and V(r) is the potential the particle feels. Properties of the wave function: 1) The wave function is the solution of the Schrödinger equation. ) The wave function contain all information of the system 3) The wave function is always continuous and if the potentials is continuous then the derivative is continuous. 4) The square of the wave function f(r) describe the particles local probability The point 4) to make sense the wave function need to be finite everywhere f(r) < And of the wave function is defined in region W the integral of its square have to be finite W f(r) d 3 r < If the integral is finite the wave function can be normalized W f(r) d 3 r = 1

7 Quantum mechanics formalism One of the largest problem in QM is that common quantities like position and momentum are rather complex. We can define operators for position and momentum and then compute the expectation values for these operators. Operators Position operator: x ˆ = x The expectation value of a particle: x = Æx æ= W f*(r) x f(r) d 3 r = Æf x fæ If we use wave function of, f(r) = A exp(-kx ) Æx æ = A - exp(-kx )x exp(-kx )dx = A - x exp(-kx )dx = 0 Momentum operator: pˆ = -i ħ d /dx it s expectation value is Æ p æ= -i ħ W f*(r) d f(r)/dx d 3 r = Æf pˆ fæ If we use wave function of, f(r) = A exp(-kx ) Æ p æ = -i ħ A - exp(-kx )dx = 0 exp(-kx )d exp(-kx )/dx dx = i ħ k A - x A free particle wave function is f(x) = A exp(ikx) and it s momentum is

8 p = -i ħ A W exp(-ikx) d exp(ikx)/dx d x = -i ħ A W exp(-ikx) ik exp(ikx) d x = ħk A W exp(-ikx)exp(ikx) d x = ħk note that A W exp(-ikx)exp(ikx) d x = 1 (Normalizing the wf) So the free particles momentum is consistent with the de Broglie law! Energy operator: Ĥ = p /m + V(r) This is the left hand side of the Schrödinger equation. Angular momentum operator: In classical mechanics the angular momentum is r x p so the corresponding operator is = = = = Uncertainty principle One of the central properties in quantum mechanics is the uncertainty principle. The uncertainty principle says that two quantities that do not commute (like position and momentum) cannot be measured simultaneously accurately, Dx = Æ x æ - Æ xæ Dx Dp ħ/, DE Dt ħ/ In fact there is nothing mysterious in the uncertainty principle. It is just a consequence of the statistical nature of the QM. In more generally

9 [A,B] = AB - BA = i C, DA DB <C> / If [A,B]=0 then the quantities A and B can be measured simultaneously accurately. The difficult thing in QM is that even the probabilities are continuous all measured quantities are discrete. Solutions of the Schrödinger equation Next few solutions of the Schrödinger equation are shown. These should be known from the previous course. Constant potential The simplest case is the constant potential V(x) = V -h m d f n (x)/dx = (E n V) f n (x) The equation can be written as -d f n (x)/dx = k f n (x), where k = m (E n V)/ h This differential equation have solution of f(x) = A exp(ikx) + B exp(-ikx) = A sin(kx) + B cos(kx) Thus, the solution is a wave. Note that in this case the energy is not quantized. Particle in a box A more realistic problem is a particle in an infinitely deep box. In 1-dim the systems is the following:

0 L Inside the box the Schrödinger equation is the same as before: -h m d f n (x)/dx = E n f n (x), 0 < x < L (.1) But one need to be more careful in the borders. There the potential is infinite.

10 0 L Inside the box the Schrödinger equation is the same as before: -h m d f n (x)/dx = E n f n (x), 0 < x < L (.1) But one need to be more careful in the borders. There the potential is infinite. Let s first set the potential to value M at O and L. -h m d f n (0,L)/dx + M f n (0,L) = E f n (0,L) When M the f n (0,L) 0 to keep E finite. So we get a boundary conditions to f n (x), f(0) = f(l) = 0 so the f n (x) = A sin(kx), k = nπ/l, n=1,,.. The wavefunction is a sin wave.

11 The energy is: E n = h k m = ( ml h np ) = ( ml hn ) 8 n=1,,.. It should be normalised A L 0 sin( kx)sin( kx) dx = A L/ = 1 A = / L The final wave function is f n (x)= / L sin(kx), k = nπ/l, n=1,,.. Here we have quantized energy. The smallest possible energy (the Zero point energy) is not = 0, it is h /8mL The different wave functions are orthogonal L 0 f ( x) f ( x) dx = 0 if n m, ( x) f ( x) dx = 1 n m L 0 f when n = m This orthogonality is a general property of Schrödinger equations. n m Probability density The probability to find a particle in quantum mechanical system is proportional to Φ(r). In the case of two lowest states of particle in box: f 1 (x) = A sin(px/l) and f (x) = A sin(px/l). The probability is

12 f 1 (x) = sin (px/l)/l and f (x) = sin (px/l)/l Particle in two dimensional box The 1 dim case is rather easy to generalize to dimensions. Now the box is (0 < x < Lx, 0 < y < Ly). The Schrödinger equations is -h M( f n,m (x,y)/ x + f n,m (x,y)/ y )= E n,m f n,m (x,y), 0 < x < L x, 0 < y < L y This may look a bit difficult but it is easy to solve. Let as look a more general separable differential equation: H(x,y) = h 1 (x) + h (y), where h 1 (x)= ħ + ( ), choose v(x)=0

13 The wave function can be solved with a trial function f n,m (x,y) = f n (x)f m (y) This leads to -h M (f m (y) f n (x)/ x + f n (x) f m (y)/ y )= (E n + E m )f n (x)f m (y) This has solution only when -h M -h M f n (x)/ x = E n f n (x) f m (y)/ y = E m f m (y) (Proof: divide both sides with f n (x)f m (y. The right hand side is constant thus 1/f n (x)* f n (x)/ x and 1/f m (y)* f m (y)/ y have to be constants, since the first depend only on x and the second only on y.) Now the -dim wave function is: f n,m (x,y) = / L x Ly sin(k x x)*sin(k y y), where

14 E n,m =(n /L x + m /L y ) h /8M n,m = 1,,... 0 < x < L x, 0 < y < L y, k x = nπ/l x, k y = mπ/l y If Lx=Ly=L then E n,m =(n + m )h /8ML Now we can have a case where to wave functions can have the same energy, like n=1, m= and n=, m=1. This is called degeneracy. Now we can choose any (normalized) linear combination of the degenerate wave functions as a wave function. f A = a1f 1 + af fb b1f 1 + bf = if E 1 = E Particle in three dimensional box The -dim can be generalized to 3-dim easily: f n,m,k (x,y,z) = / L L L sin(k x x)*sin(k y y)*sin(k z z), ja x y z

15 E n,m =(n /L x + m /L y + k /L z ) h /8M n,m,k = 1,,... 0 < x < L x, 0 < y < L y, 0 < z < L z, k x =nπ/l x, k y =mπ/l y, k z =kπ/l z Note that now we have 3 quantum numbers. This is a general result. Particle in finite box The infinite box is not realistic so a more realistic model is particle in a finite box. -L/ L/ Again inside the box the Schrödinger equation is -h m d f n (x)/dx = E n f n (x), -L/ < x < L/ And outside -h m d j n (x)/dx = (E n V)j n (x), x < -L/, x > L/ Here the V is the depth of the box. The solution inside is f n (x) = A sin(kx) + B cos(kx), k = me h And outside j n (x) = C exp(kx) + D exp(-kx), k = m( V - E) h Because the wave functions need to be integrable j n (x) = C exp(kx), x < -L/ j n (x) = D exp(-kx), x > L/

16 Now the coefficients A,B,C,D and energy E can be solved from continuation equations: f n (-L/) = j n (-L/): C exp(-lk/) = -A sin(kl/) + B cos(kl/) f n (-L/) = j n (-L/): C k exp(-lk/) = A k cos(kl/) + B k sin(kl/) f n (L/)= j n (L/): D exp(-lk/) = A sin(kl/) + B cos(kl/) f n (L/)= j n (L/): - D k exp(-lk/) = A k cos(kl/) - B k sin(kl/) -L / - j ( x) j ( x) dx + n n L / f ( x) f ( x) dx + n n n -L / L / j ( x) j ( x) dx = 1 There is no closed solutions of these equations. When E < V the energy can be solved numerically. Once the E is known the wave functions are known. n

17 As can be seen the solutions are close to the infinite box case. Exercise: What is the solutions when E > V. One do not need to solve the equations. Harmonic oscillator One of the most important model system is the harmonic oscillator. It is an approximation of any local minima. In our case, the most important application is vibrational spectroscopy but we will come to that later. The potential in the Schrödinger equation is ½ kx -h m d f n (x)/dx + 1/ kx f n (x)= E n f n (x), (harm) The general solutions has form H n (x)exp(-ax ). (Proof: try this form to eq. (harm).) The first solution is

18 f 0 (x) = A 0 exp(-ax /) a = mk h And higher energy solutions are f 1 (x) = A 1 y exp(-y /) y = a 1/ x f (x) = A (4y -)exp(-y /) f 3 (x) = A 3 y(8y -1)exp(-y /) f n (x) = A n H n (y) exp(-y /), n=0,1,,.. Here H n (x) is Hermite s polynomial. The normalization is A n = 1 n n! a Ł p ł 1/ 4, A 0 = a Ł p ł 1/ 4, A 1 = 1 a Ł p ł 1/ 4, And the energies are E n = (n + 1/)h w, w = k m The positions and momentum expectation values are <x> n = 0, <x > n = <p> n = 0, <p > n = h m k h mk (n+1/) (n+1/)

19 Now Dx Dp = ħ(n+1/) ħ/ the uncertainty principle is valid and the n=0 is the most accurate state QM allows. (here Dx = Æ x æ - Æxæ ) Also the potential and kinetic energy expectation values can be computed, <V> = k<x >/ = 1/ h k m (n+1/) <E kin> = <p >/m = 1/ h k m(n+1/) D Rotation The next quantum system of interest is rotation. In two dimensions the Schrödinger equation is -h m( F n (x,y)/ x + F n (x,y)/ y ) = E n F n (x), But this do not take into account the rotational motion. We need to go to the cylindrical coordinates, where the coordinates are the distance (r = constant = r 0 ) and angle ϕ. / x + / y = 1/r 0 / f Now the Schrödinger equation is -h r m 0 F n (f)/ f = E n F n (f) This is easy to solve F m (f) = A exp(-im f). Now we need still the boundary conditions. The wave function need to be same after π rotation, F m (f+π) =F m (f). Thus m is an integer, m=0,±1,±, etc. Note these is variable (angle) so there is on quantum number, m. The normalization p A exp( imf) exp( -imf) df = A p = 1, A = 1/ p 0 And the energy E m = h m / mr 0

20 3D Rotation The more realistic system is the 3 dim rotation. The Schrödinger equation is more complex and it is - h mr 0 Ø 1 Y( q, f) 1 Œ sinq + º sinq q Ł q ł sin q Y( q, f) ø = EY( q, f) f œ ß The boundary conditions are now Y(q,f)= Y(q+p,f) ja Y(q,f)= Y(q,f+p). The solution can be written as Y(q,f)=Q(q)F(f) Øsinq Q( q ) Œ sinq + b sin º Q( q ) q Ł q ł ø œ ß 1 F( f) q = - F( f) F( f) = -const * F( f), constant = m, f f = constant, F(f) = A exp(imf),q(q) is a polynom of sinq, cosq Q00(q) = 1, Q10(q) = cosq, Q11(q) = sinq, Q0(q) = 3cos q-1, Q1(q) = sinq cosq, Q(q) = 3sin q, etc. Now there are two angles q, f so there are two quantum numbers, l,m l which are l = 0,1,,3,.., m l = -l,-l+1,..l-1,l The energy is h l( l + 1) h l( l + 1) E l = = mr I 0 Note that energy do not depend on quantum number m. Angular momentum We can also use angular momentums to describe the rotation. The Schrödinger equation is

ˆ L Y ( q, f) = h l( l + 1) Y l, m l, m ( q, f) In spherical coordinates the angular momentum operators are ˆ L x ˆ L y Lˆ z Ø ø = -ihœ- sinf - cotq cosf œ º q f ß Ø ø = -ihœ- cosf - cotq sin f œ º q

21 ˆ L Y ( q, f) = h l( l + 1) Y l, m l, m ( q, f) In spherical coordinates the angular momentum operators are ˆ L x ˆ L y Lˆ z Ø ø = -ihœ- sinf - cotq cosf œ º q f ß Ø ø = -ihœ- cosf - cotq sin f œ º q f ß Ø ø = -ihœ œ º f ß We can compute the commutators [Lx,Ly] = iħlz, [Ly,Lz] = iħlx, [Lz,Lx] = iħly This mean that we cannot measure all the three angular momentums simultaneously. But we can show that [L,Lz] = 0, so the length of L and one component can be measured. Thus we know the L cones but not the precise vectors.

22 Ylm(q,f) Let s look the Ylm in more details: Y 00 (q,f) = 1/ 4 p, Y 10 (q,f) = 3/ 4p cos(q), Y 1±1 (q,f)= 3/ 8p sin(q)exp(±if), Y 0 (q,f)= 5 /16p (3cos (q)-1), Y ±1 (q,f)= 15/8p sin(q)cos(q)exp(±if), Y ± (q,f)= 15/3p sin (q)exp(±if), jne. Many of these functions are complex so they are not so easy to handle but because the energy do not depend on the m quantum number we can rearrange the functions to real functions: P x = (Y 11 +Y 1-1 )/ = P y = (Y 11 -Y 1-1 )/ i = d xz = (Y 1 +Y -1 )/ = 3/ 4p sin(q)cos(f), 3/ 4p sin(q)sin(f), 15/ 4p sin(q)cos(q)cos(f) d yz = (Y 1 -Y -1 )/ i = d xy = (Y -Y - )/ i = 15/ 4p sin(q)cos(q)sin(f) 15/16p sin (q)sin(f) d x-y = (Y +Y - )/ = 15/16p sin (q)cos(f) These should be familiar. These are the p- and d-orbitals which you have seen many times. Note that the p- and d-orbitals are wave functions!!

The Hydrogen atom The most important single system from point of view of chemistry. It is the only atom which Schrödinger equation we can solve exactly.

23 The Hydrogen atom The most important single system from point of view of chemistry. It is the only atom which Schrödinger equation we can solve exactly. Here the potential energy is V(r) = - e /4p e 0 r, And the Schrödinger equation is ),, ( ),, ( 4 ),, ( sin 1 ),, ( sin sin 1 ),, ( 0 f q f q pe f f q q q f q q q q f q m r E r r e r r r r r r r Y = Y - œ ß ø Œ º Ø Y + ł Ł Y + ł Ł Y - h This is not simple equation but it is a spherical problem and the wave functions can be written as

24 Y ( r, q, f) =R(r)Y lm (q,f), In the equation m is the reduced mass of electron (1/m=1/m e +1/M p, m e mass of electron, M p mass of proton). Now we can write equation for the radial part R(r) h Ø R( r) ø h e - r + l( l + 1) - R( r) = ER( r), mr Œ r r œ º Ł łß Ł mr 4pe0r ł l = 0,1,... The l(l+1) part comes from the angular functions. Note that the radial wave function depend on two quantum numbers n and l. The quantum number n is called the main quantum number, the l is the angular quantum number and m is the magnetic quantum number. These quantum numbers are connected: n = 1,,3, l = 0,1,,3,,n-1 main quantum number angular quantum number

25 m = 0,±1,±,±3,,±l magnetic quantum number The few lowest radial wave functions are n=1,l=0 R 10(r) = (1/a 0) 3/ exp(-r/a 0) n=,l=0 R 0(r) = 1/ 8(1/a 0) 3/ (-r/a 0)exp(-r/a 0) n=,l=1 R 1(r) = 1/ 4(1/a 0) 3/ r/a 0 exp(-r/a 0) n=3,l=0 R 30(r) = /81 3(1/a 0) 3/ (7-18r/a 0+r /a 0 )exp(-r/3a 0) n=3,l=1 R 31(r) = 1/9 6(1/a 0) 3/ (6r/a 0-r /a 0 )exp(-r/3a 0) n=3,l= R 3(r) = 4/81 30(1/a 0) 3/ r /a 0 exp(-r/3a 0) here a0 is the Bohr radius a 0= e 0h /pm ee = Å. This is the natural length scale in QM. It is common to mark the angular quantum numbers with letters: s (l=0), p (l=1), d (l=), f (l=3) Below all s-functions have maxima at origin. The higher l functions are =0 at origin (why?). The first l-radial function (1s, p, 3d, 4f) do not have node outside of origin. The next ones have one node (s, 3p, 4d) and next ones (3s, 4d) have two nodes.

26 These functions need to be joined with the angular funcitons n=1,l=0,m=0 j 100(r) = 1/ p(1/a 0) 3/ exp(-r/a 0) n=,l=0,m=0 j 00(r) = 1/4 p(1/a 0) 3/ (-r/a 0)exp(-r/a 0) n=,l=1,m=0 j 10(r) = 1/4 p(1/a 0) 3/ r/a 0 exp(-r/a 0)cos(q) n=,l=1,m=±1 j 1±1(r) = 1/8 p(1/a 0) 3/ r/a 0 exp(-r/a 0)sin(q) exp(±if) n=3,l=0,m=0 j 300(r) = 1/81 3p(1/a 0) 3/ (7-18r/a 0+r /a 0 ) exp(-r/3a 0) n=3,l=1,m=0 j 310(r) see Engel s book n=3,l=1,m=±1 j 31±1(r) see Engel s book n=3,l=1,m=0 j 30(r) see Engel s book

27 n=3,l=,m==±1 j 3±1(r) see Engel s book n=3,l=,m==± j 3±(r) see Engel s book The low index wave functions are rather simple 1s wave function Below p, 3p, dxy and dx-y

28 Shell structure Last important thing is that the wave functions for different main quantum number do not overlap much. This has an important consequences in chemistry. It basically explain why the periodic table is as it is.

29 Below the radial probability r j nlm is drawn Energies of the hydrogen atom The energy of the hydrogen atom is 4 me E n = - = -.179*10-18 J/n = ev/n e 8 0 h n

30 This can be measured very accurately (7 digits) and it is very strong proof that the quantum mechanics is correct. We can for example see the difference of the spectra of hydrogen and deuterium. Remember that the effective mass is 1/m=1/m e +1/M p, (m e mass of electron, M p mass of proton or mass of deuterium.) The difference is not big, ca. 0.1 % but the accuracy is much better. Also the 1/n behavior was known in 19 th century (that is before QM) but the QM predict the correct behavior.

The Hydrogen Atom. Dr. Sabry El-Taher 1. e 4. U U r

The Hydrogen Atom. Dr. Sabry El-Taher 1. e 4. U U r The Hydrogen Atom Atom is a 3D object, and the electron motion is three-dimensional. We ll start with the simplest case - The hydrogen atom. An electron and a proton (nucleus) are bound by the central-symmetric