11/2/2023 PHY 412 Introductory Quantum Mechanics continue from last lecture…….Particle in a One-Dimensional Box Normalization of the wave function Normalization means that the integral of the square of the wavefunction (probability density) over all space is equal to one. As we mentioned earlier, the quantity finding the particle in the small volume Lecture 06: The Schrödinger Equation ,Particle in a Box and The Postulates of Quantum Mechanics ∗ gives the probability of . Since the particle must be found inside the box, the total probability of finding the particle inside the box must equal 1. ∗ =1 Because the wave function doesn’t include imaginary numbers, we have ∗ = , and because the particle is limited to move in one direction, we have = . In addition, because the particle is limited to move inside the box between = 0 and = , we can put the integration limits between 0 and . 2 Particle in a One-Dimensional Box Particle in a One-Dimensional Box Using these facts, we get = ψ∗ But we have sin 2 Substituting for 2 = − sin 2 4 = sin = sin =1 ⇒ using the relation sin = − 2 2 = And when substitute for sin 2 4 = 2 − sin 4 , 3 =1 ℎ = =1 1 2 2 = in the equation of the wave function, we get we get 2 sin 0 = 1 4 = 0 for any value of . therefore, , we get = −0− 2 sin And this is the final form of the normalized wave function of the particle in a one dimensional box. 4 Particle in a One-Dimensional Box Particle in a One-Dimensional Box Orthogonality of wave functions Using this realtion This is an important mathematical characteristic of Eigenfunctions. This mathematical rule indicates that when a number of Eigenfunctions are different solutions with different eigenvalues of a given equation, these Eigenfunctions must be orthogonal. This rule is expressed mathematically as: =0, 2 = 2 sin 2 sin sin = sin − sin + − 2( − ) 2( + ) we get = ≠ 1 2 ( − ) sin ( − ) − ( + ) sin ( + ) =0 Because and are real numbers, their sum or difference is also a real number, and sin = 0 when = 0, 1, 2, 3, … ,therefore the above integral must equal zero. This means that the two wave function equations and for the particle in the one dimensional box are orthogonal. It must also be noticed that the orthogonality conditions apply only when the eigenfunctions are real solutions to the equation and not approximate solutions. For a particle in a one dimensional box = sin sin sin 5 6 1 11/2/2023 Particle in a One-Dimensional Box Particle in a One-Dimensional Box Energy and wave function of a particle in a box 2. The highest probability of finding the particle is in the middle of the distance between two nodes. The highest probability of finding the particle for = 1 is at /2, and for = 2 is at /4 and 3 /4, and so on. For a particle moving in a one dimensional box, we have shown that = 2 and sin = ℎ 8 3. The orthogonality of the wave functions is clear. From the figure we can see that We noticed from the graph that: = 2 take zero at the same 1. Both and certain points. These points are called nodes. At the nodes, the wave function of the particle vanishes and there is zero probability of finding the particle. The number of nodes for a given value of equals ( − 1) and the number of nodes increases as then value of increases. This is in consistence with the de Broglie hypothesis ( = ℎ/ ). As the kinetic energy increases the wave length decreases and the number of nodes increases. =− − − =− − − The integration over all values of for the two equations between = 0 and = /2 is equal but different in sign for the same integral between = /2 and = . This means that the integration∫ = 0 is equal to zero. ⇓ 7 Particle in a One-Dimensional Box 8 Particle in a One-Dimensional Box 4. is inversely proportional to and : as the mass of the particle and or the size of the box get smaller the higher the energy of the particle. Correspondence principle “classical mechanics emerges from quantum mechanics at high quantum numbers.” Note that in the particle in one dimensional box problem when the size of the box becomes very large or when the mass of the particle gets very large, the quantity becomes very large compared to ℎ ( ≫ ℎ ) and the energy of the system becomes continuum and we go back from quantum physics to classical physics. This is always the case. When sizes get larger, the quantum physics problem converges with classical physics. This is called the correspondence principle. So it might be said that the classical physics laws are special case of quantum chemistry laws. 9 Correspondence principle The most probable position of the particle is near the = 1 state but that the center of the box for the probability density becomes more uniformly distributed as increases. The probability density for higher energy levels (e.g. at = 20) is fairly uniformly distributed from 0 to . the particle tends to behave classically in the limit of → ∞ each position large quantum numbers (as is equally probable). | | → The probability distribution of a particle in a box. 0 → 10 Particle in a 1-D Box Model and the Real World Particle in a 1-D Box Model and the Real World The particle in a box model can be applied to electrons moving freely ( electrons) in a molecule. According to the Pauli exclusion principle each energy level can hold only two electrons (with opposite spins), and so the four electrons fill the first two levels. The first excited state of this system of four electrons is that which has one electron elevated from the = 2 state to the = 3 state. The energy needed to make a transition from the = 2 state to the = 3 state is Consider butadiene, H2C=CH-CH=CH2, which has four Butadiene has an absorption band at 217 nm for the 1st Although butadiene simplicity that the four electrons. → ∗ transition. is not a linear molecule, we will assume for electrons in butadiene move along a straight line. The length of this straight line can be estimated as 2 C=C bond (2 x 135 pm) + C- C bond (154 pm) + the distance of a carbon atom radius at each end (2 x 77.0 pm= 154pm), giving a total distance of 578 pm (5.78 Å or 578 x 10-12 m). The allowed electronic energies are given by = is the electron mass. ℎ 8 ∆ = = ℎ (3 − 2 ) 8 the mass of the electron is 9.109 x 10-31 kg, and the length of the box is taken to be 578 pm, or 578 x 10-12 m. ∆ = = 1, 2, 3, … 11 − ℎ ∆ 1 ℎ : = 1 × 10 HOMO: Highest Occupied Molecular Orbital LUMO: Lowest Unoccupied Molecular Orbital 12 2 11/2/2023 Particle in a 1-D Box Model and the Real World ∆ = (6.626 × 10 J. s ) 5 8(9.109 x 10−31 kg )(578 × 10 ) = 9.02 x 10−19 J and = (6.626 × 10 J. s )(2.998 × 10 9.02 x 10−19 J = 2.20 × 10 1 ( 220 / ) Postulates of Quantum Mechanics Any theory is usually built on a number of postulates. A postulate is a theoretical assumption that has no prove but reasonable and found to be consistent with observation. A postulates can play a key rule in deriving the theory. The following are some of the quantum mechanics postulates: First postulate The state of a system is fully described by its state function or wave function (often written as wavefunction) , that depends on the coordinates , , of the particle(s) and on time . ) ℎ : = 1 × 10 For a system consisting of described by a wave function The causes of discrepancy between experimental value (217 ) and calculated value (220 ) is the approximations that have been made. 13 ≡ ( 1, where 1, and 1, 2, and 4, 5, particles then this system can be fully which is given by 2, 3, 4, … , 3 , ) 3 are the spatial coordinates ( 1, 1, 1) and 6 are the spatial coordinates ( 2, 2 and so on. Whereas is the time. for particle 2, 2) for particle 14 Postulates of Quantum Mechanics Postulates of Quantum Mechanics When the function is independent of time the function is said to be stationary wave function. Fortunately, many applications of quantum mechanics to chemistry use stationary wavefunction. There are requirements for a wave function to be acceptable. These requirements are: The probability of finding particle lies in the volume element , where refers to the region between and + , and + , and and + is given by ∗ Now, because the particle has to be found some where in space, we must have ∗ ∗ found in one space at a given time. b. The wave function and its first derivative must be continuous since a particle can not transfer from one space to another without passing in the path connecting them. c. The wave function must be finite. =1 The notation "all space" here means that we integrate over all possible values of , , and . The quantity a. The wave function must be single valued since a particle can be is called the probability density. d. The wave function must be quadratically integrable. The wave function that fulfil these requirements is said to be well behaved wave function. 15 16 Postulates of Quantum Mechanics Postulates of Quantum Mechanics Second postulate To every measurable physical property (observable) in classical mechanics there is a linear Hermitian operator in quantum mechanics. The value of the physical property for a system can be found when the corresponding operator operates on its wave function. Physical observables and their corresponding quantum operators. The linear and Hermitian operator have been defined earlier when we discussed operators. However we add here that a Hermitian operator also fulfil these requirements: Classical mechanics Name Symbol = ( = ( )∗ − ℏ( ̂ Kinetic energy ℏ − 2 Total energy For all well behaved functions. )∗ Angular momentum 17 Multiply by Linear momentum and ∗ Operation ̂ Position ( ) Potential energy ∗ Quantum mechanics Symbol = + ( ) ( ) = + + ̂ ( + + + Multiply by ( ) − ℏ 2 ) ) ( ) + + + ( ) − ℏ( − ) − ℏ( − ) − ℏ( − ) 18 3 11/2/2023 Postulates of Quantum Mechanics Postulates of Quantum Mechanics How to derive an operator of a property? To derive the operator of a physical property, follow the next steps: a. Write the equation of the desired property using the classical physics laws in terms of the coordinates, momentum and time. b. Leave time and coordinates with no change. c. When using the cartesian coordinates, replace the momentum by the operator Now, we make the replacement , where q ≡( x, y, z ) i q pˆ x i pˆ x2 2 T= 2 x 2 Likewise 2 2 and pˆ z2 2 2 y 2 z And substitution in the operator of the kinetic energy, we get pˆ 2y 2 Example: If we want to calculate the kinetic energy of the system, the corresponding operator is and from classical physics we have x 2 2 2 2 2 2 Tˆ 2m x 2 y 2 z 2 2m The operator is called the Laplacian operator (read as “del squared”). 2 2 2 1 2 m 2 v 2 Px + Py + Pz mv = = 2 2m 2m ≡ + + 19 20 Postulates of Quantum Mechanics Applications of Postulates of Quantum Mechanics Third postulate If is an eigenfunction of the operator , then in any measurement of the observable associated with the operator , the only values that will ever be observed are the eigenvalues , which satisfy the eigenvalue equation Lets, for example, operate on the wave function for the particle in one dimensional box by the operator of the linear momentum in the direction, = = −ℏ Fourth postulate If is not an eigenfunction of the operator , we still can calculate the average or expected value of that property using the relation =−ℏ expected value = = ∗ ∫ If the wave function is a normalized , then the average value of the observable corresponding to is given by What if we operate by cos ? 2 d d2 pˆ x2 i 2 2 dx dx ∗ = 2 sin It is clear that the function here is not an eigenfunction of the operator . ∗ ∫ 2 21 Applications of Postulates of Quantum Mechanics 2 =−ℏ = ℏ = ℏ It is clear that the function and the eigenvalue is ℏ 2 cos a 2 d 2 nx nx pˆ x sin sin i dx dx a a a 0 a sin = Applications of Postulates of Quantum Mechanics Because the function is not an eigenfunction of the operator , we can calculate the expected value of as the fourth postulate states. ∫ ∗ ∗ = = ∫ ∗ sin 2 =−ℏ 22 2 2ℏ2 i 2 is an eigenfunction of the operator . Using the relation 23 2 n a nx nx sin cos dx a a 0 a a sin bx cos bx dx sin 2 bx 2b 24 4 11/2/2023 Applications of Postulates of Quantum Mechanics 2 nπ 2a Pˆx i a a nπ ⇒ a The importance of the particle in a 3-D box model arises from the idea that it shows the technique used to solve Schrodinger equation when it is a function of more than one variable. 2 nx sin a 0 2 nπ 2a sin 2 nπ sin 2 0 0 i a a nπ Therefore, the expectation value of This technique is called the separation of variables in which the total equation is separated into as many equations as there variables. is zero. For the particle in three dimensional box, the wave function is a function of the three variables , and = ( , , ) It was found from the early discussion that: n 2 2 2 p x2 a2 Taking the square root of this equation gives px n 2 2 2 a2 Particle in a Three-Dimensional Box Using the separation of variables technique, This equation can be divided into three equations, equation for each variable, and the original equation will be the product of these three equations. n a , , = Thus for a large number of measurements of , we will have half of the results with = ℏ/ , and the other half with = − ℏ/ . Therefore, the average value of the results will be zero. 25 The solution of the equation of each variable can be found separately, and the solution of the original equation can then be found. 26 Particle in a Three-Dimensional Box Particle in a Three-Dimensional Box For the particle in three dimensional box, suppose we have a particle of mass moving inside a box of dimensions , and . Now, we want to solve the Schrodinger equation for this system The potential energy inside the box is zero and infinity any where else outside the box. In this case, the particle is confined to lie within a box. Note that solving the Schrödinger equation means finding the right form of the Hamiltonian ( ), the wave function ( ) and the energy ( ) of the system. , , , , =0 =∞ 0< 0< 0< ℎ < < < = We know that when the potential energy is zero, the Hamiltonian in three dimensions , and is given by h2 2 2 2 h2 Hˆ 2 2 2 2 2 2 8 m x y z 8 m ℎ Now the equation we want to solve would be The potential term can be treated by boundary conditions (i.e., infinite potential implies that the wavefunction must be zero there). 27 Particle in a Three-Dimensional Box = and , we can write = and = 28 Likewise, we can write To solve this equation, we use the separation of variables technique. We will try to make the function , , a product of three functions , , and where each function of the three functions is a function of only one variable that is separate of the other two variables. , , = = is independent of h2 2 2 2 ( x, y, z ) E ( x, y, z ) 8 2 m x 2 y 2 z 2 Particle in a Three-Dimensional Box h2 2 2 2 ( x, y, z ) E ( x, y, z ) 8 2 m x 2 y 2 z 2 Now, because 29 = when we substitute by this into the total equation, we get ℎ − + + = 8 If we divide both sides be , we get ℎ 1 1 1 − + + = 8 We note that each one of the three terms inside the square brackets depends only on one variable and totally independent of the other two. If the particle moves only on the coordinate , then the energy of the other two directions is constant and we can solve for the energy in the direction ( ). 30 5 11/2/2023 Particle in a Three-Dimensional Box Particle in a Three-Dimensional Box In this case we can write The previous equation is just similar to the equation of a particle in one dimensional box, therefore, the solution is known and given by ℎ − 1 = 8 Similarly, when the particle is moving in the other directions ℎ 1 ℎ 1 − = − = 8 8 or . + + = Now, we can solve each equation of the above three equations separately. If we start with the equation in the x direction, it can be written in the form ℎ − = 8 31 Particle in a Three-Dimensional Box ( , , )= sin + = For the case of a cubic box = ℎ + 2 sin = = 2 , = 1, 2, 3, … ℎ 8 , = 1, 2, 3, … = ), if = = = 1 then, is normalized and so is the total function. ∗ = = 3. The three states + = 1·1·1 =1 , , and = Particle in a Three-Dimensional Box References This means that it is possible to have different states with different quantum numbers taking the same energy. This is what is called degeneracy. Some other examples are: English References: = ℎ 8 32 33 = = 1, 2, 3, … and 2. Each function , , = , the total energy is give by 8 = , = 8 And this is called the zero point energy for the particle in a cubic box. This level, , is said to be nondegenerate. = 1, 2, 3, … = 1, 2, 3, … = 1, 2, 3, … + sin ℎ 8 ℎ The total energy is given as ℎ = 8 = Notice the following: 1. For a cubic box where ( = = 1, 2, 3, … = 1, 2, 3, … = 1, 2, 3, … sin sin Particle in a Three-Dimensional Box The total wave function equation is then given as 8 2 Likewise, It is also clear that = = are equal in energy 6ℎ = = 8 34 :اﻟﻣراﺟﻊ اﻟﻌرﺑﯾﺔ McQuarrie, Donald A. “Quantum Chemistry.” 2nd Ed., University Science Books, Mill Valey, CA, 1983. = 11ℎ = 8 Levine, Ira N. “Quantum Chemistry.” 7th Ed., Pearson Education, 2014. = 9ℎ = 8 Sherrill, C. D. “A Brief Review of Elementary Quantum Chemistry.” 2001 http://vergil.chemistry.gatech.edu/notes/quantrev/quantrev.html Atkins, P. W. “Molecular Quantum Mechanics.”, 2nd Ed., Oxford University Press. Oxford 1983. Undergraduate Quantum Chemistry by Jussi Eloranta (2018). Degeneracy – different wave functions with the same energy. Reflects symmetry of the box. When = Cockett, Martin; Doggett, Graham, “Maths for Chemists.” RSC publishing, 2012. Cunningham, Allan; Whelan, Rory “Maths for Chemists.” University of Birmingham, University of Leeds 2014. = , the lowest levels have the following degeneracy factors: 35 36 6