Solution to the Particle-in-a-Box Problem in Quantum Mechanics, Essays (high school) of Electrical Engineering

Download Solution to the Particle-in-a-Box Problem in Quantum Mechanics and more Essays (high school) Electrical Engineering in PDF only on Docsity! Atomic and Molecular Quantum Theory Course Number: C561 9 Particle-in-a-box (PIB) 1. Consider a linear poly-ene. 2. The electrons are completely delocalized inside the poly-ene, but cannot leave the molecular framework. 3. Let us approximate this system by a one-dimensional box, of length L. The potential energy of the electrons inside the polyenes can be approximated by the figure below. 4. We will assume that the electron feels the same potential at all points inside the molecular framework. (An approximation that actually works pretty well and explains results in many physical systems.) 5. The particle-in-a-box is “toy” problem, but it is the starting point for many important modern- day ideas such as: (a) A fundamental understanding of resonance on polyenes is possible through PIB. (b) Frontier molecular orbital theory, which is the starting point for the Woodward Hoff- man rules in organic chemistry, is very easily explained using PIB. (c) Quantum dots, wells, and wires become very much accessible. (d) And the Thomas Fermi functional in Density Functional Theory (DFT), a very pow- ereful modern-day quantum chemistry method, has its roots in PIB. 6. So PIB is of fundamental importance. 7. Once we study the PIB, we will then derive a conceptually appealing formalism based on “scattering” to write down solutions directly for such problems (without calculus)!! 8. The box edges have infinite repulsive potentials to keep the electrons inside the molecular framework. But inside the molecular framework the electrons are completely free to move as they should be on account of resonance in poly-enes. So, V (x) = 0, 0 ≤ x ≤ L V (x) = ∞, x < 0 and x > L (9.1) 9. In region II, the Hamiltonian for the system is: H = p2 2m + V = − h̄2 2m ∂2 ∂x2 (9.2) (note that the potential is zero). Chemistry, Indiana University 35 c©2003, Srinivasan S. Iyengar (instructor) Atomic and Molecular Quantum Theory Course Number: C561 10. We would like to solve the time-independent Schrödinger Equation to obtain the wavefunc- tion for the system inside the box (in region II): HIIψ(x) = Eψ(x) − h̄2 2m ∂2 ∂x2 ψ(x) = Eψ(x) ∂2 ∂x2 ψ(x) + 2mE h̄2 ψ(x) = 0 (9.3) If we make the substitution 2mE h̄2 = k2, ∂2 ∂x2 ψ(x) + k2ψ(x) = 0 (9.4) To obtain the solution to this equation, we need to solve for ψ(x). The function should have a form such that, when differentiated twice, it gives the function back multiplied by a constant. The exponential function has this property and hence lets guess the solution to this equation as exp{sx}. Substituting this solution into the equation above leads to: s2ψ(x) + k2ψ(x) = 0 (9.5) which leads to: s = ±ık (9.6) which gives two solutions: exp{ıkx} and exp{−ıkx}. Chemistry, Indiana University 36 c©2003, Srinivasan S. Iyengar (instructor) Atomic and Molecular Quantum Theory Course Number: C561 16. But ψ(x) = 0 at x = L. This means: B sin{kL} = 0 (9.14) and since B cannot be zero (there would be no wavefunction if that were the case!!) sin{kL}, which can happen when {kL} has the values 0, π, 2π, 3π · · ·, or simply when kL = nπ, or: k = √ 2mE h̄ = nπ L (9.15) where n = 0, 1, 2, · · · , (whole numbers) which leads to ψ(x) = B sin{nπx L } (9.16) 17. The wavefunction must be normalized. Therefore ∫ L 0 B2 sin2{nπx L }dx = 1 (9.17) If we make the substitution nπx L = y, then dx = L nπ dy and the integral can be rewritten as B2 L nπ ∫ nπ 0 sin2{y}dy = 1 (9.18) which leads to B = √ 2 L (9.19) Thus the solution to ψ(x) is ψ(x) = √ 2 L sin{nπx L } (9.20) 18. From Eq. (9.15) E = n2π2h̄2 2mL2 = k2h̄2 2m (9.21) Chemistry, Indiana University 39 c©2003, Srinivasan S. Iyengar (instructor) Atomic and Molecular Quantum Theory Course Number: C561 19. What do we have? (a) Lets recap the particle in a box wavefunction and energy: ψn(x) = √ 2 L sin{nπx L } (9.22) En = n2π2h̄2 2mL2 (9.23) (b) Even at n = 1, which is the lowest energy state, the particle has a finite energy. This the “zero point energy”. As the length of the box gets smaller this energy gets larger!!! It also goes up as the mass decreases. (Note: n = 0 is not physically meaningful. Why?) (c) The solution is oscillatory. (Trigonometric functions are oscillatory.) Hence it contains points inside the box where the probability is zero. These points are called nodes. For example, ψ2(x) has a node at x = L/2. (d) Since n can only take on some values (n = 1, 2, 3, · · · ,) the energy expression in Eq. (9.21) can have only some discrete values. Thus the energy is quantized. (e) n can take on only some values due to the boundary conditions. See Points (15) and (16). (f) In fact, it is in general true that boundary conditions enforce quantization. (g) Notice further that for different values of n: 〈ψn |ψn′〉 = ∫ L 0 ψn(x)ψn′(x)dx = ∫ L 0 2 L sin{nπx L } sin{n ′πx L } = δn,n′ (9.24) where we have chosen to label the wavefunctions for the particle in a box in Eq. (9.20) using the “quantum number” n. Also, δn,n′ is the Kronecker delta: δn,n′ = 1 for n = n′ and δn,n′ = 0 for n 6= n′. Homework: Prove the above relation using trigonometric identities for Sine? (h) This means {ψn(x)} form an orthonormal set. (In fact these wavefunctions also form a complete set, in the sense that they satisfy the resolution of the identity (do you remember what that is, refresh!!). We will prove this for a general case later in the course.) Chemistry, Indiana University 40 c©2003, Srinivasan S. Iyengar (instructor) Atomic and Molecular Quantum Theory Course Number: C561 20. Homework: (a) Sketch out the first five eigenstates of the particle in a box. Do you notice any symmetry about the center of the box? Comment. (Hint: In general, there are two different kinds of functions, based on symmetry about the center. Do you see this? Comment.) (b) As n increases the difference between neighboring energy levels increases. Prove this statement. Hint: Calculate (En+1 − En). What happens to this expression as n in- creases? How about when L goes to infinity? (c) Write down the probability distribution ρn(x) = |ψn(x)|2. How does this change as n increases? Comment on the oscillatory nature of this function with increasing n. (Does it get more or less oscillatory?) Sketch out ρn(x) for n = 1, 2, 3, 4. (d) I have a detector through which I look into the box to locate the position of the particle. My detector, however, has a restriction. It has a least count of a. This means I cannot measure the probability of the particle with infinite precision. At any point x that I look at using the detector, I end up measuring the following quantity: ∫ x+a/2 x−a/2 ρn(x ′)dx′ (9.25) Which is the total probability inside the interval [x− a/2, x+ a/2]. (Do you see this? This detector is similar to an electron microscope!) Lets assume for this problem that a = L/5. Plot the value of the function above for n=1,2,3,4,5,11,21,31,51,501. (Note: The integral above will be a function of x after integration.) Describe what you see. Comment on the generality of your result. That is, is this description always valid for any a, however small (but never zero). 21. If you got the correct answer for the above problem, you have just derived what is known as the Bohr’s correspondence rule between quantum mechanics and classical mechanics. For larger n you should see little change in the probability across the box. Hence the particle becomes more and more “classical-like” (i.e. devoid of the nodes which are a manifestation of quantization and wave-like behavior) as we increase n. 22. This is actually an important result. In the early 20th century many great minds sat down to “make-sense” out of quantum mechanics. Bohr was one of them. Dirac was another and so was Heissenberg. These guys tried to see if there were certain limits where quantum theory would approach classical mechanics. In this homework you have arrived at one such general limit above. Chemistry, Indiana University 41 c©2003, Srinivasan S. Iyengar (instructor)