Quantum Mechanics for Scientists and Engineers Sangin Kim Advanced Computational Electromagnetics Lab redkid@yonsei.ac.kr Opt. 5th, 2016 Hello, my name is sangin kim from ACEM labatory.

Quantum Mechanics for Scientists and Engineers Outline Quantum Mechanics for Scientists and Engineers Particles in a finite potential wells Solving for the eigenenergies Harmonic oscillator eigensolutions Today, I’m going to present of the lecture 7. this is the my outline. First, I’ll explain to solve in a finite potential wells and then solve the harmonic oscillator eigensolutions.

Particle in a finite potential well Assumption Potential barriers as V0, with 0 potential energy at the bottom of the well The thickness of the well is Lz Solution Sinusoidal in the middle We can solve inside the well itself. First, we are going to choose the height of the barrier here to be V0 on either side of potential well. And at the bottom, here is 0 for potential energies. The thickness of the well will be Lz. Now we will choose the position origin in the center of the well. If there is an eigenergy E for which there is a solution, we know that form the solution has take sinusoidal in the middle and exponentially decaying on either side. Exponentially decaying on either side

Particle in a finite potential well For some eigenenergy E with and Solution with region condition for for for So, for some eigenenergy E, the k magnitude in the middle of the well would be the square root of 2mE over ha bar squared. And similarly, for the exponential decay constant on either side, we would have kappa equal to the square root of 2m, V0 minus E over ha bar squared. So for z less than minus Lz over 2 or greater than Lz over 2, we choose the solution is a decaying exponential. In the middle of well, we’re going to have some sort of sinusoidal or cosinusoidal solution. The constants is arbitrary. A,B,F,G constant value

Particle in a finite potential well Apply to the boundary conditions From continuity of the wavefunction at Writing Gives Similarly at Now we need to apply the boundary conditions to solve for the unknown coefficients constants A,B,F and G. from continuity of the wavefunction at z equal Lz over 2, the wave fuction at Lz over 2 is same just on the right-hand side of the barrier and the left-hand side of the barrier. The functions are writing simple notation, then the equation for continuity of the wave gives FXL equal ASL plus BCL. Similarly, applying at z equal minus LZ over 2, the equation gives GXL equal minus ASL plus BCL. Gives

Particle in a finite potential well Apply to the boundary conditions From continuity of the derivative at Gives Similarly at Gives So, we have four relations Now, the wavefunction must satisfy from continuity of the derivative. We’re similarly going to get two equation using the boundary conditions at z= LZ over 2 and z = -Lz over 2. the blue boxes is the acquired results. So we have four relations from continuity of wavefunctions.

Particle in a finite potential well Adding and subtracting process Adding and Subtracting and As long as Next, we can add or subtract this equations. Just adding the two equations, so that would give us 2BCL equal F plus G times XL. Similarly, substracting the two equations, that would give us 2BSL equal kappa over k times F plus G times XL. Now, as long as F is not equal to minus G, so anything other than the condition F equals minus G, then we can divide one by another. And we get the tangent equation.

Particle in a finite potential well Adding and subtracting process Subtracting and Adding and As long as Inversely, the first two equations process to subtracting and the second two equations process to adding. as long as F is not equal to G, then we can divide one by another. And we get the minus cotangent equation.

Particle in a finite potential well The only possibilities First case, , Note from Because and cannot both be 0. Hence in the well we have We can get the wavefunction with the constant condition. First, let’s look at the condition F is equal to G, in the possible case of tangent function. Note from this equation that we had here and this one that we had there that, if F is equal to G, this is 0 and this is 0. so the right-hand side of these equations are 0. the left-hand side must become 0. so the constant A must be 0 because the sin and cosin function cannot both be 0. So, in the well, we have the wavefunction as the cos function. This function is an even state. which is an even function

Particle in a finite potential well The only possibilities Second case, , Note from Because and cannot both be 0. Hence in the well we have And second, let’s look at the condition F is equal to minus G, in the possible case of - cotangent function. Similarly, previous case, these equations are 0, so the right-hand side of these equations are 0. the left-hand side must become 0. so the constant B must be 0. in the well, we have the wavefunction as the sin function. As you can see, this function is an odd state. which is an odd function

Solving for the eigenenergies Dimensionless units Use the energy of the first level in the “infinite” potential well Leading to a dimensionless Also, a dimensionless barrier height So, the parameters can be changed by Now, we can solve for the eigenergies. We’re going to change to dimensionless units. The energy unit we’re going to use is this one in the first level infinite potential well. So instead of energies E we’re going to epsilon. This is actually a dimensionless quantity. Also, instead of potential V0, we’are going to nu zero. Also, in terms of these energy units, example k and kappa, change to the epsilon and nu zero notation instead of E and Vzero.

Solving for the eigenenergies Dimensionless units Consequently, So, becomes or Similarly, becomes Consequently, this and this parameter also change this notations. So the tangent and minus cotangent function become to the this one and this one. or

Solving for the eigenenergies Graphical solution Choose a specific well depth and plot the curve Adding the curves Then, we can draw the figure of functions. First we plot the curve root nuzero minus epsilon graph. Increasing the nu zero value, the yellow graph increases the figure like this. If the nu zero is 8, the curve draw like this. We can add the curves of tangent and minus cotangent functions. The red line is tangent and blue line is minus cotangent function.

Solving for the eigenenergies Graphical solution Choose a specific well depth and plot the curve Adding the curves The solution of eigenenergies are the values of epsilon at the intersections of this curve, the square root curve, and these two curves. So, one of those solutions is for epsilon equal to 0.663, and we could get that off this graph. Second solution would be with the contangent, and that occur at 2.603 energy units. And the third solution occur at 5.609. The solutions are the values of at the intersections of functions

Solving for the eigenenergies Solutions These are the solutions for a well depth Note that they are all lower energies than the corresponding solutions for the infinitely deep well of the same width So these are the actual solutions. This is first solution and second and third. And there are only three such solutions in a well of this height and depth. Note that they are all lower energies than the corresponding solutions for the infinitely deep well of the same width.

Harmonic oscillator Mass on a spring A simple spring will have a restoring force F acting on the mass M From Newton’s second law Many kinds of quantum mechanical systems can be based on the simple harmonic oscillator, at least as a good first approximation. We remember that we have a mass M on a spring, and we are going to have a restoring force on that mass. The restoring force F is minus K times y which K is spring constant. from newton’s second law, the F equals M times a is minus K times y. And we get the this equation where omega square is K over M. we’re going to use small m, because we like that for our quantum mechanical particles. where

Harmonic oscillator Potential energy Harmonic oscillator equation The potential from the restoring force F is Harmonic oscillator equation The potential energy corresponding to restoring force F is something we can work out. The potential will be the integral from 0 to z of the force times the distance. Substituting the F, it gives the 1 over 2 Kz squared. And it shows the a parabola function. So that’s the potential energy we’re going to be able to put into our equations to solve this quantum mechanical problem. And it is the same thing as 1/2m omega squared times z squared. So, we can put this potential energy into a 슈레딩거 equation. For convenience, we define a dimensionless distance unit so the unit of distance is to be 시. So the 슈레딩거 equation becomes like this. We define a dimensionless distance unit

Harmonic oscillator Harmonic oscillator equation One specific solution to this equation is with a corresponding energy Solutions of the form where is some set of functions still to be determined Substituting into the equation Now, one specific solution to this equation is proportional to the exponential of minus 시 squared over 2. and this would have a corresponding energy of ha bar omega over 2. and then, the solutions of the 사이 forms like this. This set of functions here is just a set of functions still to be determined. Then we substitute it back into our 슈레딩거 equation. the 슈레딩거 equation changes the form like this. it turns out the solutions to this equation are actually quite well know. They are technically what are called Hermite polynomials This is the defining differential equation for the Hermite polynomials

Harmonic oscillator Harmonic oscillator equation Solutions to differential equation exist provided That is, The allowed energy levels are equally spaced separated by an amount These solutions here exist provide, and only if, the energy satisfies this equation. Therefore, energy is satisfied this equation. The allowed energy levels are equally spaced separated by an amount ha bar times omega. Like the potential well, there is a zero point energy. Even if we go to n equals 0, we’re still left with an ha bar over 2 times omega. Like the potential well, there is a “zero point energy” here

Harmonic oscillator Hermite polynomials Harmonic oscillator solutions Hermite polynomials are presented by Note they are either odd or even Hermite polynomials satisfy “recurrence relation” Harmonic oscillator solutions Normalizing The first hermite polynomials are constant value. The second one is a linearly varying function. The third one is a kind of parabola function. Note they are either odd or even, for example, this terms are odd and this terms are even. and hermite polynomials satisfy the recurrence relation. So we finds the Hermite functions using the previous terms. And we normalize the harmonic oscillator solutions. The An is root , 시 is root. Therefore be in the dimensionless form, this set of eigensolutions of the harmonic oscillator equation. So we’ve solved this problem for eigenfunctions and eigenenergies.

Harmonic oscillator Harmonic oscillator eigensolutions Let’s just look at what they look like. These are the energy levels, these orange dashed lines. Here’s the eigenfunctions. It’s basically a Gaussian.

Harmonic oscillator Classical turning points The intersections of the parabola and the dashed lines give the “classical turning points” where a classical mass of that energy turns round and goes back downhill The intersections of the parabola and the dashed lines give the classical turning points which a classical mass of that energy turns round and goes back downhill. So this is called classical turning points.

Summary We can solve the wavefunction to a particle in a finite potential well Accordance with constant conditions, the results in the well shows that two case We can solve the eigenenergies using the graphical solution. Using the harmonic oscillator, we can solve the harmonic oscillator eigensolutions. First case, Second case, Intersection of We can solve the wavefunction a finite potential well case. Accordance with a constant conditions, we can solve the wavefunction cos or sin in the middle of well. And using the dimensionless units, we solve the eigenenergies using the graphical solutions intersection of this and these functions. Finally, the using the harmonic oscillator model, we can solve the harmonic oscillator eigensolutions.

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