### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Introduction to Quantum Mechanics I PHY 567

Syracuse

GPA 3.64

### View Full Document

## 11

## 0

## Popular in Course

## Popular in Physics 2

This 5 page Class Notes was uploaded by Clement Bernier on Wednesday October 21, 2015. The Class Notes belongs to PHY 567 at Syracuse University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/225636/phy-567-syracuse-university in Physics 2 at Syracuse University.

## Popular in Physics 2

## Reviews for Introduction to Quantum Mechanics I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/21/15

Lecture 9 Chemical Bonding revisited We can use this matrix formulation of QM to return to the problem of un derstanding the QM origin of the chemical bond Suppose we are looking at the H ion which consists of just one electron shared between two protons If these two protons are a long apart we can envisage the elec tron being attached to one or the other giving two physical states 1 gt and 2 gt These will be our base states This truncation of the space of all states to just two will be suf cient to discuss simple properties of the ground state of the system Imagine taking the two protons in nitely far apart then we expect no overlap lt 1 H 2 gt 0 the local Hamiltonian cannot cause an electron to jump from one state to another In this limit we can trivially solve the Schroedinger equation for the two state system we nd 1 gt exp iHHth Thus H11 E0 is to be interpreted the energy an electron would have in a single hydrogen ion By symmetry it is equal to H22 the energy if the electron lived on the other proton As we put the two protons closer and closer together we nd a nonzero chance for the electron to jump from one proton to another this is represented by a nonzero matrix element between the two states lt 1 H 2 gt 0 A Thus the two state system we must solve takes the form aim we 2 Hit 0 02 where cl lt 1311 gt etc Lets 100k for solutions where both amplitudes 0102 have the same time dependence exp iEt7z Thus we need to solve the matrix eigenvalue problem H c Ec The possible eigenvalues turn out to be E Eg A with corresponding eigenvectors e l gt 2 gt Thus the ground state is lowered in energy with respect to a single hydogren atom and its amplitude is an even function of the two separate amplitudes In order to get the true ground state we must use a state in which the electron is equally split between the two protons This energy will decrease the distance between the two protons is decreased ie we assume that the magnitude of A increases the distance decreases hence chemical bonding Notice also that if we start out with the electron located on atom 1 gt it will not remain so the presence of the two time dependent factors for the eigenstates which oscillate at different frequencies ensures that at some later time it will contain an admixture of the second state ie there will be some probability that the electron would be found on the second atoml Suppose we had two distinct atoms trying to share an electron in this picture It is to modify the calculation to handle the situation when H11 g H22 For small A the new energies are A2 E H 11lam H22 A2 E H I 22 JH H22 This is typically much smaller than the splitting for equal energies and ex plains why single electron bonding in nonsymmetric molecules is not very common A more typical situation is the two electron chemical bond in which two electrons are shared by the two atoms The hydrogen molecule furnishes a nice example The situation can still be modelled approximately a two state system corresponding to the physical situation in which electron a is around the rst proton and electron b is around the second and Vice versa As before the base states for these two situations have the same energy by symmetry but before there is a possibility of hopping or exchange of the two electrons between the two protons The mathematics is identical to the single electron problem and so we expect that the allowed energies of the combined system are split by this quantum mechanical hopping and we pre dict that the energy of the ground state is lowered a result This decrease in ground state energy is accentuated for small interproton separations and results in a chemical bond the covalent bond If we admit two more base states into the picture those corresponding to the two electrons being on one or the other of the two atoms if they are dissimilar we can can allow for innic banding in this picture Thus by extensions of these ideas we can start to understand the quantum mechanical basis for chemistry Emergence of wave mechanics Consider a one dimensional crystal with lattice sites i 1N spaced 1 apart A quantum particle can move on this lattice and we can adopt a set of basis states to describe this motion follows 33139 gt corresponds to the particle being located at site i 311 gtZltei11gt ci gt 139 Writing down the Schroedinger equation for this system we nd do my Hijcj where lt 6131 gt is the amplitude for the particle to be located at site i the component of the state vector on the basis vector gm gt The matrix elements Hij are just Hij lt gt and measure the component of H cj gt on the basis vector gei gt Remember that the hamiltonian measures the change in a vector under an in nitessimal amount of time We expect that for very short times the only state a particle can hop to from lattice site i are its neighbours i 1 and i 1 Furthermore we expect that the probability for going in either direction is the same Thus we might guess HEEI39 gt gt fl 1391 gt fl i1 gt where A and V are some constants Furthermore we expect that for a free particle the vector corresponding to a uniform probability distribution for the particle is time independent This identi es V a simple function of the potential energy of the particle We thus have found Hy mam V61 where Aid is a discrete form of the operator restricted to the crystal lattice Furthermore we can rewrite this matrix equation in the suggestive form 2 a AaA V Em 7 3 In the limit in which a gt 0 with N gt 00 the sums become integrals and Lava gt 6i j where i and j can now be thought of continuous valued positions Thus this equation will be the same the fundamental equation of wave mechanics if we require 7amp2 A 7 a 2m In this limit can clearly adopt a normalization for the eigenvectors in which 2 1 1 which just goes over into the usual form we 1 when it is realized that lt 6191 gt is just the probability ampli tude for nding the particle at lattice site an Notice that the eigenvectors then take the form of Dirac delta functions gt Thus we can see that wave mechanics is just one representation of QM in which we focus on measurements of position and expand all vectors on a basis corresponding to the eigenvectors of the position operator On a nite lattice this is a nite dimensional matrix problem and a gt 0 it goes over to a in nite dimensional matrix problem Differential operators like LET being just convenient representations of in nite dimensional matrices Indeed you can see that the space of all functions satis es all the require ments of a vector space the sum of two functions is a function One can de ne Operators and scalars just complex numbers and most importantly one can de ne the notion of a dot product where summation over components is replaced by integration over a continuous valued index the position x The most delicate remaining issue consists of completeness that any function can be expanded out as a sum over a set of basis functions The boundary conditions play a crucial role here if the wavefunction is subject to suitable boundary conditions this postulate may also be satis ed eg Fourier series and the in nite square well Such functions are also square integrable that is their dot product is nite The latter is necessary for a probability interpretation of the theory Functions satisfying these requirements are said to live in a Hilbert space after the famous mathematician Band structure Suppose we use this lattice Schroedinger equation to model the situation in a crystal in which an additional constant potential energy is acquired when an electron is at a lattice site V V0 It is to see that 61km is a solution of the time independent Schroedinger equation with energy 2 E V0 1 coska for any C from 7ra to zero for an in nitely long crystal Thus the possible energies form a band with energies ranging from V0 to V0 Clearly V0 is the energy an electron would have on a isolated lattice site its original atomic energy level this is split into a band of allowed energies in the crystal Furthermore a similar process will be true for all the original atomic levels leading to a sequence of bands of allowed energies separated by nite energy gaps This structure is a crucial component to the arguments which are used to explain all metallic insulator and semiconductor behavior Electrons ll these stationary states up to some maximum energy the socalled Fermi energy there is a principle called the Pauli exclusion principle that prevents electrons from being in the same state A lled band is inert for conduction purposes it has an equal number of left and right traveling electron waves and no free states available to take more electrons such those responsible for carrying an electric current A partially lled band however has available electron states for carrying current and such a substance is a metal An insulator has a lled band and hence cannot use electrons to carry heat or electric current A semiconductor has a lled band but a small bandgap to the next empty band hence at room temperature a small number of electrons can become thermally excited to this new band and carry current Furthermore the holes left behind in the nearly lled band can also carry current this substance will be a semiconductor 0

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.