### 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

## 19

## 0

## Popular in Course

## Popular in Physics 2

This 10 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 19 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 12 Symmetry and conservation laws in QM Imagine computing the time variation of the expectation value of some ob servable corresponding to the operator A We have amgt 8t dltAgt6ltII AEIlgt ltII A dt 8t From the Schroedinger equation we have dltAgt df m lt x11 HA AH gxp gt Thus any operator which commutes with the Hamiltonian will have an ex pectation value in any state which does not vary with time it is conserved Furthermore we know from our discussion of angular momentum that eigen states of the hamiltonian can also be chosen to be eigenstates of the operator A if it commutes with H Thus they can be labelled with the corresponding eigenvalue of A This quantum number will not then change with time There are three basic operators we know that satisfy classical conservation laws the energy momentum and angular momentum In QM the associated operators will hence commute with the Hamiltonian in the rst case this is of course triviall We might ask the questiom is there any fundamental principle of nature which might guarantee that these classical conservation laws also hold true in QM The answer is yes and it has to do with symmetry Consider the operator 0 exp zap7 By expanding this operator in powers of p the momentum operator it can be seen that this operator effects a translation of the wavefunction by the distance a If the system is translation invariant this does not change the energy of the state This will be true if the Hamiltonian of the system is invariant under translation and it follows that 0 must commute with H This in turn means that p will commute with H and hence will be conserved This invariance of the hamiltonian is called a symmetry and we see the intimate connection between symmetry and conservation laws A similar result is true for angular momentum and symmetry under rotations Perturbation Theory Suppose we have solved the Schroedinger equation for some potential V0 and now want to find the eigenvectorsvalues for another potential which differs from V0 by a small amount Perturbation theory is a tool for nding approximations to the latter knowing the eigenvectorsvalues for V0 Writing H Hg AV We will take A to be a small number We write gt gt Hie gt EnE3AE If we substitute this ansatz into the Schoedinger equation and equate powers of A we nd Vszl gt Hggzp gt Eggug gt E 3l gt Taking the dot product with lt we nd E lt 3 V 3 gt This is the rst order shift in the energy To nd the eigenvector to first order in A we expand it gt Zom z gt m Notice that we do not need to include any term en in this expression such a term is not 0A Thus we nd 2w mew gt ltv EM gt m Taking the dot product with gt we nd E mean lt waxest gt Notice that the denominator is never zero unless there is degeneracy Fine Structure of Hydrogen The spectrum of hydrogen is in leading approximation given by that of the nonrelativistic Schoedinger equation with Coulomg potential However the true Hamiltonian of the system should treat the electron relativistically Since terms are supressed by 0U2C2 relative to the leading term their er 39ects can be computed using perturbation theory Using the relativistic formula T p22m p48m302 the lowest order correction to the kinetic energy is 1 2 E1 lt 2921 gt 8m302 Now p219 gt 2mE V 1L gt Hence 1 7 27m E1 E 471 3 27m l 12 Notice that this breaks the original degeneracy of the levels and makes them ldependent a 6347ng There is a further contribution to the energy of the electron coming from relativity Imagine the electron at rest with the proton circling around it Such a charge will generate a magnetic eld which interacts with the electron s spin in the we described earlier 1 1 E32Enalt 7gta2lt7gt 739 739 This yields V ysB Notice that the magnetic eld will be proportional to the orbital angular momentum vector of the electron 1 1 ifL mo 73 Ba The factor V is given by 9 Y i 2m Thus we expect a lowest order contribution to the energy of the form from this spin Orbit term to be 2 E1 8 1 lt 7731 gt 87mg NRC2T3 Now this term does not commute with L and S separately so these two are not separately conserved However V does commute with J L 3 Thus states of the hydogen atom should be labelled by the conserved quantum numbers 72 and j Furthermore LS J2 S2 L2 Thus the eigenvalues of this new term are proportional to jj 1 ss 1 ll 1 This allows us to write the electron energy shift 1 2 E3 1 W 1 mo W 12l 1 Combining this result with the correction from the kinetic energy we nd the nal ne structure formula 136 a2 n 3 7 1 7 7 Em 712 712 12 4 Degenerate Perturbation Theory If the unperturbed states are degenerate in energy then ordinary perturbation theory fails unless the two degenerate states have lt italM gt 0 which occurs if the perturbation commutes with the original Hamiltonian this is the case for the ne structure calculation above Thus we must nd some other to handle the problem Suppose that Wu gt and M gt are two orthogonal degenerate states with energy E0 Notice that any linear combi nation of states will also be an eigenstate of the unperturbed Hamiltonian Typically V will break this degeneracy The clue to hwo to proceed can be seen from the formula for the coefficients 8 The two degenerate states will start to dominate in this sum and we may analyze the system a simple 2state system by including only those two states in the equation for the new state vector We will nd H0111 V E01131 E1 0 4 with 13gt W we nd by taking appropriate dot products that a lt 3 V 3 gt 6 lt w iv m gt aEl This is just a simple matrix problem where the matrix elements are just taken with respect to the original unperturbed wavefunetions 0 Lecture 12 Symmetry and conservation laws in QM Imagine computing the time variation of the expectation value of some ob servable corresponding to the operator A We have amgt 8t dltAgt6ltII AEIlgt ltII A dt 8t From the Schroedinger equation we have dltAgt df m lt x11 HA AH gxp gt Thus any operator which commutes with the Hamiltonian will have an ex pectation value in any state which does not vary with time it is conserved Furthermore we know from our discussion of angular momentum that eigen states of the hamiltonian can also be chosen to be eigenstates of the operator A if it commutes with H Thus they can be labelled with the corresponding eigenvalue of A This quantum number will not then change with time There are three basic operators we know that satisfy classical conservation laws the energy momentum and angular momentum In QM the associated operators will hence commute with the Hamiltonian in the rst case this is of course triviall We might ask the questiom is there any fundamental principle of nature which might guarantee that these classical conservation laws also hold true in QM The answer is yes and it has to do with symmetry Consider the operator 0 exp zap7 By expanding this operator in powers of p the momentum operator it can be seen that this operator effects a translation of the wavefunction by the distance a If the system is translation invariant this does not change the energy of the state This will be true if the Hamiltonian of the system is invariant under translation and it follows that 0 must commute with H This in turn means that p will commute with H and hence will be conserved This invariance of the hamiltonian is called a symmetry and we see the intimate connection between symmetry and conservation laws A similar result is true for angular momentum and symmetry under rotations Perturbation Theory Suppose we have solved the Schroedinger equation for some potential V0 and now want to find the eigenvectorsvalues for another potential which differs from V0 by a small amount Perturbation theory is a tool for nding approximations to the latter knowing the eigenvectorsvalues for V0 Writing H Hg AV We will take A to be a small number We write gt gt Hie gt EnE3AE If we substitute this ansatz into the Schoedinger equation and equate powers of A we nd Vszl gt Hggzp gt Eggug gt E 3l gt Taking the dot product with lt we nd E lt 3 V 3 gt This is the rst order shift in the energy To nd the eigenvector to first order in A we expand it gt Zom z gt m Notice that we do not need to include any term en in this expression such a term is not 0A Thus we nd 2w mew gt ltv EM gt m Taking the dot product with gt we nd E mean lt waxest gt Notice that the denominator is never zero unless there is degeneracy Fine Structure of Hydrogen The spectrum of hydrogen is in leading approximation given by that of the nonrelativistic Schoedinger equation with Coulomg potential However the true Hamiltonian of the system should treat the electron relativistically Since terms are supressed by 0U2C2 relative to the leading term their er 39ects can be computed using perturbation theory Using the relativistic formula T p22m p48m302 the lowest order correction to the kinetic energy is 1 2 E1 lt 2921 gt 8m302 Now p219 gt 2mE V 1L gt Hence 1 7 27m E1 E 471 3 27m l 12 Notice that this breaks the original degeneracy of the levels and makes them ldependent a 6347ng There is a further contribution to the energy of the electron coming from relativity Imagine the electron at rest with the proton circling around it Such a charge will generate a magnetic eld which interacts with the electron s spin in the we described earlier 1 1 E32Enalt 7gta2lt7gt 739 739 This yields V ysB Notice that the magnetic eld will be proportional to the orbital angular momentum vector of the electron 1 1 ifL mo 73 Ba The factor V is given by 9 Y i 2m Thus we expect a lowest order contribution to the energy of the form from this spin Orbit term to be 2 E1 8 1 lt 7731 gt 87mg NRC2T3 Now this term does not commute with L and S separately so these two are not separately conserved However V does commute with J L 3 Thus states of the hydogen atom should be labelled by the conserved quantum numbers 72 and j Furthermore LS J2 S2 L2 Thus the eigenvalues of this new term are proportional to jj 1 ss 1 ll 1 This allows us to write the electron energy shift 1 2 E3 1 W 1 mo W 12l 1 Combining this result with the correction from the kinetic energy we nd the nal ne structure formula 136 a2 n 3 7 1 7 7 Em 712 712 12 4 Degenerate Perturbation Theory If the unperturbed states are degenerate in energy then ordinary perturbation theory fails unless the two degenerate states have lt italM gt 0 which occurs if the perturbation commutes with the original Hamiltonian this is the case for the ne structure calculation above Thus we must nd some other to handle the problem Suppose that Wu gt and M gt are two orthogonal degenerate states with energy E0 Notice that any linear combi nation of states will also be an eigenstate of the unperturbed Hamiltonian Typically V will break this degeneracy The clue to hwo to proceed can be seen from the formula for the coefficients 8 The two degenerate states will start to dominate in this sum and we may analyze the system a simple 2state system by including only those two states in the equation for the new state vector We will nd H0111 V E01131 E1 0 4 with 13gt W we nd by taking appropriate dot products that a lt 3 V 3 gt 6 lt w iv m gt aEl This is just a simple matrix problem where the matrix elements are just taken with respect to the original unperturbed wavefunetions 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

#### "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!"

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### 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.