Physics 70007, Fall 2009 Answers to Final Exam

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Physics 70007, Fall 2009 Answers to Final Exam"

Transcription

1 Physics 70007, Fall 009 Answers to Final Exam December 17, Quantum mechanical pictures a Demonstrate that if the commutation relation [A, B] ic is valid in any of the three Schrodinger, Heisenberg, interaction/dirac pictures, then it is valid in all pictures. Consider the three statements [A S, B S ] ic S, [A H, B H ] ic H, [A I, B I ] ic I 1 where the subscript S, H, I on the operators indicates the Schrodinger, Heisenberg, or interaction picture respectively. We need to show that any one of these statements implies the other two. One standard way of doing this, which ensures that we don't do any unnecessary work, is to show that the rst statement implies the second, the second implies the third, and the third implies the rst. This is the strategy we will pursue. We need the relations between operators in the various pictures. The Heisenberg-picture operators are given in terms of the Schrodinger-picture operators as A H e iht/ A S e iht/ with analogous relations for B and C. We have taken the initial time t 0 equal to zero without any loss of generality. Next we relate the Schrodinger and interaction pictures: A I e ih0t/ A S e ih0t/ 3 We dene the interaction picture as usual, by splitting up the full Hamiltonian H into a solvable part H 0 plus a perturbation V. Let us invert this relation to nd A S in terms of A I : A S e ih0t/ A I e ih0t/ 4 Finally we relate the Heisenberg and interaction pictures: we invert to solve for A S, and then substitute the result into 3. The result is A I e ih0t/ e iht/ A H e iht/ e ih0t/ 5 With these preliminaries done, we can proceed to our proofs. First we assume that [A S, B S ] ic S ; then we use : [A H, B H ] A H B H B H A H e iht/ A S e iht/ e iht/ B S e iht/ e iht/ B S e iht/ e iht/ A S e iht/ e iht/ A S B S e iht/ e iht/ B S A S e iht/ e iht/ [A S, B S ] e iht/ e iht/ ic S e iht/ ic H. 1

2 Next we assume [A H, B H ] ic H and use 5: [A I, B I ] A I B I B I A I e ih0t/ e iht/ A H e iht/ e ih0t/ e ih0t/ e iht/ B H e iht/ e ih0t/ e ih0t/ e iht/ B H e iht/ e ih0t/ e ih0t/ e iht/ A H e iht/ e ih0t/ e ih0t/ e iht/ A H B H e iht/ e ih0t/ e ih0t/ e iht/ B H A H e iht/ e ih0t/ e ih0t/ e iht/ [A H, B H ] e iht/ e ih0t/ e ih0t/ e iht/ ic H e iht/ e ih0t/ ic I. Finally we assume [A I, B I ] ic I and use 4: [A S, B S ] A S B S B S A S e ih0t/ A I e ih0t/ e ih0t/ B I e ih0t/ e ih0t/ B I e ih0t/ e ih0t/ A I e ih0t/ e ih0t/ A I B I e ih0t/ e ih0t/ B I A I e ih0t/ e ih0t/ [A I, B I ] e ih0t/ e ih0t/ ic I e ih0t/ ic S. b Calculate the time dependence of the position operator x H t of a free particle in the Heisenberg picture. We need to solve the Heisenberg equation of motion for x H t: d dt x H t 1 i [x, H] H 6 where operators without a subscript are in the Schrodinger picture, and the Hamiltonian is H p /m for a free particle. The needed commutator is ] [x, H] [x, p 1 [ x, p ] 1 m m m i p i p m using a known result from previous homework Sakurai problem 1.9 to evaluate the commutator. The equation of motion becomes d dt x H t 1 m p H t 7 and we see that we need to know the time evolution of p H. Its equation of motion is d dt p H t 1 i [p, H] H 1 i [p, p m since p commutes with any function of itself. Thus p H is just a constant in time: ] H 0 p H t p H 0 p 0 8 The initial condition for a Heisenberg-picture operator is that it equals the Schrodinger operator at the initial time t 0, which we took equal to zero. Using 8, we can trivially integrate the dierential equation 7 and apply the initial condition x H 0 x 0, to nd x H t x 0 + p 0 m t

3 which is our desired result.. Particle in one-dimensional potential a Show that the energy eigenstates of a simple harmonic oscillator obey the following property under the x x parity transformation: x n 1 n x n The parity operator P takes x to x, and is dened in such a way that P 1 and that P is Hermitian see Sakurai for details; thus x x P 9 Now we take as given that the ground state of the simple harmonic oscillator has even parity: P we saw this from our explicit solution for the energy wavefunctions. The higher eigenstates are found from the ground state by n 1 n! a n 0 11 where a is the usual raising operator: a mω x ip mω How does this operator behave under parity? Since position and momentum both invert: P xp x P pp p then clearly a inverts as well: P a P a, or equivalently P a a P 1 That is, P anti-commutes with a. Now we put together our results 9,10,11,1: 1 x n x a n 0 n! 1 x P a n 0 n! 1n x a n 0 P n! 1n x a n 0 n! 1 n x n b Use the above result to obtain the energy spectrum and eigenstates of a quantum mechanical particle trapped in the following potential: V x if x < 0 and V x 1 mω x if x > 0. 3

4 In the region x > 0, the particle experiences a harmonic oscillator potential, so its wavefunction will be a linear combination of simple harmonic oscillator energy eigenfunctions. In particular, any eigenfunction for our problem must be an eigenfunction of the ordinary SHO. However, since the potential is innity for x < 0, the particle is absolutely forbidden to be there: the wavefunction is forced to be zero. And since ψ must be continuous even where the potential changes by an innite amount, it must be the case that ψ 0 + lim x 0 + ψ x 0 which acts as a boundary condition which the wavefunction in the x > 0 region must satisfy. What does this condition do? For the even-parity SHO eigenfunctions, there is nothing forcing ψ 0 to be zero, and in fact it can be shown from the exact form of the wavefunction given in Appendix A.4 of Sakurai that it will always be non-zero. Therefore, the even-parity solutions do not satisfy the boundary conditions and can be eliminated. But, the odd-parity solutions satisfy ψ x ψ x; in particular, letting x 0, we nd that ψ 0 0. These solutions do satisfy the boundary condition, and hence are allowed to be eigenfunctions of our current potential. There is a small adjustment we have to make to the normalization of these eigenfunctions: as is, they are normalized so that ˆ dx ψ 1 where the integral goes over the full range of x, from to. But now ψ is set to zero for negative values of x; so the eective range of integration shrinks to 0 x <, and over this range we have ˆ 0 dx ψ 1 To restore normalization, we have to scale up ψ by a factor of. What, then, happens to the energy of the odd-parity functions, as given by ˆ dx ψ Hψ E ˆ dx ψ ψ E ˆ 0 dx ψ On the one hand, ψ is scaled up by a factor of, but on the other hand the range of integration shrinks to half of the original. These factors cancel each other, with the consequence that the total energy of the state remains the same. So let us dene n p + 1, where p 0, 1,,... and thus n is odd. Then the eigenfunctions for our potential are given by 0, x 0 ψ p x x n, x 0 where n is an eigenstate of the original simple harmonic oscillator. The energy of the state ψ p x is ω n + 1 ω p Angular momentum operators: general properties. [ ˆ a Show that L, ˆL ] i 0. Here i x, y, z}. I will omit the hats from the operators, and show all sums explicitly rather than using the Einstein summation convention. 4

5 First let us establish a general relation for any operators A, B, C, which we will use throughout this entire problem: [A, BC] ABC BCA ABC BAC + BAC BCA [A, B] C + B [A, C] So, since L k L kl k, we have [ L, L i ] k [L k L k, L i ] k [L i, L k L k ] k [L i, L k ] L k + L k [L i, L k ] Next we use the fundamental property satised by angular momenta, namely [L i, L k ] l iɛ ikll l : [ ] L, L i i ɛ ikl L l L k + L k L l k l Now notice that we are summing over indices k and l, and what we are summing is a product of one factor which is anti-symmetric in k, l and a second factor which is symmetric in k, l. We can instantly conclude that the sums give zero, the reason being that the summands with k l cancel in pairs for example, the term with k, l 1, cancels k, l, 1, since ɛ i1 ɛ i1 while L 1 L + L L 1 L L 1 + L 1 L, while the terms with k l are zero all by themselves e.g. ɛ i11 0. This is another technique we will be using repeatedly in this problem. [ ] So L, L i 0: we are done. b Demonstrate that [L i, x k ] iɛ ikl x l. First we use the denition of orbital angular momentum, L x p, to write L i ɛ ijl x j p l j notice how we avoid using the index k which is already being used in the problem statement. Improperly reusing an index is a frequent source of error in this type of computation. Then [L i, x k ] ɛ ijl [x j p l, x k ] ɛ ijl [x k, x j p l ] j l j l ɛ ijl [x k, x j ] p l + x j [x k, p l ] j l ɛ ijl 0 + x j i δ kl j l l i j ɛ ijk x j i j i l ɛ ikj x j ɛ ikl x l. c Similarly, show that [L i, p k ] i ɛ ijl p l. 5

6 The method of proof is very similar to that of part b: [L i, p k ] ɛ ijl [x j p l, p k ] ɛ ijl [p k, x j p l ] j l j l ɛ ijl [p k, x j ] p l + x j [p k, p l ] j l ɛ ijl i δ kj p l + 0 j i ɛ ikl p l. l l d Finally, show that [ L i, x ] 0 and that [ L i, p ] 0. Proceeding directly: [ Li, x ] [L i, x k x k ] k [L i, x k ] x k + x k [L i, x k ] k Use the result of part b: [ Li, x ] i ɛ ikl x l x k + x k x l k l As in part a, we are summing over the product of anti-symmetric and symmetric; thus [ L i, x ] 0. Similarly using part c: [ Li, p ] [L i, p k p k ] k [L i, p k ] p k + p k [L i, p k ] k i ɛ ikl p l p k + p k p l k l Angular momentum operators in spherical coordinates. Start from the expression for the gradient operator in spherical coordinates r, θ, φ: er r + e θr θ + e φ r φwhere e α is the unit vector in direction α, and calculate the following operators all results should be given in spherical coordinates : a the angular momentum operator L: We use the denition L r p r i ; then we write r e r r, and use the given expression for and the fact that e r, e θ, e φ form an orthogonal triad with e r e θ e φ, e θ e φ e r, e φ e r e θ 6

7 so L i e r r e r r + e θ i e φ θ e θ φ r θ + e φ r φ [I will do part c before part b] c all Cartesian components L α α x, y, z} of L: The Cartesian components are given by e.g. L x e x L, so we need to be able to express the Cartesian unit vectors in terms of spherical coordinates and spherical unit vectors. To do this, we recall what the unit vector for a given coordinate means: it is a vector of length 1, in the direction of greatest increase for that coordinate i.e. in the direction of the gradient of that coordinate: so x e x x For reference, we write out the transformation between spherical and Cartesian: x r cos φ y r sin φ z r cos θ So we calculate: x e r r + e θ r θ + e φ r φ r cos φ e r cos φ + e θ cos θ cos φ e φ sin φ It is easy to check that this vector already has length 1; so this is also e x. Therefore, L x e x L i e r cos φ + e θ cos θ cos φ e φ sin φ i cos θ cos φ φ sin φ θ e φ θ We repeat this for the next component: y e r r + e θ r θ + e φ r φ r sin φ This vector has length 1, and so equals e y. So L y e y L e r sin φ + e θ cos θ sin φ + e φ cos φ i e r sin φ + e θ cos θ sin φ + e φ cos φ i cos θ sin φ φ + cos φ θ e φ θ e θ φ e θ φ 7

8 Finally we work through z: z e r r + e θ r θ + e φ r φ r cos θ e r cos θ e θ Again this is unit and so equals e z. Then L z e z L i e r cos θ e θ i φ. e φ θ e θ φ b the angular momentum squared operator L : We know that L L x + L y + L z; so now that we have expressed the individual components of L as dierential operators in spherical coordinates, our strategy will be to apply L to a generic function f r, θ, φ: L f L xf + L yf + L zf First we calculate a useful derivative: d cos θ cos θ cos θ dθ sin 1 θ sin θ With this done, we compute each term in turn: L xf i cos θ cos φ φ sin φ θ cos θ f f cos φ sin φ φ θ i cos θ cos φ sin φ f f cos φ sin φ f sin θ φ sin θ + cos θ cos φ sin φ φ + cos θ cos φ sin θ φ + cos θ cos φ } f φ θ + sin φ f θ L yf i cos θ sin φ φ + cos φ θ cos θ f f sin φ + cos φ φ θ i cos θ cos φ sin φ f sin θ φ + cos θ sin φ sin θ + cos φ sin φ f sin θ φ cos θ cos φ sin φ f φ + cos θ sin φ } f φ θ + cos φ f θ L zf i f φ Adding these, we nd L f i 0 f cos φ + θ cos sin φ + sin φ + 1 f θ } +0 f φ θ + cos φ + sin φ f θ i 1 f sin θ φ + cos θ f i 1 f sin θ φ + 1 } θ + f θ f } θ θ 8 f θ φ + cos θ f θ + cos θ cos φ sin φ cos θ cos φ sin φ cos φ + sin φ f θ f φ θ f φ θ

9 so we conclude that L i 1 sin θ φ + 1 }. θ θ d the raising and lowering operators L ±. We calculate directly from the denition of L ± : L ± L x ± il y i cos θ cos φ φ sin φ θ ± i cos θ } sin φ φ + cos φ θ i cos θ } cos φ ± i sin φ φ ± i cos φ ± i sin φ θ i cos θ } e±iφ φ ± ie ±iφ θ i e ±iφ cos θ } φ ± i θ. 5. The rotation group. Rotations R in three dimensions form a non-commutative three-parameter three dimensional group. Show that rotations around a single axis form a commutative, one-parameter group: a Give the rotation matrix R z φ that corresponds to a rotation around the z axis with angle φ. If we consider active rotations i.e. rotations of the object rather than the axes, then the rotation matrix is given by [ cos φ ] sin φ sin φ cos φ Working this problem in terms of passive rotations is OK as well just switch the o-diagonal terms of the matrix, throughout the entire problem. b Calculate the generator A z lim φ 0 φ R z φ. The derivative of R z φ is given by sin φ cos φ cos φ sin φ As φ 0, we have cos φ 1 and sin φ 0, so the generator is 0 1 A z. 1 0 c Show that R z φ exp [φa z ]. First we show that A z I: A z

10 So now, let n be an even number; then we can write n p where p is an integer, and so A n z A p z I p 1 p I We can obtain a corresponding result for odd n p + 1, by multiplying through by A z : A p+1 z 1 p A z Then we compute directly: exp [φa z ] I n0 p0 p0 1 n! φa z n φ p p! Ap z + p0 1 p φ p p! + A z φ p+1 p + 1! Ap+1 z p0 1 p φ p+1 p + 1! I cos φ + A z sin φ cos φ 0 0 sin φ + 0 cos φ sin φ 0 cos φ sin φ sin φ cos φ R z φ. d Show all group properties of rotations around the z axis: d.1 existence of the identity element 1 z ; We take the binary operation of the group to be ordinary matrix multiplication. Then as we know, the identity element for matrix multiplication is the identity matrix And if we let φ 0, then R z φ 0 is exactly the identity matrix. Thus the group does possess an identity element, as required. Actually φ πn where n is any integer, will also work; if we are concerned with the uniqueness of the identity element, we should restrict the allowed range for the group parameter to e.g. 0 φ < π. d. closure: R z φ 1 R z φ is also a rotation around z; Calculating directly: R z φ 1 R z φ cos φ1 sin φ 1 cos φ sin φ sin φ 1 cos φ 1 sin φ cos φ cos φ1 cos φ sin φ 1 sin φ cos φ 1 sin φ sin φ 1 cos φ sin φ 1 cos φ + cos φ 1 sin φ sin φ 1 sin φ + cos φ 1 cos φ cos φ1 + φ sin φ 1 + φ sin φ 1 + φ cos φ 1 + φ R z φ 1 + φ where we have used the standard angle addition formulas for sin and cos. This shows explicitly that R z φ 1 R z φ is also a rotation around the z-axis, and so the group is closed with respect to the group operation. 10

11 If we restrict the range of φ to 0 φ < π as in the previous part, then R z φ 1 + φ is altered to R z φ 1 + φ mod π, which is completely equivalent since cos α + πn cos α and sin α + πn sin α for any α and integer n. d.3 associativity: R z φ 1 R z φ R z φ 3 R z φ 1 R z φ R z φ 3, Using the closure property demonstrated above, showing associativity for group multiplication becomes easy: we have and R z φ 1 R z φ R z φ 3 R z φ 1 + φ R z φ 3 R z φ 1 + φ + φ 3 R z φ 1 R z φ R z φ 3 R z φ 1 R z φ + φ 3 R z φ 1 + φ + φ 3 which are manifestly equal. Essentially, associativity for the group elements reduces to associativity of addition of ordinary numbers. d.4 commutativity: R z φ 1 R z φ R z φ R z φ 1. In particular, show that R z φ 1 R z φ R z φ 1 + φ. We have already shown the lemma in d., demonstrating closure. Using the lemma, we have and R z φ 1 R z φ R z φ 1 + φ R z φ R z φ 1 R z φ + φ 1 R z φ 1 + φ so the group operation is commutative: that is, the group is Abelian. We see that commutativity of the group operation reduces to commutativity of addition of ordinary numbers. 6. Representations of the rotation group. a The rotation operator around the z axis is dened by the equation ˆD z α φ φ + α. Assume that all derivatives of the ket φ with respect to the angle variable φ exist. ˆD z α exp α φ exp i αl z where Lz i φ. Show that To express φ + α as the result of some operator upon φ, we simply expand φ + α in a Taylor series around φ: φ + α φ + α d α d φ + dφ! dφ φ +... note: we are committing a very common abuse of notation by using φ to represent both the general variable and the particular value of the variable at which we are evaluating the ket and derivatives of the ket. At any rate, we can then write φ + α I + α d dφ + α d! dφ +... φ exp α φ φ 11

12 so we can identify ˆD z α exp α φ. Then remembering that L z i φ i φ which we derived in problem 4c, we have also ˆD z α exp i αl z. b The rotation operator with angle φ around an arbitrary axis ˆn is given by [ ij ˆD nφ exp ] nφ. Show that J ˆD nφ jm j j + 1 ˆD nφ jm. In other words, this means that ˆD nφ jm is still an eigenket of J. Hint: the commutation rules you proved between J and J α in [problem 3a] will be very useful here. [ ] We showed in problem 3a that J, J α 0 for any given component α of J; then, by linearity, [ J, J n] 0 for any constant vector n as well. So J will also commute with any function of J n, in particular with ˆD nφ. So J ˆD nφ jm ˆD nφ J jm ˆD nφ j j + 1 jm j j + 1 ˆD nφ jm. Thus ˆD nφ jm is an eigenket of J with eigenvalue j j + 1, and so is itself a state with denite j. c The matrix elements of the rotation operator with angle φ around the axis n, i.e. the n + 1- dimensional representation of the rotation group are D j j m m nφ j m jm ˆD nφ. Use the results obtained in the previous part of this problem to argue that D j j m m is diagonal in j: D j j m m nφ δ j,j Dj m m nφ. We saw in the previous part that ˆD nφ jm has denite j; then we can write this state as m c m jm : i.e. a linear combination of states with dierent m but the same j. So then, D j j m m nφ j m jm ˆD nφ m c m j m jm But the various states jm are orthonormal, so for any overlap matrix element j m jm to be non-zero, we must have j j. Thus the whole sum will also be zero unless j j, and so we are done. d Show that for small rotation angles ɛ D j m m nɛ δ m,m iɛ x + ɛ y j m j + m + 1δm,m+1 iɛ x ɛ y j + m j m + 1δm,m 1 iɛ z mδ m,m. 1

13 We write nɛ ɛ x e x + ɛ y e y + ɛ z e z ; then for ɛ x, ɛ y, ɛ z 1 we can write ˆD nɛ [ ] i ɛx J x + ɛ y J y + ɛ z J z exp 1 i ɛ xj x + ɛ y J y + ɛ z J z Since J ± J x ± ij y, we have J x 1 / J + + J and J y 1 /i J + J i / J + J, so we rewrite the above as ˆD nɛ 1 i ɛx J + + J iɛ y J + J + ɛ z J z 1 1 iɛx + ɛ y J + + iɛ x ɛ y J + iɛ z J z Then so using D j j m m nɛ j m 1 1 iɛx + ɛ y J + + iɛ x ɛ y J + iɛ z J z jm J + jm j m j + m + 1 j, m + 1 J jm j + m j m + 1 j, m 1 J z jm m jm we nd D j j m m nɛ δ j jδ m m iɛ x + ɛ y j m j + m + 1δj jδ m,m+1 iɛ x ɛ y j + m j m + 1δj jδ m,m 1 iɛ z mδ j jδ m m δ j jd j m m nɛ Thus, factoring out δ j j from the right-hand side, we obtain the desired expression for D j m m nɛ. 13

Angular Momentum Algebra

Angular Momentum Algebra Angular Momentum Algebra Chris Clark August 1, 2006 1 Input We will be going through the derivation of the angular momentum operator algebra. The only inputs to this mathematical formalism are the basic

More information

Physics 342 Lecture 26. Angular Momentum. Lecture 26. Physics 342 Quantum Mechanics I

Physics 342 Lecture 26. Angular Momentum. Lecture 26. Physics 342 Quantum Mechanics I Physics 342 Lecture 26 Angular Momentum Lecture 26 Physics 342 Quantum Mechanics I Friday, April 2nd, 2010 We know how to obtain the energy of Hydrogen using the Hamiltonian operator but given a particular

More information

Angular momentum. Quantum mechanics. Orbital angular momentum

Angular momentum. Quantum mechanics. Orbital angular momentum Angular momentum 1 Orbital angular momentum Consider a particle described by the Cartesian coordinates (x, y, z r and their conjugate momenta (p x, p y, p z p. The classical definition of the orbital angular

More information

Lecture 19 (Nov. 15, 2017)

Lecture 19 (Nov. 15, 2017) Lecture 19 8.31 Quantum Theory I, Fall 017 8 Lecture 19 Nov. 15, 017) 19.1 Rotations Recall that rotations are transformations of the form x i R ij x j using Einstein summation notation), where R is an

More information

Massachusetts Institute of Technology Physics Department

Massachusetts Institute of Technology Physics Department Massachusetts Institute of Technology Physics Department Physics 8.32 Fall 2006 Quantum Theory I October 9, 2006 Assignment 6 Due October 20, 2006 Announcements There will be a makeup lecture on Friday,

More information

Sample Quantum Chemistry Exam 2 Solutions

Sample Quantum Chemistry Exam 2 Solutions Chemistry 46 Fall 7 Dr. Jean M. Standard Name SAMPE EXAM Sample Quantum Chemistry Exam Solutions.) ( points) Answer the following questions by selecting the correct answer from the choices provided. a.)

More information

Addition of Angular Momenta

Addition of Angular Momenta Addition of Angular Momenta What we have so far considered to be an exact solution for the many electron problem, should really be called exact non-relativistic solution. A relativistic treatment is needed

More information

Total Angular Momentum for Hydrogen

Total Angular Momentum for Hydrogen Physics 4 Lecture 7 Total Angular Momentum for Hydrogen Lecture 7 Physics 4 Quantum Mechanics I Friday, April th, 008 We have the Hydrogen Hamiltonian for central potential φ(r), we can write: H r = p

More information

8.05 Quantum Physics II, Fall 2011 FINAL EXAM Thursday December 22, 9:00 am -12:00 You have 3 hours.

8.05 Quantum Physics II, Fall 2011 FINAL EXAM Thursday December 22, 9:00 am -12:00 You have 3 hours. 8.05 Quantum Physics II, Fall 0 FINAL EXAM Thursday December, 9:00 am -:00 You have 3 hours. Answer all problems in the white books provided. Write YOUR NAME and YOUR SECTION on your white books. There

More information

QM and Angular Momentum

QM and Angular Momentum Chapter 5 QM and Angular Momentum 5. Angular Momentum Operators In your Introductory Quantum Mechanics (QM) course you learned about the basic properties of low spin systems. Here we want to review that

More information

The 3 dimensional Schrödinger Equation

The 3 dimensional Schrödinger Equation Chapter 6 The 3 dimensional Schrödinger Equation 6.1 Angular Momentum To study how angular momentum is represented in quantum mechanics we start by reviewing the classical vector of orbital angular momentum

More information

The Simple Harmonic Oscillator

The Simple Harmonic Oscillator The Simple Harmonic Oscillator Asaf Pe er 1 November 4, 215 This part of the course is based on Refs [1] [3] 1 Introduction We return now to the study of a 1-d stationary problem: that of the simple harmonic

More information

9 Angular Momentum I. Classical analogy, take. 9.1 Orbital Angular Momentum

9 Angular Momentum I. Classical analogy, take. 9.1 Orbital Angular Momentum 9 Angular Momentum I So far we haven t examined QM s biggest success atomic structure and the explanation of atomic spectra in detail. To do this need better understanding of angular momentum. In brief:

More information

Chemistry 432 Problem Set 4 Spring 2018 Solutions

Chemistry 432 Problem Set 4 Spring 2018 Solutions Chemistry 4 Problem Set 4 Spring 18 Solutions 1. V I II III a b c A one-dimensional particle of mass m is confined to move under the influence of the potential x a V V (x) = a < x b b x c elsewhere and

More information

in terms of the classical frequency, ω = , puts the classical Hamiltonian in the form H = p2 2m + mω2 x 2

in terms of the classical frequency, ω = , puts the classical Hamiltonian in the form H = p2 2m + mω2 x 2 One of the most important problems in quantum mechanics is the simple harmonic oscillator, in part because its properties are directly applicable to field theory. The treatment in Dirac notation is particularly

More information

Physics 606, Quantum Mechanics, Final Exam NAME ( ) ( ) + V ( x). ( ) and p( t) be the corresponding operators in ( ) and x( t) : ( ) / dt =...

Physics 606, Quantum Mechanics, Final Exam NAME ( ) ( ) + V ( x). ( ) and p( t) be the corresponding operators in ( ) and x( t) : ( ) / dt =... Physics 606, Quantum Mechanics, Final Exam NAME Please show all your work. (You are graded on your work, with partial credit where it is deserved.) All problems are, of course, nonrelativistic. 1. Consider

More information

Ch 125a Problem Set 1

Ch 125a Problem Set 1 Ch 5a Problem Set Due Monday, Oct 5, 05, am Problem : Bra-ket notation (Dirac notation) Bra-ket notation is a standard and convenient way to describe quantum state vectors For example, φ is an abstract

More information

8.04 Spring 2013 March 12, 2013 Problem 1. (10 points) The Probability Current

8.04 Spring 2013 March 12, 2013 Problem 1. (10 points) The Probability Current Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem. (0 points) The Proaility Current We wish to prove that dp a = J(a, t) J(, t). () dt Since P a (t) is the proaility of finding the particle in the

More information

UNIVERSITY OF MARYLAND Department of Physics College Park, Maryland. PHYSICS Ph.D. QUALIFYING EXAMINATION PART II

UNIVERSITY OF MARYLAND Department of Physics College Park, Maryland. PHYSICS Ph.D. QUALIFYING EXAMINATION PART II UNIVERSITY OF MARYLAND Department of Physics College Park, Maryland PHYSICS Ph.D. QUALIFYING EXAMINATION PART II August 23, 208 9:00 a.m. :00 p.m. Do any four problems. Each problem is worth 25 points.

More information

Summary: angular momentum derivation

Summary: angular momentum derivation Summary: angular momentum derivation L = r p L x = yp z zp y, etc. [x, p y ] = 0, etc. (-) (-) (-3) Angular momentum commutation relations [L x, L y ] = i hl z (-4) [L i, L j ] = i hɛ ijk L k (-5) Levi-Civita

More information

= a. a = Let us now study what is a? c ( a A a )

= a. a = Let us now study what is a? c ( a A a ) 7636S ADVANCED QUANTUM MECHANICS Solutions 1 Spring 010 1 Warm up a Show that the eigenvalues of a Hermitian operator A are real and that the eigenkets of A corresponding to dierent eigenvalues are orthogonal

More information

Chemistry 532 Practice Final Exam Fall 2012 Solutions

Chemistry 532 Practice Final Exam Fall 2012 Solutions Chemistry 53 Practice Final Exam Fall Solutions x e ax dx π a 3/ ; π sin 3 xdx 4 3 π cos nx dx π; sin θ cos θ + K x n e ax dx n! a n+ ; r r r r ˆL h r ˆL z h i φ ˆL x i hsin φ + cot θ cos φ θ φ ) ˆLy i

More information

Page 404. Lecture 22: Simple Harmonic Oscillator: Energy Basis Date Given: 2008/11/19 Date Revised: 2008/11/19

Page 404. Lecture 22: Simple Harmonic Oscillator: Energy Basis Date Given: 2008/11/19 Date Revised: 2008/11/19 Page 404 Lecture : Simple Harmonic Oscillator: Energy Basis Date Given: 008/11/19 Date Revised: 008/11/19 Coordinate Basis Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page

More information

QMI PRELIM Problem 1. All problems have the same point value. If a problem is divided in parts, each part has equal value. Show all your work.

QMI PRELIM Problem 1. All problems have the same point value. If a problem is divided in parts, each part has equal value. Show all your work. QMI PRELIM 013 All problems have the same point value. If a problem is divided in parts, each part has equal value. Show all your work. Problem 1 L = r p, p = i h ( ) (a) Show that L z = i h y x ; (cyclic

More information

Prob (solution by Michael Fisher) 1

Prob (solution by Michael Fisher) 1 Prob 975 (solution by Michael Fisher) We begin by expressing the initial state in a basis of the spherical harmonics, which will allow us to apply the operators ˆL and ˆL z θ, φ φ() = 4π sin θ sin φ =

More information

Physics 221A Fall 1996 Notes 14 Coupling of Angular Momenta

Physics 221A Fall 1996 Notes 14 Coupling of Angular Momenta Physics 1A Fall 1996 Notes 14 Coupling of Angular Momenta In these notes we will discuss the problem of the coupling or addition of angular momenta. It is assumed that you have all had experience with

More information

Rotations in Quantum Mechanics

Rotations in Quantum Mechanics Rotations in Quantum Mechanics We have seen that physical transformations are represented in quantum mechanics by unitary operators acting on the Hilbert space. In this section, we ll think about the specific

More information

( ) in the interaction picture arises only

( ) in the interaction picture arises only Physics 606, Quantum Mechanics, Final Exam NAME 1 Atomic transitions due to time-dependent electric field Consider a hydrogen atom which is in its ground state for t < 0 For t > 0 it is subjected to a

More information

PHY4604 Introduction to Quantum Mechanics Fall 2004 Final Exam SOLUTIONS December 17, 2004, 7:30 a.m.- 9:30 a.m.

PHY4604 Introduction to Quantum Mechanics Fall 2004 Final Exam SOLUTIONS December 17, 2004, 7:30 a.m.- 9:30 a.m. PHY464 Introduction to Quantum Mechanics Fall 4 Final Eam SOLUTIONS December 7, 4, 7:3 a.m.- 9:3 a.m. No other materials allowed. If you can t do one part of a problem, solve subsequent parts in terms

More information

PHYS 508 (2015-1) Final Exam January 27, Wednesday.

PHYS 508 (2015-1) Final Exam January 27, Wednesday. PHYS 508 (2015-1) Final Exam January 27, Wednesday. Q1. Scattering with identical particles The quantum statistics have some interesting consequences for the scattering of identical particles. This is

More information

Physics 137A Quantum Mechanics Fall 2012 Midterm II - Solutions

Physics 137A Quantum Mechanics Fall 2012 Midterm II - Solutions Physics 37A Quantum Mechanics Fall 0 Midterm II - Solutions These are the solutions to the exam given to Lecture Problem [5 points] Consider a particle with mass m charge q in a simple harmonic oscillator

More information

Reading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom.

Reading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom. Chemistry 356 017: Problem set No. 6; Reading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom. The H atom involves spherical coordinates and angular momentum, which leads to the shapes

More information

d 1 µ 2 Θ = 0. (4.1) consider first the case of m = 0 where there is no azimuthal dependence on the angle φ.

d 1 µ 2 Θ = 0. (4.1) consider first the case of m = 0 where there is no azimuthal dependence on the angle φ. 4 Legendre Functions In order to investigate the solutions of Legendre s differential equation d ( µ ) dθ ] ] + l(l + ) m dµ dµ µ Θ = 0. (4.) consider first the case of m = 0 where there is no azimuthal

More information

Physics 215 Quantum Mechanics 1 Assignment 5

Physics 215 Quantum Mechanics 1 Assignment 5 Physics 15 Quantum Mechanics 1 Assignment 5 Logan A. Morrison February 10, 016 Problem 1 A particle of mass m is confined to a one-dimensional region 0 x a. At t 0 its normalized wave function is 8 πx

More information

Physics 505 Homework No. 8 Solutions S Spinor rotations. Somewhat based on a problem in Schwabl.

Physics 505 Homework No. 8 Solutions S Spinor rotations. Somewhat based on a problem in Schwabl. Physics 505 Homework No 8 s S8- Spinor rotations Somewhat based on a problem in Schwabl a) Suppose n is a unit vector We are interested in n σ Show that n σ) = I where I is the identity matrix We will

More information

1. Matrix multiplication and Pauli Matrices: Pauli matrices are the 2 2 matrices. 1 0 i 0. 0 i

1. Matrix multiplication and Pauli Matrices: Pauli matrices are the 2 2 matrices. 1 0 i 0. 0 i Problems in basic linear algebra Science Academies Lecture Workshop at PSGRK College Coimbatore, June 22-24, 2016 Govind S. Krishnaswami, Chennai Mathematical Institute http://www.cmi.ac.in/~govind/teaching,

More information

Quantum Dynamics. March 10, 2017

Quantum Dynamics. March 10, 2017 Quantum Dynamics March 0, 07 As in classical mechanics, time is a parameter in quantum mechanics. It is distinct from space in the sense that, while we have Hermitian operators, X, for position and therefore

More information

B. Physical Observables Physical observables are represented by linear, hermitian operators that act on the vectors of the Hilbert space. If A is such

B. Physical Observables Physical observables are represented by linear, hermitian operators that act on the vectors of the Hilbert space. If A is such G25.2651: Statistical Mechanics Notes for Lecture 12 I. THE FUNDAMENTAL POSTULATES OF QUANTUM MECHANICS The fundamental postulates of quantum mechanics concern the following questions: 1. How is the physical

More information

Lecture 45: The Eigenvalue Problem of L z and L 2 in Three Dimensions, ct d: Operator Method Date Revised: 2009/02/17 Date Given: 2009/02/11

Lecture 45: The Eigenvalue Problem of L z and L 2 in Three Dimensions, ct d: Operator Method Date Revised: 2009/02/17 Date Given: 2009/02/11 Page 757 Lecture 45: The Eigenvalue Problem of L z and L 2 in Three Dimensions, ct d: Operator Method Date Revised: 2009/02/17 Date Given: 2009/02/11 The Eigenvector-Eigenvalue Problem of L z and L 2 Section

More information

Second quantization: where quantization and particles come from?

Second quantization: where quantization and particles come from? 110 Phys460.nb 7 Second quantization: where quantization and particles come from? 7.1. Lagrangian mechanics and canonical quantization Q: How do we quantize a general system? 7.1.1.Lagrangian Lagrangian

More information

Angular Momentum in Quantum Mechanics.

Angular Momentum in Quantum Mechanics. Angular Momentum in Quantum Mechanics. R. C. Johnson March 10, 2015 1 Brief review of the language and concepts of Quantum Mechanics. We begin with a review of the basic concepts involved in the quantum

More information

26 Group Theory Basics

26 Group Theory Basics 26 Group Theory Basics 1. Reference: Group Theory and Quantum Mechanics by Michael Tinkham. 2. We said earlier that we will go looking for the set of operators that commute with the molecular Hamiltonian.

More information

Problem 1: A 3-D Spherical Well(10 Points)

Problem 1: A 3-D Spherical Well(10 Points) Problem : A 3-D Spherical Well( Points) For this problem, consider a particle of mass m in a three-dimensional spherical potential well, V (r), given as, V = r a/2 V = W r > a/2. with W >. All of the following

More information

( ) = 9φ 1, ( ) = 4φ 2.

( ) = 9φ 1, ( ) = 4φ 2. Chemistry 46 Dr Jean M Standard Homework Problem Set 6 Solutions The Hermitian operator A ˆ is associated with the physical observable A Two of the eigenfunctions of A ˆ are and These eigenfunctions are

More information

Computational Spectroscopy III. Spectroscopic Hamiltonians

Computational Spectroscopy III. Spectroscopic Hamiltonians Computational Spectroscopy III. Spectroscopic Hamiltonians (e) Elementary operators for the harmonic oscillator (f) Elementary operators for the asymmetric rotor (g) Implementation of complex Hamiltonians

More information

Quantum Mechanics Solutions. λ i λ j v j v j v i v i.

Quantum Mechanics Solutions. λ i λ j v j v j v i v i. Quantum Mechanics Solutions 1. (a) If H has an orthonormal basis consisting of the eigenvectors { v i } of A with eigenvalues λ i C, then A can be written in terms of its spectral decomposition as A =

More information

Angular Momentum - set 1

Angular Momentum - set 1 Angular Momentum - set PH0 - QM II August 6, 07 First of all, let us practise evaluating commutators. Consider these as warm up problems. Problem : Show the following commutation relations ˆx, ˆL x ] =

More information

3 Angular Momentum and Spin

3 Angular Momentum and Spin In this chapter we review the notions surrounding the different forms of angular momenta in quantum mechanics, including the spin angular momentum, which is entirely quantum mechanical in nature. Some

More information

C/CS/Phys C191 Particle-in-a-box, Spin 10/02/08 Fall 2008 Lecture 11

C/CS/Phys C191 Particle-in-a-box, Spin 10/02/08 Fall 2008 Lecture 11 C/CS/Phys C191 Particle-in-a-box, Spin 10/0/08 Fall 008 Lecture 11 Last time we saw that the time dependent Schr. eqn. can be decomposed into two equations, one in time (t) and one in space (x): space

More information

Lecture-XXVI. Time-Independent Schrodinger Equation

Lecture-XXVI. Time-Independent Schrodinger Equation Lecture-XXVI Time-Independent Schrodinger Equation Time Independent Schrodinger Equation: The time-dependent Schrodinger equation: Assume that V is independent of time t. In that case the Schrodinger equation

More information

Lecture 12. The harmonic oscillator

Lecture 12. The harmonic oscillator Lecture 12 The harmonic oscillator 107 108 LECTURE 12. THE HARMONIC OSCILLATOR 12.1 Introduction In this chapter, we are going to find explicitly the eigenfunctions and eigenvalues for the time-independent

More information

Appendix: SU(2) spin angular momentum and single spin dynamics

Appendix: SU(2) spin angular momentum and single spin dynamics Phys 7 Topics in Particles & Fields Spring 03 Lecture v0 Appendix: SU spin angular momentum and single spin dynamics Jeffrey Yepez Department of Physics and Astronomy University of Hawai i at Manoa Watanabe

More information

Quantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras. Lecture - 21 Square-Integrable Functions

Quantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras. Lecture - 21 Square-Integrable Functions Quantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras Lecture - 21 Square-Integrable Functions (Refer Slide Time: 00:06) (Refer Slide Time: 00:14) We

More information

G : Quantum Mechanics II

G : Quantum Mechanics II G5.666: Quantum Mechanics II Notes for Lecture 5 I. REPRESENTING STATES IN THE FULL HILBERT SPACE Given a representation of the states that span the spin Hilbert space, we now need to consider the problem

More information

Implications of Time-Reversal Symmetry in Quantum Mechanics

Implications of Time-Reversal Symmetry in Quantum Mechanics Physics 215 Winter 2018 Implications of Time-Reversal Symmetry in Quantum Mechanics 1. The time reversal operator is antiunitary In quantum mechanics, the time reversal operator Θ acting on a state produces

More information

2 Canonical quantization

2 Canonical quantization Phys540.nb 7 Canonical quantization.1. Lagrangian mechanics and canonical quantization Q: How do we quantize a general system?.1.1.lagrangian Lagrangian mechanics is a reformulation of classical mechanics.

More information

QUALIFYING EXAMINATION, Part 2. Solutions. Problem 1: Quantum Mechanics I

QUALIFYING EXAMINATION, Part 2. Solutions. Problem 1: Quantum Mechanics I QUALIFYING EXAMINATION, Part Solutions Problem 1: Quantum Mechanics I (a) We may decompose the Hamiltonian into two parts: H = H 1 + H, ( ) where H j = 1 m p j + 1 mω x j = ω a j a j + 1/ with eigenenergies

More information

Solutions to exam : 1FA352 Quantum Mechanics 10 hp 1

Solutions to exam : 1FA352 Quantum Mechanics 10 hp 1 Solutions to exam 6--6: FA35 Quantum Mechanics hp Problem (4 p): (a) Define the concept of unitary operator and show that the operator e ipa/ is unitary (p is the momentum operator in one dimension) (b)

More information

Physics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics

Physics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics Physics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics We now consider the spatial degrees of freedom of a particle moving in 3-dimensional space, which of course is an important

More information

Physics 342 Lecture 27. Spin. Lecture 27. Physics 342 Quantum Mechanics I

Physics 342 Lecture 27. Spin. Lecture 27. Physics 342 Quantum Mechanics I Physics 342 Lecture 27 Spin Lecture 27 Physics 342 Quantum Mechanics I Monday, April 5th, 2010 There is an intrinsic characteristic of point particles that has an analogue in but no direct derivation from

More information

Statistical Interpretation

Statistical Interpretation Physics 342 Lecture 15 Statistical Interpretation Lecture 15 Physics 342 Quantum Mechanics I Friday, February 29th, 2008 Quantum mechanics is a theory of probability densities given that we now have an

More information

The Sommerfeld Polynomial Method: Harmonic Oscillator Example

The Sommerfeld Polynomial Method: Harmonic Oscillator Example Chemistry 460 Fall 2017 Dr. Jean M. Standard October 2, 2017 The Sommerfeld Polynomial Method: Harmonic Oscillator Example Scaling the Harmonic Oscillator Equation Recall the basic definitions of the harmonic

More information

Angular Momentum - set 1

Angular Momentum - set 1 Angular Momentum - set PH0 - QM II August 6, 07 First of all, let us practise evaluating commutators. Consider these as warm up problems. Problem : Show the following commutation relations ˆx, ˆL x = 0,

More information

Spin Dynamics Basic Theory Operators. Richard Green SBD Research Group Department of Chemistry

Spin Dynamics Basic Theory Operators. Richard Green SBD Research Group Department of Chemistry Spin Dynamics Basic Theory Operators Richard Green SBD Research Group Department of Chemistry Objective of this session Introduce you to operators used in quantum mechanics Achieve this by looking at:

More information

Physics 505 Homework No. 1 Solutions S1-1

Physics 505 Homework No. 1 Solutions S1-1 Physics 505 Homework No s S- Some Preliminaries Assume A and B are Hermitian operators (a) Show that (AB) B A dx φ ABψ dx (A φ) Bψ dx (B (A φ)) ψ dx (B A φ) ψ End (b) Show that AB [A, B]/2+{A, B}/2 where

More information

Generators for Continuous Coordinate Transformations

Generators for Continuous Coordinate Transformations Page 636 Lecture 37: Coordinate Transformations: Continuous Passive Coordinate Transformations Active Coordinate Transformations Date Revised: 2009/01/28 Date Given: 2009/01/26 Generators for Continuous

More information

Physics 741 Graduate Quantum Mechanics 1 Solutions to Midterm Exam, Fall x i x dx i x i x x i x dx

Physics 741 Graduate Quantum Mechanics 1 Solutions to Midterm Exam, Fall x i x dx i x i x x i x dx Physics 74 Graduate Quantum Mechanics Solutions to Midterm Exam, Fall 4. [ points] Consider the wave function x Nexp x ix (a) [6] What is the correct normaliation N? The normaliation condition is. exp,

More information

ECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals

ECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals ECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals Laporte Selection Rule Polarization Dependence Spin Selection Rule 1 Laporte Selection Rule We first apply this

More information

For a system with more than one electron, we can t solve the Schrödinger Eq. exactly. We must develop methods of approximation, such as

For a system with more than one electron, we can t solve the Schrödinger Eq. exactly. We must develop methods of approximation, such as VARIATIO METHOD For a system with more than one electron, we can t solve the Schrödinger Eq. exactly. We must develop methods of approximation, such as Variation Method Perturbation Theory Combination

More information

P3317 HW from Lecture and Recitation 7

P3317 HW from Lecture and Recitation 7 P3317 HW from Lecture 1+13 and Recitation 7 Due Oct 16, 018 Problem 1. Separation of variables Suppose we have two masses that can move in 1D. They are attached by a spring, yielding a Hamiltonian where

More information

Quantum Physics III (8.06) Spring 2006 Solution Set 4

Quantum Physics III (8.06) Spring 2006 Solution Set 4 Quantum Physics III 8.6 Spring 6 Solution Set 4 March 3, 6 1. Landau Levels: numerics 4 points When B = 1 tesla, then the energy spacing ω L is given by ω L = ceb mc = 197 1 7 evcm3 1 5 ev/cm 511 kev =

More information

PHY 407 QUANTUM MECHANICS Fall 05 Problem set 1 Due Sep

PHY 407 QUANTUM MECHANICS Fall 05 Problem set 1 Due Sep Problem set 1 Due Sep 15 2005 1. Let V be the set of all complex valued functions of a real variable θ, that are periodic with period 2π. That is u(θ + 2π) = u(θ), for all u V. (1) (i) Show that this V

More information

Rotational motion of a rigid body spinning around a rotational axis ˆn;

Rotational motion of a rigid body spinning around a rotational axis ˆn; Physics 106a, Caltech 15 November, 2018 Lecture 14: Rotations The motion of solid bodies So far, we have been studying the motion of point particles, which are essentially just translational. Bodies with

More information

8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology Wednesday April Exam 2

8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology Wednesday April Exam 2 8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology Wednesday April 18 2012 Exam 2 Last Name: First Name: Check Recitation Instructor Time R01 Barton Zwiebach 10:00 R02

More information

Harmonic Oscillator with raising and lowering operators. We write the Schrödinger equation for the harmonic oscillator in one dimension as follows:

Harmonic Oscillator with raising and lowering operators. We write the Schrödinger equation for the harmonic oscillator in one dimension as follows: We write the Schrödinger equation for the harmonic oscillator in one dimension as follows: H ˆ! = "!2 d 2! + 1 2µ dx 2 2 kx 2! = E! T ˆ = "! 2 2µ d 2 dx 2 V ˆ = 1 2 kx 2 H ˆ = ˆ T + ˆ V (1) where µ is

More information

arxiv:quant-ph/ v1 29 Mar 2003

arxiv:quant-ph/ v1 29 Mar 2003 Finite-Dimensional PT -Symmetric Hamiltonians arxiv:quant-ph/0303174v1 29 Mar 2003 Carl M. Bender, Peter N. Meisinger, and Qinghai Wang Department of Physics, Washington University, St. Louis, MO 63130,

More information

Page 712. Lecture 42: Rotations and Orbital Angular Momentum in Two Dimensions Date Revised: 2009/02/04 Date Given: 2009/02/04

Page 712. Lecture 42: Rotations and Orbital Angular Momentum in Two Dimensions Date Revised: 2009/02/04 Date Given: 2009/02/04 Page 71 Lecture 4: Rotations and Orbital Angular Momentum in Two Dimensions Date Revised: 009/0/04 Date Given: 009/0/04 Plan of Attack Section 14.1 Rotations and Orbital Angular Momentum: Plan of Attack

More information

MATH325 - QUANTUM MECHANICS - SOLUTION SHEET 11

MATH325 - QUANTUM MECHANICS - SOLUTION SHEET 11 MATH35 - QUANTUM MECHANICS - SOLUTION SHEET. The Hamiltonian for a particle of mass m moving in three dimensions under the influence of a three-dimensional harmonic oscillator potential is Ĥ = h m + mω

More information

1 Commutators (10 pts)

1 Commutators (10 pts) Final Exam Solutions 37A Fall 0 I. Siddiqi / E. Dodds Commutators 0 pts) ) Consider the operator  = Ĵx Ĵ y + ĴyĴx where J i represents the total angular momentum in the ith direction. a) Express both

More information

An operator is a transformation that takes a function as an input and produces another function (usually).

An operator is a transformation that takes a function as an input and produces another function (usually). Formalism of Quantum Mechanics Operators Engel 3.2 An operator is a transformation that takes a function as an input and produces another function (usually). Example: In QM, most operators are linear:

More information

PHYS Quantum Mechanics I - Fall 2011 Problem Set 7 Solutions

PHYS Quantum Mechanics I - Fall 2011 Problem Set 7 Solutions PHYS 657 - Fall PHYS 657 - Quantum Mechanics I - Fall Problem Set 7 Solutions Joe P Chen / joepchen@gmailcom For our reference, here are some useful identities invoked frequentl on this problem set: J

More information

CHAPTER 8 The Quantum Theory of Motion

CHAPTER 8 The Quantum Theory of Motion I. Translational motion. CHAPTER 8 The Quantum Theory of Motion A. Single particle in free space, 1-D. 1. Schrodinger eqn H ψ = Eψ! 2 2m d 2 dx 2 ψ = Eψ ; no boundary conditions 2. General solution: ψ

More information

Quantum Mechanics in Three Dimensions

Quantum Mechanics in Three Dimensions Physics 342 Lecture 21 Quantum Mechanics in Three Dimensions Lecture 21 Physics 342 Quantum Mechanics I Monday, March 22nd, 21 We are used to the temporal separation that gives, for example, the timeindependent

More information

16.1. PROBLEM SET I 197

16.1. PROBLEM SET I 197 6.. PROBLEM SET I 97 Answers: Problem set I. a In one dimension, the current operator is specified by ĵ = m ψ ˆpψ + ψˆpψ. Applied to the left hand side of the system outside the region of the potential,

More information

Quantum Physics II (8.05) Fall 2004 Assignment 3

Quantum Physics II (8.05) Fall 2004 Assignment 3 Quantum Physics II (8.5) Fall 24 Assignment 3 Massachusetts Institute of Technology Physics Department Due September 3, 24 September 23, 24 7:pm This week we continue to study the basic principles of quantum

More information

Angular momentum & spin

Angular momentum & spin Angular momentum & spin January 8, 2002 1 Angular momentum Angular momentum appears as a very important aspect of almost any quantum mechanical system, so we need to briefly review some basic properties

More information

Coherent states of the harmonic oscillator

Coherent states of the harmonic oscillator Coherent states of the harmonic oscillator In these notes I will assume knowlege about the operator metho for the harmonic oscillator corresponing to sect..3 i Moern Quantum Mechanics by J.J. Sakurai.

More information

129 Lecture Notes More on Dirac Equation

129 Lecture Notes More on Dirac Equation 19 Lecture Notes More on Dirac Equation 1 Ultra-relativistic Limit We have solved the Diraction in the Lecture Notes on Relativistic Quantum Mechanics, and saw that the upper lower two components are large

More information

Quantum Mechanics II Lecture 11 (www.sp.phy.cam.ac.uk/~dar11/pdf) David Ritchie

Quantum Mechanics II Lecture 11 (www.sp.phy.cam.ac.uk/~dar11/pdf) David Ritchie Quantum Mechanics II Lecture (www.sp.phy.cam.ac.u/~dar/pdf) David Ritchie Michaelmas. So far we have found solutions to Section 4:Transitions Ĥ ψ Eψ Solutions stationary states time dependence with time

More information

Sommerfeld (1920) noted energy levels of Li deduced from spectroscopy looked like H, with slight adjustment of principal quantum number:

Sommerfeld (1920) noted energy levels of Li deduced from spectroscopy looked like H, with slight adjustment of principal quantum number: Spin. Historical Spectroscopy of Alkali atoms First expt. to suggest need for electron spin: observation of splitting of expected spectral lines for alkali atoms: i.e. expect one line based on analogy

More information

Waves and the Schroedinger Equation

Waves and the Schroedinger Equation Waves and the Schroedinger Equation 5 april 010 1 The Wave Equation We have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form

More information

Hilbert Space Problems

Hilbert Space Problems Hilbert Space Problems Prescribed books for problems. ) Hilbert Spaces, Wavelets, Generalized Functions and Modern Quantum Mechanics by Willi-Hans Steeb Kluwer Academic Publishers, 998 ISBN -7923-523-9

More information

PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 10: Solutions

PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 10: Solutions PHYS851 Quantum Mechanics I, Fall 009 HOMEWORK ASSIGNMENT 10: Solutions Topics Covered: Tensor product spaces, change of coordinate system, general theory of angular momentum Some Key Concepts: Angular

More information

ψ( ) k (r) which take the asymtotic form far away from the scattering center: k (r) = E kψ (±) φ k (r) = e ikr

ψ( ) k (r) which take the asymtotic form far away from the scattering center: k (r) = E kψ (±) φ k (r) = e ikr Scattering Theory Consider scattering of two particles in the center of mass frame, or equivalently scattering of a single particle from a potential V (r), which becomes zero suciently fast as r. The initial

More information

Angular Momentum. Classical. J r p. radius vector from origin. linear momentum. determinant form of cross product iˆ xˆ J J J J J J

Angular Momentum. Classical. J r p. radius vector from origin. linear momentum. determinant form of cross product iˆ xˆ J J J J J J Angular Momentum Classical r p p radius vector from origin linear momentum r iˆ ˆj kˆ x y p p p x y determinant form of cross product iˆ xˆ ˆj yˆ kˆ ˆ y p p x y p x p y x x p y p y x x y Copyright Michael

More information

( ). Expanding the square and keeping in mind that

( ). Expanding the square and keeping in mind that One-electron atom in a Magnetic Field When the atom is in a magnetic field the magnetic moment of the electron due to its orbital motion and its spin interacts with the field and the Schrodinger Hamiltonian

More information

WKB Approximation in 3D

WKB Approximation in 3D 1 WKB Approximation in 3D We see solutions ψr of the stationary Schrodinger equations for a spinless particle of energy E: 2 2m 2 ψ + V rψ = Eψ At rst, we just rewrite the Schrodinger equation in the following

More information

d 3 r d 3 vf( r, v) = N (2) = CV C = n where n N/V is the total number of molecules per unit volume. Hence e βmv2 /2 d 3 rd 3 v (5)

d 3 r d 3 vf( r, v) = N (2) = CV C = n where n N/V is the total number of molecules per unit volume. Hence e βmv2 /2 d 3 rd 3 v (5) LECTURE 12 Maxwell Velocity Distribution Suppose we have a dilute gas of molecules, each with mass m. If the gas is dilute enough, we can ignore the interactions between the molecules and the energy will

More information

Lecture 4: Equations of motion and canonical quantization Read Sakurai Chapter 1.6 and 1.7

Lecture 4: Equations of motion and canonical quantization Read Sakurai Chapter 1.6 and 1.7 Lecture 4: Equations of motion and canonical quantization Read Sakurai Chapter 1.6 and 1.7 In Lecture 1 and 2, we have discussed how to represent the state of a quantum mechanical system based the superposition

More information

Tensor Operators and the Wigner Eckart Theorem

Tensor Operators and the Wigner Eckart Theorem November 11, 2009 Tensor Operators and the Wigner Eckart Theorem Vector operator The ket α transforms under rotation to α D(R) α. The expectation value of a vector operator in the rotated system is related

More information