Quantum Mechanics. December 17, 2007

Transcription

1 Quantum Mechanics Based on lectures given by J.Billowes at the University of Manchester Sept-Dec 7 Please me with any comments/corrections: jap@watering.co.uk J.Pearson December 7, 7 Contents Review. Postulate Postulate Hermitian Operators have Real Eigenvalues Eigenfunctions are Orthonormal Eigenfunctions Form a Complete Set Postulate Postulate D QM for Finite Wells, Potential Barriers & Tunelling 5. Infinite Well Finite Square Well Finite Well in 3D Quantum Mechanical Tunnelling Examples of QM Tunnelling Orbital Angular Momentum 3. Comments & Introduction Commutation Relations Compatibility of Physical Variables i

2 3..3 Comments on the Uncertainty Principle Degeneracy Commutation Relations for Orbital Angular Momentum ˆL, ˆL z Operators in Spherical Polars Spherical Harmonics Ladder Operators Axial Symmetry of L z Eigenstates What is Constant c in ˆL + φ m = c φ m+? Spin 3 4. Commutation Relations Ladder Operators Dirac Notation Matrix Representations in QM Introduction Matrix Representations of Ŝx, Ŝy and Ŝz Matrix Representation for Spin Raising Operators Measuring a Spin Component The Stern-Gerlach Experiment Precession in Magnetic Fields Solutions to TDSE Quantum Beats Classical Precession of Spin QM Description of Precession Addition of Angular Momenta 4 6 The Hydrogen Atom Review Spin-Orbit Coupling and Fine Structure ii

3 6.. Magnitude of Spin-Orbit Splitting Zeeman Effect Weak Applied Field Strong Applied Field Time Independant Perturbation Theory 49 A Dirac Notation and Vectors I A. Basis Vectors I A. Basis Eigenvectors II A.. Spin Angular Momentum II A.. Orbital Angular Momentum IV B Finding Eigenvalues/Vectors For [L x ], Spin VI iii

4 Review Here we review the 4 postulates of quantum mechanics.. Postulate Every dynamical system can be described by a wavefunction from which all possible predictions about the physical properties can be obtained: ψ(x, y, z, t) (.) The wavefunction is normalised: ψ ψ dτ = (.) From the wavefunction ψ, we are able to determine all dynamical variables; such as momentum, energy, angular momentum..... Postulate Every dynamical variable may be represented by a Hermitian operator, whose eigenvalues represent the possible results of a measurement on the wavefunction ψ. Immediately after the measurement, the wavefunction of the system will be identical to the eigenfunction corresponding to the eigenvalue obtained in the measurement. Consider an operator ˆQ. There will be an eigenvalue equation: ˆQφ n = q n φ n (.3) Where q n are eigenvalues: if q n is measured, then the state is definitely in the state φ n. The definition of a Hermitian operator is: if the following integral holds: f ˆQg dτ = g ˆQ f dτ (.4) Then ˆQ is said to be Hermitian; if f, g are well behaved functions of position, which go to zero at infinity. Here are the properties of Hermitian operators:.. Hermitian Operators have Real Eigenvalues Proof of Real Eigenvalues Now, starting with an eigenvalue equation we may take the complex conjugate of everything, to give: ˆQφ n = q n φ n, (.5) ˆQ φ n = q nφ n. (.6)

5 Now, if we multiply (.5) by φ n, we have: φ n ˆQφ n = φ nq n φ n. (.7) Taking the integral over all space gives: φ n ˆQφ n dτ = q n φ nφ n dτ. (.8) If we similarly multiply (.6) by φ n, we have; and writing down the corresponding integral: φ ˆQ n φ n dτ = qn φ n φ n dτ. (.9) Now, due to the condition that ˆQ is a Hermitian operator, we apply its condition: φ ˆQφ n n dτ = φ ˆQ n φ n dτ, (.) the RHS of which we notice is the integral in the LHS of (.6). Hence, we have that: φ n φ n dτ = q n φ nφ n dτ (.) q n qn = q n. (.) Hence showing that q n are real... Eigenfunctions are Orthonormal This says that: φ mφ n dτ = δ nm (.3) Proof of Orthonomality Starting from the standard eigenvalue equation: ˆQφn = q n φ n, where ˆQ is some Hermitian operator. Now, we multiply both sides by some φ m, and integrate: φ ˆQφ m n = φ mq n φ n (.4) φ ˆQφ m n dτ = φ mq n φ n dτ (.5) Now, we use a property of the Hermiticity of the operator ˆQ: We write the LHS of (.5) as: φ ˆQφ m n dτ = φ ˆQ n φ m dτ (.6) Now, the RHS of (.6) can be expressed using the standard eigenvalue equation: φ ˆQ n φ m dτ = φ n qmφ m dτ (.7)

6 Now, by working backwards, the RHS of (.7) is equal to the RHS of (.5): φ n qmφ m dτ = φ mq n φ n dτ (.8) Hence, we have that: q m φ n φ m dτ = q n φ mφ n dτ (.9) (qm q n ) φ n φ m dτ = (.) Now, using another property of Hermiticity: eigenvalues of a Hermitian operator are real. Hence, we have that qm = q m. Hence, (.5) is: (q m q n ) φ n φ m dτ = (.) Hence, so long as m n, we have that φ n φ m dτ = (.) And the only thing that this proof shows about the m = n case is that φ n φ m dτ (.3)..3 Eigenfunctions Form a Complete Set Any arbitrary function can be described as a linear combination: ψ = n a n φ n (.4) Where the expansion coefficients a n are found from the overlap integral: a m = φ mψ dτ (.5) Proof of Completeness a m = φ mψ dτ (.6) = φ m an φ n dτ (.7) = a n φ mφ n dτ (.8) = a n δ mn (.9) = a m. (.3) 3

7 .3 Postulate 3 The operators are: ˆr = r (.3) ˆP x = i h (.3) x ˆP = i h (.33) All other operators can be deduced from these; and they bear the same functional relation as their classical counterparts: E = P m + V (.34) Ĥ = h m + V (.35) L = r P (.36) ˆL = i h(r ) (.37).4 Postulate 4 When a measurement of the physical property, represented by ˆQ, is carried out on ψ, the probability that the result will be q m will be: a m (.38) e.g Let ψ = a φ +a φ +a 3 φ , where φ n are eigenfunctions of ˆQ. The measurement forces ψ to collapse into one of the eigenfunctions of ˆQ, φ m, say, and will return the value q m, with probability a m. Now, as ψ ψ dτ = (i.e. normalisation of the wavefunction), this implies that: a + a + a =, (.39) the sum of all probabilities of the different outcomes is equal to unity. This allows us to define an expectation value ˆQ, which is the average value of many measurements on ψ: ˆQ = ψ ˆQψ dτ ( ) ( ) = a mφ m ˆQ a n φ n dτ m = a mφ mq n a n φ n dτ = a ma n q n φ mφ n dτ = a ma n q n δ mn = a na n q n n = n a n q n 4

8 Which is back to the definition of an average, with a n being the probability of measuring the value q n. We have made use of the eigenvalue equation: ˆQan φ n = a n ˆQqn = a n q n φ n. D QM for Finite Wells, Potential Barriers & Tunelling We will find that quantum fluctuations in energy allow for violation of conservation of energy for a short time, according to the HUP: E t h.. Infinite Well Review of the infinite square well. We have a potential V = inside a well which sits on a x a, and V = outside the well. The Schrodinger equation inside the well has the form: Where the eigenfunction solution is of the form: due to the boundary condition that u (±a) =. With wavefunction and probability distributions looking like: h d u m dx = Eu (.) u (x) = A cos kx + A sin kx (.) k me h, (.3) Figure : Wavefunctions and probability distributions for the infinite square well. Notice that the wavefunctions are all confined to the region within the box: a x a. 5

9 . Finite Square Well Here, we have the situtation that we have 3 regions: u for x < a, where V = V ; u for a x a, where V = ; and u 3 for x > a where V = V. So, we have a potential of V = inside, and V = V outside a box of size a. The energy of the particle is E, where E < V give rise to bound states. Now, we expect that due to symmetry, u and u 3 should have the same form. The Schrodinger equation, in region (inside the box) is: Hence, we have the solution: h d u m dx = Eu u = A cos kx + B sin kx k = me h In region (to the left of the box), we have a Schrodinger equation looking like: So, we have solution: h d u m h m dx + V u = Eu d u dx = (V E)u u = Ce µx + De µx µ = m(v E) h Note that µ is real, if V > E. So, the task is to find relations between the coefficients A, B, C, D. boundary conditions of the problem: continuity. At x = a, the wavefunctions To do this, we look at the u ( a) = u ( a) And their derivatives du dx = du x= a dx x= a Match up so. Similarly, at x = a, the wavefunctions u (a) = u 3 (a) And their derivatives du dx = du 3 x=a dx x=a 6

10 Notice, infact, we have that the wavefunction must decay for x ±, so we may write the seperate wavefunctions for all 3 regions: u = Ce µx u = A cos kx + A sin kx u 3 = De µx Now, at x = a, we apply the continuity boundary conditions: And at x = a: A cos ka + B sin ka = De µa (.4) Ak sin ka + Bk cos ka = µde µa (.5) A cos ka B sin ka = Ce µa (.6) Ak sin ka + Bk cos ka = Cµe µa (.7) Now, there is one solution with (.4) - (.5) / (.6)+ (.7), which gives us: Ak sin ka A cos ka = tan ka = µ k (C + D)e µa (C + D)e µa or D = C, A = And a second solution with (.4) + (.5) / (.6) - (.7), which gives us: cot ka = µ k or D = C, B = So, if we use the tan solution, we also have that D = C and B =. If we go with the cot solution, we also have that D = C and A =. So, taking the tan solution: tan ka = µ k (.8) With D = C, B =, the wavefunctions look like: u = Ce µx u = A cos kx u 3 = Ce µx Now, to solve the tan solution (.8), we do so graphically, by plotting y = tan ka and lines of µ k. To do this: µ = m(v E) h = mv h k µ λ k = a k ka λ mv a h 7

11 Figure : The graphical method for solving (.8). Notice that the lines of tan ka only cross lines of particular λ at discrete points. So, for a shallow well, we have V is small, hence λ is small, hence a small number of solutions of bound states. Note, if V =, then λ is large, hence an infinite number of bound states. Which is what has been previously derived for the infinite square well. No matter how shallow the well is, there will always be bound states. The interesting thing is, that the wavefunction seeps into classically forbidden region! This is a very odd result: remembering that the probability to find a particle somewhere is the square of its wavefunction, means that there is a finite probability to find any particle anywhere at all! Figure 3: The first two wavefunctions for bound states of the finite well. Notice the seepage of the wavefunction into classically forbidden regions. Things to note: 8

12 There is always bound states (i.e. at least one solution), no matter how small the value of λ = mv a h. As V or a increases, there are more bound solutions (e.g. bound energy levels) and lowest solution tends towards infinite square well solution. Energies are slightly lower than for the infinite well..3 Finite Well in 3D Here, we have a well of depth V, width a, starting from r = to r = a. There are two wavefunctions: inside u and outside u. Now, the Schrodinger equation has the form: ) ( h m + V (r) u = Eu (.9) We look for solutions with E < V : i.e. bound solutions. We assume spherical symmetry, hence independant of θ, φ; and the Laplacian operator looks like: u = ( r r u ) r r We try solutions of the form: u(r) = r φ(r) Hence, the TISE (.9) becomes: u = r φ r h φ m r + V φ = Eφ So, for the case V =, i.e. inside the well, we have wavefunction: And for outside, V = V : φ = A cos kr + B sin kr k = me h φ = Ce µr + De µr µ = (V E)m h Now, u = φ r must remain finite as r, so D =. Similarly, u = φ r r, so A =. Hence, we have: u = B sin kr r u = Ce µr r must remain finite as 9

13 We apply the countinuity boundary conditions: u (r = a) = u (r = a) u (r = a) = u (r = a) This easily gives: cot ka = µ k mv = h k Which we solve graphically. In a similar way to the D solution, we find that there are no solutions for ka < π. Hence, as: Then have that k a = mv a h V > π h 8ma For a (bound) solution. Basically, when we plot the wavefunctions, we find that the sin kr must have started to turn over for it to match the gradient of e µr. Hence, k has a lower limit. If the sine just about peaks where it joins the exponential, there is a lot of wavefunction outside the well, and is hence very loosely bound. An example of this is the Deuteron: H; (pn): Experimentally, we find that (V E) =.3MeV i.e. the binding energy. And that a = 5 m. There are no (bound) excited states of the deuteron - and that the first excited state is unbound by.6mev..4 Quantum Mechanical Tunnelling Consider a flux of particles of energy E = h k m approaching a barrier of width a, height V (E > V ). We assume that some flux of particles emerges on the other side. Basically, we havea zero potential setup, with a barrier of height V starting on x = to a. The momentum of incoming particles is p = hk. What are the momentum eigenfunction? ˆP x = i h x ˆP x φ = ( hk)φ φ = e ikx So, we have flux of particles incident, assume some to be reflected, and some that appear on the other side of the barrier. In the region x <, we have wavefunction u : u = Ae ikx + Be ikx

14 Where A denotes the number of incoming particles per unit length; and B being the eqivalent for reflected particles. For the region x a, we solve: d u dx = m(v E) h u u = Ce µx + De µx µ m(v E) h For x > a, we have that u 3 = F e ikx. Now, to progress, we do the standard boundary conditon match-up. At x = we have u () = u () and their derivatives: Similarly for at x = a, with u, u 3 : A + B = C + D (.) A B = µ (C D) (.) ik Ce µa + De µa = F e ika (.) Ce µa De µa = ik µ F eika (.3) Now, we want F in terms of A: i.e. we want to find out how many particles we expect to emerge from the barrier, if we know how many we are firing at it. To do so, we begin with doing the following operations, in order: (.) + (.), (.) + (.3), (.) - (.3): ( A = + µ ) ( C + µ ) D (.4) ( ik ik Ce µa = + ik ) F e ika (.5) µ ( De µa = ik ) F e ika (.6) µ And, substituting C, D from (.5) and (.6) into (.4), gives (after a bit of algebra): F A = 4iµk (iµk + µ k )e µa + (iµk µ + k )e µa e ika (.7) Now, if the tunnelling probability is small (which is what we are assuming), we neglect the e µa term in (.7). We define F A as the tunnelling probability, and we find it to be: F 6µ k A = (µ + k ) e µa = 6(V E)E V e µa Hence, the tunnelling probabilty is largely determined by the exponential decay within the barrier: exp( const width height) Where height is the difference between the energy of the particle, and the barrier height, so is V E.

15 .4. Examples of QM Tunnelling α-decay: The potential well for a preformed α-particle in the nucleus is the sum of the strong (short range) nuclear attraction, and the weaker (long range) Coulomb repulsion. We also have the exmples of the Scanning Tunnelling Microscope, and thermonuclear fusion in stars. 3 Orbital Angular Momentum 3. Comments & Introduction 3.. Commutation Relations The order of QM operators is important. Consider the difference in sequential operations on ψ: ( ˆP xˆx ˆx ˆP x )ψ = i h ( x (xψ) xi h ψ ) x ( = i h x ψ ) ( x + ψ xi h ψ ) x = i hψ ( ˆP xˆx ˆx ˆP x )ψ This result is independant of the form of ψ, so we can write: [ ˆPx, ˆx] ˆP xˆx ˆx ˆP x = i h [ Where we call ˆPx, ˆx] the commutator. Obviously, we could also write: It is evident that: And also, there are things like: [ ˆPy, ŷ] [ˆx, ˆP x ] [ = ˆPz, ẑ] = i h [ ] = ˆPx, ˆx [ˆx, ŷ] = [ŷ, ẑ] = [ẑ, ˆx] = [ [ ˆPx, ŷ] = ˆPx, ẑ] = [ If ˆQ, ˆR] =, then we say that the operators commute. It is more instructive to write: [ ] Pxi ˆ, ˆx j We also see that: [Â, ˆB] [ ˆx i, ˆx j ] = = δ ij i h [ ] = ˆB, Â

16 3.. Compatibility of Physical Variables [ If ˆQ, ˆR] =, the physical observables they represent are said to be compatible : the operators ˆQ, ˆR have a common set of eigenfunctions. Thus, a measurement of ˆQ on ψ will collapse the state into a common eigenfunction of ˆQ: φ n. A subsequent measurement of the other quantity (represented by ˆR) will have an exactly predictable result ( ˆRφ n = r n φ n ) and will leave the wavefunction unchanged. Operators of compatible observables commute: If we have that ψ = a n φ n, and if we consider: [ ˆQ, ˆR] ψ = a n ( ˆQ ˆRφ n ˆR ˆQφ n ) = a n ( ˆQr n φ n ˆRq n φ n ) = a n (r n ˆQφn q n ˆRφn ) = a n (r n q n φ n q n r n φ n ) = Hence proved; as we have assumed that ˆR and ˆQ posses common sets of eigenfunctions, the φ n Comments on the Uncertainty Principle If two operators do not commute, there is a fundamental limit on the products of the root-meansquare deviations associated with measurements of their eigenvalues. Now, Rae 4.5 proves the following statement: q r ] [ ˆQ, ˆR Where we have: q = ˆQ = ψ ( ˆQ ˆQ ) ψ dτ ψ ˆQψ dτ In later lectures, we used: q = ˆQ ˆQ For example, since [ˆx, ˆP ] x = i h, then x P x h. The UP does not imply Measurement of P x affect subsequent measurement of x. If two operators commute, then the product of their uncertainties must be zero. However, if ψ is one of the common eigenfunctions φ n, then both observables have exactly determined[ values and state stays in φ n. So, if ˆQ, ˆR] =, then ˆQφ n = q n φ n and ˆRφ n = r n φ n, we have a common set of eigenfunctions. 3

17 3..4 Degeneracy Here, if we have: ˆQφ n = qφ n ˆQφ m = qφ m Then φ n is degenerate: we have the same eigenvalue for two or more eigenfunctions. It was assumed in the proof of orthonormality, that if ˆQφ n = q n φ n, then q n q m, for all n, m. So, we had that: φ nφ m dτ = if n m q n = q m if n = m This is only the case if there is no degeneracy. Hence, a set of non-degenerate eigenfunctions are orthonormal, but a set of degenerate eigenfunctions are not nessecarily orthonormal. Orthonormality It is always possible to construct a set of orthonormal eigenfunctions, from a set of non-orthonormal eigenfunctions. Consider: ˆQφ = qφ ˆQφ = qφ Now, any linear combination of φ, φ is also an eigenfunction of ˆQ: So, if we construct: ˆQ(αφ + βφ ) = αqφ + βqφ = q(αφ + βφ ) φ = S φ φ φ φ dτ S Thus, we show that φ is orthonormal to φ : φ φ dτ = S φ φ dτ φ φ dτ = S S = Hence orthonormal. For 3 or more degenerate states, see Rae for Schmidt Orthogonalisation Procedure. 4

18 [ Compatibility If ˆQ, ˆR] =, then there exists a common set of eigenfunctions. Again, before, q n q m was assumed before. If q n = q m, a measurement result of q n would not tell you which eigenfunction the state collapsed into. Once again, let ˆQφ = qφ, ˆQφ = qφ. Then, any combination of φ = (αφ + βφ ) is an eigenfunction of ˆQ, but not necessarily ˆR: ˆRφ = (αr φ + βr φ ) const (αφ + βφ ) Nevertheless, a set of eigenfunctions can be found, that are common to both operators. Now, this happens: Measurement with ˆQ on arbitrary ψ; gives a collapse into φ n, with result q n. φ n may not be an eigenfunction of ˆR. Measurement with ˆR on φ n ; giving further collapse, into common eigenfunctions. Leaving q n unchanged, and results in value r n. All subsequent measurements with ˆQ or ˆR will always give q n, r n. In this case, q r =. An example of this is orbital angular momentum: The n = state is split into L = and L =. So, a measurement using ˆL gives an eigenvalue L(L + ) h ; so, either h or h. For the zero measurement, there is a unique, common eigenfunction of ˆL and ˆL z ; and m = being our eigenvalue. However, for the h case, we have 3 possible states when we measure with ˆL z : ones where m =,,. All further measurements will find the same eigenvalues. 3. Commutation Relations for Orbital Angular Momentum Now, from postulate 3, we can go from the classical l = r p, to the quantum mechanical: ˆL = ˆR ˆP Hence, we can write that the components of ˆL are: ˆL x = ŷ ˆP z ẑ ˆP y ˆL y = ẑ ˆP x ˆx ˆP z ˆL z = ˆx ˆP y ŷ ˆP x Where we have used standard Cartesian coordinates. Notice the cyclic nature of the coordinates. We can write that: ˆL = ˆL x + ˆL y + ˆL z 5

19 Now, to find out if there exists a common set of eigenfunctions of ˆL, ˆL x, ˆL y, ˆL z, we consider the commutation relations (where we now omit the hat): [ ˆLx, ˆL y ] = (yp z zp y )(zp x P z ) (zp x xp z )(yp z zp y ) = yp z zp x yp z xp z zp y zp x + zp y xp z zp x yp z + zp x zp y + xp z yp z xp z zp y Watch out for the order of operators! Now, remember that [ˆx, ŷ] =, and Hence, we have that: [ ˆPx, ˆP y ] = etc; but [ ˆPx, ˆx] = i h = [ ˆPy, ŷ] = [ ˆPz, ẑ]. Hence, we have that: [ ˆLx, ˆL y ] [ ˆLx, ˆL ] y zp y xp z = P y zxp z = P y xzp z = xp y zp z = yp x (P z z zp z ) + xp y (zp z P z z) [ ] = yp x ˆPz, ẑ + xp y [ẑ, ˆP ] z [ ] = (yp x xp y ) ˆPz, ẑ = ( ˆL z )( i h) = i hˆl z = i hˆl z Similarly, by cycling over indices, we have all together: [ ˆLx, ˆL ] y = i hˆl z [ ˆLy, ˆL ] z = i hˆl x [ ˆLz, ˆL ] x = i hˆl y Hence, no set of common eigenfunctions exist for any of the pairs of operators. A system in a state of definite L z (say) cannot have a definite value of L x or L y. Thus, we cannot determine the direction of L. Now, let us consider: [ ˆL, ˆL ] [ x = ˆL x, ˆL ] [ x + ˆL y, ˆL ] [ x + ˆL z, ˆL ] x [ We can immediately see that ˆL x, ˆL ] x = ˆL ˆL x x ˆL x ˆL x =, as being almost trivial. We look at: [ ˆL y, ˆL x ] = L yl x L x L y = L y (L x L y i hl z ) (i hl z + L y L x )L y = i h(l y L z + L z L y ) 6

20 Similarly: [ ˆL z, ˆL ] x = i h(l y L z + L z L y ) [ Hence, we have that ˆL, ˆL ] x =. Similarly then for the other indices. Thus, there exists a common set of eigenfunctions of ˆL and ˆL x. There exists a different set of common eigenfunction of ˆL and ˆL y ; as well as for ˆL and ˆL y. Hence, three different sets of common eigenfunctions. By convention, we work with the last set, in ˆL and ˆL z. We can always describe a state which is an eigenfunction of ˆL y (say) by a linear combination of the ˆL z eigenfunctions; as we shall see later in the course. 3.3 ˆL, ˆL z Operators in Spherical Polars Once again, we start with ˆL = ˆR P = i h(r ). Now, we write del in spherical polars thus: = ˆr r + ˆθ r θ + ˆφ r sin θ φ Hence, if we do the cross-product, with r = r in spherical polars, we get that: ( ˆL = ( i h) ˆφ θ ˆθ ) sin θ φ Now, we have the transformation of the unit vector in the z direction is z = r cos θ ˆθ sin θ. Hence: And, with much more effort, we find that: ˆL = h [ sin θ ˆL z = z ˆL = i h φ θ ( sin θ ) + θ sin θ ] φ 3.3. Spherical Harmonics The eigenfunctions of ˆL are called spherical harmonics Y (θ, φ). To find them, we use separation of variables: Y (θ, φ) = Θ(θ)Φ(φ) From this procedure (found in Rae pp46-49), we find that: Θ = l a l cos l θ l =,,,... 7

21 With conditions on a l (such as avoiding infinities, so the series must terminate). And we also find that: Φ(φ) = e imφ (note the normalisation: π Φ dφ = ). With m l. So, we have that m is constrained: m = +l, l,...,,..., l So that we have l + values of m for each l. Hence, we find that solutions have the form: Then these eigenvalues are found: Y l,m (θ, φ) = norm associated Legendre P olynomial e imφ = ( ) m [ (l + )(l m )! 4π(l + m )! ] m P l (cos θ)e imφ ˆL Y l,m = l(l + ) h Y l,m l =,,,... ˆL z Y l,m = m hy l,m m = l,...,,..., l 3.4 Ladder Operators We can find the eigenvalues and develop a complete set of eigenfunctions of the ˆL, ˆL z (i.e. commuting) operators; by using ladder operators : ˆL + ˆL x + iˆl y (3.) ˆL ˆL x iˆl y (3.) Thus, these operators are non-hermitian, and do not represent physical observables; But they change the eigenfunctions Y l,m to const Y l,m±, and are so sometimes called raising and lowering operators. So, the first thing to do is to establish commutation relations. We initally mulitply the two operators together: ˆL + ˆL = (ˆL x + iˆl y )(ˆL x iˆl y ) = ˆL x + ˆL y iˆl x ˆLy + iˆl y ˆLx = ˆL x + ˆL [ y i ˆLx, ˆL ] y [ Now, we know that ˆLx, ˆL ] y = i hˆl z, and ˆL x + ˆL y = ˆL ˆL z. Hence: We see similarly, that: ˆL + ˆL = ˆL ˆL z + hˆl z ˆL ˆL+ = ˆL ˆL z hˆl z (3.3) 8

22 Hence, we have the commutation relation: [ ] L+ ˆ, L ˆ = ˆL + ˆL ˆL ˆL+ = hˆl z Similarly, we have: [ ] ˆLz, L+ ˆ = [ ˆLz, ˆL ] [ x + i ˆLz, ˆL ] y (3.4) [ ] ˆLz, L+ ˆ [ ] Similarly, we find that ˆLz, L ˆ = hˆl. Now, consider the eigenvalue equation: = i h(ˆl y iˆl x ) (3.5) = hˆl + (3.6) = hˆl + (3.7) ˆL z φ = βφ So that we have that β is some eigenvalue (= m h), for the eigenfunction φ. Now, operate on this whole equation with ˆL + : ˆL + ˆLz φ = β ˆL + φ (3.8) Now, from (3.7), we have that ˆL z ˆL+ ˆL + ˆLz = hˆl + ˆL + ˆLz = ˆL z ˆL+ hˆl +. Thus, we put this into (3.8): (ˆL z ˆL+ hˆl + )φ = β ˆL + φ (3.9) ˆL z (ˆL + φ) = (β + h)(ˆl + φ) (3.) Similarly: ˆL z (ˆL φ) = (β h)(ˆl φ) Hence, we see that (ˆL + φ) and (ˆL φ) are (unnormalised) eigenfunctions of ˆL z, with eigenvalues (m + ) h and (m ) h. Now, consider ˆL φ = αφ - we know that α = l(l+) h. Now, if we operate on this, from the left, with both ladder operators: ˆL + ˆL φ = αˆl + φ ˆL ˆL φ = αˆl φ Now, we also know that ˆL commutes with ˆL x, ˆL y, so therefore ˆL commutes with ˆL +, ˆL ; hence, we can change the order to: ˆL (ˆL + φ) = α(ˆl + φ) ˆL (ˆL φ) = α(ˆl φ) Which are just eigenvalue equations. So therefore, both ˆL + and ˆL are eigenfunctions of the ˆL operator, with the same eigenvalue α. 9

23 Now, imagine we project the L eigenfunction onto the z axis; to give L z. Now, the length of each L line is just L = α. We can use the ladder operators to move from one L z eigenfunction to the next. If the maximum L z eigenvalue/functions are β, φ, and the minimum β, φ ; then operating on the maximum with the raising operator will produce zero; similarly for the lowering and minimum. So, we have that β, β < α, and ˆL z φ = β φ. Now, we suppose that: Now, operating on (3.) with ˆL and then use (3.3): ˆL + φ = (3.) ˆL φ = (3.) ˆL ˆL+ φ = (ˆL ˆL z hˆl z )φ = (3.3) = (α β hβ )φ = (3.4) α = β (β + h) (3.5) Hence, what we have shown is that α, which is the eigenvalue of ˆL is the same as β (β + h), where β is the maximum eigenvalue of ˆL z. We can do a similar operation on (3.), to give: α = β (β h) Which is equivalent. Thus, these two equations give a symmetry property: β = β Now, as β changes by unit of h in raising and lowering operations, it follows that: β β = n h Where n is an integer. If arrangement is symmetrical about L z = state, then n is even = l (say). Thus it follows that: Notice: α = l(l + ) h β = m h l m +l ˆL x = (ˆL + + ˆL ) ˆL y = i (ˆL + ˆL ) 3.4. Axial Symmetry of L z Eigenstates Here, we show that ˆLx = & ˆLy =. Now, since we have the definitions: ˆL + ˆL x + iˆl y ˆL ˆL x iˆl y

24 We can write: ˆL x = (ˆL+ + ˆL ) And similarly for ˆL. Now, let φ m be an eigenstate of ˆL z. For this state, we write: ˆLx = φ ˆL m x φ m dτ = φ m(ˆl + + ˆL )φ m dτ = c φ mφ m+ + c φ mφ m dτ = Where the last two steps are done using ˆL + φ m = c φ m+ and using the fact that the eigenfunctions are orthonormal. We similarly show that ˆLy =. Therefore, a state φ m has axial symmetry about z axis, since no preferred x or y projection What is Constant c in ˆL + φ m = c φ m+? We want to find the constant c in the eigenequation: ˆL + φ m = c φ m+ We need this relationship: f ˆL g dτ = = f(ˆl x iˆl y )g dτ g(ˆl x iˆl y)f dτ Which we are able to do via the definition of Hermitian operators. Now, using the normalised wavefunction φ m, we have: c φ m+φ m+ dτ = (ˆL + φ m )(ˆL + φ m ) dτ = φ ˆL m ˆL+ φ m dτ = φ m(ˆl ˆL z hˆl z )φ m dτ = l(l + ) h m h m h c = l(l + ) h m h m h From (3.3). Thus, as φ m+ φ m+ dτ =, as φ m normalised: ˆL + φ m = h l(l + ) m(m + ) φ m+

25 Similarly: ˆL φ m = h l(l + ) m(m ) φ m Notice, from these, we immediately see that: ˆL + φ m=+l = ˆL φ m= l = Therefore, to combine the knowledge we just gained: ˆL ± φ m = h l(l + ) m(m ± ) φ m±

26 4 Spin This appears naturally, from the Dirac formalism of relativistic quantum mechanics, but has no classical analogue. There is no differential operator for spin, but we can use the same algebra as for orbital angular momentum. An electron is a spin particle, so s = h, and the possible values of the projection of spin onto the z-axis are m s = ± h. We have spin angular momentum operators: Ŝ = (Ŝx, Ŝy, Ŝz) Ŝ = Ŝ x + Ŝ y + Ŝ z 4. Commutation Relations We use the orbital angular momentum analogues, to write: [Ŝx, Ŝy] = i hŝz [Ŝ, Ŝz] = With another two of each, which are similar. By convention, we work with common set of eigenfunctions for Ŝ, Ŝz. We call the eigenvalues α, β: Ŝ z α = + hα Ŝ z β = hβ So that α corresponds to a spin-up state, and β to spin-down. We also have: Ŝ α = ( ) + h α = 3 4 h α Ŝ β = 3 4 h β 3 So, we have that S = 4 h, and the projection onto the z axis is ± h. Now, a general spin wavefunction χ (e.g. spin polarised along the x axis), may be expressed as a linear combination of the Ŝ, Ŝz eigenfunctions: χ = aα + bβ Where a, b are coefficients which determine relative populations and relative phases of α, β space. We also have that a + b = ; i.e. normalisation. A bit of notation: When we refer to either α or β, we ll use χ ms. Thus: χ + χ α β 3

27 4.. Ladder Operators Here, we define thus: Ŝ + Ŝx + iŝy (4.) Ŝ Ŝx iŝy (4.) Hence, we have similar results as for orbital angular momentum ladder operators: Ŝ x = (Ŝ+ + Ŝ ) Ŝ y = i (Ŝ+ Ŝ ) Ŝ + χ ms = s(s + ) m s (m s + ) hχ ms+ Ŝ χ ms = s(s + ) m s (m s ) hχ ms Now, as for the α state we have m s = +, and for the β state m s = ; and that s = ; hence, we see that: So then, what are the eigenfunctions of Ŝx? Consider: Ŝ + α = Ŝ α = hβ Ŝ + β = hα Ŝ β = Ŝ x α = (Ŝ+ + Ŝ )α = hβ Ŝ x β = (Ŝ+ + Ŝ )β = hα So, we see that α, β are not individually eigenfunctions of Ŝx, but, adding the two results: Ŝ x (α + β) = h(α + β) Therefore, we find that (α + β) is an eigenfunction: the combination is, but seperately neither are. Hence, the normalised eigenfunction of Ŝx is (α + β), with eigenvalue h. Notice, subtracting the results gives a different eigenfunction: Ŝ x (α β) = β(β α) = (α β) 4

28 Hence, the only other eigenfunction is (α β), with eigenvalue h. Similarly, for Ŝy, the eigenfunctions and eigenvalues are: (α + iβ) (α iβ) + h h 4. Dirac Notation Simple shorthand that can be used for all aspects of quantum mechanics, but particularly for spinspace, where we need a scheme to handle objects equivalent to χ χ dτ, where we dont know the space we are integrating over. So, we write: Wavefunction φ as φ, a state vector ket ; Complex conjugate φ as φ a bra ; Normalisation φ φ dτ = as φ φ = ; Hydrogen wavefunction φ nlm (r, θ, φ) as n, l, m ; Spherical harmonics Y lm (θ, φ) as l, m ; Expectation value ˆQ = φ ˆQφ dτ as φ ˆQ φ ; Orthonormality φ nφ m dτ = δ nm as φ n φ m = δ nm or n m = δ nm ; Expansion coefficients ψ = a n φ n, where a m = φ mψ dτ as ψ = a n φ n where a m = φ m ψ. We shall see that a ket φ is a column vector, and that the bra φ is related to the ket by being a row vector, whose elements are the complex-conjugates of the ket: a φ = b c φ = (a, b, c ) In the same way that a vector must be expressed in terms of some basis (for example, the e i, e j, e k unit vectors in Cartesian coordinates), a ket φ must be expressed in terms of some basis functions. We can also see that if φ is some orthonormal vector, then φ φ =, as aa + bb + cc =. We shall thus see (this is all used later on) that the expectation value of some operator Â, represented by some matrix A can be written φ A φ ; thus: φ A φ = (a, b, c ) a a a 3 a a a 3 a 3 a 3 a 33 Which can obviously be expanded out to a scalar quantity. a b c 5

29 4.3 Matrix Representations in QM 4.3. Introduction Let us work with a complete set of eigenfunctions φ n. We call this a basis: a coordinate system for wavefunctions. Consider the eigenvalue equation of an operator ˆQ: ˆQψ = qψ Where ψ is an eigenfunction of ˆQ. We can express ψ in our chosen basis: ψ = n a n φ n Hence, the eigenvalue equation becomes: a n ˆQφn = q n n Now, if we multiply from the left by φ m, and integrate: a n φ ˆQφ m n dτ = q n n a n φ n a n φ mφ n dτ = q n a n δ mn Now, let us define: = qa m φ m ˆQφ n dτ Q mn Hence, we have that: Q mn a n = qa m n Which is just standard matrix multiplication! The dimension of the matrix [Q] (which has elements Q mn ), is the number of eigenfunctions in the basis. This is known as a matrix eigenvalue equation. Expanded out, this is: Q Q... Q Q.... a a. = q The eigenvalue solutions q n are identical in the matrix representation, and in ˆQψ n = q n ψ n. For matrix equations, we have that eigenvalues are the roots of the determinant: Q q Q... Q Q q... =. 6 a a.

30 Where we will get the same number of eigenvalues as the dimension of the matrix. Each solution (each eigenvalue) has an associated eigenvector: a a. Equivalent to the eigenfunction: ψ n = a φ + a φ + a 3 φ Where we will obviously have n running up to the dimension of the matrix. The matrix [Q] is known as an Hermitian Matrix. That is, [Q] is the Hermitian conjugate of [Q ]: [Q ] [ Q ] The transpose of the complex conjugate. With Q mn = Q nm. We can also see how this works from the integral defintions: Q mn = φ ˆQφ m n dτ Q mn = φ ˆQφ n m dτ Q mn = φ ˆQ n φ m dτ Which is the same as the definition of an Hermitian operator: φ n ˆQφm dτ = φ ˆQ m φ n dτ Hence, as we have that [Q] is an Hermitian matrix, we can therefore say that it has real eigenvalues. Now, if we had chosen to work in terms of the basis of eigenfunctions of ˆQ, so that we have ˆQφ n = q n φ n : Q mn = φ ˆQφ m n dτ = q n δ mn That is, the matrix [Q] will be diagonal, with entries q i : q... a q... a = q.. Thus, q a = q a =... q = q i. Therefore, eigenvectors will be:..... With eigenvalues q, q,... The elements of [Q] therefore depend on our choice of basis (our eigenfunction set). The elements a a. 7

31 of [Q] contain all the relations we need in order to know everything about ˆQ. In many cases, we will have an infinite set of eigenfunctions, and matrices of infinite order are generally very hard to handle! They are however, very useful in angular momentum problems, as we have a limited set of eigenfunctions: Y lm (θ, φ) has l + eigenfunctions. In the case of intrinsic spin, we have no possibility to represent spin operators by differential operators, so we can use matrix operators instead Matrix Representations of Ŝx, Ŝy and Ŝz Matrix Representation of Ŝz We choose a basis formed from the eigenfunctions of Ŝz: α, β. Now, continuing with the notation of [S z ] for a matrix, we can write: ( ) S S S S Now, we recall that the element Q ij is generated via: Q ij = φ ˆQφ j i dτ = φ j ˆQ φ i Thus, we have: Now, we also know: Ŝ z α = hα S = α Ŝz α Ŝ z β = hβ Thus: S = h α α = h Via orthogonality. Similarly, we see: S = β Ŝz β = h β β = h For the non-diagonal entries: S = α Ŝz β = h α β = = S 8

32 Where we have used the fact that two eigenfunctions are orthogonal. Hence, we can write: ( + [S z ] = h ) h This is usually written as: [S z ] = h ( = hσ z Where σ z is known as a Pauli spin matrix. You need to be careful about the factor of h in front of the matrix: it is technically inside the operator matrix, but its written outside; it does not cancel off in operations. If we can to find the eigenvalues, we need to solve: ( ) ( ) ( ) h a a = q (4.3) a Which we do by bringing the column on the RHS over: ( h q ) ( ) a h q = a Which is solved by setting the determinant of the matrix to zero, and solving the characteristic equation for the eigenvalue q: ( ) ( h q h ) q = Therefore, the eigenvalues are q = ± h, which we already knew, but were able to recover from the matrix representation. To find the eigenvectors, we stick each eigenvalue in turn, into (4.3), and solve for a, a. So, for q = + h, and multiply the matrix out: Thus, the eigenvector is: a ha + a = ha a ha = ha ) a = a = α = ( Notice, we could have made a anything, but we chose as we want orthonormal normalised eigenvectors. If we do the same for q = h, we find: ( ) β = ) 9

33 Matrix Representation of Ŝx We have previously shown that: Ŝ x = (Ŝ+ + Ŝ ) So therefore, we can write that Ŝ x α = h β Ŝ x β = h α Hence, we write the matrix, and compute its elements: ( ) S S [S x ] = S S ( α Ŝ = x α α Ŝx β β Ŝx α β Ŝx β ( = h ) h = ( ) h ) = hσ x Where we have notices things like: α Ŝx α = h α β = β Ŝx α = h β β = h To find the eigenvectors, we must solve ( h h ) ( a a ) ( a = q a ) For each eigenvalue q = ± h. Thus, we have: α x = ( ) β x = ( ) The factor of outfront is to keep the vectors having length one. 3

34 Matrix Representation of Ŝy Again, we have shown that: Ŝ y = i (Ŝ+ Ŝ ) So, we have that: Ŝ y α = i β = i β Ŝ y β = i α And we notice that: α Ŝx β = = i hi α α h And similar, untill we get to being able to write all the elements of the matrix: [S y ] = ( ) i h i = hσ y And, upon solving for the eigenvectors: α y = ( ) i β y = ( ) i Other Uses We can also demonstrate the commutation relations by matrix multiplication. For example: [Ŝx, Ŝy] = [S x ][S y ] [S y ][S x ] Being careful of the rubbish notation! So, we multiply the spin matrices together: [Ŝx Ŝy], = 4 h (σ x σ y σ y σ x ) = 4 h [( = 4 h [( i i = ( ) i 4 h i = i ( ) h = i h σ z ) ( i ) i ( i i ) ( i i )] ) ( )] 3

35 Similarly, we can also find Ŝ in matrix representation: Ŝ = Ŝ x + Ŝ y + Ŝ z = 4 h (σ x σ x + σ y σ y + σ z σ z ) = 3 ( ) 4 h Which is the unity( matrix. ) Any spin a vector is an eigenfunction of b Ŝ : ( 3 4 h ) ( a b ) = 3 ( a 4 h b ) Notice, if s =, then 3 4 h = s(s + ) h Matrix Representation for Spin- We can find the matrices for spin- operators (L x, L y, L z, L +, L, L ), in, for example, a basis spanned by the eigenfunctions of L and L z. I have left off hats! This basis is denoted by m, where m m z = ±,. So, if we define m z = =, m z = = and m z = = as our basis, we can compute the elements of [L x ]: [L x ] = Where the elements can be computed; L ij = m i ˆL x m j L L L 3 L L L 3 L 3 L 3 L 33 = m i ˆL + m j + m i ˆL m j ˆL x ˆL x ˆL x [L x ] = ˆL x ˆL x ˆL x ˆL x ˆL x ˆL x Be careful, as we have defined m = m z =, m 3 = m z = etc. We have used the ladder operator representation for ˆL x = (ˆL + + L ). We also need the previous relations: ˆL + m = h l(l + ) m(m + ) m + ˆL m = h l(l + ) m(m ) m 3

36 We also have that l =. So, to compute a few elements: L = ˆL x = ˆL + + ˆL = + h = L = ˆL x Continuing for all other elements, we find: = ˆL + + ˆL = + h = h [L x ] = h And, if we want eigenvalues, we must find the roots of: q q q = q(q ) ( q) = Hence, the roots are either q =, ±. However, q is already in units of h/. Thus, we have that q =, ± h; which is again a result we already had! The eigenvectors for q = + h are found by solving the system h a b c = + h a b c b = a (a + c) = b b = c Thus, the (normalised) eigenvector is: 33

37 So, this is the eigenvector of the + h eigenvalue, of the x-component of angular momentum, with respect to the z-basis. Hence, in Dirac notation, this is: m x = + = m z = + + m z = + m z = Now, in Dirac notation, we have the interpretation that a m = b c We can thus write the interpretation of m : m = (a b c ) Thus, when we write an object like m m, we actually have; m m = (a b c ) a b c = aa + bb + cc = Which is done via standard matrix multiplication. We hence can say that m m is like the dot product. Example Suppose we have a state initially in the m z = eigenstate. Compute ˆL x and ˆL x. So, we start on L x. We know that m z = = : L x = m z = ˆL x m z = = h ( ) = h ( ) = L x = 34

38 Now for ˆL x : ˆL x = m z = ˆL x ˆLx m z = = h ( ) = h = h ( ) Hence, we can write down the overall uncertainty in ˆL x : ˆL x = ˆL x ˆL x = h Example We know: Find the eigenvectors of Ŝx. Thus, we must solve the equation: ( h h [S x ] = h ) ( a b ( ) ) ( a = q b Eigenvalues are the roots of the determinant, and can easily be shown to be q = + h, q = h. Now, for the q eigenvalue, to find the corresponding eigenvector, we must solve: ( ) ( ) h a = ( ) a b h b Easily resulting in a = b. Thus, the eigenvector v is: Which, in Dirac notation, is the same as: v = ( m x = + = m z = + + m z = ) ) Similarly, for q = h, we end up with v = ( ) Or: m x = = m z = + m z = 35

39 And, to show that these two eigenvectors are orthogonal (which they should be), we write: ( ) ( ) v v = Thus the eigenvectors are orthogonal. = Raising Operators If we want the matrix representation of the raising operator Ŝ+, we write: Ŝ + = Ŝx + iŝy = h(σ x + iσ y ) = h [( ( ) = h ) ( i + i i )] We can use this on (say) β: ( h Which again, is a result we already knew. Ŝ + β = ) ( ) ( ) = h = hα 4.4 Measuring a Spin Component Charged particles with intrinsic spin, or orbital angular momentum have a magnetic moment µ. For some classical current loop, where a current I is going round an area A, we have µ = IA; where the area vector points according to the right-hand-rule. For some classical electron orbit, we have that the charge flowing per unit time is I = e T = eω π, where the period T π ω. If the area of the orbit is A = πr, then we easily have that: µ = eωr π Now, we also have that l h = rp = m e ωr. Hence: µ = l h eω m e ω = l he m e = g l lµ B 36

40 It is actually more correct to write this as: µ z = g l l z µ B Just in terms of notation, l z = m z. Where we have defined the Bohr Magneton µ B, and the gyromagnetic ratio g l = for electrons. So, in vector form, we have: µ B e h m e µ = g l lµ B Where µ z is the projection of µ along the z-axis, and is always in the same direction as l. Now, the electron also has an intrinsic magnetic moment - like a bar magnet - along the direction of s: µ s = g s sµ B Simple Dirac theory (apparentyl) yields g s =, and we use the approximation µ s µ B. The nuclear magneton is pretty much identical, except it uses the mass of the proton. Hence, spin effects due to the nucleus are tiny, as m p >> m e : µ N = e h m p If we remember the classical gyroscope, if it has angular momentum, it does not fall over, but precesses about the vertical axis (direction of gravity). Thus, a magnetic moment µ would align itself with an applies magnetic field B (like a compass needle), except if the particle also has spin, when the particle will then precess about B The Stern-Gerlach Experiment The experiment has the following setup: An oven evaporates silver atoms which are emitted in all direction, but are collimated by some aperture. This beam of silver atoms then passes through an inhomogeneous magnetic field. The valence electrons spin will interact with the field, and will produce a discrete pattern on the screen: atoms with different spins leave the field in different directions. Now, as force = -grad PE, we hence have: F = +µ B z = µ z B z z Hence, a beam of atoms is deflected either up or down, depending on the sign component of µ along the z-axis: m z = + or m z =. So, an unpolarised beam will be in the state: 5% α + 5% β 37

41 Where α = m z = +, and β = m z =. Now, suppose we have selected the α component, via a Stern-Gerlach type-experiment, we now have the state: % α Which is actually half of the initial state. Now, if we want to then measure S x, we need to know how to describe the α eigenstate in terms of eigenstates of S x. We have previously done this: m x = + = ( α + β ) m x = = ( α β ) Hence: α = ( m x = + + m x = ) Now, suppose we further select the m x = + state using another magnet, we then get half of what enters the second magent, which was half of what entered the first magnet. 4.5 Precession in Magnetic Fields 4.5. Solutions to TDSE These give the time-evolutioon of the wavefunction for a particle in a potential V (r): ] [ h m + V (r) ψ(r, t) = i h ψ(r, t) t If V (r) is not a function of time, we can apply the standard separation of variables technique. Let ψ(r, t) = φ(r)t (t), and substitute into the above TDSE, and divide by ψ: h m φ φ + V (r) = i h T T t = const = E Where we have called the constant E, say. So, we can separate the two halfs of the equation, so looking at the time equation: i h T t = ET Which has solutions of the form: T (t) = const e iet/ h Sometimes, using E = hω, this is written: T (t) = const e iωt And the spatial part of the TDSE looks like: ] [ h m + V (r) φ = Eφ 38

42 Which has eigenvalues E n for eigenfunctions φ n. Any arbitrary state ψ at t = could be described by: ψ = n a n φ n Then we now know how the state evolves in time: ψ(t) = n a n e iωnt φ n 4.5. Quantum Beats Consider an initial state which is a linear combination of energy states, for example: ψ = φ + φ Where, for this example, φ, φ are real functions. So we can write the time evolution version of the wavefunction: ψ(t) = φ e iω t + φ e iω t = e iω t (φ e iωt + φ ) Where we have defined ω ω ω. We can compute the probability distribution function via: ψ ψ = (φ e iωt + φ )(φ e iωt + φ ) = φ + φ + φ φ (e iωt + e iωt ) = φ + φ + φ φ cos ωt The interference between the two phase factors creates a time-dependance of the probability distribution that has the appearance of motion Classical Precession of Spin Consider a magnetic dipole moment µ in a uniform magnetic field B along the z-axis. The torque on such a dipole is given by: Γ = µ B Γ acts to align the dipole with the magnetic field, and acts perpendicular to the plane containing both. If the dipole also has orbital angular momentum L, we also have: We find that (a) solution is given by: dl dt = Γ L x = L sin θ cos ωt L y = L sin θ sin ωt L z = L cos θ 39

43 Where we have that θ is a constant, measured from the z-axis, and ωt the precession speed. Hence, we have described a precession, where: ω = µb L Is called the Larmor precession frequency. Note, this is the case for simple precession only! Notice, if µ = g l L h µ B Then the Larmor frequency is: ω = g lµ B B h QM Description of Precession If we have the Hamiltonian due to magnetic field: Ĥ mag = µ B = ˆµ z B = g l ˆLz h µ BB Eigenfunctions of which are l, m l, and ˆL z l, m l = m l h l, m l ; we hence see that energy eigenvalues are: E m = g l m l µ B B = + hωm l Where: ω = g l µ B B h So, to summarise, we have a Hamiltonian due to the applied magnetic field Ĥmag = µ B, with eigenstates l, m l, and eigenvalues g l m l µ B B. Now, as Ĥmag is not a function of time, energy is conserved in the system. Hence, m l is a constant of motion. Thus, the state m l = + (say) is a stationary state. L points with equal probability everywhere on the surface of the cone that the spin marks out. The time evolution of the stationary state l, m l is given by: ψ(t) = e iemt/ h l, m l = e iωm lt l, m l Where ω is the previous Larmor frequency. The phase factor (e iωm lt ) cannot be directly measured by any experiment: we cannot observe any precession if the system is in an eigenstate l, m l. However, consider a spin L = i h, polarised along the x-axis at time t = in a field B along z. Now, for reference: [L x ] = h 4

44 And, we have already seen that: ψ(t = ) = m x = + = m z = + + m z = + m z = Which is found from finding the normalised eigenvectors of [L x ], where the fraction out-front is from the normalisation. The time evolution of the state is found by multiplying each term by the phase factor e iωm lt : Where: ψ(t) = e iωt m z = + + m z = + eiωt m z = ω = g l µ B B h We are now able to observe interference between the different phase factors, which will give us quantum beats. So, let us compute ˆL x = ψ(t) ˆL x ψ(t) : ˆL x = h (e iωt,, e iωt) = h (e 4 iωt,, e iωt) = h ( e iωt + e iωt + e iωt + e iωt) 4 = h cos ωt e iωt + e iωt Thus, we have that ˆL x will sometimes be found along the + h axis, and other times along the h axis. Similarly, we can find ˆL y = h sin ωt, and ˆL z = =constant. Which is what we would expect for Larmor precession. Hence, we see that putting an atom in a magnetic field has the effect of splitting the energy levels evenly, by hω. If the magnetic field is non-homogeneous, then non-even splitting. Note: The classical analogy does not generally work; if states m are not equally spaced in energy, then each spin appears to precess with a mixture of frequencies simultaneously. e iωt e iωt 5 Addition of Angular Momenta We consider the vector addition of angular mometa. In this example, they will be spin and orbital angular momentum, but the algebra below applies equally well to other types of angular momenta (e.g. spin-spin). The vector equation is: J = L + S (5.) 4

45 Or, with a particular component: J z = L z + S z J z is now a constant of motion, whereas L z, S z are fuzzy; and only have the restriction of adding to J. Squaring J: Or, where everything is an operator: J J = J = L + S + L S Ĵ = ˆL + Ŝ + ˆL Ŝ Ĵ z = ˆL z + [ Ŝz It can be easily shown that Jˆ, ˆL ] [ ] = Jˆ, Ŝ =, thus there exists a common set of eigenfunctions. But, Ĵ does not commute with either ˆL z or Ŝz due to the ˆL Ŝ term in Ĵ. By writing: ˆL Ŝ = ˆL x Ŝ x + ˆL y Ŝ y + ˆL z Ŝ z We can show: [(ˆL Ŝ), ˆL z ] [ ] (ˆL Ŝ), Ŝz = i h (ˆLx Ŝ y Ŝx ˆL ) y = i h (ˆLx Ŝ y Ŝx ˆL ) y Which, upon adding, results in : [ ] (ˆL Ŝ), Ĵz = [ ] Therefore, we have been able to show that Jˆ, Ĵz =. The operators Ĵ, Ĵz, ˆL, Ŝ have common sets of eigenstates, but are not eigenstates of ˆL z, Ŝz. We write states as: j, j z, l, s So, in an isolated system, the total angular momentum J is a constant of motion, as is j z = m l +m s. We will sometimes use m j = j z. All the usual angular momentum relations apply to Ĵ and Ĵz, as they did to ˆL, Ŝ, ˆL z, Ŝz: [Ĵx, Ĵy] = i hĵz Ĵ = Ĵ x + Ĵ y + Ĵ z [ ] Jˆ, Ĵx = Ĵ ± = Ĵx ± iĵy = ˆL ± + Ŝ± Ĵ j, m j = j(j + ) h j, m j Ĵ z j, m j = m j h j, m j Ĵ ± j, m j = h j(j + ) m j (m j ± ) j, m j ± Ĵ ± j, m j = ±j = 4