Spin ½ (Pages 1-12 are needed)

Transcription

1 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Spin ½ (Pages - are needed) Recall that in the H-atom solution, we showed that the fact that the wavefunction ψ (r) is singlevalued requires that the angular momentum quantum number be integer: l = 0,,.. However, operator algebra allowed solutions l = 0, /,, 3/, Eperiment shows that the electron possesses an intrinsic angular momentum called spin with l = ½. B convention, we use the letter s instead of l for the spin angular momentum quantum number : s = ½. The eistence of spin is not derivable from non-relativistic QM. It is not a form of orbital angular momentum; it cannot be derived from L= r p. (The electron is a point particle with radius r = 0.) Electrons, protons, neutrons, and quarks all possess spin s = ½. Electrons and quarks are elementar point particles (as far as we can tell) and have no internal structure. However, protons and neutrons are made of 3 quarks each. The 3 half-spins of the quarks add to produce a total spin of ½ for the composite particle (in a sense, makes a single ). Photons have spin, mesons have spin 0, the delta-particle has spin 3/. The graviton has spin. (Gravitons have not been detected eperimentall, so this last statement is a theoretical prediction.) Spin and Magnetic Moment We can detect and measure spin eperimentall because the spin of a charged particle is alwas associated with a magnetic moment. Classicall, a magnetic moment is defined as a vector µ associated with a loop of current. The direction of µ is µ perpendicular to the plane of the current loop (right-hand-rule), and the magnitude is µ= ia = iπ r. The connection between orbital angular momentum (not spin) and r magnetic moment can be seen in the following classical model: Consider a particle with i mass m, charge q in circular orbit of radius r, speed v, period T. q π r qv qv qvr i i =, v = i = µ = ia = ( π r ) = T T πr π r m, q r qvr q angular momentum = L = p r = m v r, so v r = L/m, and µ= = L. m So for a classical sstem, the magnetic moment is proportional to the orbital angular momentum: q µ = L (orbital). m The same relation holds in a quantum sstem. In a magnetic field B, the energ of a magnetic moment is given b E = µ B = µ B (assuming B = Bˆ ). In QM, L = m. Writing electron mass as m e (to avoid confusion with the magnetic quantum number m) and q = e we have e e µ = m, where m = l.. l. The quantit µ B is called m m e e the Bohr magneton. The possible energies of the magnetic moment in B = Bˆ is given b Eorb = µ B = µ B Bm. For spin angular momentum, it is found eperimentall that the associated magnetic moment is twice q as big as for the orbital case: µ= S(spin) (We use S instead of L when referring to spin angular m e momentum.) This can be written µ = m = µ B m. The energ of a spin in a field is m e

2 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Espin = µ B Bm (m = ±/) a fact which has been verified eperimentall. The eistence of spin (s = ½) and the strange factor of in the gromagnetic ratio (ratio of µ to S ) was first deduced from spectrographic evidence b Goudsmit and Uhlenbeck in 95. Another, even more direct wa to eperimentall determine spin is with a Stern-Gerlach device, (This page from QM notes of Prof. Roger Tobin, Phsics Dept, Tufts U.) Stern-Gerlach Eperiment (W. Gerlach & O. Stern, Z. Phsik 9, (9). F= µ ib = µ i B ( ) B F = ˆ µ Deflection of atoms in -direction is proportional to -component of magnetic moment µ, which in turn is proportional to L. The fact that there are two beams is proof that l = s = ½. The two beams correspond to m = / and m = /. If l =, then there would be three beams, corresponding to m =, 0,. The separation of the beams is a direct measure of µ, which provides proof that µ = µ B m The etra factor of in the epression for the magnetic moment of the electron is often called the "gfactor" and the magnetic moment is often written as µ = gµ Bm. As mentioned before, this cannot be deduced from non-relativistic QM; it is known from eperiment and is inserted "b hand" into the theor. However, a relativistic version of QM due to Dirac (98, the "Dirac Equation") predicts the eistence of spin (s = ½) and furthermore the theor predicts the value g =. A later, better version of relativistic QM, called Quantum Electrodnamics (QED) predicts that g is a little larger than. The g-factor has been carefull measured with fantastic precision and the latest eperiments give g = (±76 in the last two places). Computing g in QED requires computation of a infinite series of terms that involve progressivel more mess integrals, that can onl be solved with approimate numerical methods. The computed value of g is not known quite as precisel as eperiment, nevertheless the agreement is good to about places. QED is one of our most well-verified theories. Spin Math Recall that the angular momentum commutation relations [L ˆ, L ˆ ˆ ˆ ˆ ] = 0, [L i, L ] = i L k (i,, and k cclic) were derived from the definition of the orbital angular momentum operator: L = r p. The spin operator S does not eist in Euclidean space (it doesn't have a position or momentum vector associated with it), so we cannot derive its commutation relations in a similar wa. Instead we boldl postulate that the same commutation relations hold for spin angular momentum: [ Sˆ, Sˆ ˆ ˆ ˆ ] = 0, [ Si, S] = i Sk. From these, we derive, ust a before, that 3 Ŝ sms = s(s ) sms = sms ( since s = ½ ) 4 Ŝ sm s = m s sm s = ± sm s ( since m s = s,s = /, / ) B

3 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Notation: since s = ½ alwas, we can drop this quantum number, and specif the eigenstates of Ŝ, and Ŝ b giving onl the m s quantum number. There are various was to write this: spin up ( ) χ α > > 0 χ± = s, ms >= ms > 0 spin down ( ) χ β > > These states eist in a D subset of the full Hilbert Space called spin space. Since these two states are eigenstates of a Hermitian operator, the form a complete orthonormal set (within their part of Hilbert space) a and an, arbitrar state in spin space can alwas be written as χ = a b = and the b normaliation gives: χχ = a b =.. Note that: ( 0) = =, 0 similarl: =, = = 0 If we were working in the full Hilbert Space of, sa, the H-atom problem, then our basis states would be m sms. n is another degree of freedom, so that the full specification of a basis state requires 4 quantum numbers without n. (More on the connection between spin and space parts of the state later.) [Note on language: throughout this section I will use the smbol Ŝ (and Ŝ, etc) to refer to both the observable ("the measured value of Ŝ is /") and its associated operator ("the eigenvalue of Ŝ is /"). The matri form of S () and S in the m basis can be worked out element b element. (Recall that for an operator A, ˆ A = m Aˆ n mn ˆ 3 ˆ 3 S = δss' δmsm s', S = δss' δ ˆ msm s', S 4 4 = 0, etc. Sˆ ˆ ˆ = δss' δmsm s', S = δss' δmsm s', S = 0, etc. Then in the matri notation one finds: ( Ŝ Sˆ Sˆ αα αβ α 0 ) α α β = = = Sˆ Sˆ β α β β βα ββ 0 0 = 0 and 3

4 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc 3 0 ( Ŝ ) = 4 0 Operator equations can be written in matri form, for instance, 0 Ŝ = = We are going ask; what happens when we make measurements of S, as well as S and S?, (using a Stern- Gerlach apparatus). Will need to know: What are the matrices for the operators S and S? These are derived from the raising and lowering operators: ˆ ( ˆ ˆ Sˆ ˆ ˆ S S S ) = S is = Sˆ = Sˆ isˆ ˆ S = Sˆ Sˆ ( ) i To get the matri forms of Ŝ and Ŝ, we need a result: Ŝ± s,ms = s(s ) m s(ms ± ) s, ms ± For the case s = ½, the square root factors are alwas or 0. For instance, s = ½, m = / gives 3 s(s ) m(m ) = =. Consequentl, leading to Then: ( ) ( )( ) Sˆ =, Sˆ = 0 and Sˆ =, Sˆ = 0, ( ) S = 0, S =, etc. Sˆ Sˆ 0 0 Ŝ = = = Sˆ Sˆ and 0 0 ( Ŝ ) = 0 Notice that S, S are not Hermitian. ˆ ˆ ˆ ˆ S = S S and S = S ˆ S ˆ ields Using ( ) i( ) 0 0 i = = 0 i 0 ( Sˆ ) ( Sˆ ) These are Hermitian, of course. H.W. Check the following table: α β α β Ŝ Ŝ Ŝ 3 4 α α β 3 4 β β α Ŝ Ŝ Ŝ i β 0 β i α α 0 4

5 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Eample: Find the epectation value for the Hamiltonian b are constants. Answer: Use the epression; Sˆ = Sˆ ˆ ˆ S S We can find: And Then ˆ ˆ ˆ ˆ ˆ ˆ ˆ 3 ˆ Hˆ = a( S S S 3 S ) bs = as as bs { ˆ ˆ ˆ } Hˆ s, m = as 3 as bs s, m s s { } = as ( s ) 3 am bm s, m 3 { } 4 4 ˆ ˆ ˆ ˆ = a 3 a bm s, m = bm s, m Hˆ = as ( S S ) bs, where a and s s s s s s s s, m Hˆ s, m = bm s, m s, m = bm s s s s s s One-electron sstem The Hamiltonian p Z H o = m r has the uncoupled wave function, m, s, ms =, m s, ms which identif the angular and spin parts of the wave function. m is the proection quantum number associated with and m s is the proection quantum number associated with s satisfies the relations: ' ' ˆ ', m, s', m L, m, s, m = ( ) δ δ δ δ ' s s ss ' ' mm mm s s m s m Lˆ m s m = m δ δ δ ' ' ',, ',,,, ' s s ss ' ' mm mm s s m s m Sˆ m s m = s s ' ' ',, ',,,, ( ) ' s s ss ' ' mm mm s s m s m Sˆ m s m = m δ δ δ ' ' ',, ',,,, ' s s s ss ' ' mm mm s s ' ' ' ' δ δ δ δ Aslo, the wave function, s,, m in LS-coupling has similar relations: ' ˆ ', s', ', m L, s,, m = ( ) δ δ δ δ In which J = L S, and ' ss' ' ' m m ' ˆ = δ' δss' δ ' δ ' m m ' ˆ = δ' δss' δ ' δ ' mm ' ˆ = δ' δss' δ ' δ ' m m ', s', ', m S, s,, m s( s ) ', s', ', m J, s,, m ( ) ', s', ', m J, s,, m m δ δ 5

6 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc J = J J J = L Sˆ L Sˆ = L Sˆ L Sˆ L Sˆ L Sˆ, ˆ ˆ ˆ ˆ ˆ ˆ ˆ. ˆ ˆ ˆ Note that, s,, m representation. are not eigenfunctions of L ˆ or S ˆ., s,, m Lˆ = ( Lˆ Lˆ )/ i, Lˆ = ( Lˆ Lˆ )/ ˆ ˆ ˆ ˆ ˆ = ˆ ˆ ˆ ˆ ˆ = LL L L L LL L L L Lˆ l, m = l( l ) m( m ± ) l, m ± ± Jˆ = Jˆ ± ijˆ ˆ ˆ ˆ ± i ϕ L± L ± il =± e ± i θ Lˆ = i φ Lˆ cosθ sinθ φ = sinθ sinθ θ θ sin θ φ are said to be in the coupled ± ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ =. ˆ ˆ ˆ ˆ ˆ ˆ = = J J J J L S LS L S L S L S L S Jˆ, ˆ ˆ, ˆ, ˆ ˆ, ˆ, ˆ ˆ ˆ ˆ ˆ J i J J J i J J J = = = ij J J = ij ˆ, J m >= ( ), m > ˆ ˆ J, m >= m, m > ; J, m >= m m, > Jˆ, m >= ( ) m ( m ± ), m ± > ± Jˆ, Jˆ Jˆ, ˆ, ˆ ˆ, ˆ, ˆ ˆ J J J J J = = = J ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ J, J J, J J, J,, ˆ = = = J J = J J = 0, 6

7 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc - Two spin particles Addition of Angular momentum 7

8 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc 8

9 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc s = s = sˆ s m = m s m i i i i i i ˆ i i i = i( i ) i i, =, (I) s s m s s s m i smsm sm sm (II) sˆ s m s m = m s m s m ˆ = ( ) s sms m s s sms m sˆ s m s m = m s m s m sˆ sms m = s( s ) sms m ( ) sˆ sms m = sˆ sˆ sms m ( ˆ ) ( ˆ ) ( ) ( ) = s s m s m s s m s m = m s m s m m s m s m = ( m m) smsm = m s m s m m = m m ( ) ( ) ( ) ( ) ˆ = ˆ ˆ = ˆ ˆ ˆ ˆ ˆ ˆ s s s s s s s s s ( III ) ( IV ) ( V ) = α α χ 0 S = = triplet states( Smmetric, Ortho or Even) β α α β = β β χ = 00 = A singlet states (Antismmetric, Para or Odd) β α α β 9

10 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc ( αβ βα ) = ( )( αβ βα ) Sˆ sˆ sˆ ( sˆ ) ( sˆ ) ( sˆ ) ( sˆ ) = β α α β α β β α ( βα αβ αβ βα ) = = 0 H.W. ( ) ( ) sˆ ˆ ˆ ˆ s s ( ˆ ˆ ˆ ˆ s s s s s ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ S = s s = s s s s = s s s s s s s s = H.W. check the following ˆ 0, ( ) S ψ = ψ ψ = α β βα ( αβ βα ) = ( αβ ) ( αβ βα ) = ( βα ) ( ) 3 3,, sˆψ = ψ sˆψ = ψ sˆ sˆ ψ = ψ sˆ sˆ 0 sˆ sˆ 0 ( α β βα ) = ( α β βα ) Sˆ ( ),. Sˆ 0 ( αβ βα ) = ( αβ βα ) 0

11 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Q: What is the configuration for the p -orbital ( =) for the electron in the Hdrogen atom in LSJ-coupling scheme? Answer: The wave function of the Hdrogen atom can be given b: Ψtotal R n( r) Y, m ( θϕχ, ) ± = n,, m s, ms = n,, m, s, ms = n,, s,, m Where =, s =, = ±, m =,,,. Here we have two cases; First case at: ma And it has four degenerate states. Second case at: m,,, = s = = = min = s = = m =, And it has two degenerate states. 3 Start with the highest value ma =, so coupled uncoupled, m m, m 33 ' = Y α =,, Using the relation: J ˆ ±, m = ( ) m( m ± ), m ±, one finds in the coupled representation: J ˆ 33, = ( ˆ ˆ ), S (*) LHS of (*) implies: / J ˆ , ' = ( ) ( ), ' 3, = ' (A) And the RHS of (*) implies: ( Lˆ ˆ ), ˆ, ˆ S = L S, / / = ( ) ( ) 0, ( ) ( ), 0,, = (B) Equate the equations (A) and (B), we have: 3, ' = 0,, = Y,0α Y,β What is the last equation mean? The last equation indicates that the eigen state m, is a linear combination of the eigen states ls, m, m. l s s

12 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc 3, ' = 0,, = Y α Check our epression: , m c ml, ms c ml, ms Is c c =? Is J = L s? Is m = ml ms? H.W. Prove the following 3, ' =, 0, = Y α , ' =, = Y, β Y,,0 β Y,0, These are the last two states for the value ma =. Note that the degenerac is 3 d 3/ = = 4 3, ' For the second case, start with the maimum one,, with min = and we will suppose that it take the linear combination form:, ' = c 0, c, = cy,0α cy,β From the normaliation we have ',, ' = c c = And from the orthogonalit with the state 3, ' we have 3 ',, ' = c c = 0 c = c 3 3 From both, we have c =± 3 Finall, we reach the relation:, ' = 0,, = Y α Using the lowering operator, we can have: Y,0,, ' = 0,, = Y β Y β,0, α β

13 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc a Suppose we measure S on a sstem in some state χ =. Postulate sas that the possible b results of this measurement are one of the S eigenvalues: / or /. Postulate 3 sas the probabilit of finding, sa /, is ( ) a Prob(find /) = χ = 0 = b. Postulate 4 sas that, b as a result of this measurement, which found /, the initial state χ collapses to. But suppose we measure S? (Which we can do b rotating the SG apparatus.) What will we find? Answer: one of the eigenvalues of S, which we show below are the same as the eigenvalues of S : / or /. (Not surprising, since there is nothing special about the -ais.) What is the probabilit that we find, sa, S = /? To answer this we need to know the eigenstates of the S operator. Let's () call these (so far unknown) eigenstates and () () (Griffiths calls them χ and we find these? Answer: We must solve the eigenvalue equation: S χ = λ χ, 0 where λ are the unknown eigenvalues. In matri form ( Sˆ ) = and 0 a 0 a a ( Sˆ ) = λ b = which gives, 0 b b 0 / a a λ / a = λ which can be rewritten as / 0 b b / λ b = 0. In linear algebra, this last equation is called the characteristic equation. λ / λ / This sstem of linear equations onl has a solution if Det = = 0. So / λ / λ ( ) λ = λ = ± / 0 / As epected, the eigenvalues of S are the same as those of S (or S ). Now we can plug in each eigenvalue and solve for the eigenstates: 0 a a 0 a a = a = b ; = 0 b b 0 b b a = () () So we have = and = () χ b. () ). How do Now back to our question: Suppose the sstem in the state =, and we measure S. What is the 0 probabilit that we find, sa, S = /? Postulate 3 gives the recipe for the answer: () () ( ) Prob(find S = /) = = = = / 0 3

14 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc Question for the student: Suppose the initial state is an arbitrar state What are the probabilities that we find S = / and /? a χ = b and we measure S. Let's review the strangeness of Quantum Mechanics. () Suppose an electron is in the S = / eigenstate =. If we ask: What is the value of S? Then there is a definite answer: /. But if we ask: What is the value of S, then this is no answer. The sstem does not possess a value of S. If we measure S, then the act of measurement will produce a definite result and will force the state of the sstem to collapse into an eigenstate of S, but that ver act of measurement will destro the definiteness of the value of S. The sstem can be in an eigenstate of either S or S, but not both. HW Check the following: ( Sˆ ) 0 = 0 Eigen-values smbol Eigen states 0 0 ( Sˆ ) 0 = 0 Eigen-values smbol Eigen states 0 = 0 = { } α β 0 = 0 = { } α β 4

15 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc = i 0 i ( Sˆ ) 0 Eigen-values smbol Eigen states 0 = i i 0 = { α i β} 0 = i i 0 = { α i β} 5

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18 Prof. Dr. I. Nasser Phs- 55 (T-) October 3, 03 Spin3.doc = = Pauli spin matrices Often written: S = σ, Where 0 0 i 0 σ =, σ =, σ = 0 i 0 0 are called the Pauli spin matrices and the have the following properties: σ = σ = σ =, Tr( σ ) = 0, det σ =, { } i i σ, σ = σ σ σ σ = δ, ( i, ) = (,, ) i i i i 8