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Lecture 33: Quantum Mechanical Spin Phy851 Fall 2009 Intrinsic Spin • Empirically, we have found that most particles have an additional internal degree of freedom, called ‘spin’ • The Stern-Gerlach experiment (1922): • Each type of particle has a discrete number of internal states: – 2 states --> spin _ – 3 states --> spin 1 – Etc…. Interpretation • It is best to think of spin as just an additional quantum number needed to specify the state of a particle. – Within the Dirac formalism, this is relatively simple and requires no new physical concepts • The physical meaning of spin is not wellunderstood • Fro Dirac eq. we find that for QM to be Lorentz invariant requires particles to have both anti-particles and spin. • The ‘spin’ of a particle is a form of angular momentum Spin Operators • Spin is described by a vector operator: r r r r S = S x ex + S y e y + S z ez • The components satisfy angular momentum commutation relations: [ S x , S y ] = ihS z [ S y , S z ] = ihS x [ S z , S x ] = ihS y • This means simultaneous eigenstates of S2 and Sz exist: S 2 = S x2 + S y2 + S z2 S 2 s, ms = h 2 s ( s + 1) s, ms S z s, ms = hm s, ms Allowed quantum numbers • For any set of 3 operators satisfying the angular momentum algebra, the allowed values of the quantum numbers are: j ∈ {0, 12 ,1, 32 , K} m j ∈ {− j ,− j + 1, K , j} • For orbital angular momentum, the allowed values were further restricted to only integer values by the requirement that the wavefunction be single-valued • For spin, the quantum number, s, can only take on one value – The value depends on the type of particle – S=0: Higgs – s=1/2: Electrons, positrons, protons, neutrons, muons,neutrinos, quarks,… – s=1: Photons, W, Z, Gluon – s=2: graviton ms ∈ {− s,− s + 1, K , s} Complete single particle basis • A set of 5 commuting operators which describe the independent observables of a single particle are: r 2 R, S , S z – Or equivalently: 2 2 R , L , Lz , S , S z • Some possible basis choices: r {r , s, ms r {p, s, ms } } {r , l, m , s, m } {n, l, m , s, m } l s l s • When dealing with a single-particle, it is permissible to drop the s quantum number Intrinsic Magnetic Dipole Moment • Due to spin, an electron has an intrinsic magnetic dipole moment: ge e r r µe = − S 2me – ge is the electron g-factor – For an electron, we have: g e = 2.0023193043622 ± 0.0000000000015 – The is the most precisely measured physical quantity • For most purposes, we can take ge≈ 2, so that e r r µe = − S me • For any charged particle we have: r gq r µ= S 2M Each particle has a different g-factor Hamiltonian for an electron in a magnetic field • Because the electron is a point-particle, the dipole-approximation is always valid for the spin degree of freedom • Any `kinetic’ energy associated with S2 is absorbed into the rest mass • To obtain the full Hamiltonian of an electron, we must add a single term: e r r r H →H+ S ⋅ B(R) me r r 2 r e r r r 1 r H= P + e A( R) − e Ö ( R) + S ⋅ B( R) 2me me [ ] Uniform Weak Magnetic Field as a perturbation • For a weak uniform field, we find e B0 P2 (Lz + 2S z ) H= + 2me 2me • With the addition of a spherically symmetric potential, this gives: e B0 P2 (Lz + 2S z ) H= + V ( R) + 2me 2me – If the zero-field eigenstates are known H 0 n, l, ml = E0,n n, l, ml – The weak-uniform-field eigenstates are: {n, l, m , m } l s H n, l, ml , ms = En ,ml ,ms n, l, ml , ms En ,ml ,ms = E0,n + µ B B0 (ml + 2ms ) µB = eh 2me ‘Bohr Magneton’ Wavefunctions • In Dirac notation, all spin does is add two extra quantum numbers • The separate concept of a ‘spinor’ is unnecessary • Coordinate basis: r {r , s, ms r 2 R, S , S z – Eigenstate of } • Projector: I= s ∑ r r ∫ d r r , s , ms r , s , ms 3 ms = − s V • Wavefunction: r r ψ ms (r ):= r , s, ms ψ Spinor Notation: r r ψ ms (r ):= r , s, ms ψ • We think of them as components of a length 2s+1 vector, where each component is a wavefunction • Example: s=1/2 r ψ ↑ (r ):= r ψ ↓ (r ):= r 1 r , s, 2 ψ r r , s,− 12 ψ • Spinor wavefunction definition: r r r ψ ↑ (r ) ψ 12 (r ) [ψ ](r ):= r = r ψ ↓ (r ) ψ − 12 (r ) Note that descending order is unusual • If external and internal motions are not entangled, we can factorize the spinor wavefunction: r c↑ r [ψ ](r ):= ψ (r ) c↓ c↑ c↓ Is then a pure spinor Schrödinger's Equation • We start from: d ih ψ = H ψ dt • Hit from left with with r r , ms r d r ih r , ms ψ = r , ms H ψ dt • Insert the projector s r r r d r 3 ′ ′ ′ ih r , ms ψ = ∑ ∫ d r r , ms H r , ms r ′, m′ ψ dt m′s = − s V r P2 [[I ]]+ [[V ]]( R) H= 2M • Let: – For s=1/2: 1 0 0 1 [[I ]]= r r r V↑↑ (r ) V↑↓ (r ) [[V ]](r ) = r r V↓↑ (r ) V↓↓ (r ) 2 r r r r d h 2 ih [ψ ](r ) = − ∇ [ψ ](r )+ [[V ]](r )[ψ ](r ) dt 2M Example: Electron in a Uniform Field r r r r d h2 2 ih [ψ ](r ) = − ∇ [ψ ](r )+ [[V ]](r )[ψ ](r ) dt 2M r r r 2 ( ) ( ) ( ) ψ r ψ r ψ r 1 0 d h ∂ ↑ ↑ ↑ 2 ih ∇ − iµ B B0 r = − r + µ B B0 r dt ψ ↓ (r ) 2me ∂φ ψ ↓ (r ) 0 − 1ψ ↓ (r ) • This is just a representation of two separate equations: r h2 2 r r d ∂ ih ψ ↑ (r ) = − ∇ − iµ B B0 ψ ↑ (r )+ µ B B0ψ ↑ (r ) dt ∂φ 2me r h2 2 r r d ∂ ih ψ ↓ (r ) = − ∇ − iµ B B0 ψ ↓ (r )− µ B B0ψ ↓ (r ) dt ∂φ 2me • We would have arrived at these same equations using Dirac notation, without ever mentioning ‘Spinors’ Pauli Matrices r r r 2 ( ) ( ) ( ψ r ψ r ψ r 1 0 ↑ ) d ↑ h ∂ ↑ 2 ih ∇ − iµ B B0 r = − r + µ B B0 r ( ) ( ) ( ψ r ψ r ψ r dt ↓ 2me ∂φ ↓ 0 − 1 ↓ ) • Here we see that we have recovered one of the Pauli Matrices: 1 0 σ z = 0 − 1 • The other Pauli matrices are: 0 1 σ x = 1 0 0 − i σ y = i 0 • Then in the basis of eigenstates of we have: r Sz h r S= σ 2 r h2 2 r d ∂ ih [ψ ](r ) = − ∇ + µ B B0 − i + σ z [ψ ](r ) dt ∂φ 2M • If we only care about spin dynamics: e r r d i [c ]= σ ⋅ B[c ] dt 2me Two particles with spin • How do we treat a system of two particles with masses M1 and M2, charges q1 and q2, and spins s1 and s2? – Basis: r r r1 , s1 , ms1 ; r2 , s2 , ms 2 – Wavefunction: r r r r ψ s1 ,ms1 , s2 ,ms 2 (r1 , r2 ) := r1 , s1 , ms1 ; r2 , s2 , ms 2 ψ – Hamiltonian w/out motional degrees of freedom: q1 r r r q2 r r r H =− S1 ⋅ B( R1 ) − S 2 ⋅ B( R2 ) M1 M2 – Hamiltonian w/ motional degrees of freedom: r r r 1 r q1 r r r H= P1 − q1 A( R1 ) + q1Φ ( R1 ) − S1 ⋅ B( R1 ) 2M 1 M1 r r r 1 r q2 r r r P2 − q2 A( R2 ) + q2 Φ ( R2 ) − S 2 ⋅ B( R2 ) 2M 2 M2 ( ) ( ) Example #1 • A spin _ particle is in the ↑ state with respect to the z-axis. What is the probability of finding it in the ↓-state with respect to the x-axis? • Let: ψ = ↑z {↑ ,↓ } z z • In the basis, the operator for the x-component of spin is: 0 1 σ x = 1 0 • By symmetry, σx must have eigenvalues +1 and -1 • The eigenvector corresponding to -1 is defined by: σ x ↓x = − ↓x Example #1 continued: 0 1 σ x = 1 0 σ x ↓x = − ↓x ↓ x = −σ x ↓ x = −1 ↑z ↓x = − ↑z σ x ↓x = − ↓z ↓x • This implies that: ↓x 1 = ↑z − ↓z 2 ( P = ↓x ↑z 2 1 = 2 ) Example #2 • Two identical spin-1/2 particles are placed in a uniform magnetic field. Ignoring motional degrees of freedom, what are the energy-levels and degeneracies of the system? • States: {↑↑ , ↑↓ , ↓↑ , ↓↓ } – Z-axis chosen along B-field • Hamiltonian: gqB0 (S1z + S 2 z ) H =− 2M • Basis states are already eigenstates: hgqB0 H ↑↑ = − ↑↑ 2M H ↑↓ = H ↓↑ = 0 hgqB0 E1 = − ; d1 = 1 2M hgqB0 H ↓↓ = ↓↓ 2M hgqB0 E3 = ; d3 = 1 2M E2 = 0; d 2 = 2