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 use the properties of the Pauli matrices that σ i = I and {σ i σ j } = 0 provided i j n σ) = n x σ x + n y σ y + n z σ z )n x σ x + n y σ y + n z σ z ) = n x σ x + n y σ y + n z σ z + n x n y σ x σ y + σ y σ x ) + n y n z σ y σ z + σ z σ y ) + n z n x σ z σ x + σ x σ z ) = n x + n y + n z)i = I is The unitary operator which generates a rotation by ϕ about the axis n in spinor space U = e iϕ n S/ h b) Show that U = cos ϕ/ + in σ sin ϕ/ First we note that n S/ h = n σ/ Second we expand the exponential in a Taylor series and note that since n σ) = I even terms in the series will involve the identity matrix and odd terms will involve n σ to the first power U = e iϕ n S/ h = e iϕ/)n σ) i n = I n! ϕ/)n + in σ) n=04 = I cos ϕ/ + in σ) sin ϕ/ n=35 i n n! ϕ/) n c) Show that UσU = nn σ) n n σ) cosϕ + n σ) sinϕ You might find it helpful to remember that a product of two Pauli matrices gives the identity matrix or plus or minus i times the third Pauli matrix according to the formula σ i σ j = δ ij + iǫ ijk σ k
Physics 505 Homework No 8 s S8- where the summation convention is assumed and where we have just used in place of I UσU = cos ϕ/ + in σ sin ϕ/)σcosϕ/ in σ sin ϕ/) = cos ϕ/ σ + i cos ϕ/ sinϕ/n σ σ σ n σ) + sin ϕ/ n σ σ n σ At this point we just grind out the i th component of each term The first term is just cos ϕ/ σ i For the second term we need n σ σ i σ i n σ = n j σ j σ i σ i n j σ j = n j [σ j σ i ] = in j ǫ jik σ k = iǫ ijk n j σ k = i n σ) i and the middle term is sin ϕ n σ) i For the third term we need n σ σ i n σ = n j σ j σ i n k σ k = n j n k σ j σ i σ k = n j n k δ ji + iǫ jil σ l )σ k = n i n σ + in j n k ǫ jil δ lk + iǫ lkm σ m ) = n i n σ n j ǫ jil ǫ lkm n k σ m = n i n σ n j ǫ jil n σ) l = n i n σ + ǫ ijl n j n σ) l = n i n σ + n n σ)) i The second term looks like what we want but we still have more work to do on the first and third terms We combine the two terms and use the identities cos ϕ/ = +cos ϕ)/ and sin ϕ/ = cos ϕ)/ to get σ + nn σ) + n n σ))/ + cos ϕσ nn σ) n n σ))/) We now use the vector identity A B C) = BA C) CA B) In the first term above σ + nn σ) + n n σ))/ = σ + nn σ) + nn σ) σn n))/ = nn σ)
Physics 505 Homework No 8 s S8-3 In the second term we have cos ϕσ nn σ) n n σ))/) = cos ϕσ nn σ) nn σ) + σn n))/ = cos ϕσ nn σ)) = cos ϕn n σ)) Putting everything together we get the desired result: UσU = nn σ) n n σ) cos ϕ + n σ) sinϕ d) Consider the special case n = e z and discuss infinitesimal rotations and In this case U + iσ z δϕ/ = + iδϕ/ 0 ) 0 iδϕ/ σ = UσU σ e x σ y δϕ + e y σ x δϕ The rotation about the z-axis carries a little bit of the x-component of the Pauli operator into the y-direction and a little bit of the y-component into the negative x-direction just as what happens with an ordinary vector under rotation e) For a spinor χ = α+ α ) calculate the transformed spinor χ for n = e z ) χ α+ = Uχ = cos ϕ/ + iσ z sin ϕ/) α ) 0 0 = cos ϕ/ + i sin ϕ/ = 0 α+ cos ϕ/ + iα + sin ϕ/ α cos ϕ/ iα sin ϕ/ )) ) α+ 0 α ) α = + e +iϕ/ ) α e iϕ/
Physics 505 Homework No 8 s S8-4 f) What happens to the spinor when ϕ = π? The spinor changes sign! A rotation of 4π is required to bring the spinor back to its original state! This is a weird property of fermions However expectation values involve χ so the sign cancels out! Spin precession a) The spinors = ) 0 = ) 0 are the eigenfunctions of σ z They are an orthonormal complete basis in which σ z is diagonal In this same basis what are the eigenfunctions spinors) of σ x and σ y? The eigenvalues are ± so the expression for the eigenspinors of σ x is 0 0 ) ) a = ± b ) a b which gives b = ±a In order that the spinors be normalized we take a = b = / Then the eigenspinors of σ x are x = ) x = ) For σ y 0 i i 0 ) ) a = ± b ) a b which gives b = ±ia Again we must take a = b = / and the eigenspinors of σ y are y = ) i y = ) i Now we consider a spin / particle in a uniform magnetic field For definiteness we suppose it s an electron so the Hamiltonian for the spin degree of freedom is H = ge mc S B e mc S B
Physics 505 Homework No 8 s S8-5 b) The state of the system is represented by the spinor χ = i h t χ+ χ ) What is the equation of motion for χ? After writing down the equation for the general case specialize to the case B = Be z Your results should depend on the frequency ω = eb/mc which will turn out to be the precession frequency It s just the Schroedinger equation also known as the Pauli equation in this case) χ+ χ ) = e ) ) mc S B χ+ χ+ = µ χ B σ B = µ χ B When B = Be z the equation of motion becomes i t χ+ χ ) = µ BB h B z B x + ib y ) ) 0 χ+ = ω ) ) 0 χ+ 0 χ 0 χ ) ) B x ib y χ+ B z where ω = µ B B/ h = eb/mc) is the frequency which will turn out to the precession frequency) c) In the case with B = Be z and in the Heisenberg representation what is the time dependence of the spin operator St)? In the Heisenberg representation χ ds z dt = ī h [H S z] = 0 so S z t) = S z 0) For the other two components it s more convenient to work with S x ±is y d dt S x ± is y ) = ī h [H S x ± is y ] = ieb mc h [S z S x ± is y ] = ieb mc h i hs y is x ) = ± ieb mc S x ± is y ) This means or S x t) ± is y t) = e ±iωt S x 0) ± is y 0)) S x t) = cos ωt S x 0) sin ωt S y 0) S y t) = sin ωt S x 0) + cos ωt S y 0)
Physics 505 Homework No 8 s S8-6 d) In the case with B = Be z and in the Schroedinger representation what is the time dependence of the spinor χt)? The argument of the exponential is χt) = e iht/ h χ0) iht h = iebt mc h S z = iωt h S z = iϕn S/ h provided we identify ϕ = ωt and n = e z This form is exactly the form for the unitary operator discussed in problem so the time dependence of χ is χt) = ) χ+ t) = cos ωt/) χ t) = e iωt/ 0 0 e +iωt/ e = iωt/ ) χ + 0) e +iωt/ χ 0) ) 0 0 + i sin ωt/) 0 0 ) ) χ+ 0) χ 0) )) ) χ+ 0) χ 0) This makes sense The up and down states are the stationary states of the Hamiltonian with energy hω/ for the up state The spin in the up state is aligned with B but the magnetic moment is anti-aligned so the energy is higher than in the down state which has energy hω/ e) At t = 0 the spin is aligned along the x axis What is the probability of getting h/ in a measurement of S z at time t The initial state must be to within a phase!) χ0) = x = ) From part e) we find the state at time t and project that onto z to find the amplitude for going to spin up along z and then square it to get the probability e P z ) = 0) iωt/ / ) e +iωt/ =
Physics 505 Homework No 8 s S8-7 which shouldn t be a surprise If the spin is pointed along the x axis it s equally likely to be up or down if a measurement is made along the z axis When the spin is in a magnetic field in the z-direction it precesses about the z-axis which does not change its z-component or the probability distribution of its z-component) f) At t = 0 the spin is aligned along the x axis What is the probability of getting h/ in a measurement of S x at time t We solve this part the same way except we project onto x P x ) = / / e ) iωt/ / ) e +iωt/ so the spin is in fact precessing with a frequency ω! = cos ωt/ = + cosωt))/ The solutions to parts e) and f) have been based on the Schroedinger representation In the Heisenberg representation we could ask about the expectation value of S x t) Recall that the state does not change We use S x t) from part c) S x t) = / / ) 0 he iωt / he iωt / 0 ) ) / / = h cos ωt which is consistent with the idea that the spin is precessing around the z-axis which means its projection on the x-axis oscillates between ± h/ 3 Magnetic Resonance This problem continues from where problem ended We have an electron in a uniform magnetic field along the z-axis In addition we have a small time variable field along the x-axis: B x = B p cos ωt where ω is determined by the field in the z-direction ω = eb/mc We suppose that at t = 0 the electron has spin down along the z-axis The experimental picture is that the magnetic moment of the electron is aligned with the field so the spin is anti-aligned) and the probe field is turned on at t = 0 What happens? In particular what is S z t)? Use the interaction representation to solve this problem Hint: The probe field is oscillating along the x-axis However you can think of it as the sum of two fields each rotating in the xy-plane with angular velocity ω One rotates in the positive direction and one rotates in the negative direction When you transform the Hamiltonian due to the probe field to the interaction representation one of these will wind up stationary and the other will wind up rotating at ω Make an argument for why the ω term can be ignored and then proceed with the stationary term
Physics 505 Homework No 8 s S8-8 The additional term in the Hamiltonian is V t) = eb p mc S x cos ωt = ω p S x cos ωt In the interaction picture the operators evolve just as they did in the Heisenberg picture so we can use problem part c) for the St) In particular S z t) = S z 0) so that was easy! Transforming the extra term in the Hamiltonian requires a bit more work First as mentioned in the hint we can write the extra magnetic field as the sum of B + + B with Then the extra term is V + + V with B ± = B p cosωt e x ± sin ωt e y ) V ± = eb p mc S x cos ωt ± S y sin ωt) Now we must transform this extra term to the Heisenberg representation based on the Hamiltonian without V Again we use problem c) and replace S x t) by its expression in terms of S x 0) and S y 0) and similarly for S y t) So V I± t) = ω p cos ωts x 0) cos ωt sinωts y 0) ± sin ωts x 0) ± cos ωt sinωts y 0) ) V I+ t) = ω p S x 0) V I t) = ω p cos ωts x 0) sin ωts y 0)) The integral of the Hamiltonian with time gives the phase of the wave function V I is oscillating about 0) twice as fast as anything else so it essentially never has time to build up an appreciable phase Thus we ignore it and continue with V I+ t) which is just a constant The state evolves as states in the Schroedinger representation but with the interaction Hamiltonian rather than the full Hamiltonian i h t χ = V I+t)χ χt) = e iv I+t)/ h χ0) = e iω p e x S0)t/ h χ0) Again we make use of problem to write this as χt) = cos ω p t/) + iσ x 0) sin ω p t/))χ0)
Physics 505 Homework No 8 s S8-9 The state at t = 0 is spin down along the z-axis so ) ) ) χ+ cos ωp t/ i sin ω = p t/ 0 = i sin ω p t/ cos ω p t/ χ ) i sin ωp t/ cos ω p t/ So we have found how the state evolves due to the probe field and we know how S z evolves due to the main field and we can calculate S z ) ) h/ 0 i sin ωp t/ S z t) = +i sin ω p t/ cos ω p t/ ) = h 0 h/ cos ω p t/ cos ω pt After all that we have a very simple answer! The spin flips up and down with frequency ω p Physically the Heisenberg representation in this case corresponds to a frame rotating with the precessing spin Since we made the probe field oscillate with the same frequency as the rotation it split into two pieces one that s stationary in the rotating frame and one that s rotating like mad and has no net effect Since the field is stationary in the rotating frame it causes the spin to precess about this field in the rotating frame and it oscillates between aligned and anti-aligned You have just discovered the principle of magnetic resonance imaging which generally uses nuclear spins but the idea is the same) A sample is placed in a strong field this causes the spins to precess Because the environment of the spins depends not only on the applied field but the field due to neighbors the actual precession frequency will vary with chemical environment The probe field is scanned in frequency slowly compared to ω and ω p ) and there will be an absorption and emission of energy by the sample as the resonant frequency is passed The strength of the signal depends on the density of aligned spins A lot can be learned about the sample by studying how the spins precess and flip! 4 Combining angular momentum Two electron spins S and S can be summed to produce a total angular momentum J = S + S The states j ms s can be expanded in the states s m s m Determine the expansion Addition of two spin / angular momenta produces a spin system with j = and m = 0 + the triplet) and a spin system with j = 0 and m = 0 So Recall that + / / = / / / / L ± l m = h ll + ) mm ± ) l m ± We operate with J = S + S to get + ) ) 0 / / = /)/ + ) /)/ ) / / / / + /)/ + ) /)/ ) / / / /
Physics 505 Homework No 8 s S8-0 or 0 / / = / / / / + / / / / ) If we operate with J again we find Finally / / = / / / / 0 0 / / = / / / / / / / / ) We find this by noting that it must be orthogonal to 0 / / and it must contain the same m and m states that add up to 0 Comment: we will learn later that the wave functions for Fermions must be antisymmetric that is change sign) when any two Fermions are exchanged The singlet state above is anti-symmetric so this must be combined with a symmetric state for the spatial wave function to have a suitable overall state Similarly the triplet state must be combined with an anti-symmetric spatial wave function