QM1 Problem Set 1 solutions Mike Saelim

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1 QM Problem Set solutions Mike Saelim If you find any errors with these solutions, please me at a We can assume that the function is smooth enough to Taylor expand: fa = n c n A n The rest is a straightforward evaluation, using the property A a = a a : a fa a = n c n a A n a = n c n a a n a = n c n a n a a = fa δ aa b Since the operator can be written as a diagonal matrix of its eigenvalues, A = a a a a, the determinant is simply the product of all these eigenvalues The trace of the logarithm of the operator is Tr ln A = a a ln A a = a ln a, so taking the exponential of that gives exptr ln A = exp a ln a = a a = det A c e ia is unitary if e ia e ia =, where is the identity operator To show this, let s decompose the exponentials into their spectral decompositions The spectral decomposition of any function fa is a fa a a, which can be reasoned out from what we proved in part a You can also compute it out directly by replacing A with its spectral decomposition in the Taylor expansion of fa So, e ia = a e ia = a e ia a a e ia a a = a e ia a a Putting these together, e ia e ia = a = a e ia e ib a a b b = a b a a = e ia e ib a δ ab b b Alternatively, given that we showed e ia = e ia, we could just say that e ia e ia = e ia e ia = e 0 =, but we would have to specify that we can only combine the two exponentials naively because their arguments commute e X e Y = e X+Y is not true if [X, Y 0

2 I find it easiest to start with the right-hand side and work to the left-hand side of this equation We can even use our friends the Taylor expansion and spectral decomposition again fu AU = n c n U AU n = n c n U AUU AUU AU U AU = U n c n A n U = U fau 3 a Calculating e iˆn σφ involves using the Taylor expansion of the exponential, which will necessitate calculating ˆn σ One way to do this is by brute force If we let ˆn = n x, n y, n z, nz n ˆn σ = x in y n x + in y n z ˆn σ n = x + n y + n z n x + n y + n = z 0 since ˆn is a unit vector Thus, factors of ˆn σ n will simply alternate between the identity matrix and ˆn σ The other way to do this is to cheat sorta and use the identity you prove in part b Either way, ˆn σ = ˆn ˆn + i σ ˆn ˆn = e iˆn σφ = + iˆn σφ ˆn σ φ i 3! ˆn σ3 φ 3 + = + iˆn σφ φ 3! iˆn σ3 φ 3 + = cos φ + iˆn σ sin φ where the bold signifies the x identity matrix b The quickest way to prove this identity is to use two identities that define the Pauli matrices: [σ i, σ j = k iε ijk σ k {σ i, σ j } = δ ij Then, a σ b σ = ij = ij a i b j σ i σ j = a i b j {σ i, σ j } + [σ i, σ j ij a i b j δ ij + iε ijk σ k = a b + i σ a b k 4 a Again with the Taylor expansions: de xb = d + xb + x B + 3! x3 B 3 + = B + xb + x B 3 + = B e xb

3 Or, this can also be done with the definition of the derivative, extended to functions of operators: de xb = lim [ex+ɛb e xb = e xb lim ɛ [eɛb ɛ 0 = e xb lim = B e xb [ ɛb + ɛ B + but one must be careful to specify that you can split e x+ɛb into e xb e ɛb only because xb and ɛb commute b This will require us to use the definition of the derivative, as well as a Taylor expansion of the operators Ax and Bx: dab = lim [Ax + ɛbx + ɛ AxBx [ = lim Ax + ɛ dax + Oɛ Bx + ɛ dbx + Oɛ AxBx [ = lim AxBx + ɛ dax Bx + ɛaxdbx + Oɛ AxBx = A B + AB c This relies on the identity in part b Since AA =, d AA = da A + A da = 0 = da da = A A 5 Photon Filters The linear polarizer can be represented by a matrix that picks out the component of the photon s state vector that is parallel to the axis rotated by α from the x-axis In the x-y basis, it takes the x and y components of the input state vector, and spits out the x and y components of the output state vector We can do this by rotating the state vector by α, picking out its x-component, and rotating the result back by α: P xy α = cos α sin α sin α cos α cos α sin α cos = α sin α cos α sin α cos α sin α cos α sin α What does the matrix look like if we have it take in and spit out R and L components instead? One way is to transform Pα xy with the matrices that transform the components of the state vector The prompt gives us the transformation for the basis states: R x = U U = i L y i 3

4 However, the components of the state vector will actually transform with U : ξ = x ξ x ξ y = ξ y x ξ y U x U = R ξ y R ξ L L = ξ R ξ L = ξx ξ y U ξr = = U ξx ξ y ξ L So, to transform P α into a matrix that transforms between R-L components, we use the U matrix that transforms x-y components into R-L components, and its inverse U T that transforms R-L components into x-y components: P RL α = U Pα xy U T = i i = cos α sin α cos α sin α cos α sin α e iα e iα i i Another way, which pretty much does the same thing but avoids the confusion over what matrix to use, involves playing with bras and kets Notice that P α = α α where α = cos α x + sin α y in the x-y basis We can transform this to the R-L basis: α = cos α R + L + sin α i R L = e iα R + e iα L Using this α, P α gives the same result We can do a similar transformation from Q xy to Q RL to account for the behavior of the quarterwave plate Q xy = 0 0 i = Q RL = U Q xy U T = + i i = e iπ/4 i + i Now let s consider the combination QP π/4 Q Plugging and chugging, Q P π/4 Q xy = i i Q P π/4 Q RL 0 = 0 0 So, we can summarize the effects of this system in a table: Input Output Transmission probability = Output x R i y R R 0 0 L R e iπ/4 e iπ/4 e iπ/4 This filter blocks right-circularly polarized light, and turns left-circularly polarized light into rightcircularly polarized light 4

5 Indeed, this fits with what we would expect Q P π/4 Q to do We ve set the fast and slow axes of our quarter wave plates to be parallel to the x and y axes, so light linearly polarized parallel to the x and y axes will be unaffected by the first quarter-wave plate: the plate cannot induce a shift between two components if only one is present Then, it is halved in intensity by the 45 linear polarizer, and turned into circularly-polarized light by the second quarter-wave plate The first quarter-wave plate will also shift left-circularly polarized light into linearly polarized light parallel to the axis of the linear polarizer, while it shifts right-circularly polarized light into linearly polarized light perpendicular to that axis 6 Sakurai 7 Since we can decompose any state into a linear combination of eigenstates of A, α = a c a a, let s consider these operators effects on eigenstates first a For some eigenstate a, A a a = a a a a a which equals 0 because of the term a = a Since this holds for all the eigenstates, the operator will return 0 for any state b For some eigenstate a, A a a a a = a a a a a a a a a We have two possible cases: either a = a and all the terms will be, or a a, and one term will have 0 in the numerator So, A a a a = a a, a a is the projection operator Note that this is only possible if the states are nondegenerate, because if two different states have the same eigenvalue, the product will have 0/0 for one of its terms c The null operator, a S z a = will return 0 for either state The projection operator for S z h, a h/ S z + h S z h, S z a h a = S z h, will return h/ when acting on h/ and 0 when acting on h/ 5