Quantum Mechanics (Symmetry in Quantum Mechanics)

4 Symmetry in Quantum Mechanics 4.1 (a) Assuming that the Hamiltonian is invariant under time reversal, prove that the wave function for a spinless nondegenerate system at any given instant of time can always be chosen to he real. (b) The wave function for a plane-wave state at 10 is given by a complex function, Why does this not violate time-reversal invariance? (a) Suppose that in) in a nondegenerate energy eigenstate. Then If Theta|n rangle= Theta H|n rangle=E 0 |n rangle Rightarrow Theta|n rangle= e ^ (i*phi) |n rangle e^ i(E n / hbar Theta|n rangle=e^ i( E n t hbar + delta) |n rangle=e^ i( 2E_{n}*t hbar +i) |n,t 0 =0;i rangle Rightarrow Theta[ int d^ 3 x| vec x rangle langle vec x |]|n,t 0 =0;t rangle= e ^ (i((2pi_{m}*z)/n + t)) [ int d^ 3 x| vec x rangle langle vec x ]]|n,t n =0;t rangle Rightarrow int d^ 3 x langle vec x |v,t 0 =0;i rangle^ * | vec x rangle= int d^ 8 ,xe^ i( 2E_{x}*z hbar + phi) langle vec x |n,t 0 =0;t rangle| vec x rangle Rightarrow phi a ^ a ( vec x ,t)= e ^ (i((x*x_{a}*t)/b + t)) * phi_{a} ( vec x ,t). (4.1) So if we choose at any instant of time real. delta=- 2E - j hbar the wave function will be (b) In the case of a free particle the Schrödinger equation is p^ 2 2m |n rangle=E|n rangle Rightarrow- hbar^ 2 2m vec nabla phi_{u}(x) = E*phi_{u}(x) (4.2) The wave functions same eigenvalue (1) E = E/(2m) and phi_{n}' * (x) =e^ i vec E mathcal E /k correspond to the and so there is degeneracy since these correspond to different state | vec p rangle and |- vec p rangle So we cannot apply the previous result. 4.2 Let phi ( vec p ) the momentum-space wave function for state (o), that is, phi ( vec p )= langle vec p | alpha rangle. Is the momentum-space wave function for the 4. SYMMETRY IN QUANTUM MECHANICS 95 time-reversed state Theta | alpha rangle given by phi ( vec p ), (7), phi ^ * ( vec p ^ * ) s or phi ^ * (- vec p ^ * )? Justify your answer. In the momentum space we have | alpha rangle= int d^ 9 p^ prime langle vec p | alpha rangle| vec p rangle Rightarrow| alpha rangle= int d^ 0 p' * phi ( vec p )| vec p rangle Rightarrow Theta| alpha rangle= int d^ 3 mu' * Theta \ langle vec p [ alpha rangle| vec p ^ prime rangle]= int d^ 3 p^ prime langle vec rho ( alpha)^ * Theta| vec p rangle. (4.3) For the momentum it is natural to require langle alpha | vec p | alpha rangle=- langle tilde alpha | vec p | hat alpha rangle Rightarrow langle hat alpha | Theta vec p * Theta ^ - 1 | hat alpha rangle Rightarrow Theta hat p Theta^ -1 =- vec p | So Theta vec p | vec p rangle^ (4.4) - vec p*Theta | vec p rangle Rightarrow Theta| vec p rangle=|- vec p ^ prime rangle (4.5) up to a phase factor. So finally Theta| alpha rangle= int d^ prime p' langle vec p ^ prime | alpha rangle^ * |- vec p ^ prime rangle= int d^ prime p^ prime langle- vec p ^ prime | alpha rangle^ * | vec p ^ prime rangle Rightarrow langle vec mu|Theta| * alpha rangle= tilde phi ( vec p ^ prime )= langle- vec p ^ prime | alpha rangle^ * = phi^ * (- vec mu ^ prime ). (4.6) 4.3 Read section 4.3 in Sakurai to refresh your knowledge of the quantum mechanics of periodic potentials. You know that the en-ergybands in solids are described by the so called Bloch functions psi a, k fullfilling, Valr+a)=() where a is the lattice constant, labels the band, and the lattice momentum & is restricted to the Brillouin zone [(- pi) / a, pi / a] Prove that any Bloch function can be written as, (4.4) where the sum is over all lattice vectors R_{i} (In this simble one di-mensional problem R_{i} = i*sigma_{i} but the construction generalizes easily to three dimensions.). The functions phi_{pi} are called Wannier functions, and are impor-tant in the tight-binding description of solids. Show that the Wan-nier functions are corresponding to different sites and/or different bands are orthogonal, de prove int dx phi m ^ * (x - R_{i}) * phi_{u} (x-R j ) sim delta 0 delta mn Hint: Expand the dot psi_{n} s dot u Bloch functions and use their orthonor-mality properties. The defining property of a Bloch function psi n, k (x) is (+0)=() (4.7) We can show that the functions En (R) satisfy the same relation Σα-Σ-(-)(R) R R_{i} - q =R j e^ i hat k k sum R j phi n (x-R j )e^ i hat k R j (4.8) which means that it is a Bloch function (2) - Σ2 - (4.9) The last relation gives the Bloch functions in terms of Wannier functions. To find the expansion of a Wannier function in ternes of Bloch functions we multiply this relation by c ^ (- i * k*R_{i}) and integrate over k Rightarrow int - pi f4 ^ pi*fu dke^ -ik hat R t phi n,k (x)= sum R 1 phi n (x-K i ) int - pi f4 ^ pi fu e^ ik( overline R t - overline R j ) dk 4.10) 4. SYMMETRY IN QUANTUM MECHANICS 97 But N-d 2sin/(RR) R-R = (RR) = delta_{o} * (2pi)/n (4.11) (4.12) (4.13) where in the last step we used that RRna, with n ∈ Z. So integrate 1 dk from (- pi) / n to pi / n * e ^ (- i * k*R_{z}) psi eta,k (x)= sum R i phi w (x-R i ) delta ij 2 pi a Rightarrow dot phi u hat x -R i )= a 2 pi int - tau/n ^ tau/ pi e^ -ikR i dot psi u,k (x)dk So using the orthonormality properties of the Bloch functions int dx phi n ^ * (x - R_{i}) * phi_{n}(x - R_{j}) = (2) (2) = (z)(z)drdkdk (2) 2г 4.4 Suppose a spinless particle is bound to a fixed center by a potential V ( vec x ) so assymetrical that no energy level is degenerate. Using the time-reversal invariance prove langle vec t rangle=0 for any energy eigenstate. (This is known as quenching of orbital angular momemtum.) If the wave function of such a nondegenerate eigenstate is expanded as sum l sum m F lm (r) * Y_{l} ^ (- m) * (theta, phi) ,