Physics 305 Summary of Week 1 SchrodingerÄ and Heisenberg picture We are all familiar with the SchrÄodinger equation, that tells us how a state ªs depends on time j i d ih¹ ª = H ª : (1) dtj si j si Here H is the hamiltonian, in general some given function of the position and momentum. One can write the formal solution to the SchrÄodinger equation as ¡ i tH ª = e h¹ ª : (2) j si j hi where ª is time-independent: it coincides with ª at t = 0. We call ªs the SchrÄodinger j hi j si j i wavefunction and ªh the Heisenberg wavefunction. We can introduce a complete or- thonormal basis ofjeigenstatesi of H H n = En n ; n m = ±nm (3) j i j i h j i and decompose ª as j hi ªh = X cn n cn = n ªh : (4) j i n j i h j i 2 The cn's are (time-independent) complex numbers such that cn = 1. The time- Pn dependent SchrÄodinger wavefunction can now be expressed as j j i ¡ h¹ tEn ªs = X cne n : (5) j i n j i There are now two equivalent ways of describing time evolution in quantum mechanics. In the SchrÄodinger picture, states are time-dependent and satisfy the SchrÄodinger equation (1). Operators As in the SchrÄodinger picture, such as the position and momentum operator and functions thereof, are time-independent. Their expectations values, de¯ned as A = ª A ª , do depend on time, via h i h sj sj si d i A = [H; A] : (6) dth i h¹ D E 1 In the Heisenberg picture, on the other hand, the states ªh are time-independent and related to the SchrÄodinger states via eqn (2). The time-depj endencei in this formulation is contained in the operators Ah, which satisfy the equation of motion d i A = [H; A ] (7) dt h h¹ h The Heisenberg operators are related to the time-independent SchrÄodinger operators via i tH ¡ i tH Ah = e h¹ Ase h¹ (8) This relation combined with (1) implies that expectation values in the Heisenberg picture, de¯ned as A = ª A ª , are identical to those de¯ned in the SchrÄodinger picture h i h hj hj hi i tH ¡ i tH ª A ª = ª e h¹ A e h¹ ª = ª A ª (9) h hj hj hi h hj s j hi h sj sj si So it is clear that both formalisms are indeed equivalent: in both pictures, the expectation value of operators satisfy the Ehrenfest equation of motion (6). In the SchrÄodinger picture, the time-dependence comes from the wavefunctions, in the Heisenberg picture from the operators. Charged particle in an electro-magnetic field To write the hamiltonian of a charged particle in an electro-magnetic ¯eld, we introduce the electro-magnetic vectorpotential : A~(~x; t) and potential : Á(~x; t) : (10) The B~ and E~ ¯eld are obtained from these potentials via 1 @A~ B~ = ~ A E~ = ~ Á (11) r £ ¡r ¡ c @t Given A~ and Á, the hamiltonian of a particle with charge e and mass m is given by 1 e H = (p~ + A~(~x; t))2 + eÁ(~x; t) : (12) 2m c 2 The form of this hamiltonian is ¯xed by the requirement that the resulting equations of motion take the correct form in terms of the Lorentz force. It is convenient to work in the Heisenberg picture. The position and momentum operators satisfy d~x i @H = [ H; ~x ] = (13) dt h¹ @p~ dp~ i @H = [ H; p~ ] = ; (14) dt h¹ ¡ @~x where we used the canonical commutation relations h¹ [p ; x ] = ± k; l = 1; 2; 3: (15) k l i kl The above equations are just the Hamilton equations of classical mechanics, except that ~x and p~ are now operators, rather than just numbers. For the time-derivative of the position operator we ¯nd d~x 1 e = (p~ + A~): (16) dt m c So velocity and momentum are no longer in the same direction. The above relation can be rewritten as e d~x p~ = m~v A~ ; ~v (17) ¡ c ´ dt With a bit more work { taking the time-derivative of (16) and using the second Hamilton equation (14) and relations (11) { one derives that d2x e m = eE~ ~v B~ : (18) dt2 ¡ ¡ c £ This is the Lorentz force equation. Charged particle in a constant magnetic field We now consider the case of a constant magnetic ¯eld. Without loss of generality, we can choose the B-¯eld in the direction of the z-axis: B~ = ( 0; 0; B) (19) with B a constant. For the corresponding vector potential, we choose A~ = ( 1 By; 1 Bx; 0) : (20) ¡ 2 2 3 You can verify that A~ and B~ are related via (11). Plugging (20) into (12), we obtain the hamiltonian 1 e 2 e 2 2 H = (px By) + (py + Bx) + p : (21) 2m³ ¡ 2c 2c z´ This can be rewritten as: 1 p2 H = (p2 + p2) + 1 m!2(x2 + y2) + !L + z (22) 2m x y 2 z 2m with Lz = xpy ypx (23) ¡ the angular momentum in the z direction, and eB ! ! = = L : (24) 2mc 2 !L is the well-known Larmor frequency. Note that this hamiltonian does not look invariant under translations in the (x; y) plane, although the physics should be translation invariant. We will explain this later. The hamiltonian (22) also looks a bit complicated, because it has many terms. Things are not as bad as they look, however. As we will see, it can be reduced to the hamiltonian of a simple harmonic oscillator. First we note that the motion in the z-direction is trivial: the hamiltonian equations of motion for z and pz read dz p dp = z ; z = 0; (25) dt m dt which tells us that the particle moves with a uniform velocity in the z-direction. From now on we will ignore this z-motion, and concentrate of the motion in the (x; y)-plane. A second helpful fact is that all the terms in H are (at most) quadratic in positions and momenta. As a result, the corresponding Hamilton equations of motion can be written as a linear equation: x px=m !y 0 ! 1=m 0 x 0 1 0 ¡ 1 0 ¡ 1 0 1 d y py=m + !x ! 0 0 1=m y B C = B 2 C = B 2 C B C : (26) dt B px C B m! x !py C B m! 0 0 ! C B px C B C B 2 ¡ C B 2 ¡ C B C @ py A @ m! y + !px A @ 0 m! ! 0 A @ py A 4 To solve these equations, we would like to ¯nd suitable linear combinations of the coordi- d nates and momenta, such that the time-derivative dt acts via a diagonal matrix. In other words, we want to diagonalize the above 4 4 matrix. We will do this in two steps. £ The ¯rst two terms of the hamiltonian are those of a simple harmonic oscillator. It is therefore natural to introduce creation and annihilation operators via h¹ y hm!¹ y x = (ax + a ) px = i (ax a ) (27) q 2m! x ¡ q 2 ¡ x h¹ y hm!¹ y y = (ay + a ) py = i (ay a ) (28) q 2m! y ¡ q 2 ¡ y These are the standard expressions for the creation and annihilation operators of a harmonic oscillator. From (15) one deduces the familiar commutation relations y y [ax; ax] = 1 ; [ay; ay] = 1: (29) Let us express the hamiltonian in terms of these new operators. A simple calculation shows that y y y y H = h!¹ (a ax + a ay + 1) + ih!¹ (a ax a ay) (30) x y y ¡ x In the ¯rst term we recognize the hamiltonion of two harmonic oscillators, and the last term is equal to !Lz. The hamiltonian can be simpli¯ed a bit further, by introducing the complex combina- tions 1 y 1 y y a§ = (ax iay) ; a§ = (a ia ) (31) p2 § p2 x ¨ y which also satisfy the standard commutation relations of creation and annihilation operators y y [a+ ; a+ ] = 1 ; [a¡ ; a¡ ] = 1: (32) The hamiltonian reduces to the very simple form y 1 H = h!¹ L(a+ a+ + 2 ) (33) y Note that H does not depend on a¡ and a¡. Using (7) and (32), we ¯nd for the equation y of motion of the creation and annihilation operators (a§; a§) a+ i!La i!L 0 0 0 a+ 0 1 0 ¡ + 1 0 ¡ 1 0 1 d a¡ 0 0 0 0 0 a¡ B y C = B y C = B C B y C (34) dt B a+ C B i!La+ C B 0 0 i!L 0 C B a+ C B y C B C B C B y C @ a¡ A @ 0 A @ 0 0 0 0 A @ a¡ A 5 So, as promised, we have diagonalized the equation of motion. It is now an easy task to obtain the energy spectrum. The following formulas are just the standard ones for a harmonic oscillator; in our case, we just have two oscillators, one with frequency !L and one with frequency 0. We introduce the number operators y y N+ = a+ a+ ; N¡ = a¡ a¡ : (35) The corresponding eigenstates are N+ n ; n¡ = n n ; n¡ ; N¡ n ; n¡ = n¡ n ; n¡ (36) j + i + j + i j + i j + i The creation and annihilation operators act on these eigenstates as follows y a n+ ; n¡ = n+ + 1 n++ 1; n¡ ; a+ n+ ; n¡ = pn+ n+ 1; n¡ (37) + j i q j i j i j ¡ i y a n+ ; n¡ = n¡ + 1 n+ ; n¡+ 1 ; a¡ n+ ; n¡ = pn¡ n+ ; n¡ 1 (38) ¡ j i q j i j i j ¡ i All these eigenstates can be obtained from the ground state by acting with the creation operators 1 y n y n n ; n = (a ) + (a ) ¡ 0; 0 (39) j + ¡ i + ¡ j i qn+! n¡! where 0; 0 is annihilated by both annihilation operators j i a 0; 0 = a 0; 0 = 0 (40) + j i ¡ j i 1 The hamiltonian (33) is just H = h!¹ L(N+ + 2 ), so its eigenvalues and eigenstates are 1 H n ; n¡ = h¹!L(n+ + ) n ; n¡ : (41) j + i 2 j + i Note that the eigenvalues are independent of n¡ , so there are in¯nitely many eigenstates with the same given eigenenergy.