Physics 305 Summary of Week 1 Schrödinger and Heisenberg

Physics 305 Summary of Week 1

Schrodinger¨ and Heisenberg picture

We are all familiar with the Schr¨odinger equation, that tells us how a state Ψs depends on time | i d ih¯ Ψ = H Ψ . (1) dt| si | si Here H is the hamiltonian, in general some given function of the position and momentum. One can write the formal solution to the Schr¨odinger equation as

− i tH Ψ = e h¯ Ψ . (2) | si | hi where Ψ is time-independent: it coincides with Ψ at t = 0. We call Ψs the Schr¨odinger | hi | si | i wavefunction and Ψh the Heisenberg wavefunction. We can introduce a complete or- thonormal basis of|eigenstatesi of H

H n = En n , n m = δnm (3) | i | i h | i and decompose Ψ as | hi

Ψh = X cn n cn = n Ψh . (4) | i n | i h | i

2 The cn’s are (time-independent) complex numbers such that cn = 1. The time- Pn dependent Schr¨odinger wavefunction can now be expressed as | |

i − h¯ tEn Ψs = X cne n . (5) | i n | i

There are now two equivalent ways of describing time evolution in quantum mechanics. In the Schrödinger picture, states are time-dependent and satisfy the Schrödinger equation (1). Operators As in the Schrödinger picture, such as the position and momentum operator and functions thereof, are time-independent. Their expectations values, defined as A = Ψ A Ψ , do depend on time, via h i h s| s| si d i A = [H, A] . (6) dth i h¯ D E

1 In the Heisenberg picture, on the other hand, the states Ψh are time-independent and related to the Schr¨odinger states via eqn (2). The time-dep| endencei in this formulation is contained in the operators Ah, which satisfy the equation of motion

d i A = [H, A ] (7) dt h h¯ h

The Heisenberg operators are related to the time-independent Schr¨odinger operators via

i tH − i tH Ah = e h¯ Ase h¯ (8)

This relation combined with (1) implies that expectation values in the Heisenberg picture, defined as A = Ψ A Ψ , are identical to those defined in the Schrödinger picture h i h h| h| hi

i tH − i tH Ψ A Ψ = Ψ e h¯ A e h¯ Ψ = Ψ A Ψ (9) h h| h| hi h h| s | hi h s| s| 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 Schr¨odinger 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 ﬁeld, we introduce the electro-magnetic

vectorpotential : A~(~x, t) and potential : φ(~x, t) . (10)

The B~ and E~ ﬁeld are obtained from these potentials via

1 ∂A~ B~ = ~ A E~ = ~ φ (11) ∇ × −∇ − c ∂t

Given A~ and φ, the hamiltonian of a particle with charge e and mass m is given by

1 e H = (p~ + A~(~x, t))2 + eφ(~x, t) . (12) 2m c

2 The form of this hamiltonian is ﬁxed 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 h¯ ∂p~ dp~ i ∂H = [ H, p~ ] = , (14) dt h¯ − ∂~x where we used the canonical commutation relations h¯ [p , x ] = δ 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 ﬁnd 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 ﬁeld. Without loss of generality, we can choose the B-ﬁeld 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 mω2(x2 + y2) + ωL + z (22) 2m x y 2 z 2m with

Lz = xpy ypx (23) − the angular momentum in the z direction, and

eB ω ω = = L . (24) 2mc 2

ωL 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 ωy 0 ω 1/m 0 x    −   −    d y py/m + ωx ω 0 0 1/m y   =  2  =  2    . (26) dt  px   mω x ωpy   mω 0 0 ω   px     2 −   2 −     py   mω y + ωpx   0 mω ω 0   py 

4 To solve these equations, we would like to ﬁnd 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 ﬁrst two terms of the hamiltonian are those of a simple harmonic oscillator. It is therefore natural to introduce creation and annihilation operators via

¯h † ¯hmω † x = (ax + a ) px = i (ax a ) (27) q 2mω x − q 2 − x ¯h † ¯hmω † y = (ay + a ) py = i (ay a ) (28) q 2mω 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

† † [ax, ax] = 1 , [ay, ay] = 1. (29)

Let us express the hamiltonian in terms of these new operators. A simple calculation shows that † † † † H = hω¯ (a ax + a ay + 1) + ihω¯ (a ax a ay) (30) x y y − x In the ﬁrst term we recognize the hamiltonion of two harmonic oscillators, and the last term is equal to ωLz. The hamiltonian can be simpliﬁed a bit further, by introducing the complex combinations

1 † 1 † † a§ = (ax iay) , a§ = (a ia ) (31) √2 § √2 x ∓ y which also satisfy the standard commutation relations of creation and annihilation operators

† † [a+ , a+ ] = 1 , [a− , a− ] = 1. (32)

The hamiltonian reduces to the very simple form

† 1 H = hω¯ L(a+ a+ + 2 ) (33)

† Note that H does not depend on a− and a−. Using (7) and (32), we ﬁnd for the equation † of motion of the creation and annihilation operators (a§, a§)

a+ iωLa iωL 0 0 0 a+    − +   −    d a− 0 0 0 0 0 a−  †  =  †  =    †  (34) dt  a+   iωLa+   0 0 iωL 0   a+   †       †   a−   0   0 0 0 0   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 ωL and one with frequency 0. We introduce the number operators

† † N+ = a+ a+ , N− = a− a− . (35)

The corresponding eigenstates are

N+ n , n− = n n , n− , N− n , n− = n− n , n− (36) | + i + | + i | + i | + i The creation and annihilation operators act on these eigenstates as follows

† a n+ , n− = n+ + 1 n++ 1, n− , a+ n+ , n− = √n+ n+ 1, n− (37) + | i q | i | i | − i

† a n+ , n− = n− + 1 n+ , n−+ 1 , a− n+ , n− = √n− n+ , n− 1 (38) − | i q | i | i | − i All these eigenstates can be obtained from the ground state by acting with the creation operators

1 † n † n n , n = (a ) + (a ) − 0, 0 (39) | + − i + − | i qn+! n−! where 0, 0 is annihilated by both annihilation operators | i a 0, 0 = a 0, 0 = 0 (40) + | i − | i

1 The hamiltonian (33) is just H = hω¯ L(N+ + 2 ), so its eigenvalues and eigenstates are

1 H n , n− = h¯ωL(n+ + ) n , n− . (41) | + i 2 | + i

Note that the eigenvalues are independent of n− , so there are infinitely many eigenstates with the same given eigenenergy. This has an immediate physical interpretation. As we will see, each eigenstate corresponds to some local wavefunction, centered around some location in the (x, y)-plane. Since the magnetic field is uniform, translating the center position of the state should not change its energy. So indeed there should be infinitely many states with the same energy.