Minimal coupling and Berry s phase

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Theodore Fields
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1 Phys460.nb 63 5 Minimal coupling Berry s phase Q: For charged quantum particles (e.g. electrons), if we apply an E&M field (E or B), the particle will feel the field. How do we consider the effect E B fields in quantum mechanics? A: Minimal coupling 5.1. Minimal coupling review on E&M (vector potential gauge transformations) In E&M, we learned that we can describe E B fields using potentials : scalar potential φ vector potential A. E = - φ - A B = A (5.1) (5.2) In addition, we learned that the potentials are NOT uniquely defined. In fact, we can change the scalar vector potential in the following way, which would NOT change E B fields thus all physics would remain the same. φ r φ' r = φ r - Λ r, t (5.3) A r A ' r = A r + Λ r, t (5.4) where Λ r, t is any arbitrary function of r t. To prove that these new potentials φ' A ' gives exactly the same E B fields as φ A, we compute the corresponding field strength E ' = - φ' - A' = - φ - Λ A + Λ - = - φ + Λ - A - Λ = - φ - A + Λ - Λ (5.5) The last two terms are identical (with opposite sign), so they cancel. E ' = - φ - A + Λ - Λ So, the E field remains the same when we change φ A to φ' A'. For the B field, = - φ - A = E (5.6) B ' = A ' = A + Λ = A + Λ (5.7)

2 64 Phys460.nb The last term contains, which is zero (The cross product for a vector with itself is always zero). B ' = A + Λ = A = B (5.8) So, the B field remains the same when we change φ A to φ' A'. As a results, we can freely to choose whether we want to use φ A as our potentials or φ' A'. This free choice is known as a gauge choice. Using a different gauge will NOT change any physics. The transformation mentioned above is known as a gauge transformation φ φ' = φ - Λ (5.9) A A ' = A + Λ (5.10) Comment: scalar vector potentials φ A are NOT measurable quantities. In E&M we can only measure E B. We cannot measure φ A. This is because we can freely choose a different gauge, which changes the value of φ A. On the other h, for any measurable quantities, we can measure its value, thus it must have a unique value. Since φ A do not have unique values, they cannot be measurable quantities. In fact, we get a more general statement along the same line of thinking: any gauge-dependent quantity (i.e. a quantity whose value changes when we choose a different gauge), can NOT be a measurable quantity Gauge transformation in quantum mechanics In classical physics, gauge transformation only changes potentials. Matters (e.g. motion of particles) don t care about gauge choice at all. This is because in classical physics, the motion of a particle (or any other objects) are described by measurable quantities (e.g. velocity, momentum, mass, etc). For measurable quantities, we know they can NOT change under gauge transformations. However, in quantum, the story is different. Because in quantum mechanics, matters are described by a non-measurable quantity: i.e., the wavefunction ψ r, t. The wavefunction is a complex function. The absolute value of the wavefunction is a measurable quantity (i.e. the square of the absolute value is the probability to find a particle at this point). However, the phase of a wavefunction is NOT measurable. As a result, (at least in principle), the phase of a wavefunction could depends on the gauge choice. In fact, this is exactly what happens in quantum mechanics. φ φ' = φ - Λ (5.11) A A ' = A + Λ ψ ψ' = ψ e i q Λ ħ (5.12) (5.13) where q is the charge of the particle. Comment: for charge neutral particles, q = 0, the wavefunction doesn't change, when we switch from φ A to φ' A'. But for any charged particles, the complex phase of the wavefunction must change, when we switch to a different gauge choice. To prove why we need to change the wavefunction when we change to a different gauge, we shall look at the Schrodinger equation. For any physics law, it should be invariant (i.e. remains the same) when we change to a different gauge (because different gauge choices describes exactly the same physics). As will be shown below, for the Schrodinger equation (which is one fundamental physics law), if we do not change the wavefunction, it would NOT remain the same after we switch to a different gauge. And the only way to ensure that the Schrodinger equation doesn t change is to change the wavefunction according to the transformation describe above Minimal coupling Now, we try to write down the Schrodinger s equation for a charged particle moving in E B fields. The way to couple the motion of a charged particle with E&B fields is known as the minimal coupling. We start from charge neutral particles with q = 0, whose Schrodinger equation is

3 Phys460.nb 65 i ħ ψ = P 2 + V(r) ψ (5.14) where P = -iħ is the momentum quantum operator. Now, if we want to describe the motion of a charged particle, we only need to do the following -i ħ -i ħ -q A (5.15) In addition, we need to do one more thing iħ iħ -q φ (5.16) In summary, what we need to do are two things: iħ iħ -q φ -i ħ -i ħ -q A (5.17) (5.18) More precisely, this are four formulas, instead of two, because the last formula is a vector formula with three components, x, y z iħ iħ -q φ -i ħ x -i ħ x -q A x -i ħ y -i ħ y -q A y -i ħ z -i ħ z -q A z (5.19) (5.20) (5.21) (5.22) When we have four formulas (one related with time three with space), special relativity told us that we can often combine them together as one four-dimensional vector formula. In special relativity, we can combine ϕ A together as a four-dimensional vector, known has the electromagnetic four-potential A μ = φ c, A x, A y, A y (5.23) In the same time, we can group together as a four-dimensional vector, known as the four-gradient t μ = 1 c, - x, - y, - y (5.24) As a result, when we study charged particles, we can start from the Schrodinger equation for a charge neutral particle. Whenever, we see a (time or spatial) derivative, we do the following substitution. i ħ μ i ħ μ -q A μ (5.25) Now, let s try the minimal coupling on the following Schrodinger equation i ħ (-i ħ )2 ψ = + V(r) ψ (5.26) After we do the minimal substitution, we get -q φ ψ = + V(r) ψ ψ - q φ ψ = + V(r) ψ (5.27) (5.28) we can move the second term on the l.h.s. to the r.h.s.

4 66 Phys460.nb ψ = + V(r) ψ + q φ ψ ψ = + V(r) + q φ ψ (5.29) (5.30) And know, we can define a new Hamiltonian (-i ħ -q A)2 H = + V(r) + q φ (5.31) The minimal coupling told us that the Hamiltonian has two modifications: 1. For the kinetic energy, the momentum is not longer just a derivative. We get an extra part q A. This part is new, it is the first time that we learned it (unless your theoretical mechanics class discussed similar topics for Hamiltonian mechanics). 2. For the potential energy, we get an additional contribution q φ. This part is NOT new NOT surprising at all. For static electrics, scalar potential φ is just the electric potential. If we have a charge particle with charge q moving in the presence of electric potential φ, it gets a potential energy q φ Verify the gauge transformation Now we prove that the Schrodinger equation with minimal coupling is invariant under the gauge transformation φ φ' = φ - Λ A A ' = A + Λ ψ ψ' = ψ e i q Λ ħ (5.32) (5.33) (5.34) We know that -q φ ψ = + V(r) ψ (5.35) Now, if we change to a different gauge, ψ', φ' A', the Schrodinger equation shall remain the same. We start from the l.h.s.. After the gauge transformation, it becomes i ħ ψ' -q φ' ψ' = i ħ -q φ' ψ' (5.36) For the first term i ħ ψ' = i ħ ψ q Λ ei ħ = i ħ e i q Λ ħ - q ħ Λ ψ ψ + i ħ ψ q Λ ei ħ = i ħ e i q Λ ħ ψ - q ψ e i q Λ ħ Λ (5.37) As a result, i ħ ψ' -q φ' ψ' = i ħ - q φ' ψ' - q Λ ψ - q φ' ψ - q φ' + Λ ψ - q φ ψ - q Λ ψ - q φ' e i q Λ ħ ψ (5.38) Now, we look a the right h side. After the gauge transformation, it becomes,

5 Phys460.nb 67 + V(r) ψ' = ψ' + V(r) ψ' (5.39) For the first term, (-i ħ -q A') 2 (-i ħ -q A') (-i ħ -q A') ψ' ψ' = (5.40) Let s start from (-i ħ -q A') ψ' = (-i ħ -q A') e i q Λ ħ ψ = -i ħ ψe i q Λ ħ - q A' ψe i q Λ ħ = -i ħ e i q Λ ħ ψ - i ħ ψ e i q Λ ħ - q A' ψe i q Λ ħ = -i ħ e i q Λ ħ ψ + q ψe i q Λ ħ Λ - q A' ψe i q Λ ħ = e i q Λ ħ [-i ħ ψ + q ψ Λ - q A' ψ] = e i q Λ ħ [-i ħ ψ - q (A' - Λ) ψ] = e i q Λ ħ [-i ħ ψ - q A ψ] = e i q Λ ħ (-i ħ -q A ) ψ (5.41) HEre, we proved that (-i ħ -q A') ψ' = (-i ħ -q A') e i q Λ ħ ψ = e i q Λ ħ (-i ħ -q A ) ψ (5.42) i.e., we can move e i q Λ ħ from the right side of (-i ħ -q A ) to its left side, but in the same time, we change A' to A. Now, lets consider the second (-i ħ -q A ) (-i ħ -q A') (-i ħ -q A') ψ' = (-i ħ -q A') e i q Λ ħ (-i ħ -q A ) ψ (5.43) Above, we proved that when we move e i q Λ ħ from the right side of (-i ħ -q A ) to its left side, we need to change A' to A. As a result, (-i ħ -q A') (-i ħ -q A') ψ' = (-i ħ -q A') e i q Λ ħ (-i ħ -q A ) ψ = e i q Λ ħ (-i ħ -q A) (-i ħ -q A ) ψ (5.44) = e i q Λ ħ (-i ħ -q A) 2 ψ Therefore, + V(r) ψ' = = e i q Λ (-i ħ -q A)2 ħ + V(r) ψ ψ' + V(r) ψ' = e i q Λ ħ (-i ħ -q A) 2 ψ + e i q Λ ħ V(r) ψ (5.45) So, the l.h.s. of the Schrodinger equation becomes i ħ -q φ' ψ' = ei q Λ ħ i ħ ψ - q φ ψ (5.46) which is the old l.h.s. with an extra factor e i q Λ ħ the r.h.s. becomes + V(r) ψ' = e i q Λ (-i ħ -q A)2 ħ + V(r) ψ (5.47) which is the old r.h.s., with an extra factor e i q Λ ħ Because we know that -q φ ψ = + V(r) ψ (5.48) We immediately know that

6 68 Phys460.nb i ħ -q φ' ψ' = + V(r) ψ' (5.49) i.e., using the new gauge, the Schrodinger equation takes exactly the same form (i.e. the physics law remains the same) Example: Lau levels quantum Hall effect D electron gas Now, we consider free electrons moving in 2D. (-i ħ )2 H = = (-i ħ x) 2 + (-i ħ y) 2 (5.50) Here, we have no potential energy, only kinetic energy. The eigenwavefunctions eigenenergy for this Hamiltonian are very simple. The eigen-wavefunctions are 2D plane waves ψ = ei kx x+i ky y (5.51) the corresponding eigenenergy is ϵ = P2 = P 2 x + P 2 y = ħ k x 2 + ħ k 2 y = ħ kx2 + k 2 y (5.52) Key conclusion: The eigenenergy can be any positive real number D electron gas + uniform magnetic field Now we apply a uniform B field (B is along the z-direction). As will shown below, the eigen-energy will now become discrete energy levels (similar to a harmonic oscillator). Here, eigen-energy can only some discrete values. For an uniform B field along z, we can represent it using the following scalar vector potential φ = 0 (5.53) A = (0, B x, 0) (5.54) It is easy to verify that E = - φ - A = 0 (5.55) B = A = x y z x y z = A x A y A z x y z x y z 0 B x 0 = ( x B x) z = B z (5.56) Note: This choice of φ A are not the unique choice. We can choose any gauge, but the final result would be the same, since our Schrodinger equation doesn t depends on gauge choices. Now the Hamiltonian changes into H = (-i ħ x -q A x ) 2 + (-i ħ y -q A y ) 2 The static Schrodinger equation is = (-i ħ x) 2 + (-i ħ y -q B x) 2 (-i ħ x ) 2 ψ(x, y) + (-i ħ y -q B x) 2 ψ(x, y) = ϵ ψ(x, y) (5.58) First, we notice that (5.57)

5.2. Example: Landau levels and quantum Hall effect

5.2. Example: Landau levels and quantum Hall effect 68 Phs460.nb i ħ (-i ħ -q A') -q φ' ψ' = + V(r) ψ' (5.49) t i.e., using the new gauge, the Schrodinger equation takes eactl the sae for (i.e. the phsics law reains the sae). 5.. Eaple: Lau levels quantu