5.2. Example: Landau levels and quantum Hall effect

Transcription

1 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 Hall effect D electron gas Now, we consider free electrons oving in D. (-i ħ ) H = = (-i ħ ) + (-i ħ ) (5.50) Here, we have no potential energ, onl kinetic energ. The eigenwavefunctions eigenenerg for this Hailtonian are ver siple. The eigen-wavefunctions are D plane waves ψ = ei k +i k (5.51) the corresponding eigenenerg is ϵ = P = P + P = ħ k + ħ k = ħ k + k (5.5) Ke conclusion: The eigenenerg can be an positive real nuber D electron gas + unifor agnetic field Now we appl a unifor B field (B is along the z-direction). As will shown below, the eigen-energ will now becoe discrete energ levels (siilar to a haronic oscillator). Here, eigen-energ can onl soe discrete values. For an unifor B field along z, we can represent it using the following scalar vector potential φ = 0 (5.53) A = (0, B, 0) (5.54) It is eas to verif that E = - φ - A t = 0 (5.55) B = A = z z = A A A z z z 0 B 0 = ( B ) z = B z (5.56) Note: This choice of φ A are not the unique choice. We can choose an gauge, but the final result would be the sae, since our Schrodinger equation doesn t depends on gauge choices. Now the Hailtonian changes into H = (-i ħ -q A ) + (-i ħ -q A ) The static Schrodinger equation is = (-i ħ ) + (-i ħ -q B ) (-i ħ ) ψ(, ) + (-i ħ -q B ) ψ(, ) = ϵ ψ(, ) (5.58) First, we notice that (5.57)

2 Phs460.nb 69 [H, -i ħ ] = 0 (5.59) This is because H doesn t contain an, so it coutes with -i ħ. This coutator iplies that we can find coon eigenstates of H -i ħ (-i ħ ) ψ(, ) + (-i ħ -q B ) ψ(, ) = ϵ ψ(, ) (5.60) -i ħ ψ(, ) = k ħ ψ(, ) (5.61) The solution of the second equation is eas to find ψ(, ) = f () e i k (5.6) If we put this solution back to the first equation, we get (-i ħ ) f () ei k + (-i ħ -q B ) We ultipl both sides with e-i k f () e i k = ϵ f () e i k (5.63) (-i ħ ) f () + e (-i ħ -i k -q B ) f () e i k = ϵ f () (5.64) (-i ħ ) f () + f () e (-i ħ -i k -q B ) e i k = ϵ f () (5.65) In the second ter on the l.h.s., we cannot sipl cancel e -i k e i k. This is becoe one of the is before, while the other is after. We know that we cannot switch e i k, because does not coute with. Here, we utilize the fact that -i ħ e i k = k ħ e i k (-i ħ ) f () + f () e (ħ k -i k - q B ) e i k = ϵ f () (5.66) (-i ħ ) f () + (ħ k - q B ) f () = ϵ f () (5.67) (-i ħ ) (-i ħ ) How, we define + (ħ k - q B ) f () = ϵ f () (5.68) q B - ħ k + q B f () = ϵ f () (5.69) 0 = ħ k q B (5.70) k = q B (-i ħ ) + k ( - 0) f () = ϵ f () (5.7) This equation is the static Schrodinger equation for a haronic oscillator (with equilibriu position at = 0 ). For a haronic oscillator, we know that the eigenenerg is ϵ n = ħ ω (n + 1/) (5.73) (5.71)

3 70 Phs460.nb where ω = k / Here, our k is q B, so we found that ω = k / = q B = q B (5.74) This frequenc is known as the cclotron frequenc the eigenenergies are ϵ n = ħ ω (n + 1/) = q ħ B n + 1 (5.75) The eigenenerg can no longer take arbitrar positive values. Onl soe discrete values are not allowed. Each of these discrete energ level is known as a Lau level cclotron frequenc The cclotron frequenc has a siple phsical eaning. In classical echanics, if we have a charged particle oving in a unifor B field, the trajector is a circle. This is because the Lorentz force is alwas perpendicular to velocit, which eans that the acceleration is alwas perpendicular to the velocit, i.e. a circular otion. If the velocit of the particle is v, the Lorentz force is F = q v B (5.76) For a circular otion, the acceleration is a = v r (5.77) We know that F = a, thus q v B = v r q B = v r (5.78) (5.79) We found hat neither v nor r is a universal nuber. However, their ratio is independent of v or r. Once we fied the B field, because q/ can never change for a particle (ignore relativistic effects), v/r is alwas a constant. Q: what is v/r for a circular otion? A: it is the angular frequenc ω ω = q B This is the cclotron frequenc. Quantu echanics told us, once we have a periodic otion with angular oentu ω, if it is a haronic oscillator, we shall epect eigenenerg ϵ = ħ ω (n + 1/). This is eactl what happened here for Lau levels.. (5.80) 5.3. Eaple: particles oving around a solenoid. Consider a charged particle oving around a solenoid.

4 Phs460.nb 71 For an infinite long solenoid, there is B field inside the solenoid, but the B field outside the clinder is zero. Now consider a charged particle with charge q. We put the particle on a ring, which circles around the solenoid, i.e. the particle can onl ove on this circular ring. B changing the electric current that passes thought the solenoid, we can change the field inside the solenoid, but the B field outside will alwas reain zero. Now, we ask the question whether the otion of the charged particle will change when we change the B field inside the solenoid. In classical echanics, the otion of the particle does NOT depend on the solenoid, because the B field is alwas zero for an point on the ring. In classical echanics, if B = 0, it eans that the particle cannot see the B field thus the solenoid doesn t atter at all, as far as the particle is concerned. However, in quantu echanics, this is not the case. It turns out that the particle does know the agnetic field inside the solenoid, although the field strength for an point on the ring is zero vector potential In quantu echanics, the agnetic field couples to the otion of a particle via vector potential A. This is ver different fro classical echanics, where what atters is B, instead of A. If the otion of the particle onl depends on B, then the solenoid doesn t atter, because B = 0 for the ring (this is what happens in classical echanics). If the otion of the particle relies on A, things would be different, because when B = 0, A a not be zero. Q: what is A for a solenoid A: It is not zero on the ring, if the solenoid has a nonzero current. Consider the following integral along the ring: d r A = d σ A = d σ B = (5.81) Here, we used Stokes theore the fact that the curl of A is B. is the agnetic flu inside the solenoid. Conclusion: the integral d r A equals to the agnetic flu inside the solenoid. Although B = 0 outside the solenoid, A is NOT zero (if A = 0, the integral would be zero, which is ipossible, because we know that the agnetic flu is nonzero). We can choose the gauge such that A = π r φ Here, we use the clindrical coordinate, ϕ is the unit vector that is tangential to the ring. It is eas to verif that A = (-sin φ, cos φ, 0) = π r π r - r, r, 0 = π - r, r, 0 = π - +, (5.8) +, 0 (5.83) Hailtonian

5 7 Phs460.nb H = (-i ħ -q A ) + (-i ħ -q A ) + (-i ħ z -q A z ) = -i ħ + π ħ +i = - Start with the first ter: +i i ħ - π - ħ -i + + ψ(,, z) = +i + (-i ħ z) - ħ z + +i + ψ(,, z) = ψ(,, z) + i + ψ(,, z) + i ψ(,, z) + i + + = ψ(,, z) + i + ψ(,, z) + i + = ψ(,, z) + i = ψ(,, z) + i ψ(,, z) + i + ψ(,, z) - + ψ(,, z) - i + ψ(,, z) - i ψ(,, z) - + ψ(,, z) (5.85) Siilarl, the second ter is -i + = ψ(,, z) - i ψ(,, z) + ψ(,, z) + i ψ(,, z) (5.86) As a result, ħ +i H = ħ -i = - ħ + + z - ħ i = - ħ + + z - i = - ħ ħ - i ħ π r ( - ) + 1 H ψ(,, z) = - ħ π ħ z + V r + ħ π r π ψ(,, z) - i ħ π r ( - ) ψ(,, z) + 1 π r (5.87) ψ(,, z) (5.88) Convert to the clindrical coordinate, ψ(r, φ, z) = 1 r r ψ(r, φ, z) r + 1 ψ(r, φ, z) + ψ(r, φ, z) (5.89) r r φ z Because the particle can onl ove along the ring, r z are fied thus the wavefunction onl depends on φ, independent of r z. ψ(r, φ, z) = ψ(φ). Therefore, z ψ = z ψ = 0

6 Phs460.nb 73 ψ = 1 r For the second ter, r r ψ r + 1 r ψ φ + ψ z = 1 ψ r φ = 1 r d ψ dφ (5.90) ( - ) ψ(,, z) = -z (,, 0) (,, z ) ψ(,, z) = -z r ψ(,, z) = -z r ψ(,, z) We know that ψ = ψ r = r r r r + 1 r ψ φ φ + ψ z z r ψ(,, z) = r r ψ r r + 1 r = ψ φ r φ + r ψ z r z ψ φ φ + ψ z z = r r 1 r z r ψ(,, z) = ψ φ z (r φ ) + r ψ z z (r z ) = ψ φ ( - ) ψ(,, z) = -z r ψ(,, z) = - ψ φ = - dψ dφ ψ φ φ + r r ψ z z (5.91) (5.9) (5.93) (5.94) (5.95) (5.96) Therefore, H ψ(,, z) = - ħ = - ħ 1 r ψ(,, z) - i ħ π r ( - ) ψ(,, z) + 1 d ψ dφ + i ħ d ψ π r d φ + 1 π r ψ π r ψ(,, z) (5.97) The Schrodinger equation H ψ(φ) = E ψ(φ) (5.98) So we have - ħ 1 r d ψ dφ + i ħ d ψ π r d φ d ψ dφ + i d ψ d φ + π r ψ = E r - d ψ dφ + i d ψ d φ = E r - ħ ψ = E ψ(φ) (5.99) ħ ψ (5.100) ψ (5.101) Define β = E r ϵ = - β ħ So, - d ψ dφ + i β dψ dφ = ϵ ψ d ψ dφ - i β dψ dφ + ϵ ψ = 0 (5.10) (5.103) The solution for this second order differential equation is ψ(φ) = A e i λ φ -λ ψ + λ β ψ + ϵ ψ = 0 (5.104) (5.105)