Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 7

a) Division of the SE with h̄ω_{0}/2 yields

[−\frac{h̄}{mω_{0}} \frac{d^{2}}{dy^{2}} + \frac{mω_{0}}{h̄} y^{2}] ϕ = εϕ.

Substitutions y^{2} = h̄ξ^{2}/mω_{0} and dy^{2} = h̄dξ^{2}/mω_{0} turns the equation into the one given in the exercise.

b) Let’s start by writing down the SE:

[−\frac{h̄^{2}}{2m} (∇ + \frac{ieA}{h̄})^{2} + \frac{1}{2}mω_{0}^{2}y^{2}] ψ = Eψ.

With ∇ = x̂∂/∂x + ŷ∂/∂y, A = −yBx̂, and ψ = exp(ikx)ϕ_{n}(y), the SE becomes

[−\frac{h̄^{2}}{2m} (ik − i\frac{eyB}{h̄} )^{2} − \frac{h̄^{2}}{2m} \frac{∂^{2}}{∂y^{2}} + \frac{1}{2}mω_{0}^{2}y^{2}] ϕ_{n}(y) = Eϕ_{n}(y).

We note that

\frac{h̄^{2}}{2m} · \frac{1}{l_{B}^{2}} = \frac{h̄^{2}}{2m} · \frac{eB}{h̄} = \frac{1}{2}h̄\frac{eB}{m} = \frac{1}{2}h̄ω_{c}.

Thus, if we divide the SE with this factor, we obtain

[l_{B}^{2} (k − \frac{eyB}{h̄} )^{2} − l_{B}^{2} \frac{∂^{2}}{∂y^{2}} + m^{2}ω_{0}^{2}l_{B}^{2}y^{2}/h̄^{2}] ϕ_{n}(y) = εϕ_{n}(y).

On the left hand side, the first term is (κ − η)^{2} = (η − κ)^{2}, since κ = kl_{B} and η = y/l_{B} = yl_{B}/l_{B}^{2} = yl_{B}eB/h̄. The second term is obviously ∂^{2}/∂η^{2}. The third term is

m^{2}ω_{0}^{2}l_{B}^{4}η^{2}/h̄^{2} = m^{2}ω_{0}^{2} \frac{h̄^{2}}{e^{2}B^{2}} η^{2}/h̄^{2} = \frac{ω_{0}^{2}}{ω_{c}^{2}} η^{2} = α^{2}η^{2}.

Consequently, the equation for ϕ_{n}(η) is

[−\frac{d^{2}}{dη^{2}} + (η − κ)^{2} + α^{2}η^{2}] ϕ_{n}(η) = εϕ_{n}(η).

c) We write the two terms inside the brackets as

(η − κ)^{2} + α^{2}η^{2} = (1 + α^{2})η^{2} − 2κη + κ^{2}
= (1 + α^{2}) (η − \frac{κ}{1 + α^{2}} )^{2} + κ^{2} − \frac{κ^{2}}{1 + α^{2}}
= (1 + α^{2}) (η − \frac{κ}{1 + α^{2}} )^{2} + \frac{α^{2}κ^{2}}{1 + α^{2}}

Hence, the equation for ϕ_{n}(η) can be written

[−\frac{d^{2}}{dη^{2}} + (1 + α^{2}) (η − \frac{κ}{1 + α^{2}} )^{2}] ϕ_{n}(η) = (ε − \frac{α^{2}κ^{2}}{1 + α^{2}} ) ϕ_{n}(η).

Comparison with the dimensionless equation in a) shows that this is an equation for a 1D harmonic oscillator, centered at η_{c} = κ/(1 + α^{2}), and with a redefined energy eigenvalue ε − α^{2}κ^{2}/(1 + α^{2}). In addition, the factor 1 + α^{2} = 1 + ω_{0}^{2}/ω_{c}^{2} = Ω^{2}/ω_{c}^{2} in front of the quadratic term on the left hand side means that this equation comes from a harmonic oscillator equation where the potential energy is not mω_{c}^{2}y^{2}/2, but rather mΩ^{2}y^{2}/2. Hence, we know that the dimensionless energy on the right hand side actually equals (n + 1/2)h̄Ω/(1/2)h̄ω_{c}. Now, we are ready to transform back to original variables. First, the center position of the harmonic oscillator wave functions:

y_{c} = l_{B}η_{c} = l_{B}κ/(1 + α^{2})
= l_{B}^{2}kω_{c}^{2}/Ω^{2} = L_{B}^{2}k

Next, the energy spectrum:

(n + \frac{1}{2})h̄Ω = \frac{1}{2}h̄ω_{c} (ε − \frac{α^{2}κ^{2}}{1 + α^{2}} )
= E − \frac{1}{2}h̄\frac{eB}{m} \frac{ω_{0}^{2}}{Ω^{2}} k^{2} \frac{h̄}{eB}
= E − \frac{h̄^{2}k^{2}}{2M_{B}}

Therefore,

E = E_{n,k} = (n + \frac{1}{2})h̄Ω + \frac{h̄^{2}k^{2}}{2M_{B}} .

The presence of the confining potential V (y) has turned the Landau levels into ”Landau bands”: Electrons at the Fermi level distribute their energy E_{F} between discrete levels (n + 1/2)h̄Ω and a continuous ”free particle like” part h̄^{2}k^{2}/2M_{B}, suggesting electron transport along the 1D channel.

d) Since A has no y–component and the functions ϕ_{n}(y − L_{B}^{2}k) are real, we have j_{y} = 0. The x–component, however, is not zero:

j_{x} = \frac{h̄}{m}Im (e^{−ikx}ϕ_{n} \frac{∂}{∂x}e^{ikx}ϕ_{n}) − \frac{eyB}{m} ϕ_{n}^{2}

= (\frac{h̄k}{m} − \frac{eyB}{m} ) ϕ_{n}^{2}
= ω_{c} (l_{B}^{2}k − y) ϕ_{n}^{2}(y − L_{B}^{2}k).

We see that j_{x} > 0 if y < l_{B}^{2}k and j_{x} > 0 if y > l_{B}^{2}k.

e) The total probability current along the channel is

I_{x} = ∫ dyj_{x}(y) = ω_{c} ∫ dy(l_{B}^{2}k − y)ϕ_{n}^{2}(y − L_{B}^{2}k).

We take the hint given in the exercise and use the property

ϕ_{n}^{2}(y − L_{B}^{2}k) = ϕ_{n}^{2}(L_{B}^{2}k − y),

which must be true, since the harmonic oscillator functions have a certain parity, either ϕ_{n}(x) = ϕ_{n}(−x) or ϕ_{n}(x) = −ϕ_{n}(−x). Hence, if we rewrite I_{x} slightly,

I_{x} = ω_{c} ∫ dy [−(y − L_{B}^{2}k) − (L_{B}^{2} − l_{B}^{2})k] ϕ_{n}^{2}(y − L_{B}^{2}k),

we see that the first term gives an odd integrand which integrates to zero, whereas the second term is simply the constant −ω_{c}(L_{B}^{2} − l_{B}^{2})k multiplied by the normalization integral

∫ dyϕ_{n}^{2}(y − L_{B}^{2}k) = 1.

Hence, the total probability current is

I_{x} = ω_{c}(l_{B}^{2} − L_{B}^{2})k = … = \frac{h̄k}{m} \frac{ω_{0}^{2}}{Ω^{2}} .

Limiting cases:

1. I_{x}(ω_{0} → 0) = 0

2. I_{x}(ω_{c} → 0) = \frac{h̄k}{m}

These limits are both quite reasonable: When ω_{0} → 0, we have an infinite 2DEG, without any borders, and the electrons are simply spinning around without getting anywhere. In other words, I_{x} = 0. When ω_{c} → 0, the magnetic field is turned off, and we simply have electrons moving along the 1D channel with velocity v_{x} = h̄k/m.