15D (Quantum Mechanics)

i ℏ \frac{\partial ψ}{\partial t} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ + U ψ, - i ℏ \frac{\partial ψ ^{*}}{\partial t} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ^{*} + U ψ^{*} .

(1)

Because the potential

U

is real, its two contributions cancel when differentiating the probability density

ρ = ψ^{*} ψ

. Therefore

\frac{\partial ρ}{\partial t} = ψ^{*} \frac{\partial ψ}{\partial t} + ψ \frac{\partial ψ ^{*}}{\partial t} = \frac{i ℏ}{2 m} (ψ^{*} \nabla^{2} ψ - ψ \nabla^{2} ψ^{*}) = - \nabla \cdot [- \frac{i ℏ}{2 m} (ψ^{*} \nabla ψ - ψ \nabla ψ^{*})] .

(2)

The expression in square brackets is the probability current

J

, so

\frac{\partial ρ}{\partial t} + \nabla \cdot J = 0 .

(3)

This probability continuity equation says that probability can leave a region only through the current across its boundary.

b

 0  0

i

 0  0

Solution

 0  0

Write the spatial factors of the stationary wavefunction as

ψ (x, t) = e^{- i Et /ℏ} ⎩ ⎨ ⎧ e^{ik x} + R e^{- ik x}, C e^{k x} + D e^{- k x}, T e^{ik x}, x < - a, - a < x < a, x > a,

(1)

where

k = \frac{2 m E}{ℏ} .

(2)

Since

U_{0} - E = E

, the decay constant inside the potential barrier is also

k

. Continuity of the wavefunction and its first derivative at

x = - a

and

x = a

gives four linear equations. Solving them yields

T = \frac{e ^{- 2 ika}}{c o s h ( 2 ka )} .

(3)

Consequently the transmitted wave is

ψ_{tr} (x, t) = \frac{exp ( i [ k ( x - 2 a ) - Et /ℏ ] )}{cosh ( 2 ka )}, x > a,

(4)

and its probability density is

ρ_{tr} (x, t) = ∣ ψ_{tr} ∣^{2} = sech^{2} (2 ka) .

(5)

This is finite square barrier transmission at half barrier height.

ii

 0  0

Solution

 0  0

For a plane wave

A e^{i (k x - Et /ℏ)}

, the probability current is

J = \frac{ℏ k}{m} ∣ A ∣^{2} .

(1)

The incident amplitude is one and the transmitted amplitude is

T

, so

\frac{J _{tr}}{J _{in}} = ∣ T ∣^{2} = sech^{2} (2 ka) .

(2)

This is the transmission probability for quantum tunnelling. In the stationary state,

\partial ρ / \partial t = 0

, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

\frac{∣ J _{refl} ∣}{J _{in}} = 1 - sech^{2} (2 ka) = tanh^{2} (2 ka) .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 15D (Quantum Mechanics)

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