QM of a particle in a 1D Dirac delta potential
A particle in a Dirac delta potential
and scattering states^[yet to be dirac normalized]
where
Proof
We begin with bound states (i.e.
). Letting πΈ < 0 the SchrΓΆdinger equation for π = β β 2 π πΈ / β reads π₯ β 0 π 2 π π π₯ 2 = π 2 π which has the general solution
. Applying the boundary conditions π ( π₯ ) = Λ π΄ π π π₯ + Λ π΅ π β π π₯ and continuity of π ( Β± β ) = 0 we conclude π . Integrating the complete SchrΓΆdinger equation over π ( π₯ ) = π΅ π β π | π₯ | gives ( β π , π ) πΈ β« π β π π ( π₯ ) π π₯ = β β 2 2 π β« π β π π 2 π π π₯ 2 π π₯ β πΌ β« π β π πΏ ( π₯ ) π ( π₯ ) π π₯ = β β 2 2 π [ π 2 π π π₯ 2 ] π₯ = π π₯ = β π β πΌ π ( 0 ) and taking the limit
gives π β 0 β 2 π πΌ β 2 π΅ = π π π π₯ ( 0 + ) β π π π π₯ ( 0 β ) = 2 π π΅ hence
and π = β π πΌ / β 2 πΈ = β β 2 π 2 2 π = β π πΌ 2 2 β 2 Normalization yields
1 = | π΅ | 2 β« β β β π β 2 π | π₯ | π π₯ = 2 | π΅ | 2 β« β 0 π β 2 π π₯ π π₯ = | π΅ | 2 π hence
. π΅ = β π = β π πΌ / β For scattering states (
), let πΈ β₯ 0 . The SchrΓΆdinger equation for π = β 2 π πΈ / β thence becomes π₯ β 0 π 2 π π π₯ 2 = β π 2 π it follows
π ( π₯ ) = { Λ π΄ π π π π₯ + Λ π΅ π β π π π₯ π₯ β€ 0 Λ πΉ π π π π₯ + Λ πΊ π β π π π₯ π₯ β₯ 0 where continuity requires
. The derivatives are π΄ + π΅ = πΉ + πΊ π π π π₯ ( 0 β ) = π π β π΄ β π΅ ) π π π π₯ ( 0 β ) = π π ( πΉ β πΊ ) hence
β 2 π πΌ β 2 ( π΄ β π΅ ) = π π ( πΉ β πΊ β π΄ + π΅ ) which may be reΓ€rranged to
πΉ β πΊ = π΄ ( 1 + 2 π π½ ) β π΅ ( 1 β 2 π π½ ) where
π½ = π πΌ / β 2 π
Properties
- The reflection and transmission coΓ«fficients (regardless of which side the particle enters) for scattering states are
which do not depend on the sign of
Proof of 1β2
Since the velocities of the particle are equal on either side of the potential, it is sufficient to compare the coΓ«fficients of the unnormalized scattering state. Since the setup is symmetric, without loss of generality let the particle scatter from the left, so
. Thus πΊ = 0 corresponds to the incident wave amplitude, π΄ to the reflected wave, and π΅ to the transmitted wave. πΉ π΅ = π π½ 1 β π π½ π΄ πΉ = 1 1 β π π½ π΄ thus
π = | π΅ | 2 | π΄ | 2 = π½ 2 1 + π½ 2 = 1 1 + ( 2 β 2 πΈ / π πΌ 2 ) π = | πΉ | 2 | π΄ | 2 = 1 1 + π½ 2 = 1 1 + ( 2 π πΌ 2 / 2 β 2 πΈ ) as claimed by ^P1. Note that these are unchanged for negative
. πΌ First we compute the necessary derivatives
π π π₯ π ( π₯ ) = β π πΌ β { π πΌ β 2 π π πΌ π₯ / β 2 π₯ β€ 0 β π πΌ β 2 π β π πΌ π₯ / β 2 π₯ β₯ 0 = ( π πΌ β 2 ) 3 / 2 ( Ξ ( β π₯ ) π π πΌ π₯ / β 2 β Ξ ( π₯ ) π β π πΌ π₯ / β 2 ) π 2 π π₯ 2 π ( π₯ ) = ( π πΌ β 2 ) 3 / 2 ( β πΏ ( π₯ ) ( π π πΌ π₯ / β + π β π πΌ π₯ / β ) + π πΌ β 2 Ξ ( β π₯ ) ( π π πΌ π₯ / β 2 + Ξ ( π₯ ) π β π πΌ π₯ / β 2 ) ) = ( π πΌ β 2 ) 3 / 2 ( β 2 πΏ ( π₯ ) + π πΌ β 2 π β π πΌ | π₯ | / β 2 ) where
is the Heaviside function. Thus Ξ β¨ π | Λ π₯ | π β© = π πΌ β 2 β« β β β π₯ π β 2 π πΌ | π₯ | / β 2 β __ β __ β o d d π π₯ = 0 β¨ π | Λ π₯ 2 | π β© = π πΌ β 2 β« β β β π₯ 2 π β 2 π πΌ | π₯ | / β 2 β __ β __ β e v e n π π₯ = 2 π πΌ β 2 β« β 0 π₯ 2 π β 2 π πΌ π₯ / β 2 π π₯ = 2 π πΌ β 2 ( 2 ! ) ( β 2 2 π πΌ ) 3 = β 4 2 π 2 πΌ 2 β¨ π | Λ π | π β© = β π β β« β β β π ( π₯ ) π π π₯ π ( π₯ ) π π₯ = β π β π πΌ β 2 β« β β β ( Ξ ( β π₯ ) β Ξ ( π₯ ) ) π ( π₯ ) 2 β ____ β ____ β o d d π π₯ = 0 β¨ π | Λ π 2 | π β© = β β 2 β« β β β π ( π₯ ) π 2 π π₯ 2 π ( π₯ ) π π₯ = β ( π πΌ β ) 2 β« β β β π β π πΌ | π₯ | / β 2 ( β 2 πΏ ( π₯ ) + π πΌ β 2 π β π πΌ | π₯ | / β 2 ) π π₯ = ( π πΌ β ) 2 [ 2 β π πΌ β 2 2 β« β β β π β 2 π πΌ | π₯ | / β 2 π π₯ ] = ( π πΌ β ) 2 [ 2 β π πΌ β 2 β 2 2 π πΌ ] = ( π πΌ β ) 2 [ 2 β π πΌ β 2 β 2 π πΌ ] = ( π πΌ β ) 2 proving ^P2.
Footnotes
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2018. Introduction to quantum mechanics, Β§2.5.2, pp. 63ff. β©