The Delta-Function Potential - Physics &...
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The Delta-Function Potential As our last example of one-dimensional bound-state solutions, let us re-examine the finite potential well: and take the limit as the width, a, goes to zero, while the depth, V 0 , goes to infinity keeping their product aV 0 to be constant, say U 0 . In that limit, then, the potential becomes: ( ) ( ) x U x V δ 0 − = and we can have a sense that if there is at least one bound state of this potential, it should look (as drawn above) like the ground state of the finite square well. Let’s examine the TISE for this potential: x V(x) x = -a/2 x = a/2 V = -V 0 x V(x) x = -a/2 x = a/2 V = -V 0 x V(x) x V(x)
Transcript of The Delta-Function Potential - Physics &...
The Delta-Function Potential As our last example of one-dimensional bound-state solutions, let us re-examine the finite potential well: and take the limit as the width, a, goes to zero, while the depth, V0, goes to infinity keeping their product aV0 to be constant, say U0. In that limit, then, the potential becomes: ( ) ( )xUxV δ
0−=
and we can have a sense that if there is at least one bound state of this potential, it should look (as drawn above) like the ground state of the finite square well. Let’s examine the TISE for this potential:
x
V(x)
x = -a/2 x = a/2
V = -V0
x
V(x)
x = -a/2 x = a/2
V = -V0
x
V(x)
x
V(x)
( )
( )
otherwise2
0at x2
2
2
22
02
22
02
22
ψψ
ψψδψ
ψψδ
ψψ
Exm
ExUxm
ExUxm
EH
=∂∂−
==−∂∂−
⇒=⎥⎦
⎤⎢⎣
⎡−
∂∂−
⇒=
h
h
h
)
Let’s first take a look at the region outside of x = 0. Here, we can rewrite the TISE as:
h
h
h
mE
mEx
Exm
2
22
222
2
2
22
−=
=−
=∂∂
⇒=∂∂−
κ
ψκψψ
ψψ
Here, kappa is real since the energy is negative. This simple differential equation has solutions we know: ( ) xx BeAex κκψ += −
For negative x (left side of the potential), ψ will blow up as x goes to negative infinity, so A must be zero. On the right side the same thing happens and there, the constant B must be zero. So, we have:
( )⎪⎩
⎪⎨⎧
><
= − 00
xAexBe
x x
x
κ
κ
ψ
At x = 0, the wavefunction must be continuous, so A = B and we have:
( )⎪⎩
⎪⎨⎧
≥≤
= − 00
xAexAe
x x
x
κ
κ
ψ
Now, we must use the information at x = 0. In the region near x = 0, the TISE gives us:
( ) ψψδψ ExUxm
=−∂∂−
02
22
2h
we integrate both sides with respect to x over an infinitesimally small region around the delta potential, say from –ε to +ε:
( )
( ) ( ) ( )
( ) ( ) ( ) ( )∫
∫∫∫
−−
−−−
=−⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂−
∂∂−
⇒=−∂
∂−
ε
εεε
ε
ε
ε
ε
ε
ε
ψψψψ
ψψδψ
dxxEUxx
xx
m
dxxEdxxxUdxx
xm
02
2
0
2
02
22
h
h
Now, let’s let ε go to zero. Then the right hand side of the equation will go to zero since ψ is finite and we integrate it over a zero width. This gives us:
( ) ( ) ( )02
20 ψψψ
εε h
mUxx
xx −
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂−
∂∂
−
Now, from above (and letting epsilon go to zero):
( )
( )
( )
( ) κψ
κψ
κψ
κψ
κ
κ
Axx
eAxx
Axx
eAxx
x
x
=∂
∂
⇒=∂
∂<
−=∂
∂
⇒−=∂
∂>
−
−
0
0
0, for x
0, for x
So that
( ) ( ) ( )02
2 20 ψκκκψψ
εε h
mUAAA
xx
xx −
=−=−−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂−
∂∂
−
Now, ψ(0) = A, so:
20
20
22
h
hmU
AmU
A
=
⇒−
=−
κ
κ
So that there is one, and only one allowed energy level:
2
20
20
2
2
h
hh
mUE
mEmU
−=
⇒−
==κ
To be complete, we can find the constant A via the normalization condition:
( )
h
0
2
0
222 12
mUA
AdxeAdxx x
==
⇒=== ∫∫∞
−∞
∞−
κ
κψ κ
Then there is one and only one bound state, and one energy eigenvalue:
( ) 2
200
22
0
hhh
mUEe
mUx x
mU
−==− ,ψ
