0 Bound State O
Attractive delta-function type potential Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: November 26, 2013)
We discuss the 1D barrier problem in the presence of an attractive Dirac-delta function-type potential. This provides a simple model system for the bound state and transmission and reflection. When the energy of the particle is negative, we have a bound state. When the energy of the ;particle is positive, there occurs the transmission and reflection for the incident particle. For the bound state (the energy of the particle is negative), the wave function should be either an even function or an odd function, since the delta function potential is an even function. This property is discussed in terms of the parity operator.
1. Bound state (E<0) in a delta function potential
V x Bound state
O x
E <0
Fig. Bound state of the system (having the negative energy E<0) in the presence of an attractive delta-function type potential.
We now consider a attractive potential of the form of Dirac delta function,
V (x) V0 (x) .
We find a bound state for the particle energy E (<0). The Schrödinger equation of this system is given by
2 d 2 (x) V (x) (x) E (x) . 2m dx2 0
Note that the unit of V0 is [erg cm], but not [erg]. Integrating this equation between - and +. Letting approach 0, we show that the derivative of the eigenfucntion (x) presents a discontinuity at x = 0.
2 d 2 (x)dx V (x) (x)dx E (x)dx , 2 0 2m dx or
2 d (x) 2 d (x) ( ) ( ) V (0) (x)dx E (x)dx , x x 0 2m dx 2m dx
d (x) d (x) 2m ( ) x ( ) x 2 V0 (0) 2 (0) , (boundary condition) dx dx
We use the parameter as
mV0 2 .
The unit of is [cm-1]. Then the boundary condition is given by
(i) ( ) ( ) .
d (x) d (x) (ii) | | 2 (0) . dx x dx x
We assume that the energy E is negative (bound state). The Schrödinger equation is given by
2 d 2mE 2 2 (x) 2 (x) (x) , dx where
2 2mE 2 (>0).
For x<0, the solution for (x) is given by
(.x) Aex
For x>0,
(.x) Bex
Since the potential energy V(x) is an even function, the wave function should be either even or odd function. When the wave function is an odd function, the wave function should take a zero at the origin.
(i) The case when the wave function is an even function of x
2mE A B , 2 , or
2 2 2 mV0 E 2 , (bound state). 2m 2 with
mV0 2 .
The unit of a is [cm-2]. From the normalization condition, we have
(.x) e x
(ii) The case when the wave function is an odd function of x
A B 0 .
So there is no trivial solution.
((Note)) Probability current density The probability current density is defined by
1 j Re[ *(x) (x)]. x m i x
Suppose that the wave function is a real function of x. Then jx should be equal to zero as shown above for the case of bound state case). In fact, the wave function is given by a real function.
2. Green's function: modified Helmholtz equation We solve the problem (the bound state) using the Green function (see the APPENDIX). We start with the Schrodinger equation for the bound state
2 d 2 2 2 (x) V (x) (x) E (x) (x) , 2m dx 2 0 2m where
2 2 E (<0), 2m
2 d 2 2m ( 2 ) (x) Lx (x) 2 V0 (x) (x) f (x) , dx where
2m f (x) 2 V0 (x) (x) 2 (x) (x) , and
mV0 2 .
Using the 1D Green’s function (modified Helmholtz), the solution of this equation can be given by
(x) G(x, ) f ( )d 1 e|x | 2 ( ) ( )d 2 (0)e|x| where
1 G(x, ) e|x | . (Green’s function: modified Helmholtz equation). 2
At x = 0, we get
(0) (0) .
Then we have
2 2 , E . 2m
The normalized wave function is
(x) e|x| .
3. Transmission of a "delta function" potential barrie for E>0
V x
a1 a2 b1 E >0 O x
Fig. The transmission and reflection of free particle (with the positive energy E) in the presence of an attractive delta-function type potential.
For x<0, we assume that
ikx ikx I a1e b1e , where k is defined by
2k 2 E (>0). 2m
The probability current density is
J k( a 2 b 2 ) . I m 1 1
For x>0, we assume that
ikx II a2 e .
The probability current density is
J k a 2 . II m 2
From the continuity of the wave function at x = 0;
a1b1 a2 . (1)
From the condition
d (x) d (x) ( ) ( ) 2 (0) , dx x dx x we get
ika2 ik(a1 b1) 2a2 , or
2 a b (1 i )a , (2) 1 1 k 2
Then we have
a (1 i )a , b i a . 1 k 2 1 k 2
The transmissivity is
2 2 k a2 2 m a2 1 k T 2 2 2 2 . 2 k a1 k a1 1 m k 2
The reflectivity is
2 2 2 2 k b1 2 m b1 a2 b1 R 2 2 2 2 2 . 2 k k a a1 a1 a2 m 1 where
T R 1.
The wave functions are obtained as
ikx ikx I a1e b1e [(1 i )eikx i eikx ]a k k 2 2 [eikx sin(kx)]a k 2
ikx II a2 e , where
mV0 V 0 . k 2mE 2E
((Mathematica)) We make a plot of the wave function in the regions I and II, where k and a are chosen appropriately.
Re y x
x
Fig. k = 5. a = 5. a2 3. Transmission and reflection of the free particle under the influence ikx of a Diract delta function type attractive potential (denoted by black). fincident Re[a1e ] ikx (green, x<0), freflected Re[b1e ] (blue, x<0), ftotal fincident freflected (red, x<0), and ikx ftransmitted Re[a2e ] (red, x>0).
4. Bound state under attractive double-delta potential We discuss the eigenvalue (the bound state) problem for the Schrodinger equation for the attractive double-delta potential, V x
Bound state
O x
E <0
-V0
2 d 2 2 2 (x) V [ (x a) (x a)] (x) E (x) (x) 2m dx 2 0 2m where the potential energy is an even function of x,
V (x) V0 [ (x a) (x a)] and the energy is given by
2 2 E (<0) 2m for the bound state. Then we get
d 2 ( 2 ) (x) L (x) 2[ (x a) (x a)] (x) f (x) , dx2 x where
mV0 2 .
Using the 1D Green’s function (modified Helmholtz), the solution of this equation can be given by
(x) G(x, ) f ( )d 1 e|x | 2[ (x a) (x a)] ( )d 2 [ (a)e|xa| (a)e|xa| ] where
LxG(x, ) (x ) ,
d 2 2 G(x, ) (x ) , 2 dx
1 G(x, ) e|x | . (Green’s function: modified Helmholtz equation). 2
At x = a
(a) [ (a) (a)e2a ] .
At x = -a
(a) [ (a)e2a (a)].
We note that the wave function (x) is either an even function or an odd function of x.
(i) Even parity for the wave function
e (a) e (a) , or
1 (e2a 1) , or
1 e2a .
For simplicity we put
x 2a ,
x 1 e x . 2a
y 2.0 q=0.1
1.5 q=0.5 x q=1 1.0 y= -x 2 q y=1+e
0.5
0.0 x 0.0 0.5 1.0 1.5 2.0 2.5 3.0
Fig. Even parity solution. Plot of y and y 1 e where q is changed as a parameter. 2q 2a . q a . The two curves intersect at a non-zero finite value of . When q = 1, we have = 2.21772. y = 1.10886.
The wave function (even parity) is given by
(x) e [e|xa| e|xa| ]. e (a)
When q a = 1, we have x 2a =2.21772. For a = 1 and = 1.10886, we make a plot of this function as a function of x. Clearly this function is an even function of x.
y x y a=1
1.5
1.0
0.5
x -4 -2 2 4
(x) Fig. Plot of e as a function of x. 1. a = 1 and = 1.10886. e (a 1)
(ii) Odd parity for the wave function
o (a) o (a) or
1 (e2a 1) or
e2a 1
The value of is obtained from the intersection of two curves
y , y 1 e 2q with 2a and q a (parameter).
y 2.0 q=0.1
1.5 q=0.5 x q=1 1.0 y= 2 q y=1-e-x
0.5
0.0 x 0.0 0.5 1.0 1.5 2.0 2.5 3.0
Fig. Odd parity solution. Plot of y and y 1 e where q is changed as a parameter. 2q 2a . q a .When q>0.5, there are two solutions including = 0. When q<0.5, there is only one solution = 0. When q = 1, we have = 1.59362. y = 0.796812.
The wave function (odd parity) is given by
(x) o [e|xa| e|xa| ]. o (a)
When q a = 1, we have 2a =1.59362. For a = 1 and = 0.796812, we make a plot of this function as a function of x. Clearly this function is an odd function of x.
y x y a=1
1.0
0.5
x -4 -2 2 4
-0.5
-1.0
(x) Fig. Plot of o as a function of x. 1. a = 1 and = 1.10886. o (a 1)
______REFERENCE Gordon Baym, Lectures on Quantum Mechanics (Westview Press, 1990). Claude Cohen-Tannoudji, Bernard Diu, and Franck Laloë, Quantum Mechanics, volume I and volume II (John Wiley & Sons, New York, 1977). David. Bohm, Quantum Theory (Dover Publication, Inc, New York, 1979). Leonard Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955). A. Messiah, Quantum Mechanics, vol.I and vol.II (North-Holland, 1961). R. Dick, Advanced Quantum Mechanics Materials and Photons (Springer 2012).
APPENDIX Green’s function: modified Helmholtz equation We construct the 1D Green's function for the modified Helmholtz equation
d 2 L k 2 . x 2 dx
Suppose that
Lx f (x) .
Using the Green’s function we get
G(,x, ) f ( )d
L L G(x, ) f ( )d (x ) f ( )d f (x) . x x
Thus we need G(x, ) which satisfies
LxG(x, ) (x ) ,
d 2 k 2 G(x, ) (x ) , 2 dx 1 G(x, ) ek|x | , 2k
We consider the case at = 0,
d 2 k 2 G(x) (x) , 2 dx
1 G(q) G(x)eiqxdx , 2
1 G(x) G(q)eiqxdq , 2
1 (,x) eiqxdq 2
1 d 2 1 k 2 G(q)eiqxdq eiqxdq , 2 2 dx 2 or
1 1 q2 k 2 G(q)eiqxdq eiqxdq . 2 2
Then we have
1 1 1 (q2 k 2 )G(q) , or G(q) , 2 2 q2 k 2
1 eiqx G(x) dq . 2 q2 k 2
(i) For x > 0,
1 eiqx 1 G(x) dq ekx , 2 q2 k 2 2k C1
where C1 is in the lower-half plane (counter-clock wise). There is a simple pole of z = i k inside the contour C1. We use the Residue theorem.
y
G1
C1 x>0
ik
x -¶O ¶
(ii) For x < 0,
1 eiqx 1 G(x) dq ex , 2 q2 k 2 2k C2
where C2 is in the lower-half plane (clock-wise). There is a simple pole of z = -i k inside the contour C2. We use the residue theorem. y
-¶O ¶ x
-ik
x<0 C2
G2
In summary we have the Green's function
1 G(x) ek|x| , 2k or
1 G(x, ) ek|x | . 2k