UFABC - Física Quântica - Curso 2017.3 Prof. Germán Lugones
Aula 11
Soluções da equação de Schrödinger:
potencial degrau
2
Imaginemos um potencial com o perfil na figura abaixo, onde sua variação ocorre de maneira significativa somente num pequeno intervalo em torno de x = 0, sendo assim aproximadamente constante nos domínios I e II.
Esse potencial resulta em uma força impulsiva que atua somente num pequeno intervalo, −δ ≤ x ≤ δ. Fora desse intervalo o potencial nos domínios I e II é constante de modo que não há força resultante sobre a partícula.
Partícula em presença de um potencial degrau
•
x = 0 I II
x I II
F = d
3
No caso em que δ →0, o potencial adota a forma: V(x) = 0 para x < 0
V(x) = V0 para x > 0
As soluções da Eq. de Schrödinger são bastante diferentes dependendo de se:
- E > V0
4
Resolveremos primeiro o caso 1 (E > V0) com a partícula incidindo desde a
esquerda.
REGIÃO I: Temos x<0,V =0 e a equação a ser resolvida é:
ou E > V0 x = 0 V = ⇢ 0 x < 0 V0 , x > 0 I E x < 0 V = 0 ~2 2m d2 d x2 I (x) = E I (x) d2 d x2 I (x) = k 2 1 I (x) , k 2 1 = 2m E ~2 E > V0 x = 0 V = ⇢ 0 x < 0 V0 , x > 0 I E x < 0 V = 0 ~2 2m d2 d x2 I (x) = E I (x) d2 d x2 I (x) = k 2 1 I (x) , k 2 1 = 2m E ~2 Caso com E > V0
Nesse caso é conveniente escrever solução geral de funções oscilatórias com as formas exponenciais que representam onda incidente e onda refletida
A razão disso é porque ao se introduzir a dependência temporal multiplicando a autofunção pelo fator exponencial e−iEt/ħ = e−i!t para obter a
função de onda tem-se:
Dessa forma existem dois tipos de onda na região I: uma onda que se move para a direita com amplitude A; e, uma onda que se move para a esquerda com amplitude B.
I (x) = A eik1x + B e ik1x
= inc (x) + ref (x)
e ~iEt = e iwt
(x, t) = (x) e ~iEt = (x) e iwt
inc (x, t) = A eikx iwt ref (x, t) = B e ikx iwt
I A B II K = E V0 x > 0 V = V0 ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) I (x) = A eik1x + B e ik1x = inc (x) + ref (x) e ~iEt = e iwt (x, t) = (x) e ~iEt = (x) e iwt
inc (x, t) = A eikx iwt
ref (x, t) = B e ikx iwt
I A B II K = E V0 x > 0 V = V0 ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) I (x) = A eik1x + B e ik1x = inc (x) + ref (x) e ~iEt = e iwt (x, t) = (x) e ~iEt = (x) e iwt
inc (x, t) = A eikx iwt ref (x, t) = B e ikx iwt
I A B II K = E V0 x > 0 V = V0 ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) 5
REGIÃO II: partícula move-se para a direita com energia cinética K = E − V0
x>0, V =V0 e a equação a ser resolvida é
ou
Nesse caso é conveniente escrever a solução geral de funções oscilatórias com as exponenciais
Na região II as partículas movem-se para a direita de modo que D = 0.
I (x) = A eik1x + B e ik1x
= inc (x) + ref (x)
e ~iEt = e iwt
(x, t) = (x) e ~iEt = (x) e iwt
inc (x, t) = A eikx iwt ref (x, t) = B e ikx iwt
I A B II K = E V0 x > 0 V = V0 ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) d2 d x2 II (x) = k 2 2 II (x) , k 2 2 = 2m (E V0) ~2 II (x) = C eik2x + D e ik2x = C eik2x = tra (x) II D = 0 • I (0) = II (0) d I d x (0) = d II d x (0) d2 d x2 II (x) = k 2 2 II (x) , k 2 2 = 2m (E V0) ~2 II (x) = C eik2x + D e ik2x = C eik2x = tra (x) II D = 0 • I (0) = II (0) d I d x (0) = d II d x (0) 6
Usamos agora as condições de contorno:
Lembrando que:
as condições de contorno levam às seguintes relações: d2 d x2 II (x) = k 2 2 II (x) , k 2 2 = 2m (E V0) ~2 II (x) = C eik2x + D e ik2x = C eik2x = tra (x) II D = 0 • I (0) = II (0) d I d x (0) = d II d x (0) I (x) = A eik1x + B e ik1x II (x) = C eik2x A + B = C k1A k1B = k2C B = k1 k2 k1 + k2 A C = 2k1 k1 + k2 A B 6= 0 R T I II I II I (x) = A eik1x + B e ik1x II (x) = C eik2x A + B = C k1A k1B = k2C B = k1 k2 k1 + k2 A C = 2k1 k1 + k2 A B 6= 0 R T I II I II 7
Dessas expressões obtemos:
A solução mostra que de fato há onda refletida, i. e., B ≠ 0.
I (x) = A eik1x + B e ik1x II (x) = C eik2x A + B = C k1A k1B = k2C B = k1 k2 k1 + k2 A C = 2k1 k1 + k2 A B 6= 0 R T I II I II 8
Consideremos o feixe incidente constante e determinemos os coeficientes de reflexão R e de transmissão T.
Em ambos os domínios, I e II, as funções de onda descrevem partículas livres. A energia é conservada nos processos de reflexão e transmissão. Portanto, as frequências das funções em I e II devem ser as mesmas
inc (x, t) = A eik1x iwt ref (x, t) = B e ik1x iwt
tra (x, t) = C eik2x iwt
@ @ xj + @ @ t⇢ = 0 j = i ~ 2m ✓ ⇤ @ @ x @ @ x ⇤◆ ⇢ = ⇤ j II jII (x) = i ~ 2m ✓ ⇤ II @ @ x II II @ @ x ⇤ II ◆ = ~k2 m | II | 2 = ~k2 m | C | 2 ⇥
| C |2⇤ = [ ⇤ ] = # part´ıculas/uni. comprimento
SOLUTION
S OF
TIME-INDEPENDENT SCHRO
EDING
ER EQUATIONS
constant as in the bottom part of Figure 6-1. In the region x < 0 the wave function
is a combination of the incident traveling wave (of amplitude
A)moving to the right,
and a reflected traveling wave (of amplitude A/3) moving to the left. As the
ampli-tude of the reflected wave is necessarily smaller than that of the incident wave, the two
cannot combine to yield a pure standing wave. Their sum `Y(x,t) in that region is,
instead, something between a standing wave and a traveling wave. This is seen in the
behavior of `I'*(x,t) 11'(x,t) for x < 0, which looks like something between the pure
standing wave probability density of Figure 6-7 and the pure traveling wave
proba-bility density of Figure 6-1 in that it oscillates but has minimum values greater than
zero.
The ratio of the intensity of the reflected wave to the intensity of the incident wave
gives the probability that the particle will be reflected by the potential step back into
the region x < 0. This probability is the
reflection coefficient R.That is
R _
B*B_
k 1 — k2l
* (k1
— k2 k 1 — k 2 2 E > VA*A \k1 + k2) \k1 + k2) k 1 + k2) °
(6-40)
We see from this result that
R <1 when
E > Vo ,i.e., when the total energy of the
particle is greater than the height of the potential step. This is in contrast to the value
R =
1 when
E < V° ,that we obtained from the result of Section 6-3. Of course, the
thing that is surprising about the present result is not that
R <1, but that
R >0. It
is surprising because a classical particle would definitely not be reflected if it had
enough energy to pass the potential discontinuity. On the other hand, at a
corre-sponding discontinuity a classical wave would be reflected, as we shall discuss shortly.
Also of interest is the
transmission coefficient T,which specifies the probability that
the particle will be transmitted past the potential step from the region x < 0 into the
region x > 0. The evaluation of
Tis slightly more complicated than the evaluation
of
Rbecause the velocity of the particle is different in the two regions. According to
accepted convention, transmission and reflection coefficients are actually defined in
terms of the ratios of probability fluxes. A
probability fluxis the probability per second
that a particle will be found crossing some reference point traveling in a particular
direction. The incident probability flux is the probability per second of finding a
par-ticle crossing a point at x < 0 in the direction of increasing x; the reflected
proba-bility
flux is the probability per second of finding a particle crossing a point at x < 0
in the direction of decreasing x; and the transmitted probability flux is the probability
per second of finding a particle crossing a point at x > 0 in the direction of increasing
x. Since the probability per second that a particle will cross a given point is
pro-portional to the distance it travels per second, the probability flux is propro-portional not
only to the intensity of the appropriate wave but also to the appropriate velocity of
the particle. (A more detailed discussion of this point is given in connection with
Figure L-2 in Appendix L.) Thus, according to the strict definition, the reflection
co-efficient
R isR_ v 1 B B — B B
(6-41)
v1A*A A*A
where v 1 is the velocity of the particle in the region x < 0. Since the velocities cancel,
what remains is identical to the formula we have used previously for
R.For
T,the
velocities do
notcancel, and we have
v2 C*C
_
v2 (
2k1)2
_ T
v
1 A*A v l
I\
k l + k2 )v2 is the velocity of the particle in the region x > 0. Now
pi hk1 p2 hk2 v 1 =—=
and v2 =—_
m m
m m
SOLUTION S OF TIME-INDEPENDENT SCHRO EDING ER EQUATIONSconstant as in the bottom part of Figure 6-1. In the region x < 0 the wave function
is a combination of the incident traveling wave (of amplitude
A)moving to the right,
and a reflected traveling wave (of amplitude A/3) moving to the left. As the
ampli-tude of the reflected wave is necessarily smaller than that of the incident wave, the two
cannot combine to yield a pure standing wave. Their sum `Y(x,t) in that region is,
instead, something between a standing wave and a traveling wave. This is seen in the
behavior of `I'*(x,t) 11'(x,t) for x < 0, which looks like something between the pure
standing wave probability density of Figure 6-7 and the pure traveling wave
proba-bility density of Figure 6-1 in that it oscillates but has minimum values greater than
zero.
The ratio of the intensity of the reflected wave to the intensity of the incident wave
gives the probability that the particle will be reflected by the potential step back into
the region x < 0. This probability is the
reflection coefficient R.That is
R _
B*B_
k 1 — k2l
* (k1
— k2 k 1 — k 2 2 E > VA*A \k1 + k2) \k1 + k2) k 1 + k2) °
(6-40)
We see from this result that
R <1 when
E > Vo ,i.e., when the total energy of the
particle is greater than the height of the potential step. This is in contrast to the value
R =
1 when
E < V° ,that we obtained from the result of Section 6-3. Of course, the
thing that is surprising about the present result is not that
R <1, but that
R >0. It
is surprising because a classical particle would definitely not be reflected if it had
enough energy to pass the potential discontinuity. On the other hand, at a
corre-sponding discontinuity a classical wave would be reflected, as we shall discuss shortly.
Also of interest is the
transmission coefficient T,which specifies the probability that
the particle will be transmitted past the potential step from the region x < 0 into the
region x > 0. The evaluation of
Tis slightly more complicated than the evaluation
of
Rbecause the velocity of the particle is different in the two regions. According to
accepted convention, transmission and reflection coefficients are actually defined in
terms of the ratios of probability fluxes. A
probability fluxis the probability per second
that a particle will be found crossing some reference point traveling in a particular
direction. The incident probability flux is the probability per second of finding a
par-ticle crossing a point at x < 0 in the direction of increasing x; the reflected
proba-bility
flux is the probability per second of finding a particle crossing a point at x < 0
in the direction of decreasing x; and the transmitted probability flux is the probability
per second of finding a particle crossing a point at x > 0 in the direction of increasing
x. Since the probability per second that a particle will cross a given point is
pro-portional to the distance it travels per second, the probability flux is propro-portional not
only to the intensity of the appropriate wave but also to the appropriate velocity of
the particle. (A more detailed discussion of this point is given in connection with
Figure L-2 in Appendix L.) Thus, according to the strict definition, the reflection
co-efficient
R isR_ v 1 B B — B B
(6-41)
v1A*A A*A
where v 1 is the velocity of the particle in the region x < 0. Since the velocities cancel,
what remains is identical to the formula we have used previously for
R.For
T,the
velocities do
notcancel, and we have
v2 C*C
_
v2 (
2k1)2
_ T
v
1 A*A v l
I\
k l + k2 )v2 is the velocity of the particle in the region x > 0. Now
pi hk1 p2 hk2 v 1 =—=
and v2 =—_
m m
m m
SOLUTION S OF TIME-INDEPENDENT SCHRO EDING ER EQUATIONSconstant as in the bottom part of Figure 6-1. In the region x < 0 the wave function is a combination of the incident traveling wave (of amplitude A) moving to the right,
and a reflected traveling wave (of amplitude A/3) moving to the left. As the ampli-tude of the reflected wave is necessarily smaller than that of the incident wave, the two cannot combine to yield a pure standing wave. Their sum `Y(x,t) in that region is, instead, something between a standing wave and a traveling wave. This is seen in the behavior of `I'*(x,t) 11'(x,t) for x < 0, which looks like something between the pure standing wave probability density of Figure 6-7 and the pure traveling wave proba-bility density of Figure 6-1 in that it oscillates but has minimum values greater than zero.
The ratio of the intensity of the reflected wave to the intensity of the incident wave gives the probability that the particle will be reflected by the potential step back into the region x < 0. This probability is the reflection coefficient R. That is
R _
B*B _ k 1 — k2l
* (k1
— k2 k 1 — k 2 2 E > VA*A \k1 + k2) \k1 + k2) k 1 + k2) ° (6-40)
We see from this result that R < 1 when E > Vo , i.e., when the total energy of the
particle is greater than the height of the potential step. This is in contrast to the value
R = 1 when E < V° , that we obtained from the result of Section 6-3. Of course, the
thing that is surprising about the present result is not that R < 1, but that R > 0. It is surprising because a classical particle would definitely not be reflected if it had enough energy to pass the potential discontinuity. On the other hand, at a corre-sponding discontinuity a classical wave would be reflected, as we shall discuss shortly. Also of interest is the transmission coefficient T, which specifies the probability that
the particle will be transmitted past the potential step from the region x < 0 into the region x > 0. The evaluation of T is slightly more complicated than the evaluation
of R because the velocity of the particle is different in the two regions. According to
accepted convention, transmission and reflection coefficients are actually defined in terms of the ratios of probability fluxes. A probability flux is the probability per second
that a particle will be found crossing some reference point traveling in a particular direction. The incident probability flux is the probability per second of finding a par-ticle crossing a point at x < 0 in the direction of increasing x; the reflected proba-bility flux is the probability per second of finding a particle crossing a point at x < 0
in the direction of decreasing x; and the transmitted probability flux is the probability per second of finding a particle crossing a point at x > 0 in the direction of increasing x. Since the probability per second that a particle will cross a given point is pro-portional to the distance it travels per second, the probability flux is propro-portional not only to the intensity of the appropriate wave but also to the appropriate velocity of the particle. (A more detailed discussion of this point is given in connection with Figure L-2 in Appendix L.) Thus, according to the strict definition, the reflection co-efficient R is
R_ v 1 B B — B B
(6-41)
v1A*A A*A
where v 1 is the velocity of the particle in the region x < 0. Since the velocities cancel, what remains is identical to the formula we have used previously for R. For T, the
velocities do not cancel, and we have
v2 C*C
_
v2 ( 2k1 )2_ T
v 1 A*A v l
I\
k l + k2 )v2 is the velocity of the particle in the region x > 0. Now
pi hk1 p2 hk2 v 1 =—= and v2 =—_ m m m m SOLUTION S OF TIME-INDEPENDENT SCHRO EDING ER EQUATIONS
constant as in the bottom part of Figure 6-1. In the region x < 0 the wave function is a combination of the incident traveling wave (of amplitude A) moving to the right,
and a reflected traveling wave (of amplitude A/3) moving to the left. As the ampli-tude of the reflected wave is necessarily smaller than that of the incident wave, the two cannot combine to yield a pure standing wave. Their sum `Y(x,t) in that region is, instead, something between a standing wave and a traveling wave. This is seen in the behavior of `I'*(x,t) 11'(x,t) for x < 0, which looks like something between the pure standing wave probability density of Figure 6-7 and the pure traveling wave proba-bility density of Figure 6-1 in that it oscillates but has minimum values greater than zero.
The ratio of the intensity of the reflected wave to the intensity of the incident wave gives the probability that the particle will be reflected by the potential step back into the region x < 0. This probability is the reflection coefficient R. That is
R _
B*B _ k 1 — k2l
* (k1
— k2 k 1 — k 2 2 E > VA*A \k1 + k2) \k1 + k2) k 1 + k2) ° (6-40)
We see from this result that R < 1 when E > Vo , i.e., when the total energy of the
particle is greater than the height of the potential step. This is in contrast to the value
R = 1 when E < V° , that we obtained from the result of Section 6-3. Of course, the
thing that is surprising about the present result is not that R < 1, but that R > 0. It
is surprising because a classical particle would definitely not be reflected if it had enough energy to pass the potential discontinuity. On the other hand, at a corre-sponding discontinuity a classical wave would be reflected, as we shall discuss shortly. Also of interest is the transmission coefficient T, which specifies the probability that
the particle will be transmitted past the potential step from the region x < 0 into the region x > 0. The evaluation of T is slightly more complicated than the evaluation
of R because the velocity of the particle is different in the two regions. According to
accepted convention, transmission and reflection coefficients are actually defined in terms of the ratios of probability fluxes. A probability flux is the probability per second
that a particle will be found crossing some reference point traveling in a particular direction. The incident probability flux is the probability per second of finding a par-ticle crossing a point at x < 0 in the direction of increasing x; the reflected proba-bility flux is the probability per second of finding a particle crossing a point at x < 0
in the direction of decreasing x; and the transmitted probability flux is the probability per second of finding a particle crossing a point at x > 0 in the direction of increasing x. Since the probability per second that a particle will cross a given point is pro-portional to the distance it travels per second, the probability flux is propro-portional not only to the intensity of the appropriate wave but also to the appropriate velocity of the particle. (A more detailed discussion of this point is given in connection with Figure L-2 in Appendix L.) Thus, according to the strict definition, the reflection co-efficient R is
R_ v 1 B B — B B
(6-41)
v1A*A A*A
where v 1 is the velocity of the particle in the region x < 0. Since the velocities cancel, what remains is identical to the formula we have used previously for R. For T, the
velocities do not cancel, and we have
v2 C*C _ v2 ( 2k1 )2
_ T
v 1 A*A v l
I\
k l + k2 )v2 is the velocity of the particle in the region x > 0. Now
pi hk1 p2 hk2
v 1 =—= and v2 =—_
m m m m
10
So the above expression gives
T _ k 2 (2k 1 )2 4kik 2
k2)2 = k2)2 E > Vo (6-42) cn
k 1 (k1 + k2) (k1 + k2) cD
It is easy to show by evaluating R and T from (6-40) and (6-42) that o>
R + T = 1 (6-43)
This useful relation is the motivation for defining the reflection and transmission co-efficients in terms of probability fluxes.
The probability flux incident upon the potential step is split into a transmitted flux and a reflected flux. But (6-43) says their sum equals the incident flux; i.e., the proba-bility that the particle is either transmitted or reflected is one. The particle does not
vanish at the step; nor does the particle itself split at the step. In any particular trial the particle will go one way or the other. For a large number of trials, the average probability of going in the direction of decreasing x is measured by R, and the
aver-age probability of going in the direction of increasing x is measured by T.
Note that R and T are both unchanged in value if k 1 and k2 are exchanged in (6-40)
and (6-42). A moment's consideration should convince the student that this means the same values of R and T would be obtained if the particle were incident upon the
potential step in the direction of decreasing x from the region x > 0. The wave func-tion describing the mofunc-tion of the particle, and consequently the probability flux, is partially reflected simply because there is a discontinuous change in V(x), and not
because V(x) becomes larger in the direction of the incidence of the particle. The
be-havior of R and T when k 1 and k2 are exchanged involves a characteristic property
of all waves that, in optics, is sometimes called the reciprocity property. When light
passes perpendicularly through a sharp interface between media with different indices of refraction, a fraction of the light is reflected because of the abrupt change in its wavelength, and the same fraction is reflected independent of whether it is incident from one side of the interface or from the other. Exactly the same thing happens when a microscopic particle experiences an abrupt change in its de Broglie wavelength. In fact, the equations governing the two phenomena are identical in form. We see, once again, that a microscopic particle moves in a wavelike manner.
In Figure 6-11 the reflection and transmission coefficients are plotted as functions of the convenient ratio E/Vo . By evaluating k 1 and k2 in (6-40) and (6-42), we find
that these expressions for the reflection and transmission coefficients can be written in terms of the ratio as
R=1 —T— 1 — — Vo /E 2 — E > 1 (6-44) —
1 + 1Vo/ E) ^ Vo
0.5 1.0 1.5 2.0
E/V0
Figure 6-11 The reflection and transmission coefficients R and T for a particle incident
upon a potential step. The abscissa E/Vo is the ratio of the total energy of the particle to the increase in its potential energy at the step. The case k 1 = 2k2 , illustrated in Figure 6-10, corresponds to E/Vo = 1.33. THE S TE P PO TENTIAL (E NE RGY GREATER T HAN S TEP HEI GHT )
Os coeficientes R e T são definidos pelas razões
onde e
são as velocidades da partícula nas regiões I e II respectivamente. Logo:
A medida que E cresce k2 → k1 e o coeficiente R vai a zero.
Em suma, os coeficientes R e T dão as probabilidades de que uma partícula de energia E > V0 seja refletida ou transmitida.
E k2 ! k1 R
R T
E > V0
I | |2
A diferença agora é que no domínio II a equação de Schrödinger é
ou
onde o fator que multiplica ψII(x) do lado direito na equação diferencial acima
é positivo. Caso com E < V0 E < V0 II ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) , d2 d x2 II (x) = 2 2 II (x) , 2 = p 2m (V0 E) ~ , II (x) II (x) = C e 2x + D e2x limx!1 II (x) = 0 D = 0 II (x) = C e 2x E < V0 II ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) , d2 d x2 II (x) = 2 2 II (x) , 2 = p 2m (V0 E) ~ , II (x) II (x) = C e 2x + D e2x limx!1 II (x) = 0 D = 0 II (x) = C e 2x 12
Com isso a solução geral é
A condição de convergência limx→∞ψII(x) = 0 implica que D = 0. Assim, a
autofunção decresce exponencialmente.
A solução na região I tem a mesma forma que no caso anterior, portanto: E < V0 II ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) , d2 d x2 II (x) = 2 2 II (x) , 2 = p 2m (V0 E) ~ , II (x) II (x) = C e 2x + D e2x limx!1 II (x) = 0 D = 0 II (x) = C e 2x E < V0 II ~2 2m d2 d x2 II (x) + V0 II (x) = E II (x) , d2 d x2 II (x) = 2 2 II (x) , 2 = p 2m (V0 E) ~ , II (x) II (x) = C e 2x + D e2x limx!1 II (x) = 0 D = 0 II (x) = C e 2x I (x) = A eik1x + B e ik1x II (x) = C e 2x I (0) = II (0) d I d x (0) = d II d x (0) A + B = C ik1 (A B) = 2C ) A B = i 2 k1 C A B C 13
As condições de contorno
agora fornecem
Essas duas relações permitem escrever A e B em termos de C
I (x) = A eik1x + B e ik1x II (x) = C e 2x I (0) = II (0) d I d x (0) = d II d x (0) A + B = C ik1 (A B) = 2C ) A B = i 2 k1 C A B C I (x) = A eik1x + B e ik1x II (x) = C e 2x I (0) = II (0) d I d x (0) = d II d x (0) A + B = C ik1 (A B) = 2C ) A B = i 2 k1 C A B C A = C 2 ✓ 1 + i2 k1 ◆ B = C 2 ✓ 1 i2 k1 ◆ C (x) = 8 < : C 2 ⇣ 1 + i2k1⌘eik1x + C2 ⇣1 i2k1⌘e ik1x, x 0 C e 2x, x 0 (x) e ~iEt = e iwt (x, t) = (x) e ~iEt (x, t) = (x) e iwt = ( C 2 ⇣ 1 + i2 k1 ⌘ eik1x iwt + C2 ⇣1 i2 k1 ⌘ e ik1x iwt, x 0 C e 2x iwt, x 0 x 0 14
Assim, a função de onda fica:
Multiplicando de ψ(x) pelo fator e−iEt/ħ = e−i!t tem-se a função de
onda Ψ(x,t) = ψ(x) e−iEt/ħ : 15 A = C 2 ✓ 1 + i2 k1 ◆ B = C 2 ✓ 1 i2 k1 ◆ C (x) = 8 < : C 2 ⇣ 1 + i2k1⌘ eik1x + C2 ⇣1 i2k1⌘ e ik1x, x 0 C e 2x, x 0 (x) e ~iEt = e iwt (x, t) = (x) e ~iEt (x, t) = (x) e iwt = ( C 2 ⇣ 1 + i2 k1 ⌘ eik1x iwt + C2 ⇣1 i2 k1 ⌘ e ik1x iwt, x 0 C e 2x iwt, x 0 x 0 A = C 2 ✓ 1 + i2 k1 ◆ B = C 2 ✓ 1 i2 k1 ◆ C (x) = 8 < : C 2 ⇣ 1 + i2 k1 ⌘ eik1x + C2 ⇣1 i2 k1 ⌘ e ik1x, x 0 C e 2x, x 0 (x) e ~iEt = e iwt (x, t) = (x) e ~iEt (x, t) = (x) e iwt = ( C 2 ⇣ 1 + i2 k1 ⌘ eik1x iwt + C2 ⇣1 i2 k1 ⌘ e ik1x iwt, x 0 C e 2x iwt, x 0 x 0
O coeficiente de reflexão R é:
Portanto, não há partículas transmitidas: T = 1 – R = 1 – 1 = 0.
No entanto, como a função de onda é não nula em x>0 há uma probabilidade não nula de se encontrar a partícula nessa região que seria proibida pela mecânica clássica.
16 SOLU TIONS OF TIME-IND EPEND ENT SCHROEDING ER EQ UATIO NS
is obtained by equating these derivatives at x = 0. Thus we set
—k 2D(e -k2x)x0 — ik 1 A(eik1 x)x=0 — ik1B(e iktx)x=0
This yields
Adding (6-20) and (6-21) gives
Subtracting gives
^ \ k 2 /
B = —
1
- -l
We have now determined
A, B,and
Cin terms of
D.Thus the eigenfunction for the
step potential, and for the energy
E < V0 ,is
tŸ(x) = -2
(1 + ik2/kl)eiklx +
D
(1 —
ik2/k1)e'lx
x
< 0 (6-24)
De -k2 x x>-
0
The one remaining arbitrary constant,
D,determines the amplitude of the
eigen-function, but it is not involved in any of its more important characteristics. The
presence of this constant reflects the fact that the time-independent Schroedinger
equation is linear in ifr(x), and so solutions of any amplitude are allowed by the
equation. We shall see that useful results can usually be obtained without bothering
to carry through the normalization procedure that would specify
D.The reason is
that the measurable quantities that we shall obtain as predictions of the theory
con-tain
Din both the numerator and the denominator of a ratio, and so it cancels out.
The wave function corresponding to the eigenfunction is
A eikix e - iEt/h
+
Be iklx e iEt/h = Ae i(k ix-Et/h) + Bei(- klx-Et/h) x <O
W(x,t) =
De - k2x e - iEt/h
x > 0 (6-25)
Consider the region x < 0. The first term in the wave function for this region is a
trav-eling wave propagating in the direction of increasing x. This term describes a particle
moving in the direction of increasing x. The second term in the wave function for x <
0 is a traveling wave propagating in the direction of decreasing x, and it describes
a particle moving in that direction. This information, plus the classical predictions
described earlier, suggests that we should associate the first term with the incidence
of the particle on the potential step and the second term with the reflection of the
particle from the step. Let us use this association to calculate the probability that
the incident particle is reflected, which we call the
reflection coefficient R.Obviously,
R
depends on the ratio
B/A,which specifies the amplitude of the reflected part of the
wave function relative to the amplitude of the incident part. But in quantum
mechan-ics probabilities depend on intensities, such as
B*Band
A*A,not on amplitudes.
Thus, we must evaluate
Rfrom the formula
B*B =
A*
(6-26)
That is, the reflection coefficient is equal to the ratio of the intensity of the part of
the wave that describes the reflected particle to the intensity of the part that describes
the incident particle. We obtain
R — B*B
— (1 — ik2 /kl) *(1 — ik2/kl)
A*A (1 + ik 2/k i)*(1 + ik2/ki)
k2 D =A—B ^ ^\ k 2
/
A = — 1+ i(6-21)
(6-22)
(6-23)
xy
7
4\
A
^ A^ À
\o
^ THE STE P P OTENTIAL (ENERGY LESS THAN S TEP H EIGHT )Figure 6-6 Illustrating schematically the combination of an incident and a reflected wave of
equal intensities to form a standing wave. The wave function is reflected from a potential step at x = O. Note that the nodes of the traveling waves move to the right or left, but those of the standing wave are stationary.
or
R — (1
+
ik 2/k i )(1 — ik2/ki) _ 1 E < Vo (6-27)(1 — ik2/ki)( 1 + ik2/ki)
The fact that this ratio equals one means that a particle incident upon the. potential step, with total energy less than the height of the step, has probability one of being reflected—it is always reflected. This is in agreement with the predictions of classical mechanics.
Consider now the eigenfunction of (6-24). Using the relation
eiktx = cos k i x + i sin kix (6-28)
it is easy to show that the eigenfunction can be expressed as
D cos k l x—Dk2 sin k ix x<0
t//(x) = De-k2x k1 (6-29)
x>- 0
If we generate the wave function by multiplying /i(x) by a - `E:m, we see immediately
that we actually have a standing wave because the locations of the nodes do not change in time. In this problem the incident and reflected traveling waves for x < 0 combine to form a standing wave because they are of equal intensity. Figure 6-6 depicts this schematically.
In the top part of Figure 6-7 we illustrate the wave function by plotting the eigen-function, (6-29), which is a real function of x if we take D real. The wave function
can be thought of as oscillating in time according to e - iEt/J, with an amplitude whose
space dependence is given by 0(x). Here we find a feature which is in sharp contrast to the classical predictions. Although in the region x > 0 the probability density
^*^ = D* e k 2 x e +iEt^^ e -k2x e -iEtlh = D*De 2k2x (6-30)
illustrated in the bottom of Figure 6-7, decreases rapidly with increasing x, there is
a finite probability of finding the particle in the region x > 0. In classical mechanics it would be absolutely impossible to find the particle in the region x > 0 because there the total energy is less than the potential energy, so the kinetic energy p 2/2m
