Download Relativity Problem Set 9 - Solutions Prof. J. Gerton October 23, 2011

Relativity Problem Set 9 - Solutions
Prof. J. Gerton
October 23, 2011
Problem 1
(10 pts.) The quantum harmonic oscillator
(a) The Schroedinger equation for the ground state of the 1D QHO is
~2 ∂ 2
mω 2 x2
−
+
ψ(x) = E0 ψ(x).
2m ∂x2
2
(1)
Evaluating the kinetic operator gives
−
~2
~2 ∂ 2
ψ(x)
=
B (1 − 2B x2 ) ψ(x),
2m ∂x2
m
If we want the Schroedinger equation to be satisfied it must be
2
2~2 B 2 m ω 2
~ B
2
−
+
− E0 = 0.
x +
m
2
m
(2)
(3)
This is a polynomial equation, in which each coefficient of the polynomial in the
variable x must equate zero. The coefficient of x2 gives
B=
mω
.
2~
(4)
E0 =
1
~ω,
2
(5)
The other equation then gives
which is the correct form for the ground state energy.
(b) We impose the normalization condition
Z +∞
dx |ψ(x)|2 = 1,
(6)
−∞
which leads to
A
2
Z
+∞
2
dx e−2B x = 1.
−∞
1
(7)
2
To evaluate the integral we use the result for Gaussian integrals
Z +∞
√
2
dy e−y = π,
(8)
−∞
√
together with the substitution y = x 2B that leads to
Z +∞
dy −y
A2 √
2
√
A
e =√
π = 1,
2B
2B
−∞
or
A=
Problem 2
2B
π
1/4
=
m ω 1/4
π~
.
(9)
(10)
(10 pts.) Expectation values for a harmonic oscillator
(a) Since ψ1 (x) is an odd function, p̂ψ1 (x) = −i~∂x ψ1 (x) is even. Then the integrand
(ψ1 (x)) (p̂ψ1 (x)) is a product of an even function and an odd function, so it is
odd. The integral is thus zero because it is the integral of an odd function over
an even domain.
(b) With the same reasoning, the integrand (ψ1 (x)) (x̂ψ1 (x)) is odd in x thus its
integral over the even domain specified is zero.
(c) We need to compute
Z
+∞
hK̂i =
−∞
3
3
ψ1 (x) ( − β x2 ) ~ω ψ1 (x) = ~ ω.
2
4
(11)
This result is expected, because of the following fact: we know that the total
energy must be E1 = 3/2~ω, and that for the virial theorem applied to the
harmonic field we have
hKi = hV i.
(12)
So,
E1 = hKi + hV i = 2hKi,
and
hKi =
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1
3
E1 = ~ω.
2
4
(13)
(14)
PS #
3
Problem 3
(10 pts.) The Evanescent Wave
p
p
(a) The momentum in the region x > 0 is k = 2m(E − V0 )/~ =
i
p 2m|E − V0 |/~,
the last equality holding because V0 > E. We introduce κ = 2m|E − V0 |/~, so
that k = iκ and the wave function is
ψ(x) = C e−κx ,
(15)
with C a constant.
(b) The number of particles dN within the positions x and x + dx is given by
dN = |ψ(x)|2 dx.
(16)
So, in the case discussed in part (a), the probability density to find the particle
at the point x > 0 is
|ψ(x)|2 = e−2κx .
(17)
This probability, though small, is non-zero, so that there is a certain small probability that the particle be at some x > 0. This probability decays exponentially
with x, so further away from the origin the probability drops to zero.
(c) This probability is given by the reflection coefficient,
p − k 2 p − iκ 2
=
R = p + iκ .
p + k
Now, computing this quantity gives
p − iκ 2 p − iκ p + iκ
=
R = = 1.
p + iκ p + iκ p − iκ
(18)
(19)
So, for any value of κ and p the reflection coefficient is one.
(d) The answers to (b) and (c) are in contrast, because in (b) one finds that there
is a transmitted current in x > 0 and in (c) one finds that all of the incident
particles are reflected. This ambiguity is solved by thinking of the definition for
the transmitted wave,
Jtransmitted 2
,
(20)
T = Jincident where J is a current. So, the expression for T is
T = T0 e−2κx ,
(21)
with T0 a constant. This expression holds for large x 1/κ and is particularly
good for E V0 , where T 1. So, in these cases the transmission coefficient is
much smaller than R, and energy conservation is not violated. However, in the
region x . 1/κ or for E ∼ V0 this approximation is no longer true.
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4
Problem 4
(10 pts.) The double well - Part 1
(a) Since the potential is infinite in these regions, the wave function is zero.
(b) In the regions [−a, −b] and [a, b], the potential is V (x) = 0. Energy conservation
then gives
√
p2
= E, or p = 2mE.
(22)
2m
In the region [−b, b] we have V (x) = V0 , with the condition V0 > E. Indicating
the momentum in this region with p0 , the energy conservation relation is thus
p02
+ V0 = E,
2m
or
p02
= E − V0 < 0.
2m
(23)
So, in the last equation, the right hand side is always negative, by assumption.
This means that the momentum has to be imaginary, as suggested in the text.
We write p0 = iκ, with κ a real quantity with the dimension of a momentum. If
we do so, the kinetic energy is
(iκ)2
κ2
p02
=
=−
.
2m
2m
2m
(24)
The term E − V0 , on the other hand, can be written as E − V0 = −|E − V0 |,
because we know that E − V0 < 0. Here, |x| indicates the absolute value of x.
Finally, putting together these expressions, we find the expression for κ ss
−
p
κ2
= −|E − V0 | or κ = 2m|E − V0 |.
2m
(25)
Compare the expression for κ with the expression for p above.
(c) In classical mechanics, the particle cannot stay in the region [−b, b] because this
would violate the conservation of energy constraint. In fact, since the kinetic
energy has to be always positive in classical mechanics, from KE + V (x) = E
and KE ≥ 0 we obtain
E − V (x) ≥ 0.
(26)
The expression above yields the domain x where the particle can move in the
classical picture. In our case, this domain is the union of the two regions [−a, −b]
and [a, b], but not the region [−b, b].
In quantum mechanics instead, the wave function extends in the classically forbidden region as well, so there is a certain probability to find the particle in this
region. This phenomenon is peculiar of quantum mechanics, and gives rise to
cute effects like the quantum tunneling, which explains the α-decay in nuclei or
how the tunnel diode works, among many other things.
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5
Problem 5
(10 pts.) The double well - Part 2
(a) We begin with the region [−a, −b], where the momentum is p. The wave function
is then
ψ1 (x) = C1 e−ipx + C2 eipx ,
(27)
with C1 and C2 some normalization constants to be found. In the region [−b, b],
the momentum is iκ and the wave function is then
ψ2 (x) = D1 e−i(iκ)x + D2 ei(iκ)x = D1 eκx + D2 e−κx
(28)
with D1 and D2 constants to be fixed. Finally, in the region [b, a] the momentum
is p and the wave function is
ψ3 (x) = C10 e−ipx + C20 eipx ,
(29)
with C10 and C20 some normalization constants which are different from C1 and
C2 . Summing up, the wave function is oscillating in the classically allowed region
and exponentially suppressed or enhanced in the classically forbidden region.
(b) Imposing the symmetry of the wave function ψ(−x) = ψ(x) allows us to get rid of
three of the six constants above. In the region [−b, b], which is already symmetric,
we have
ψ2 (x) = D1 eκx + D2 e−κx ,
(30)
and
ψ2 (−x) = D1 e−κx + D2 eκx .
(31)
Imposing ψ2 (−x) = ψ2 (x) in this region gives D1 = D2 , so that the wave function
is
ψ2 (x) = D1 e−κx + eκx .
(32)
We can use the function cosh(x) = (e−x + ex )/2, so that finally
ψ2 (x) = 2D1 cosh(κx).
(33)
To impose the symmetrization in the other regions, we impose that the wave
function ψ1 (x) in [−a, −b] being equal to the wave function ψ3 (−x) in [b, a], that
is, the wave function is symmetric in the symmetric domain [−a, −b] + [b, a]. We
have
ψ1 (x) = C1 e−ipx + C2 eipx , in [−a, −b],
(34)
and
ψ3 (−x) = C10 eipx + C20 e−ipx ,
in [b, a].
(35)
Imposing the condition ψ3 (−x) = ψ1 (x) gives C1 = C20 and C1 = C10 . Thus, if
we decide to get rid of the constants C10 and C20 , the symmetrization gives the
expression for the wave function in [b, a] as
ψ3 (x) = C2 e−ipx + C1 eipx ,
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(36)
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while the wave function ψ1 (x) in [−a, −b] is unchanged.
Imposing boundary conditions, we first impose that the wave function in x = −a
is zero. Thanks to the symmetrization, this already includes the condition that
the wave function being zero at x = a. At x = −a we have
0 = C1 e−ip(−a) + C2 eip(−a) = C1 eipa + C2 e−ipa .
(37)
This gives the relation
C2 = −C1 e2ipa .
(38)
We impose the continuity at x = −b. This means that the expressionψ(x) =
C1 e−ipx + C2 eipx , valid in the region [−a, −b], must be equal to the expression
ψ(x) = 2D1 cosh(κx), valid in the region [−b, b], when x = −b. Imposing this, we
have
C1 e−ip(−b) + C2 eip(−b) = 2D1 cosh(κ(−b)).
(39)
This gives the relation
C1 eipb + C2 e−ipb = 2D1 cosh(κ(−b)) = 2D1 cosh(κb),
(40)
where in the last step we use the fact that the function cosh(x) = cosh(−x), as
required by the symmetry.
Imposing now the continuity of the derivative, we have
Since
∂ψ1 (x)
= ∂ψ2 (x)∂x at x = −b.
∂x
(41)
∂ψ1 (x)
= −ip C1 e−ipx + ip C2 eipx ,
∂x
(42)
and
∂ψ2 (x)
= 2D1 κ sinh(κx),
∂x
the continuity of the derivative at x = −b yields
−ip C1 e−ip(−b) + ip C2 eip(−b) = 2D1 κ sinh(κ(−b)).
(43)
(44)
Using sinh(−x) = − sinh(x), we finally obtain the third relation
−ip C1 eipb + ip C2 e−ipb = −2D1 sinh(κb).
(45)
(c) We rewrite the three conditions below:
C2 = −C1 e2ipa
continuity at x = −a,
C1 eipb + C2 e−ipb = 2D1 cosh(κb) continuity at x = −b,
(46)
(47)
−ip C1 eipb +ip C2 e−ipb = −2D1 κ sinh(κb) continuity of the derivative at x = −b.
(48)
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Plugging the first relation in the second gives
C1 eipa eipb−ipa − e−ipb+ipa = 2D1 cosh(κb),
(49)
while plugging the first relation in the third gives
−ip C1 eipa eipb−ipa + e−ipb+ipa = −2D1 κ sinh(κb).
(50)
Now, since
eipb−ipa + e−ipb+ipa = 2 cos p(a − b),
(51)
eipb−ipa − e−ipb+ipa = −2i sin p(a − b),
(52)
and
the two relations above are
−2iC1 eipa sin p(a − b) = 2D1 cosh(κb),
(53)
−2ip C1 eipa cos p(a − b) = −2D1 κ sinh(κb).
(54)
and
Dividing these two relations gives
2C1 eipa sin p(a − b)
2D1 cosh(κb)
=
,
2p C1 eipa cos p(a − b)
−2D1 κ sinh(κb)
or
κ tan p(a − b) = −
p
.
tanh κb
(55)
(56)
(d) In terms of the given quantities, this is
tan
h√
s
i
hp
i
2mE(a − b) tanh
2m|E − V0 |b = −
E
.
|E − V0 |
(57)
This condition gives the value of the energy E, which then results quantized. The
above relation is thus a quantization requirement.
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