Download Quantum Tunneling and Transmission Coefficients and more Study notes Physics in PDF only on Docsity!
Quantum Tunneling
In this chapter, we discuss the phenomena which allows an electron to quan- tum tunnel over a classically forbidden barrier.
9 eV 10 eV
99% of time Rolls over 10 eV
1% of the time 10 eV
Rolls back
This is a strikingly non-intuitive process where small changes in either the height or width of a barrier create large changes the tunneling current of particles crossing the barrier. Quantum tunneling controls natural phenomena such as radioactive α decay where a factor of three increase in the energy released during a decay is responsible for a 10^20 fold increase in the α decay rate. The inherent sensitivity of the tunneling process can be exploited to produce photographs of individual atoms using scanning tunneling microscopes (STM) or produce extremely rapid amplifiers using tunneling diodes. It is an area of physics which is as philosophically fascinating as it is technologically important.
Most of this chapter deals with continuum rather than bound state systems. In bound state problems, one is usually concerned with solving for possible sta- tionary state energies. In tunneling problems, one has a continuous spectrum of possible incident energies. In these problems we are generally concerned with solving for the probability that an electron is transmitted or reflected from a given barrier in terms of its known incident energy.
Quantum Current
Tunneling is described by a transmission coefficient which gives the ratio of the current density emerging from a barrier divided by the current density incident on a barrier. Classically the current density J~ is related to the charge
density ρ and the velocity charge velocity ~v according to J~ = ρ ~v. Its natural to relate the current density ρ with the electron charge e and the quantum PDF(x) according to ρ = eΨ∗(x)Ψ(x). It is equally natural to describe the velocity by ˆp/m where (in 3 dimensions) ˇp = −i¯h(∂/∂x) → −i¯h∇~. Of course ∇~ is an operator which needs to operate on part of ρ. Recalling this same issue from our discussion on Quantum Measurement we expect:
J^ ~ ∼ e mΨ
This form isn’t totally correct but fairly close as we shall see.
The continuity equation which relates the time change of the charge density to the divergence of the current density, provides the departure point for the proper derivation of the quantum current.
∂ρ ∂t +^
∇ ·~ J~ = 0 (2)
By integrating both sides of the continuity current over volume (d^3 x) and using Gauss’s theorem, one can show that the continuity equation is really just an elegant statement of charge conservation or the relationship between the rate of change of the charge within a surface and the sum of the currents flowing out of the surface. ∂ ∂t
V
d^3 x ρ +
V
d^3 x ∇ ·~ J~ = 0
0 = ∂Q∂t +
S
d~a · J~ = ∂Q∂t + Iout (3)
Rather than talking about the charge and current current density; one often removes a factor of e and talks about the probability density (PDF = ρ) and probability current. We know that the probability density is given by just ρ = Ψ∗Ψ and can get a formula like the continuity equation by some simple, but
In computing the current using Eq. (9) one must consider both the time dependence as well as the space dependence. In order to produce a non-vanishing current density, the wave function must have a position dependent phase. Otherwise, the phase of Ψ(x, t) will be the same as the phase of ∇~Ψ(x, t) and therefore Ψ∗^ ∇~Ψ will be real. The current density for an electron in a stationary state of the form Ψ(x, t) = ψ(x) exp (−iωt) is zero since the phase dependence has no spatial dependence. This makes a great deal of sense since the PDF of a stationary state is time independent which indicates no charge movement or currents. A combination of two stationary states with different energies such as: Ψ(x, t) = a ψ 1 (x) exp (−iω 1 t) + b ψ 2 (x) exp (−iω 2 t) will have a position dependent phase , an oscillating PDF, and a non-zero current density which you will explore in the exercises.
To reinforce the idea that a position dependent phase is required to support a quantum current, consider writing the wave function in polar form ψ(x) = |ψ(x)| exp(iφ(x)) where we have a real modulus function |ψ(x)| and a real phase function φ(x). Using the chain rule it is easy to show that:
J^ ~ = ¯h m|ψ(x)|
(^2) ∇~φ(x) (10)
Hence the quantum current is proportional to the gradiant of the phase – a constant phase implies no current.
A particularly simple example of a state with a current flow is a quantum traveling wave of the form: Ψ(x, t) = a exp (ikx − iωt). Direct substitution of this form into Eq. (9) or (10) gives us:
J^ ~ = (a∗a) ¯hk m xˆ^ (11)
A Strategy For Solving Tunneling Problems
We will limit ourselves to one-dimensional tunneling through a various po- tential barriers. An important consequence of working in one dimension is that
the current must be the same at every point along the x-axis since there is no where for the charges to go. We can insure this automatically by using a single, stationary state wave function corresponding to a particle with a definite energy to describe the current flow everywhere. Let us see why this works. In one dimension, the (probability) continuity equation becomes:
∂ ∂t {Ψ
∗Ψ} + ∂J
∂x = 0^ ,^
∂t {Ψ
∗Ψ} = 0 for a stationary state → ∂J ∂x = 0 (12)
Eq. (12) implies that J is independent of position, and since it is constructed from a stationary state wave function, Eq. (9) tells us that J is independent of time. We thus automatically get a constant current with the same value everywhere along the x axis.
How do we find the stationary state wave function? Usually we choose a constant (often zero) potential region on the left of any barriers to “start” a wave function of the form ψ = exp(ikx) + r exp(−ikx) where k =
2 mE/¯h. We think of the exp(ikx) piece as the incident wave which travels along the positive x axis and the r exp(−ikx) piece as the reflected wave. One can show† that the total current in this zero potential region (or any other region) is J = ¯hk/m − |r|^2 ¯hk/m. We can think of the total current as the algebraic sum of the incident and reflected currents where each contribution is computed by Eq. (11). One then uses continuity of the wave function and derivative continuity to find the unknown r coefficient. The result of the calculation is generally expressed by a reflection coefficient R ≡ |r|^2 which is equivalent to R = |Jr|/|Ji| where Jr is the reflected current due to r exp(−ikx) piece and Ji is the incident current due to the exp(ikx) piece. Our goal is to calculate R as a function of E which determines the k value we start with.
If it turns out that R < 1, there will be a net current at x = +∞ (and everywhere else) which we will call the transmitted current or Jt. This will be a
† In homework you show that the interference between the incident and reflected parts of the wave function carries no current which is far from obvious without an explicit calculation
∂x e
ik 1 x+ ∂ ∂x r e
−ik 1 x (^) = ∂ ∂x t e
ik 2 x|x=0 → ik 1 −ik 1 r = ik 2 t → 1 −r = k^2 k 1 t^ (14)
Adding Eq. (13) and Eq. (14) we get
2 = 1 + k k^2 1 t → t = (^) k^2 k^1 1 +^ k 2
From 1 + r = t we can find r = 1 − (^) k^2 k^1 1 +^ k 2 → r = k k^1 −^ k^2 1 +^ k 2
We note that k 1 =
2 mE/¯h and k 2 =
2 m(E − V )/¯h which means for the step up curb: k 1 > k 2 and r > 0. If the curb were a step down curve r < 0.
We turn next to the R and T coefficients. These are not the relative ampli- tudes t and r, but rather are the ratio of the currents carried by the transmitted or reflected waves over the incident wave. Following Eq. (11) we have:
T = (t
∗t) k 2 /m k 1 /m =^
4 k 1 k 2 (k 1 + k 2 )^2
R = (r
∗r) k 1 /m k 1 /m =
(k 1 − k 2 )^2 (k 1 + k 2 )^2
Using algebra you can show from Eq. (17) and Eq. (18) that T + R = 1 as we expect. The current would not be conserved if we had (incorrectly) written the transmission coefficient as the just ratio of the transmitted over incident squared moduli (T = t∗t) rather than the correct expression T = t∗t k 2 /k 1. The formula for the reflection and transmission coefficients for a light wave passing from air to glass is exactly the same as Eq.(17) and Eq. (18) which one gets from classical electrodynamics.
We can use k 1 =
2 mE/¯h and k 2 =
2 m(E − V )/¯h to write the transmis- sion and reflection coefficients in terms of the dimensionless ratio E/V.
T = 4
E/V
E/V − 1
E/V +
E/V − 1
) 2 ,^ R^ =
E/V −
E/V − 1
E/V +
E/V − 1
The below figure shows a sketch of the R and T coefficient as a function of E/V.
through step barrier
Reflection and Transmission
1. E/V
T
T
R
1 R
The above plot suggests that once E < V there is 100 % reflection and 0 % transmission. This case is formally discussed in one of the exercises, but is reasonably easy to understand. If E < V , x > 0 becomes a classically forbidden region with an exponential wave function of the form ψ = te−βx^ = |t| eiδ^ e−βx.
ei k x
V=
E
V
x=
ψ = t e βx
The current J~ = (¯h/m)Im
Ψ∗^ ∇~Ψ
must vanish in this region since the com- plex phase is a constant (δ) independent of x. Informally there is no transmission
different complex phase between these two contributions, the complex phase will develop a position dependence and the classically forbidden region will carry a current, implying there will be a current in the x < 0 region as well and hence R < 1, T > 0 and there will be a current at infinity.
Another way of looking at this is based on the fact that the current according to Eq. (9) is proportional to the wave function as well as its derivative. If there is just a classically forbidden region past x > 0, the wave function will die out to zero and there will be no possibility of a non-zero ψ at infinity. Since, J ∝ ψ∗^ ∂ψ∂x according to Eq. (9), this means there can be no quantum current at x → ∞ and thus no net current anywhere. If , however, we restore the potential back to ground, we can “catch” the dying wave function before it totally decays away, and have thus have ψ 6 = 0 at infinity and thus a current everywhere.
We have crudely sketched the PDF = ψ∗ψ of the electron in the limit of low transmission. We can estimate the transmission coefficient in this limit by making use of the classically forbidden, quantum curb PDF’s discussed in the last section. The PDF’s in region #1 , #2, and #3 will be
PDF 1 ≈ 2+2 cos(2kx−δ) ; PDF 2 ≈ |c|^2 e−^2 βx^ ; PDF 3 = |teikx|^2 = |t|^2 (20b)
The region #1 PDF is approximate since we set |r| = 1 whereas |r| is slightly less than 1 in the low transmission limit. The region #2 PDF is approximate since we threw away the exp(βx) piece that must be present in region #2 to convey the current but it should be small in this limit. We can now estimate T = |t|^2
by matching the approximate PDF’s at x = 0 and x = a.
PDF 1 (0) = 2 + 2 cos(δ) = f (E/V ) = PDF 2 (0) = |c|^2 → |c|^2 = f (E/V )
We write PDF 1 (0) = f (E/V ) since the phase δ is a function of E/V for the classically forbidden quantum current.
PDF 2 (a) = f (E/V )e−^2 βa^ = PDF 3 (a) = |t|^2 → T = f (E/V )e−^2 βa^ (20c)
This is indeed the correct form of the exact solution when βa 1 given by Eq. (26) that we discuss in the next section. We will show that f (E/V ) is a relatively smooth function that is approximately 1 meaning that the transmission coefficient is roughly T ≈ exp − 2 βx where β =
2 m(V − E)/¯h. As we will argue later this means that the tranmitted current is very sensitive to very small (atomic scale) changes in a and the forbidden gap V − E. Here is an illustration. Consider an energy gap of V − E = 2 eV. This is typical of the work function of metals which forms the barrier preventing metal electrons from escaping into space. The β corresponding to this work function is:
β =
2 m(V − E) ¯h =
2 mc^2 (V − E) ¯hc =
2(0. 511 × 106 eV )(2 eV ) 197 eV nm = 7.^26 nm
− 1
Now consider varying the tunneling length a from a = 0. 25 nm to a = 0. 20 nm, the ratio of the tunneling current is:
T (0.20) T (0.25) =^
f (E/V ) exp(−2(7.26)(0.20)) f (E/V ) exp(−2(7.26)(0.25)) =^ e
To put this into perspective, we found that changing the tunneling length by the radius of a hydrogen atom (0.05 nm) changes the tranmission coefficient or tunneling current by 210 %. This extreme sensitivity of tunneling to distance changes on the scale of atomic dimensions forms the foundation of the STM or scanning tunneling microscope that we will describe later.
for the transmission coefficient:
T =
1 + sinh
(^2) (βa) 4(E/V )(1 − E/V )
where β =
2 m(V − E) ¯h (25)
The reflection coefficient follows from R = 1 − T. To get some insight into this we will go to various limits.
The βa 1 limit
In this limit sinh β → 12 eβa^ 1. This means that sinh^2 ( )/[ ] dominates the expression 1 + sinh^2 ( )/[ ] and Eq. (25) becomes:
T ≈ 16(E/V )(1 − E/V ) e−^2 βa^ (26)
This expression agrees with the form we deduced in Eq. (20c) in the T 1 limit.
The Delta function limit
First a few words about δ-functions in case you haven’t encountered them before. A δ function is a function with an infinitesimal width and an infinite height but a unit area. A force described as δ-function in time such as F (t) = δ(t) is known as a unit impulse which occurs at time t = 0. As you probably know from both mechanics and circuit theory; it is often relatively easy to describe a the behavior of a circuit or mechanical system to a voltage or force impulse. The same is true of quantum mechanical systems.
In quantum mechanics we often think of the a δ-function potential. We can think of this potential as a rectangular function of width w and height h in the limit that w → 0 , h → ∞ , and wh = 1 although there are many other limiting forms which approach the δ-function as well. The δ-function centered at x = 0 is written as δ(x). To shift the δ-function to the right so that it centers on xo we translate the function by subtracting xo from its argument or δ(x − xo).
o h
0 x o
f(x )
x
δ( x) δ(^ x-x o
w
)
The operational definition of the δ-function is as follows: ∫
x∈xo
dx f (x) δ(x − xo) = f (xo) (28)
In words, the integral of the product of f (x) × δ(x − xo) over any interval con- taining the point xo is just the function evaluated at xo. Its easy to see how our rectangular representation of δ(x − xo) has this property.
∫
x∈xo
dx f (x) δ(x − xo) = limw→ 0
xo∫+w/ 2
xo−w/ 2
dx h × f (x) = wh × f (xo) = f (xo) (29)
δ-function potentials in the Schr¨odinger Equation
We write the time independent Schr¨odinger Equation for the case of a δ- function potential of strength g or V (x) = g δ(x − xo). In writing this, we note that dimensions of strength g are energy × distance (eg eV · nm) since the dimensions of δ(x − xo) are distance−^1 in order that
dx δ(x − xo) = 1 (dimensionless).
−¯h^2 2 m
∂^2 ψ ∂x^2 +^ g δ(x^ −^ xo)ψ^ =^ Eψ^ (30)
If we restrict ourselves to the region in an infinitesimal neighborhood of xo, it is
have 4(E/V )(1 − E/V ) → 4 E/V since V → ∞. Hence
T =
1 +^2 ma
2 (V − E)V
4 E¯h^2
1 + ma
2 V 2
2 E¯h^2
Inserting aV = g and E = ¯h^2 k^2 / 2 m we have:
T =
1 + m
(^2) g 2 ¯h^4 k^2
Classically allowed barrier We next consider the case of a traveling wave incident on a classically allowed barrier with E > V as illustrated below.
a/ e
0 0
c + d
**-a/
k
e^ -
1 = 2m (E - V)h
x -ikx t ei k x
k = k = (^) h 2 m E h
i k x
2 m E
e
E
ik 1 x ik 1 x
V
The transmission coefficient for this barrier is:
T =
1 + sin
(^2) (k 1 a) 4(E/V )(E/V − 1)
where k 1 =
2 m(E − V ) ¯h (35)
The real difference between this case and the classically forbidden case is the use of c exp (ik 1 x) + d exp (−ik 1 x) rather than c exp (βx) + d exp (−βx) for the wave function in the 0 < x < a region. Essentially the exponential argument β → ik 1. We note that we can get to Eq. (35) from the forbidden T in Eq. (31) by the substitution sinh iβ → i sin k 1 which describes how a hyperbolic sine of an imaginary number is related to the usual sine.
We note from the form of Eq. (35), that we have perfect tranmission whenever k 1 a = nπ , n = 1, 2 , .... This condition is equivalent to n(λ 1 /2) = a where λ 1
is the electron wavelength in the barrier region. Here is a crude sketch of the
transmission coefficient as a function of E/V :
1 E/V
T
The phenemona of 100 % tranmission through a barrier at specific magic en-
ergies or wavelengths is often called transmission resonance. Examples occur
in both atomic and nuclear physics. In atomic physics , one has the Ramsauer
effect (discovered 1908) where noble gas atoms become nearly transparent to sev- eral volt electrons of of specific energies. A very similar phenomena, known as
“size resonance”. occurs for several MeV neutrons which can pass transparently
through the nucleus at resonant energies.
Transmission resonance at magic wavelengths also occurs in reflections of
electromagnetic waves from thin films as shown below. We have angled the incident ray a bit for clarity but we will discuss the case of normal incidence.
Rate (sec -1)
Po
0.3 0.
1010
1
10
10 -20 U^238
212
( MeV -1/2) Q
1
George Gamow in the 1930’s proposed a quantum tunneling explanation for nuclear α decay. In this model, the α particle is initially bound by the strong in- teraction within a nuclear well created by the Thorium nucleus to form Uranium. The Uranium decays by having the α particle tunnel through the Coulomb bar- rier of the Thorium nucleus. As depicted below, the gap by which the tunneling is classically forbidden decreases as the energy release (Q) increases. The de- cay rate is proportional to the transmission coefficient through the barrier which depends exponentially on the energy gap.
U^23292 α Th 90228
Q = 5 MeV
V
t u^ n^ n e l
7 fm
260 MeV fm r
r
Nuclear α decay is controlled by very, small tranmission coefficients. At the other extreme, we consider the nearly free motion of conduction electrons
through a metal. Although electrons are bound within individual metal atoms on
quantum levels of the atomic discrete spectrum, they exhibit nearly free motion
in a metal crystal where there is regularly spaced lattice of ions on a spacing of
a few tenths of a nanometer.
severaltenths nm
We can think of the loosely bound valence electrons in outer orbitals as
jumping from ion to ion by quantum tunneling through fairly weak barriers owing
to the narrow width and small depth of the effective interatomic barrier. We will
discuss this in depth on our chapter on Crystals.
The Scanning Tunneling Microscope
A very impressive device which exploits the extreme sensitivity of quantum
tunneling is called the Scanning Tunneling Microscope (STM) developed by Gerd
Binnig and Heinrich Rohrer of the IBM Zurich Research Laboratory in 1981.
They won the 1986 Nobel Prize for Physics for this achievement. The basic idea
of the STM is sketched below:
