Download 1-d examples

Examples: A particle in a 1-D potential
Reading: Shankar Chpt. 5, Griffiths Chpt. 2
Our usual protocol for solving
1) Determine energy eigenfunctions
Schroedinger equation:
(in the x-basis):
by solving the time-independent
2) Expand the initial wavefunction (if known) in terms of the energy eigenbasis
3) The time evolution of the wavefunction is
4)
tells you the probability that a measurement of the
particle’s position at time t will yield a value in the range x=a to x=b.
Technical note:
If V(x) is finite everywhere, then when solving for the energy
eigenstates, both
and
will be continuous (even if the potential
V(x) itself has a discontinuity!)
If V(x) is infinite somewhere, then
derivative need not.
remains continuous, but its
Preview: Key differences between classical and quantum behavior:
E < Vmin:
Not possible quantum mechanically
Vmin < E < VL :
“BOUND STATES” --- Discrete energy spectrum (doubly degenerate);
Particle can be found past classical turning points xa, x’a (but with
exponentially decreasing probability). Ground state energy > Vmin.
VL < E < VR :
“UNBOUND (or SCATTERING) STATES” --- Energy spectrum
continuous, non-degenerate. Particle can reach –(infinity). Can be
found beyond classical turning point xb.
VL < E < VR :
“UNBOUND (or SCATTERING) STATES” --- Energy spectrum
continuous, doubly degenerate.
Example 1: Free particle (V(x)=0 everywhere)
A technical difficulty: The energy eigenstates are not normalizable (check it!) –
and hence they are unphysical!
Remedy: Ok, so apparently you can’t have a free particle with a definite energy
or momentum (as we’ll see later, according to the uncertainty principle,
if you know a particle’s exact momentum, then you know nothing about
its location – it could be anywhere in the universe, so the probability of it
being at any particular location is zero. This is why the eigenstate is
non-normalizable). But … if you have a superposition of energy
eigenstates, you can form a normalizable state vector:
By choosing the right superposition, you can construct a (physical)
wavefunction that describes a particle traveling along with a fairly welldefined momentum– it’s just that the momentum won’t be exact, since it
will contain a mixture of different energy/momentum eigenstates.
So are the pure energy/momentum eigenstates of a free particle useless?
No. They still give you much physical insight, despite the nonnormalizability. We’ll see this again in the next problem …
Example 2: Scattering from a potential step
Overview of what we’ll find:
Case 1: E > V0
Classically, a particle entering from the left would
have sufficient energy to make it over the potential
barrier, and would continue to travel rightwards (at
a slower speed). QM says that the particle has
both the possibility of making it over the barrier (as
in the classical case), as well as the possibility of
being reflected back to the left! Both are possible!
The quantum state is a superposition of both.
Case 2: E < V0
Classically, a particle entering from the left lacks
sufficient energy to make it over the potential
barrier, and would be reflected backwards (at the
same speed as it came in with). QM says that the
particle has the possibility of partially penetrating
into the classically forbidden region!
Let’s begin the calculations …
Case 1: E > V0
What does this wavefunction actually tell us, especially since (as you may
have noticed) it’s not normalizable, and hence “unphysical”? Before
answering, let’s picture the wavefunction:
Interpretation: Think of A1 as the amplitude of the incoming (incident) wave, B1
as the amplitude of the reflected wave, and A2 as the amplitude of the transmitted
wave. Even though none can be normalized, we can think of the ratio
as representing the relative probability that the wave gets reflected, and
as the relative probability that it gets transmitted to the other side of the step.
i.e., in analogy with electromagnetism, define
(If we think of a beam of particles hitting the potential step, R indicates the
fraction that are reflected, and T the fraction that are transmitted. Note that
R+T=1.)
Case 2: E < V0
This time we find R = 1, T = 0. In other words, the beam of particles is entirely
reflected, and not transmitted. This is like the classical case.
HOWEVER, unlike the classical case, the particles do partially penetrate into the
forbidden region x > 0. There is a probability of finding a particle there!
Physical applications of the step potential:
• conduction electrons in a metal
• reflection/refraction of light on glass! (Total internal reflection corrseponds
to T=0, in which case the evanescent wave corresponds to the penetration
into forbidden region!)
Remark on Quantum Tunneling
In the above example, the particle was able to penetrate into the classically
disallowed region, though not very effectively, since the wavefunction decayed
away exponentially the further you went into that region.
But what if the classicaly forbidden region has only finite thickness?
Here, wavefunction starts decaying inside barrier region, but can re-emerge
on other side --- i.e., the particle can “tunnel” across the barrier!
Tunneling occurs, for example, when the nucleus of uranium U238 undergoes
radioactive decay, emitting an α-particle (2 protons, 2 neutrons):
Likewise, in a solid-state device called a tunnel diode, electrons tunnel
through a potential barrier.
Further remarks on scattering:
1) The scattering of a wave off a step potential is an example of a more
general scattering problem. Consider the following:
Here, think of A as the amplitude of the incoming wave (when it’s at the far left,
well away from the scattering area, B as the amplitude of the reflected wave
(again, well to the left of the scattering site), and C the amplitude of the
transmitted wave (far to the right). Applying boundary conditions (like we did
in the preceding problem) will yield some linear relationship between the
amplitudes, which can be expressed as:
The matrix S is called the scattering matrix (S-matrix). The reflection
and transmission coefficients can be expressed as
2) Interestingly, when a quantum particle encounters a potential well
you will still get a reflected wave (even though classically all such
particles would make it over to the other side!) -- i.e., R ≠ 0.
3) Another way of interpreting R, T: The probability current density
This will take a little explaining … we’ll need to remind ourselves about
conservation laws.
If some physical quantity is conserved (e.g., charge, mass, etc.), then though
it may shift location from one region of space to another, the total amount of
it remains constant.
But there is also a local equivalent to this global conservation law: If charge is
conserved, then the only way the amount of charge in some particular finite
volume of space can change is if charge exits (or enters) through the surface of
that volume (i.e., it is never spontaneously generated within the volume).
Mathematically, we write
By the divergence theorem,
In which case our conservation law can be expressed as
In QM, the conserved quantity of interest here will be the total probability for
finding a specified particle somewhere in space, which always equals one:
So here,
plays the role of charge density
(except now we
interpret it as probability density). So what is the corresponding (probability)
current density
?
Let’s calculate …
The probability current density tells you the rate at which probability is
crossing a unit area per unit time:
The above is a general result. Now, suppose we have
like we had in our step-potential scattering problem. In this case, we find
So this tells us the probability current associated with the incident wave.
Similar results hold for the reflected and transmitted waves. Thus, the
deeper way of thinking about the reflection and transmission
coefficients R and T is
So even though it wasn’t really proper to talk about probabilities for nonnormalizable wavefunctions like we originally did in that problem, it is okay to
talk about the relative current densities.
Hence we have managed to slither away from the abyss of non-normalizability
with at least a modicum of our self-respect intact!
Example 3: Particle in a box
To the left and right of the box we expect the wavefunction will be fixed at zero
(i.e., there is exponential decay in classically forbidden regions, and since the
potential at the walls is infinitely large this decay will be infinitely fast.)
In the interior of the box, the energy eigenstates are given by
Instead of using complex exponentials, I could equally well have written the
solutions in terms of sines/cosines:
I’ll choose this formulation here. But we’re not done yet, since we still have to
apply our boundary conditions, i.e., that
is continuous everywhere (in
particular, at x=-L/2 and x=L/2), and we’ll also have to normalize.
(Note: the derivative of
need not be continuous, because V(x) is infinite)
There are two distinct classes of solutions to the above equations:
Case 1:
B=0 and cos(kL/2)=0
Case 2:
A=0 and sin(kL/2)=0
Converting k back to energy, we see that the
allowable energies for a particle in a box are
quantized!
These are the bound states of a particle in a box. Here’s what they look
like:
Remarks:
1) Energies are discrete, non-degenerate, non-uniform spacing.
2) Ground state energy is not zero! (i.e., it is above the bottom of the
potential well). This is a general feature of QM. Here, it can be
heuristically justified by invoking the uncertainty principle.
3) Note that the ground state eigenfunction has no nodes (disregarding the
boundaries); the first excited state has one node; the second excited
state two nodes, and so on. This is also a general feature of bound
states of particles in one-dimensional potentials. (Note that the
existence of nodes means that there are places in the box where the
particle will never be found!)
4) Note that some of the eigenfunctions are even under reflections (x →-x),
while others are odd. This is a direct consequence of the fact that the
potential itself is symmetric under reflection. (Later on we will introduce
the Parity operator to discuss this idea more systematically.)
Example: A particle in a box is initially in state
What is the probability that it is in its
ground state after time t?
Solution:
So at time t,
So the probability that the particle is in its ground state at time t is given by the
amplitude squared of the coefficient of the ground state wavefunction:
Notice that this is exactly the same as the probability it was initially (i.e., at
t=0) in its ground state! More generally, for time-independent potentials, the
probability that a particle is in its nth energy eigenstate doesn’t change in
time! That’s why the energy eigenstates are called “stationary states”.
Example: A “between a rock and a hard place” potential
Ok, that gave you some practice with a scattering state of a
delta-function potential. Next let’s consider a related bound
state problem …
Example: Same as previous, but now flip the potential barrier so it
becomes a well
This time, we’ll search for bound states (E < 0) only.
Here, a little thought will help much more than Maple or Mathematica will!
First, look at the LHS of the equation. It’s just
“k”. Plotting, it’s just a simple line at 45°
Now look at the RHS of the equation. It
just looks like this:
When, if ever, will these two curves intersect? There are two possibilities:
Observe that the slope of the exponential curve at the origin (k=0) is
Meanwhile, the slope of the straight line is one.
So the curves will intersect (case B) if
In this case, there is exactly one point of intersection (ignoring k=0). Hence,
there is a single bound state!
However, if
, then no bound state solutions exist!
Thus, we have found the criterion which determines if a bound state exists.
We could go on from here further (though we won’t). For instance, in the case
of a bound state, we could solve for the unique k value (which in turn would tell
us the bound-state energy). We could also solve for the coefficients A, B, and
D, and thus get an explicit representation of the wavefunction. (Note: to solve
for the coefficients A,B,D, we would need more than just the three boundary
equations given earlier; we would also have to demand that the wavefunction
be normalized. The boundary equations together with the normalization
condition would then uniquely determine these coefficients up to an overall
phase factor.)
Example: A Periodic Potential --- Bloch’s theory of band structure
In a crystal solid, the outer (“valence”) electrons are not attached to specific
molecules, but rather can move freely around the lattice. Such electrons thus
“see” not just the coulombic potential of a single atom, but rather an periodic
potential of the entire crystal lattice.
We will crudely model this situation as a series of delta-function potentials. (This
may seem much too simple to be at all realistic, but our model will nonetheless
capture the essence of what’s going on in real solids!)
Before proceeding, we’ll need to know the following two things:
Bloch’s theorem
If you have a periodic potential V(x+a)=V(x), then the
wavefunction has the form
(See Sect. 5.3.2 of Griffiths for details)
Periodic boundary conditions Our periodic lattice can’t in reality go on forever,
since in reality it’s composed of a finite number of molecules (e.g., N ≈1023). For
N very large, it is reasonable to assume that the effect of the edges will be
unimportant compared to what’s happening “in the bulk.” For mathematical
simplicity then, we can imagine wrapping the solid around into a circle, so that
lattice site N+1 is the same as lattice site 1. (If edges are truly unimportant, then
wrapping around the left edge of the lattice to meet the right edge shouldn’t effect
its behavior.) Doing this leads to so-called “periodic boundary conditions.” It’s a
favorite trick of physicists and mathematicians! In the present context, it implies
From Bloch’s theorem, this in turn implies
(Note that for large N (≈1023), these K values are so ridiculously closely spaced that they
are, for all practical purposes, continuous.)
Ok, now let’s solve Schrödinger's equation for a long string of delta-function
potentials (assuming periodic boundary conditions):
As usual, in-between the delta-function spikes, the Schrödinger equation
becomes
The solution in the “cell” 0 < x < a is just
The solution in the neighboring cell on the left, -a < x <0, is, by Bloch’s theorem,
Our boundary conditions at x=0 (i.e., at the boundary of these two cells) are:
1)
is continuous
2)
suffers a discontinuity of size
These conditions become
1)
2)
A little algebra shows that nontrivial solutions for A, B exist only if
(This equation determines the allowed values of k, and hence the allowed energies.)
For ease of analysis, define z ≡ ka and β ≡
, in which case we have
Now, recalling that the allowed values of K are so closely spaced as to be
essentially continuous, the LHS of the preceding equation can take on virtually
any value between -1 to 1.
Meanwhile, the RHS of the equation looks like
So the allowed z-values (for which the RHS is between -1 and 1), form bands.
This in turn means that the allowable energies of the system forms (virtually
continuous) bands, separated by gaps, as shown in the following graph:
The existence of energy bands
separated by gaps is characteristic
of periodic potentials in general (not
just the delta-function array
considered here).
If a band is completely filled with
electrons, it takes lots of energy to
excite the electrons, since they have
to jump the forbidden zone to the
next band  electric insulators!
If a band is only partially filled with
electrons  conductors!
If a band is nearly full except for
some “holes”, or if it’s nearly empty
except for a few extra electrons
(which can occur if you “dope” an
insulator)  semiconductors!