 Writing the
Hamiltonian:
The brilliance of the early quantum
mechanicians lay in substituting the operators ( e.g E_{op}, p^ or p _{op })
for the corresponding quantities of the original classical Hamiltonian, then
multiplying through by the wave function y :
H _{op} y = E_{op}_{
} y
In this way,
a drastically simplified basic quantum mechanical equation could be obtained,
which could then be expanded once one substituted the operators, i.e.:
p _{op} = i h (¶/ ¶ x)
E_{op } = i h (¶ /¶t)
So, the full
wave equation becomes:

h / 2m (¶ ^{2}/ ¶ x^{2}) y +
V(x) y = i h (¶ /¶t) y
A more common form of the above equation
for less advanced work, say as applied in Calculus Physics courses is the 1dimensional
form of the Schrodinger equation, which is not time dependent but timeindependent.:
d^{2}y/dx^{2} + 8π^{2} m_{e} /h^{2} {W  V(x)}y = 0
Where W is the
total energy of each electron so (W  V) is the kinetic energy,
i.e. W = V + [m_{e} v^{2}/2] = V + KE
so: KE = W  V = [m_{e} v^{2}/2], thence:
2 m_{e} (W  V) = [m_{e} v^{2}/2],
Then:
m_{e}
^{2 }v^{2}/ h^{2}
= 2 m_{e} (W  V)/ h^{2}
Illustrating some basic
properties of the 1dimensional Schrodinger is straightforward. The best
approach is to apply it to a specific case for which some parameters are
known. Consider then an electron of
charge e, moving freely. Such a “free particle” is not bound to any potential,
so in this specific case, we have that V(x) = 0, so:
d^{2}y/dx^{2} + 8π^{2} m_{e} /h^{2} {W} y = 0
Another interesting facet of the
Schrodinger equation refers to the superposition aspect. If we start, say, with two different initial
conditions, to obtain two waves:
y = y1(x)
and y = y2(x)
Then all solutions of the given Schrodinger wave equation are of the form:
y = A y1(x) + By2(x)
In the task of understanding quantum mechanics it is useful to see the working from actual examples. As I’ve learned in teaching the subject, students are apt to get much more out of it if several different examples of quantum systems are explored and solved.
y = y1(x)
and y = y2(x)
Then all solutions of the given Schrodinger wave equation are of the form:
y = A y1(x) + By2(x)
In the task of understanding quantum mechanics it is useful to see the working from actual examples. As I’ve learned in teaching the subject, students are apt to get much more out of it if several different examples of quantum systems are explored and solved.
To that end,
we now consider a simple QM system set up such as shown in the sketch
below where we have a beam of electrons of energy kinetic energy E incident on
a plane where there is a step potential such that E is less than V(x).
Since
the total energy E is a constant, then by classical Newtonian mechanics the
electron cannot enter the region at x > 0. We must have then:
E = p^{2}/ 2m + V(x) < V(x) or: p^{2}/
2m < 0
 h^{2}/ 2m [d^{2}y(x)/dx^{2}]
= E y(x),
x < 0
In
the other region, for which x > 0 we need:

h^{2}/
2m [d^{2}y(x)/dx^{2}]
+ V_{o} y(x) = E y(x)
These
two equations can be solved separately. An eigenfunction for the entire range
of x is then obtained by joining the two solutions together at x = 0 in such a
way to satisfy the key conditions, i.e. that dy(x) /dx must be finite
and continuous:
The
first equation is simply that for a free particle, so the general solution can
be written in the traveling wave form:
y(x) = A exp (iK1 x) + B exp (iK1 x)
Where:
K1 = Ö (2mE) / h
For
the 2^{nd} equation the general solution would be:
y(x) = C exp (iK2 x) + D exp (iK2 x)
Where:
K1 = Ö (2m(V_{o}
 E) / h for x > 0
Now,
continuity of y(x)
is satisfied if the relation:
D
exp (iK2 x) _{x=0} = A exp (iK1
x) _{x= 0} + B exp (iK1 x)_{x = 0}
Is
satisfied. If it is, then:
D
= A + B
Continuity
of the derivatives of the solutions is expected if:

K2 D exp (iK2 x) _{x=0} = i K1 A
exp (iK1 x) _{x= 0} – iK1 B exp (iK1 x)_{x = 0}
In
the latter case: iK2D/ K1 = A – B
Adding
the two equations in D, A and B:
A
= D/ 2(1 + iK2/K1)
Subtracting
gives:
B
= D/2 (1 – iK2/K1)
Then
the eigenfunction for this potential will be:
For
x £ 0:
y(x) = D/ 2(1 +
iK2/K1) exp (iK1 x) + D/2 (1 –
iK2/K1) exp (iK1 x)
And: for x > 0
y(x) = D
exp (iK2 x)
This amounts to a transmitted and reflected mode and can be interpreted using the probability flux, S(x,t) in the region x < 0. According to the de Broglie wave postulate:
p1
= mv = h K1
And the probability flux is: S(x,t) = v A*A = v
P(x,t)
But: P(x,t) = B*B so that we can thereby obtain
the expression:
S(x,t)
= v A*A  v B*B
Where
v = h K1/ m
The first term
in the S(x,t) expression originates from the probability flux flowing in
the direction of increasing x. The second or ‘Bterm’ originates from the
probability flux flowing in the opposite direction. As a result we can associate the first term
(in the region x < 0) with the incidence of the particle on the point where
the potential energy V(x) changes and the second term with the reflection of
the particle from the change in potential.
The computation of the intensity of the reflected probability flux to
the incident intensity can then be made:
R = v B*B/ vA*A = B*B/ A*A =
[(1 – iK2/K1)* (1 – iK2/K1)] / [(1 + iK2/K1)* (1 + iK2/K1)]
=
[(1 + iK2/K1)* (1  iK2/K1)]/ [(1 + iK2/K1)* (1  iK2/K1)] = 1
The
transmitted probability flux can be found by calculating S(x,t) at some point
in the region x > 0. Thus:
S(x,t)
= v2 C*C
Where
v2 = h K2 / m = p2/m
The
ratio of the intensity of the reflected flux to the incident flux is the
probability the particle will be reflected back to the region x < 0. This is
written:
R
= v1 B*B/ v1 A*A = (K1 – K2)^{2}/
(K1 + K2 )^{2}
The ratio of the intensity of the intensity of the
transmitted flux to the intensity of the incident flux is the probability that
the particle will be transmitted into the region x > 0 or:
T = v2 C*C/ v1 A*A = K2 (2 K1)^{2}/ K1 ((K1
+ K2 )^{2}
=
4K1 K2/ (K1 + K2 )^{2}
And
it is easy to show then that: R + T = 1
1) For the step potential, show that for the region x > 0 and E < V _{o}
with K1 = Ö (2m(V_{o}  E) / h
The general solution of the appropriate Schrodinger equation is:
y(x) = C exp (iK2 x) + D exp (iK2 x)
2) For the transmitted and reflected flux of the step potential, show that:
R + T = 1.
To be continued
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