## Thursday, October 2, 2014

### Introduction to Quantum Mechanical Operators (Pt.1)

1. Writing the Hamiltonian:

The brilliance of the early quantum mechanicians lay in substituting the operators ( e.g Eop, p^ or p op  ) for the corresponding quantities of the original classical Hamiltonian, then multiplying through by the wave function y :

H op y  =  Eop  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)

Eop  =  i h  ( /t)

So, the full wave equation becomes:

- h / 2m ( 2/ x2) 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 1-dimensional form of the Schrodinger equation, which is not time dependent but time-independent.:

d2y/dx2 + 8π2  me /h2   {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 + [me v2/2] = V + KE

so: KE = W - V = [me v2/2], thence:

2 me  (W - V) =  [me v2/2],

Then:

me 2 v2/ h2  = 2 me (W - V)/ h2

Illustrating some basic properties of the 1-dimensional 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:

d2y/dx2 + 8π2  me /h2   {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.

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  = p2/ 2m + V(x) < V(x)  or:  p2/ 2m  < 0

To determine the motion of the electron quantum mechanically we must find the wave function that is a solution for the total energy E < Vo. In one region the appropriate time – independent Schrodinger equation is:

- h2/ 2m [d2y(x)/dx2] =  E y(x),     x < 0

In the other region, for which x > 0 we need:

-        h2/ 2m [d2y(x)/dx2] + Vo 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 2nd equation the general solution would be:

y(x)  = C exp (iK2 x) + D exp (-iK2 x)

Where: K1 = Ö (2m(Vo -  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 ‘B-term’ 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

Problems:

1)     For the step potential, show that for the region x > 0    and E < V o

with K1 = Ö (2m(Vo -  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