Tuesday, September 2, 2014

An Introduction to Quantum Mechanics (3)


(Continued from previous section)

4. The Wave-Particle Duality & Heisenberg Microscope


     We now look in somewhat more detail at wave-particle duality as it arises in quantum mechanics.  In the particle interpretation, electrons  fired from a device such as an electron gun would not all follow the same path since the trajectory of an electron – unlike a missile- can’t be predicted from its initial state. We consider here the case of electron diffraction, whereby (based on Fig. 7) electrons are emitted from an electron gun and pass through a slit toward a detector or photographic plate onto which a diffraction pattern appears. This pattern will also coincide with an intensity distribution such as shown in Fig. 10.

In effect, the intensity distribution basically describes the probability for an individual particle (electron) to strike each of several areas designated on the photographic film. This discloses a fundamental indeterminacy that has no counterpart in Newtonian mechanics. Now, consider an electron striking at some angle q, such as indicated:


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Fig. 10: Showing electron diffraction and intensity pattern on screen.

We have, from the quantities shown:

p y/ p x = tan q  or   p y =  p x  q  (in limit of small q)

Therefore, the y-component of momentum can be as large as:

p y =  p x  (l/ a)

Where a denotes the slit width. The narrower the dimension of a the broader the diffraction pattern, and the greater D p. From Louis de Broglie’s matter wave hypothesis (already introduced into the Bohr atom, as we saw, cf. Fig. 7) :   lD = h/ p x

Therefore:

p y =  p x  (h/ p x  a)  = h/ a  or: p y a =  h

But ‘a’ represents uncertainty in electron position vertically (D y), i.e. as it passes through the slit. We can reduce D py  only by narrowing the slit width a and vice versa. Thus we get:

p y a =   D py  D y »  h

Which is one form of the Heisenberg Uncertainty Principle which states that the momentum of a quantum particle and its positions cannot simultaneously be known to the same arbitrary precision. One corollary is that to detect a particle any given detector must interact with it thereby altering the motion of the particle.

     This view is no longer taken in any literal way because we understand that the quantum measurements are statistical in nature and hence a particular measurement is the result of a vast statistical assembly. Paul Dirac, in his book Quantum Mechanics, defined the “principle of superposition” thusly[1]:


“A state of a system may be defined as a state of undisturbed motion that is restricted by as many conditions or data as are theoretically possible without mutual interference or contradiction"

     But let’s examine this in more detail. By “undisturbed motion” Dirac meant the state is pure and hence no extraneous observations are being made such that the state experiences interference effects to displace or disturb it. In the Copenhagen Interpretation, “disturbance” of mutually defined variables, say x, p or position and momentum, occurs only if:

[x, p] = -i h =  -i h/ 2p


If it were the case that [x, p] = 0, one would say the variables “commute” and hence there’s no interference. If the condition doesn’t hold, then interference exists. Hence, Dirac’s setting of an upper limit in the last portion of his definition, specifying as many conditions as theoretically possible “without mutually interfering interference.” This state is undisturbed.  We have a statistical perspective!



Fig. 11: Sketch of Heisenberg Microscope and key parameters.

    It is important to see from the preceding, how the Heisenberg Uncertainty Principle arises not just from an ad hoc assumption, but from the limits (or “tolerance thresholds”) of explicit quantities (e.g. p, x), when considered in the quantum limit. Hence, the model of the Heisenberg “microscope” provides a useful (although not practical, since it can’t actually be constructed) means of deriving the statistical principle of superposition based on an observational ansatz.

    Consider a measurement made to determine the instantaneous position of an electron by means of a microscope. In such a measurement the electron must be illuminated, because it is actually the light quanta (photon) scattered by the electron that the observer sees. The resolving power of the microscope determines the ultimate accuracy with which the electron can be located.  This resolving power is known to be approximately:

 l/ 2 sin q

Where  l is the wavelength of the scattered light and q is the half-angle subtended by the objective lens of the microscope.  Then:

Δx   =  l/ 2 sin q

In order to be collected by the lens, the photon must be scattered through any range of angle from -q to q. In effect, the electron’s momentum values range from:

+ h sin q/ l   to   -   h sin q/ l  

Then the uncertainty in the momentum is given by:

D px  =  [ h sin q/ l  -  (-   h sin q/ l)]    =   2 h sin q/ l  

Then the Heisenberg Uncertainty Principle product is:

D px   Δx   =    (2 h sin q/ l )  l/ 2 sin q   = h


4. Probability density and Expectation Values

Earlier we saw:


P = ½y (1s) y (1s) *½

Which is the probability density and a quantity we can actually measure, e.g. for the 1s state of hydrogen. Then this needs to be generalized to apply to more than one case.

Since the electron locations can’t be computed from Newtonian mechanics but more plausibly based on an analogous probability density to what  we saw above, then we can generalize and write:

P ab  =     ò ba   y(x) 2  dx

Where x is the state under consideration and this system is 1-dimensional with the probability assessed from a to b.  Note that we define the normalization condition as:

òba   y2  dx   =  1

Normalization is simply a condition stating that the particle exists at some point at all times. Thus if we had:


òba   y2  dx   =  0

The probability would not exist. The probability condition then allows us to specify the probability of observing a particle even though we cannot specify the position. The normalization then gives the probability of finding the particle in the range a <  x  < b, say in one dimension.

The wave function, y(x) satisfies the Schrodinger equation. For the simple one dimensional case we can write:

d2 y/dx2 + F(x)   y = 0

Though the wave function y(x) it self is not a measurable quantity, other measurable quantities such as the energy E and momentum of the particle can be derived from it. Also, if the wave function is known it is possible to compute the average position of the particle, known as the expectation value:

  =       ò¥-¥   x y(x) 2  dx

This expression implies the particle is in a definite state so that the probability density is time –independent.

Example Problem: Consider the 1D quantum system shown, and a particle confined therein:
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With maximum dimension L in direction +x. Find: a) the probability P ab  the particle is between x = 0 and x  = L, b) the expectation value and (c) show the energy for the particle can be quantized according to:


E n = (h2/ 8m L2) n2

Let the wave function be:
y(x)  = Ö2/ ÖL    sin (kx)


Solution:

We rewrite the wave function as:

y(x)  =   Ö2/ ÖL   [sin (px/L)]   where k = p/L

Then:

P ab  =     ò ba   y(x) 2  dx =   ò L 0   (Ö2/ ÖL )2   sin2 (px/L) dx

=    2/ L ò L 0    ½ [1 - cos (px/L)] dx

(Let q = px/L  and use:  sin2 q= ½ (1 – cos 2q))

P ab  =     2/ L [½  ò L 0    dx  - ò L 0   cos (px/L)] dx

P ab  =      1 -  1/p  sin (2px/L)] L 0   = 1 -  1/p  sin (2p)


But : sin (2p) = 0 so P ab  =      1  

The expectation value is:

  =       ò¥-¥   x y(x) 2  dx

=  2/ L [ ò L 0   x sin (px/L)2] dx

= L2/ 4  -  [x sin (2px/L)/ 4p/L - cos (2px/L)/8(p2/L2)]

= 2/L  (L2 /4)  =  L/ 2

Finding the energy:  We have the Schrodinger equation:


dy2/dx2  + K2 y = 0

where K = 
Ö [2mE] / ħ

If we examine the sketch below:



We see plots of the wave function y(x)   vs. position x (far left),  and of the probability density (middle) and the energy levels. Since we have represented the wave function by a sinusoidal function then it follows that the allowed wavelengths are those for which the length L is equal to an integral number of half wavelengths, or:

L = nl/ 2

These allowed states are called stationary states and represent standing waves (analogous to the ones seen earlier for the Bohr atom). Thus, the wavelengths of the particle are restricted by the condition:

l  = 2L/ n

Then the magnitude of the momentum p is also restricted to specific values (e.g. using p = h/ l) [2] such that:

p = h/ l =  h/ 2L/ n = nh/ 2L

The energy associated with the particle is then:

E = 1/2 mv2  =  p2/ 2m  = (nh/ 2L)2 / 2m

E= ( h2/ 8mL2) n2


(n= 1, 2, 3 etc.)

Thus the energy is quantized with the energy of the lowest energy state corresponding to n =1 so:

E1= ( h2/ 8mL2)

This least energy that the particle can have is called the “zero point energy” and means the particle can never be at rest.

Note that the above energy result can also obtained through the use of differential equations, e.g.

The probability density can be extended to 3 dimensions by writing:


P   =     ò¥- ¥  y(x) 2  dV

The quantized energy will  then be (for a 3D box, for which dV = dx dy dz):

E= ( h2/ 8mL2)[ n x 2  +  n y 2    + n z 2  ]


Problems:

1)For a 1D box, let one electron inside have the wave function:

y(x)  =   Ö2/ ÖL   [sin (2px/L)]  

Find the probability of locating the electron between x = 0 and x = L/4.

2)Use the uncertainty principle to estimate the uncertainty in momentum for a particle in a 1D box. Estimate the ground state energy using this means and compare it to the actual ground state energy.

3) The wave function for a particle confined to moving in a 1D box is given by:

y(x)  =   A [sin (npx/L)]  

Use the normalization condition on y(x)   to show the constant A is given by:  A =  Ö2/ ÖL  

4) It is known from quantum mechanics that a particle in a one dimensional potential well (such as shown in the diagram) can exist in a number of energy states. Imagine an electron confined between the boundaries x and x +  Dx, where Dx is 0.5 Angstroms.

Approximately, what is the uncertainty in the x-component of  the  momentum of the electron?

5)(a) Consider a free particle confined between two impenetrable walls at x and x + L.  What is the probability according to classical physics that the particle will be found between x and x + L/3 if no other information is given?

b) What is the probability according to quantum mechanics that the particle in its lowest energy state will be found between x and x + L/3?


c) What is the probability according to quantum mechanics that the particle in the second lowest energy state will be found between x and x + L/3?



[1] Dirac, P.A.M.: 1941, Quantum Mechanics, Oxford University Press, 11.
[2] Recall from Planck’s law: E = hc/ l and p = Ö [2mE].

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