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:

                                                                             

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) :   l_D = 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 p_y  only by narrowing the slit width a and vice versa. Thus we get:

p _y a =   D p_y  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 p_x  =  [ h sin q/ l  -  (-   h sin q/ l)]    =   2 h sin q/ l  

Then the Heisenberg Uncertainty Principle product is:

D p_x   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 ^b_a   ‖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:

 ò ^b_a  ‖y‖²  dx   =  1

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

ò ^b_a   ‖y‖²  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:

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 ^b_a   ‖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.

http://brane-space.blogspot.com/2013/05/applications-of-differential-equations.html

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].

An Introduction to Quantum Mechanics (1)

1. The Wave Model of the Atom.

Though useful, especially in terms of identifying spectral lines, Bohr’s model had its limitations. For example, it couldn’t account for how angular momentum is conserved in atoms, nor how electronic transitions originate. More seriously, it wasn’t able to deal with the problem of lost energy and why atoms don’t simply collapse.

    Consider the following dynamical picture:  as the electrons whir about the nucleus they ought to be losing energy, in the context of Bohr’s orbital model. If they lose kinetic energy over time they must spiral into the nucleus, and the atom then ceases to exist. This ought to happen in a very short time, so that most atoms in the universe cease to exist and hence the whole universe. But this isn’t observed. Why?

     The only explanation is that Bohr’s orbital model can’t be correct.  Thus was born the theoretical basis for the wave model which we mostly accept today in modern quantum mechanics.  Unlike the Bohr model, electrons don’t follow defined orbital paths but instead are referenced to regions or volumes in which they will be more or less probable.  The basic allocation of electrons, say for the hydrogen atom, is then confined to “orbitals” or regions of higher probability.

     We now look at the experimental basis provided for this model.

     Around 1926, a young French physicist named Louis de Broglie actually postulated the basis for material particles, such as electrons, acting as waves.  This was experimentally verified in the (1927) Davisson and Germer electron diffraction experiment sketched below:

 

Fig. 1: The Basic Davisson and Germer experiment

    From the experiment, with electrons moving through a potential difference V = 4,000 volts, the kinetic energy gained should be equal to the work done, or:

½ m v²   =   eV

Where m the mass of the electron is: 9.1 x 10 ^-31 kg

And the electron charge e = 1.6 x 10^-19 C

The velocity then is:

v = Ö (2eV/m)

The momentum p = mv = m Ö (2eV/m) = Ö (2eVm)

And the de Broglie wavelength is:

l_D= h/p = h/Ö (2eVm)

For a voltage V = 3,000 V one would find:

l_D =

(6.626 x 10^-34 J-s) / [(3.2 x 10^-19 C) (3000V) ( 9.1 x 10^-31 kg)]^1/2

l_D =  2 x 10^-11 m

Which is the de Broglie wavelength of the electron in this experiment.

    A first step to uncovering the wave model from Bohr’s is to examine his quantized relationship:

m vr  = nh/ 2p  = L

where L is the angular momentum.  Re-arranging:

h/ mv =  2p r/ n = l_D    or  2p r =  n l_D

Showing the circumference is scaled into n (standing) waves of wavelength l_D   as shown below:

Fig. 2. Standing wave model for the Bohr atom.

     This wave-orbiting electron atom still has a radius r,  but with waves each separated by one de Broglie wavelength , l_D.

     Thereby  an integral number of such wavelengths form the circumference of the atomic orbit, as required by  the condition:  2p r =  n l_D.

  For example, in the case of hydrogen the first three of these cloud-wave regions are shown in Fig. 3.

Fig. 3: Electron (“orbital”) clouds-regions in Hydrogen

Let us ‘zoom in’ on the more spherical n= 1 configuration, and the probability for the electron in this space as depicted in Fig. 4 below:

Fig. 4: The n = 1 electron orbital for hydrogen

    This diagram more than any other dispenses with the notion that hydrogen electron occupies a definite position. Instead, it’s confined someplace within a “cloud” or probability space (b) but that probability can be computed as a function of the Bohr radius (a_o = 0.0529 nm).  The probability P_1s for the 1s orbital is itself a result of squaring the “wave function” for the orbital.  If the wave function is defined y (1s) = 1/Öp (Z/ a_o) exp (-Zr/ a_o), and the probability function is expressed:

P = ½y (1s) y (1s) *½

     Where y (1s) * is the complex conjugate, then the graph shown in Fig. 4 is obtained. Inspection shows the probability of finding the electron at the Bohr radius is the greatest, but it can also be found at distances less than or greater than  0.0529 nm.

     We thereby see from Figs. 3-4 that Bohr’s original quantizing number, n (the "principal quantum number"), has far more meaning than simply to parse the number of standing waves for a given atom. We already see that it determines the energy of the atom, viz.

E _n =  - 13.6/ n²   

     But it also indicates the average distance of the electron from the nucleus.  Thus, de Broglie’s wavelength provides the basis for the wave-particle duality that lies at the basis of the “smeared” probabilistic atoms peculiar to modern atomic theory.

       At the bottom of Fig. 3 are the quantum numbers: n and ℓ, which are identified as: the principal quantum number, and the angular momentum quantum number, respectively.

    There are two physical meanings attendant on n: i) it determines the energy of an orbital (specifically in the H-atom), and (ii) it indicates the average distance of an electron in a particular orbital, to the nucleus, To fix ideas, I show in the accompanying diagram (Fig. 5)  a sketch of one lobe for an electron orbital associated with the (3, 2, +2) state in the Hydrogen atom. The key point is the orbital denotes an electron density associated with a probability of finding the electron in some defined space.

Fig. 5: one lobe for an electron orbital associated with the (3, 2, +2) state in hydrogen

     In the case shown one must also visualize a symmetrical lobe on the other side (making the whole orbital resemble a dumbbell) to make it complete. As one alters the set of quantum numbers the electron densities change and so do the probabilities associated with the orbit.(See Fig.6)

    Fig. 6: Further hydrogen orbitals with higher quantum numbers

     Describing orbitals using the set of given quantum numbers means knowing the numbering rules applied to each. In the case of the principal quantum number, n, we allow it to have integral (non-zero) values: 1, 2, 3, 4 etc.

     The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital. The numbering rule for l is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n -1). This means if n = 2, then ℓ can have (n- 1) = (2 -1) = 1. But if n =1, then ℓ = (n - 1) = 1 - 1 = 0.

       Note that the ℓ -quantum numbers appear more than once for any  orbital with n >1. Thus, for the n = 2 case, we have two values of ℓ occurring: one for l = 0, the other for ℓ = 1. If we go on to n= 3 there are three values of ℓ, for n = 4, four values and so on. One also finds the ℓ -value specified for lettered orbitals: s, p, d, f, g, h. The s-orbital is for ℓ =0, the p for ℓ =1, the d for ℓ = 2 and so on.

     There is no special significance to the letters (apart from the physical meaning we already gave for the angular momentum quantum number, ℓ, and they are mainly of historical import- though still retained, for example, in chemistry. (By extension, one also often hears the term "atomic shell" used in chemistry). A collection of orbitals under the same value of n is called a "shell". Thus, for n = 4, we have ℓ =0, ℓ = 1, ℓ =2, ℓ = 3 so comprising the collection of orbitals: s, p, d and f.

Lastly, there is the magnetic quantum number, usually designated m _ℓ (subscript the same as the angular momentum quantum number) because it is contingent upon it. This quantum number describes the orientation of the orbital in 3-D space. For a given angular momentum quantum number, ℓ, we have integral values specified as follows:

m _ℓ = - ℓ, (-ℓ +1)....0......( ℓ - 1), + ℓ

Note the above set of m _ℓ numbers is given as a SERIES, e.g. starting with (-ℓ) and terminating at + ℓ. Look at the simplest example for ℓ = 0, then:

m _ℓ  = 0.    (Since all terms are zero)

What about ℓ = 1?

Then: m _ℓ = -1, 0, 1

What about ℓ = 2?

We have:

m _ℓ = -2, -1, 0, 1, 2


As a general rule then, we can use the formula:

N(m _ℓ) = {(2 x l) + 1}

to give the total number of m _ℓ   numbers.

In any problems to do with identifying electron shell structure- configuration, it is well to bear in mind the Pauli Exclusion Principle to make sure the electrons are distributed so that no two electrons have the same set of quantum numbers. It is instructive to study the table below to see how the principal quantum number, n, changes with ℓ, and the subshell as well as the symbol for the primary electron shell.  The  Table shown below shows the respective Shell and Subshell Symbols and associated Quantum Numbers.

n  =	SHELL	ℓ   =	Subshell
1	K	0	s
2	L	1	p
3	M	2	d
4	N	3	f
5	O	4	g
6	P	5	h

In working out assorted problems, preparation of a schematic energy level diagram associated with the state of the system can also be of immense value.

Problems:

(1) Find the de Broglie wavelength  l_D  of a proton subject to 4000 V in a Davisson-Germer type experiment.

 (Take m _p = 1.7  x 10 ^-27  kg)

(2) Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, ℓ, m _ℓ, m _s  for each of the electrons in O16.

(Hint: The  m _s or spin quantum number, which we will encounter in the next section, has either +1/2 or -1/2 value. )