# Chapter 3. Quantum Mechanics of Simple Systems

### Particle in a box, Harmonic oscillator, and rotation

We examine several systems that are analytically solvable, and gain insight into how one might solve these problems. In addition, we examine how to extend a simple one-dimensional system into multi-dimensions. Our hope is to learn about how each of the systems might have applications in chemistry.

Here is the prescription of how to solve simpler quantum problem in general.

1) Draw a model of physical system with identifying coordinates and potential energy (interaction energy).

2) Write down Hamiltonian according to the model.

3) Place Hamiltonian into Schrödinger equation.

4) Taking boundary conditions into account to solve Schrödinger equation for Ψ.

5) Solve for energy.

6) Normalize wave function.

That's it. What I'd like you to take home overall is that how the shapes of wave functions change with respect to the potential energy given, as well as spacing of quantized energy levels of a given system.

## Particle in a Box

The particle-in-a-box system is to confine a particle in an infinitely deep potential well, as shown in Figure 1.

This is expressed as the following.

 $V( x )={ 0 0≤x≤L ∞ x<0,x>0$ 1
In the regions where, x < 0 and x > 0, V = ∞ therefore, Shrödinger equation yield the wave function, Ψ, to vanish since the second derivative values should be smaller than ∞, so Ψ must compensate for infinity in the potential to obtain zero on the right-hand side.
 $− ℏ 2 2m d 2 Ψ d x 2 +( V−E )Ψ=0$ 2 $− ℏ 2 2m d 2 Ψ d x 2 +( ∞−E ) Ψ →0 =0$ 3

Figure 1. Particle in a box potential.
Above shows if the potential is infinite, no particle can exist. Let us now look at the middle region where V = 0. Since V = 0, Schrödinger equation has the form,
 $− ℏ 2 2m d 2 Ψ d x 2 =EΨ$ 4
If we rearrange Eqn. 4 to isolate second derivative, we have
 $d 2 Ψ d x 2 =− 2mE ℏ 2 Ψ$ 5
The right-hand side of Eqn. 5 is constant times the wave function. We seek a form of function that gives original function after taking second derivative. Functions such as sine, cosine, and exponential functions are the candidate.

Additionally, we want to consider the most conveniet function that vanishes at x = 0 and L. For this reason, we choose sine function as Ψ. In order to fit sine function to the interval [0,L], we use the argument of sine to be n π x / L, with n is positive integer. This way, we can fit functions such as the ones shown in Figure 2.

Figure 2. First few wave functions.
Therefore, the wave function has the form,
 $Ψ=sin⁡( nπx L )$ 6
Let's test to see if we can construct an eigenvalue equation with our choice of wave function. We want the original wave function back after second derivative operation.
 $dΨ dx = nπ L cos⁡( nπx L )$ 7 $d 2 Ψ d x 2 = d dx dΨ dx =− n 2 π 2 L 2 sin⁡( nπx L )$ 8
The second derivative gives Ψ back with some constants, thus we have eigenvalue equation.

Now, let us look at what the expression of energy is. As can be seen, the right-hand sides of Eqns. 5 and 8 are the same. Then, the constants assocated with them are equal, as well. That is,

 $− 2mE ℏ 2 =− n 2 π 2 L 2$ 9
Then, we can solve for E to obtain the expression for the energy.
 $E= n 2 π 2 ℏ 2 2m L 2$ 10
The energy of a particle with mass m confined in an infinitely deep well has energy dependent on n and L, both with square dependence. n is the quantum number that describes the state, and it is a positive integer for a given box. The energy spacing between adjacent energy levels for a give box size increases as n becomes larger, is the characteristics of the particle in a box model.

Now, we need to normalize the wave function. We need to square integrate Ψ. We can follow the procedure in the section (Eqns. 8 to 11) described in Chapter 3.

 $〈Ψ|Ψ〉= ∫ A 2 sin⁡ 2 ( nπx L ) dx=1$ 13
By looking at the table of integrals for trignometric functions, one finds
 $∫ sin⁡ 2 axdx= x 2 − 1 4a sin⁡2ax$ 14
Then, we get
 $A 2 L 2 =1$ 15 $A= ( 2 L ) 1/2$ 16
Finally, we can write the wave function to be
 $Ψ= ( 2 L ) 1/2 sin⁡( nπx L )$ 17

Figure 3 shows first three lowest energy levels and their wave functions.

Figure 3. Energy levels and wave functions

On the left axis, the energy levels are labeled in the units of n2. The lowest energy state, n = 1 state, is called ground state. Any other states are called excited states. As the quantum number increase, the number of nodes, where the wave function becomes zero, increases.

## Particle in a finite potential

Figure 4. Finite energy well.
In this potential, the well depth is no longer infinite. Therefore, we must consider each of the regions (I, II, and III). Schrödinger equation for these regions are,
 $− ℏ 2 2m ∇ 2 Ψ=EΨ 0 18 $− ℏ 2 2m ∇ 2 Ψ+ V 0 Ψ=EΨ x≤0,x≥0$ 19
The solution for Eqns. 18 and 19 are
 $Ψ II =Asin⁡kx+Bcos⁡kx$ 20 $Ψ I,III =C e Gx +D e −Gx$ 21
Here we define G2 to be,
 $G 2 = 2m( V 0 −E ) ℏ 2$ 22
Eqn. 19 can be written as
 $∇ 2 Ψ− G 2 Ψ=0$ 23
By this definition, the second derivative of Eqn. 21 gives the second term on the left-hand side.

The first term of the right-hand side of Eqn. 21 is for region I, and the last term is for region III. It can be understood by the fact that the wave function vanish at -∞ and +∞.

We determine the coefficients on the wave functions by use of some of the conditions for well-behaved wave functions. Conditions to which the wave functions must have, are:

 24 25
Let us look at x = L for the moment, and consider only the case where B = 0 from the physical argument before. Therefore, we have the following.
 $C e −Gx =Asin⁡kx$ 26 $−GC e −Gx =kAcos⁡kx$ 27
If we divide Eqn. 27 by G and add it to Eqn. 26, we have
 $Asin⁡kx− kAcos⁡kx G =0$ 28
Then, we have
 $Asin⁡kx Acos⁡kx =tan⁡kx= k G$ 29
 $tan⁡( 2mE ℏ )= ( 2mE 2m( V 0 −E ) ) 1/2$ 44
If one plots the left-hand side of Eqn. 30 for different E and also plot the right-hand side as a function of E as well. Then, the intersetions of the two functions are the eigenvalue for this problem.

## Particle in a 2-dimension box

Let us now extend the dimensinality of our system to 2D. It will be clear soon how to extend even to six dimensions. Let's consider the following potential.

Figure 4. Particle in 2D box.

We now confine a particle in two dimensions. Similar to 1D case, we set the potential energy to be infinite outside of the box and V = 0 inside. So, we have

 $V( x,y )={ 0 0≤x≤ L x and 0≤y≤ L y ∞ x<0,x>0 and y<0,y>0$ 45

Then, inside of the box, Schrödinger equation becomes

 $− ℏ 2 2m ( ∇ x 2 + ∇ y 2 )Ψ=EΨ$ 46
where ${\nabla }_{x}^{2}$ is the second derivative operator in x coordinate. Similarly, ${\nabla }_{y}^{2}$ for y coordinate.

### Separation of Variables

The general procedure of dealing with multidimensional problem is separation of variable. In this procedure, we form the total wave function by product of one-dimensional wave functions. Thus,
 $Ψ= Ψ x Ψ y = Ψ x ( x ) Ψ y ( y )$ 47
If we substitute Eqn. 20 into Eqn. 19, we get
 $− ℏ 2 2m ( ∇ x 2 + ∇ y 2 ) Ψ x Ψ y =E Ψ x Ψ y$ 34
Isolating second derivatives so it is easier to see,
 $( ∇ x 2 + ∇ y 2 ) Ψ x Ψ y =− 2mE ℏ 2 Ψ x Ψ y$ 35
Since ∇x depends only on x and ∇y for y, one has
 $Ψ y ∇ x 2 Ψ x + Ψ x ∇ y 2 Ψ y =− 2mE ℏ 2 Ψ x Ψ y$ 36
Divide both sides by Ψ,
 $∇ x 2 Ψ x Ψ x + ∇ y 2 Ψ y Ψ y =− 2mE ℏ 2$ 37
It means that we can separate Eqn. 24 into two parts as,
 $∇ x 2 Ψ x Ψ x =− 2m E x ℏ 2$ 38 $∇ y 2 Ψ y Ψ y =− 2m E y ℏ 2$ 39

### 2-D Wave functions

It means that we have separated Schrödinger equation for each dimension, for which we already know how to solve.
 $− ℏ 2 2m ∇ x 2 Ψ x = E x Ψ x$ 40 $− ℏ 2 2m ∇ y 2 Ψ y = E y Ψ y$ 41
One can clearly see E = Ex + Ey. To make clear, let me rewrite Eqn. 20, now that we have the separated functions.
 $Ψ= 2 ( L x L y ) 1/2 sin⁡( n x πx L x )sin⁡( n y πy L y )$ 42
Now, let us look into how wave function looks like. The ground state wave function in the 2-D case is given by nx = nx = 1. Both are individually the ground states of 1-D wave function. The movie below shows the amplitude of the wave function in the intervals [0,Lx] and [0,Ly]. In all cases, the red color indicates highest positive numbers and blue being the lowest negative numbers. Green is zero and closer to it.
 Above is for nx = 1, ny = 1 nx = 1, ny = 2 nx = 1, ny = 3 nx = 2, ny = 2
You can imagine that there are nx = 2 and ny = 1, similar to above-right wave function. You should consider such function by rotating the picture by 90° in the plane of the monitor.

### 2-D Energies

The total energy of the system in 2-D is given by
 $E=( n x 2 L x 2 + n y 2 L y 2 ) π 2 ℏ 2 2m$ 43
For a given box size, Lx and Ly, the quantum states are determined. As has been mentioned that the wave functions obtained by nx = 1 and ny = 2 is equivalent to the state with nx = 2 and ny = 1. The energies of these two states are also the same. Then, these states are said to be degenerate .

## Harmonic Oscillator

So, we continue to follow the prescription of how we solve simple 1-D system. Let's make a model of the system.

Harmonic oscillator is quite important system since many things oscillates back and forth in motion. Electromagnetic wave is an obvious example, but for application to chemistry, harmonic oscillator is a model vibration. You have used IR spectrometer, a great identification instrument! Right? The energy range of IR light corresponds to the motion of nuclei in the molecule. When a carbonyl group is present, you see very prominent peak in the spectrum around 1600 cm-1, and this peak is due mainly to stretching motion of C and O in the C=O bond. We'll see later how we can obtain IR spectrum from the fundamental point of view.

In Figure 5, we have a model for harmonic oscillator and its potential.

Figure 5. Harmonic Oscillator. a) Model and b) Potential.

A particle with mass m is connected to a mass-less ideal spring and move back and force in x direction. The force law comes from Hooke's law, where the force is linearly dependent on the stretching or shrinking distance.

 $F=−kx$ 44
From previous chapters, we know that F = -dV/dx, so we integrate the potential V to have
 $V( x )= 1 2 k x 2$ 45
The potential energy is now placed into Schrödinger equation.
 $− ℏ 2 2m ∇ x 2 Ψ+ 1 2 k x 2 Ψ=EΨ$ 46
The equation looks benign, but it requires a special function in order to solve it. So, I will have this done in the collapsible window. If you want to skip it for now, read on!

+ Harmonic Oscillator Solution
We need to define several variables and make y into unitless variable
 $y= ( μω ℏ ) 1/2 x$ C1 $D= 2E ℏω$ C2 $ω=2πν= ( k μ ) 1/2$ C3
Substitute Eqns. C1 to C3 into Eqn. 33, and multiply both sides by -2ℏ/ω to yield
 $−2 ℏω [ − ℏ 2 2μ ∇ x 2 Ψ+ 1 2 k x 2 Ψ ]= −2 ℏω EΨ$ C4 $[ ℏ μω ∇ x 2 − k x 2 ℏω + 2E ℏω ]Ψ=0$ C5

Now making the independent variable x into y by factoring out the (μω/ℏ) term.

 $[ ℏ μω d 2 d ( ( μω ℏ ) 1/2 x ) 2 − μω x 2 ℏ +D ]Ψ=0$ C6
 $[ d 2 d y 2 − ( μω ℏ x ) 2 +D ]Ψ=0$ C7
 $[ d 2 d y 2 − y 2 +D ]Ψ=0$ C8
The general solution has a form of Gaussian function.
 $Ψ( y )= e −β y 2 ψ( y )$ $Ψ= e − y 2 /2 ψ$ C9
We neglect the solution with the positive exponent since such function is not normalizable as it tends to infinity at y = ∞. To check the validity of the general solution, we have
 $dΨ dy =−y e − y 2 /2 ψ+ e −β y 2 /2 ψ ′$ C10 $d 2 Ψ d y 2 =[ ψ ″ −2y ψ ′ +( y 2 −1 )ψ ] e − y 2 /2$ C11
Here the primes on ψ indicates derivatives with respect to y. Substitute Eqn. C11 into Eqn. C8 gives the following.
 $[ ψ ″ −2y ψ ′ +( D−1 )ψ ] e − y 2 /2 =0$ C12
Now, we seek a solution to the terms in the square braket. It turns out (this is the kind of language I have to use, since the following is not too logically connected) that the solution is a polynomial, a power series expansion of y, to be more specific.
 $ψ= a 0 + a 1 y+ a 2 y 2 + a 3 y 3 +⋯= ∑ n=0 a n y n$ C13
Let's now try expressing the first and second derivatives of ψ. The first derivative is
 $ψ ′ = a 1 +2 a 2 y+3 a 3 y 2 +4 a 4 y 3 +⋯= ∑ n=0 n a n y n−1$ C14
Further taking a derivative for Eqn. 14, we get
 $ψ ″ =2 a 2 +6 a 3 y+12 a 4 y 2 +20 a 5 y 3 +⋯= ∑ n=0 ( n+1 )( n+2 ) a n+2 y n$ C15
Substitute Eqns. C13, C14, and C15, into the square bracket terms in Eqn. C12, i.e.
 $ψ ″ −2y ψ ′ +( D−1 )ψ=0$ C16
Then, we get
 $∑ n=0 [ ( n+1 )( n+2 ) a n+2 y n −2yn a n y n−1 +( D−1 ) a n y n ] =0$ C17
Clean this up a bit.
 $∑ n=0 [ ( n+1 )( n+2 ) a n+2 −2n a n +( D−1 ) a n ] y n =0$ C18
In order to have 0 on the left-hand side is to have the terms in the square braket to vanish. By rearranging the terms in the braket yields,
 $a n+2 = a n ( 2n+1−D ) ( n+1 )( n+2 )$ C19
It is a recursion relation if an is known, an+2 can be obtained. This is useful, but in order to have the complete set, we need to know first two sets of a's, a0 and a1.

Now, as |y| → ∞, both the exponential part as well as the polynomial parts of Eqn. C9 become infinite, as well. In order to prevent the unphysical infinity, we limit the terms of the polynomial expansion. This can be achieved by recurrsion to stop in Eqn. C19. It means that if D in Eqn. C19 is 2n + 1, then an+2 = 0. From the definition of D, we have

 $D=2n+1= 2E ℏω$ C20
Therefore, we have
 $E=ℏω( n+ 1 2 )=hν( n+ 1 2 )$ C21
From our definition of the polynomial, the integer n = 0, 1, 2, .... When E satisfies C21 for a given n, the expansion in C18 is valid. It means that the summation of Eqn. C18 only runs up to the value of particular n. Then the wave function, when satisfying Eqn. C21, has the form
 $Ψ n,even =[ a 0 + a 2 y 2 + a 4 y 4 + a 6 y 6 +⋯ a n y n ] e − y 2 /2$ C22 $Ψ n,odd =[ a 1 + a 3 y 3 + a 5 y 5 + a 7 y 7 +⋯ a n y n ] e − y 2 /2$ C23
Eqn. C22 is for even n, and Eqn. C23 is for odd n. The terms in the bracket are obtained explicitly by using recurrsion relation, Eqn. C19. Then, Eqns. C22 and C23 are rewritten as
 $Ψ n,even =[ a 0 ( 1+ 1−D 2 y 2 + ( 1−D )( 5−D ) 4! y 4 + ( 1−D )( 5−D )( 9−D ) 6! y 6 + ( 1−D )( 5−D )⋯( 2( n−2 )+1−D ) n! y n ) ] e − y 2 /2$ C24 $Ψ n,odd =[ a 1 ( 1+ 3−D 3! y 3 + ( 3−D )( 7−D ) 5! y 5 + ( 3−D )( 7−D )( 11−D ) 7! y 7 + ( 3−D )( 7−D )⋯( 2( n−2 )+1−D ) n! y n ) ] e − y 2 /2$ C25
These equations, C24 and C25, can be expressed in a more concise manner, and it is called Hermite polynomial, generally defined by
 $H n ( y )= ( −1 ) n e y 2 /2 d n e − y 2 /2 d y n$ C26
So, finally, we can express the wave function C9 in the following.
 $Ψ= e − y 2 /2 H n ( y )$ C27
Eqn. C27 satisfies Eqn. C8 and Hermite polynomial goes up to the number of terms that is defined by C26.

We make some substitution and change variables.

 $y= ( μω ℏ ) 1/2 x$ 47 $D= 2E ℏω$ 48 $ω=2πν= ( k μ ) 1/2$ 49
With these choices of variables, Schrödinger equation is read
 $[ d 2 d y 2 − y 2 +D ]Ψ=0$ 50
The general solution has a product form of Gaussian and Hermite polynomial.
 $Ψ= e − y 2 /2 H n ( y )$ 51
The energy is given by
 $E n =hν( n+ 1 2 )$ 52

The normalization procedure is the same, except the fact that it is bit more involved. So, its derivation is skipped. The normalized wave function is

 $Ψ n = ( π 1/2 2 n n! ) −1/2 H n ( y ) e − y 2 /2$ 53

The first few Hermite polynomial is listed in Table 1.

n Hn(y) En Normalized Wave Function
01 $\frac{1}{2}h\nu$
12y $\frac{3}{2}h\nu$
24y2 - 2 $\frac{5}{2}h\nu$
38y3 - 12y $\frac{7}{2}h\nu$
416y4 -48y2 + 12 $\frac{9}{2}h\nu$
532x5 - 160y3 + 120y $\frac{11}{2}h\nu$
664x6 - 480y4 + 720y2 - 120 $\frac{13}{2}h\nu$
Table 1. Harmonic oscillator solution. n = 0 to 6.

One of the most interesting aspect of the quantum mechanical solution to the harmonic oscillator model is that there is a n = 0 is quantized to be higher than the bottom of the well. This is what we call zero-point energy. So, the loweest possible energy seen in the potential can not be achieved by quantum systems! Quite intriguing! Also, you might have already noticed, but the energy spacing between two adjacent energy levels are the same. Look at the En column of Table 1. Each and every adjacent energy levels have .

### Classical turning point

As has been mentioned that when a ball is placed in a 1-D bowl (consider that there is no friction), and you let go. You can easily guess the motion of the ball goes back and forth. The point at which the ball stops moving and the momentum changes its direction, is what we call classical turning point. That's exactly what we expect, say, you stretch a spring and let go. It will bounce back to original stretched length, but not beyond. You can see that in Figure 6.

Figure 6. Classical turning point on harmonic oscillator.

Now, look at our wave function. Let us plot the ground state wave function along with its energy and the potential energy function. It is shown in Figure 7. The potential energy function represented here has mass, μ, and force constant, k, are set to 1 in Eqns. 34 - 36.

Figure 7. Harmonic oscillator zero-point energy and wave function.

In the figure, the lowest energy state (zero-point energy), red line labeled with n = 0, is superimposed on the potential energy curve (red). The blue curve is the ground state wave function. The dashed vertical lines represent the position of x = 1 and -1. The classical turning points occur at these points. Now, you might have noticed already, but the where the dashed lines meet the wave function, the value of the wave function is not zero. In fact, there are finite probability exists beyond the classical turning points. Figure 8 shows the classically allowed region is shaded for the probability distribution, | ψ |2.

Figure 8. Classically allowed region of ground state harmonic oscillator

Since we measure the probability of finding a particle is normalized to be one we can think about how much of that particle, in a probablistic sense, can be found in the classical region vs. outside of classical realm; let's call it tunneling region. The classically allowed region is nearly 85%, while the two extreme ends have about 15% together! So, 15 out of 100 oscillators might be found outside of what is expected! That's odd!

The region outside of classically allowed one is generally referred to as tunneling. There is a bunch of examples of tunneling in chemistry. I'll show you some little later.

### Harmonic oscillator as molecular vibration

As mentioned briefly above, harmonic oscillator model is used to gain information on molecular vibration.

Figure 9. Energy levels of harmonic oscillator and "Experimental" potential.

One can think about a diatomic molecule to be two balls hooked up with a spring, as shown in Figure 9. The red curve is the potential energy in harmonic oscillator model, with its quantized energy levels and the transition between n = 0 state to higher states. Similarly, in blue, the potential energy obtained from experiment is given. Also shown are the energy levels according to the experiment and the transitions from n = 0 to higher states. The energy spacing for harmonic oscillator model is, of course, constant given by , while for the blue curve, the energy spacing between adjacent levels gets closer as energy becomes higher. Despite the fact that overall shapes of the two curves are different, in the vicinity of the lowest energy points, the two curve match decently. So, only near the equilibrium bond length, harmonic oscillator model for vibration is valid. What I mean by "valid" is that the fundamental transition, that is the molecule in the n = 0 state absorbing a photon to become n = 1 state, a primary transition registered by an IR spectrometer as a peak in the spectrum, is good enough as an approximation to molecular vibration. As it can be seen in the figure, the fundamental transition for the blue curve closely matches with that of the red potential. This is not the case for higher transition, n = 0 to n = 2, for example.

This is the general shape for stretching of any chemical bond from the small bond distance to large and eventually dissociates into two atoms. At a very large distance, two atoms don't interact, therefore, the potential energy curve tapers off to asymptotic value, which can be made to be zero. The details are much rather complicated because by stretching one coordinate, other parts of the molecule is affected, and the fact changes among other things, dissociation energy and the force constant associated with the particular stretching coordinate.

Generally, what we do in the theoretical calculations to obtain the molecular geometry at equilibrium bond distance, then calculate what is known as Hessian which is the second derivative matrix of the molecular energy with respect to all coordinate. One way to think about what Hessian is, is from your undergrad calculus class. At extremum, second derivative tells you whether the function in question is concaving up or down. Right? So, the second derivative characterizes the concavity of the function at the extremum.

Also, let me approximate the potential energy curve by Talor series expansion. What I'm trying to do here is that you only have the knowledge of the bottom of the well geometry. Now you have to "feel" around the bottom of the well by moving your atoms ever so slightly in two directions, back and forth. The Taylor series expansion of the potential is,

 $V( x )= V 0 + 1 1! d V 0 dx Δx+ 1 2! d 2 V 0 d x 2 Δ x 2 + 1 3! d 3 V 0 d x 3 Δ x 3 + 1 4! d 4 V 0 d x 4 Δ x 4 +⋯$ 54
The first term on the right-hand side, V0, is the molecular geometry at the bottom of the well, the stationary point. Since V0 is at a stationary point, meaning that the gradient dV0/dx is zero. Since Δx is a small displacement, Δx3 is really really small, and this term and beyond, the expansion is terminated to have,
 $V( x )= 1 2 d 2 V 0 d x 2 Δ x 2$ 55
Here we subtracted out the V0. Do you notice something? If we equate d2V0/dx2 with k, the force constant, in Hooke's law, we have V(x) = 1/2 kx2. So, this shows that the Taylor series expansion of V at V0 truncated at second-order is the harmonic oscillator approximation to the molecular vibration. For a diatomic molecule, given the second derivative, which means you know the force constant, and by knowing the mass of the atoms we have all the energy levels and wave functions by using the Ean. 38. This also means that the transition energies are known as well.

For molecules beyond diatomic molecules, we need to transform the Cartesian coordinate to normal coordinate, which is a way to separate coordinates that are strongly bonded together, oscillating atoms are unisome and having the same frequency. Since each atom can move in x, y and z diredtion and contributing to the vibration, which means that there are 3N different frequencies are available with N being the number of atoms in a molecule. It turns out that three of 3N coordinates are taken as translational motion, and another three coordinates are for rotation for non linear molecules. The rest of the coordinates, 3N - 6 of them, are for vibrational degrees of freedom. For linear molecules, two of the rotations are equivalent, so the number of vibration becomes 3N - 5.

The normal mode analysis gives the vibrational frequencies of a molecule and for each frequency how each atom moves. The coordinate to which the atoms move in one frequency is called normal mode.

+ Normal Mode Analysis

For a molecule with more than two atoms, it is not obvious how molecule vibrates. For a diatomic molecule, there is only one coordinate for which the distance between two atoms either shrink or elongate. What happens to a molecule such as water? How does it vibrate?

This is what normal mode analysis gives. The normal mode analysis is valid only in the local region of much larger global potential energy surface. It makes sense to limit to the small region. Afterall, when the IR spectrosopy is carried out, the spectra obtained are generally in the vicinity of the minimum that the molecule was in.

The model of a molecule is a linear triatomic molecule to make the argument simpler, as shown in Figure C1.

Figure C1. Linear triatomic molecules.

When the molecule vibrates, the coordinates, x1, x2, and x3, assiciated with each mass change in time. Since the motion is assumed to be harmonic (oscillatory), we can write the motion in terms of displacement xi, where i = 1, 2, and 3, from the equilibrium in the following way.

 $x i = A i e iωt$ C1
where Ai is the amplitude, ω is the frequency and t is time. We know from Newton's law, F = ma, the acceleration of mass is the second derivative of the dispacement with respect to time. Then,
 $x ¨ i =− ω 2 A i e iωt$ C2
The double dot on x denotes second derivative with respect to time. The Hooke's law force is given as F = -kq where q in this problem is the displacement coordinate xi that is connected to a spring. The force law allows us to write down equations of motion in the following way.
 $m 1 x ¨ 1 − k 1 ( x 2 − x 1 )=0$ C3 $m 2 x ¨ 2 + k 1 ( x 2 − x 1 )− k 2 ( x 3 − x 2 )=0$ C4 $m 3 x ¨ 3 + k 2 ( x 3 − x 2 )=0$ C5
In these equations, we see the equivalence of forces from the kinetic term (first term on the left-hand side) and the potential energy terms (second term on the left). In Eqn. C4, the second term on the right-hand side the sign of the x2 and x3 is reversed, since the mass m2 is attached to the both ends of the spring, the force associated with the springs have opposite signs.

If you think what is just written is rather arbitrary, then you should consider setting up Lagrangian function, L,, which is used in dynamics often. It is set up such that L(q, $\stackrel{˙}{q}$,t) = T(q, $\stackrel{˙}{q}$, t) - V(q), with $T\left(\stackrel{˙}{q},q,t\right)=\frac{1}{2}m{\stackrel{˙}{q}}^{2}$, where $\stackrel{˙}{q}$ is the time derivative of q. Then, solve the following equation.

 $d dt ( ∂L ∂ q ˙ )− ∂L ∂q =0$ C6
We should get the same result, i.e. Eqns. C3, C4 and C5 using Eqn. C6. More details will be found this way.

Now we seek a new coordinate by transforming these relationships such that all masses vibrate with the same frequency ω. We can substitute Eqns. C1 and C2 into Eqns. C3, C4, and C5, then we divide each equation by eiωt. Then, we get

 $( − m 1 ω 2 + k 1 − k 1 0 − k 1 − m 2 ω 2 + k 1 + k 2 − k 2 0 − k 2 − m 3 ω 2 + k 2 )( A 1 A 2 A 3 )=0$ C7
The amplitude of the vibration is give as Ai, and we seek solution for. We can solve this by using the following determinant

 $| − m 1 ω 2 + k 1 − k 1 0 − k 1 − m 2 ω 2 + k 1 + k 2 − k 2 0 − k 2 − m 3 ω 2 + k 2 |=0$ C8
Before calculations becomes too hard to see for illustrative purpose, let us look at a CO2 motions in x coordinate only. We consider, then, k1 = k2 = k and m1 = m3 = mO and m2 = mC for CO2, as shown in Figure C2.

Figure C2. CO2 molecule in stretching coordinates.

So, the matrix equation C7 becomes

 $( − m O ω 2 +k −k 0 −k − m C ω 2 +2k −k 0 −k − m O ω 2 +k )( A 1 A 2 A 3 )=0$ C9
Eqn. C8 is solved as determinant such that
 $| − m O ω 2 +k −k 0 −k − m C ω 2 +2k −k 0 −k − m O ω 2 +k |=0$ C10
You can follow this link to see how the mathematics works. If we expand the determinant,
 $( − m O ω 2 +k )[ ( − m C ω 2 +2k )( − m O ω 2 +k )− k 2 ]−( −k )( −k )( − m O ω 2 +k )=0$ C11 $( − m O ω 2 +k )[ m C m O ( ω 2 ) 2 − m C ω 2 k−2 m O ω 2 + k 2 ]− k 2 ( − m O ω 2 +k )=0$ C12 $( − m O ω 2 +k )[ m C m O ( ω 2 ) 2 − m C ω 2 k−2 m O ω 2 ]=0$ C13 $ω 2 ( − m O ω 2 +k )( m C m O ω 2 − m C k−2 m O k )=0$ C14
So, we have ω2 = 0, ω2 = k/mO, and ω2 = 2k/mC + k/mO. The motion associated with each frequency, ω must be solved separately.

ω2 = 0 case.: Substituting ω2 to Eqn. C9, we get from first row, A1 = A2, and the third row similarly has A2 = A3. These are substituted into the second row, we get zero. Therefore, there is no relative motion. This is a translational motion.

ω2 = k/mO: Substitution gives A2 = 0 from the first and the third row. It means that A1 = -A3, as shown in Figure C3.

Figure C3. CO2 normal mode: Symmetric stretch.

ω2 = 2k/mC + k/mO: Substitution gives A1 = A3 from the first and third rows. The second row gives A2 = -2A1 mO/mC. So, the two oxygen atoms move in the same direction and the carbon atom moves in one direction, as shown in Figure C4.

Figure C4. CO2 normal mode: Asymmetric stretch.

For a general case, we can see from Eqn. C7 that diagonal terms contain mass and frequency2 product with force constant, while off-diagonal elements are the negative of force constants, which shows connectivity of atoms in a molecule.

## Rotations

Next model system we're going to examine is rotations. This leads to discussion of how the shapes of atomic orbitals are the way they are, and this has to do with the angular momentum of the electron in the atom. Let us start with a simpler example of a particle on a ring system.

Classically, rotational kinetic energy is described by angular momentum. For a mass m to travel around a circle with its radius r, we can see that circumference is given by

 $2πr= s ν$ 56
where s is the velocity of the mass and ν is the frequency of the mass in circular motion. If we define rotational frequency to be ω = 2πν, then ωr = s, we have,
 $T= 1 2 m s 2 = 1 2 m ( ωr ) 2 = 1 2 I ω 2 = L 2 2I$ 57
where I = mr2, and L = I ω = mr2(2πν). The angular momentum is the momentum in rotation, and determines the kinetic energy of the mass.

Angular momentum give a force perpendicular to the plane of rotation, as shown in Figure 10. This is the same force acting on the wheels of bicycle, and when the speed is too high, it is hard to turn.

Figure 10. Angular momentum of a wheel

### Hamiltonian in 3-D rotation

Before moving to relatively simple system of particle on a ring, I'd like to show how Hamiltonian looks like in all full 3-D rotation. Let me give you a new coordinate system, a spherical polar coordinate. The coordinate in relation to Cartesian system is shown in Figure 11.
 Figure 11. Spherical coordinate system The mass is attached at the end of line segment r. The coordinate system along with Cartesian coordinates are in Figure 11. The angle φ is the polar angle formed by the projection of r on the x-y plane. It starts on the positive x axis in counterclockwise fashon. The angle θ is the azimuth angle of the r to the z axis.

The Hamiltonian of 3-D rotation in spherical polar coordinate is written as the following.

 $H ^ =− ℏ 2 2m 2 r 2 sin⁡θ [ sin⁡θ ∂ ∂r ( r 2 ∂ 2 ∂ r 2 )+ ∂ ∂θ ( sin⁡θ ∂ ∂θ )+ 1 sin⁡θ ∂ 2 ∂ ϕ 2 ]+V$ 58

It looks rather formidable.

### Particle on a ring

Let us look at a simpler system, where the rotation only occurs on a ring with a fixed r. It means that we need to fix θ = π/2 (which makes sinθ = 1), and ∂/∂θ goes to zero, because it is not changing. Then, we have much more benign looking Hamiltonian,
 $H ^ =− ℏ 2 2m r 2 d 2 d ϕ 2 +V$ 59 or $H ^ =− ℏ 2 2I d 2 d ϕ 2 +V$ 60
In terms of the physical model, the following should suffice.

Figure 12. Particle on a ring system

Now place Hamiltonian (Eqn. 60) in Schrödinger equation, we have

 $− ℏ 2 2I d 2 Φ d ϕ 2 =EΦ$ 61 $d 2 Φ d ϕ 2 =− 2IE ℏ 2 Φ$ 62
The right-hand side of Eqn. 62 is a constant. We seek a function that gives a constant and the function back. One of such solutions is Φ = Aeimφ with A and m are yet to be determined.

### Boundary condition

The boundary condition for this system is called cyclic boundary condition in that the function Φ must have the same value after one revolution. It means that Φ(φ) = Φ(φ+2π). Figure 13 shows the cyclic boudary condition, where Φ(0) = Φ(2π) with the matching first derivatives, as well.

Figure 13. Φ = eimφ at φ = 0 and 2π

### Normalization

So, we assume the wave function to have the form,
 $Φ=A e imϕ$ 63
The wave function contains i, and the complex conjugation makes the integration easy.
 $∫ 0 2π Φ * Φdϕ= 1$ 64 $A 2 ∫ 0 2π e −imϕ e imϕ dϕ= A 2 ϕ| 0 2π = A 2 ( 2π )=1$ 65 $A= ( 1 2π ) 1/2$ 66

### Energy

By differenciating the wavefunction twice, you get
 $dΦ dϕ =im ( 1 2π ) 1/2 e imϕ =imΦ$ 67 $d 2 Φ d ϕ 2 = d dϕ dΦ dϕ = d( imΦ ) dϕ =im ( 1 2π ) 1/2 d e imϕ dϕ =− m 2 Φ$ 68
By comparing Eqn. 68 and Eqn. 62, we can see that
 $− m 2 =− 2IE ℏ 2$ 69
Therefore, the energy of particle on a ring system is given as
 $E= ℏ 2 m 2 2I$ 70
where m = 0, ±1, ±2, 3± ...

Figure 14. Energy levels of particle on a ring

The wave function and density for the lowest three sets are given in Figure 15. For m = 0, the amplitude is constant. For higher energy states, the number of nodes increase by two as the absolute value of m increases by one to meet the cyclic boundary condition requirement.

Figure 15. Wave functions of particle on a ring

Here is a cute illustrative example using particle on a ring system.

Example: Benzene can be thought of as a system of electron on a ring. It is rather a drastic example, it suffices for back-of-envelope type calculation to get a ballpark figure. Let's predict the longest wave length associated with the abosorption of photon in benzene. Our system looks like the following.

Figure E1. Benzene as particles on a ring system

Since there are six electrons in π-orbitals, three lowest orbitals should be occupied. The transition for absorption of the longest wave length is from HOMO-LUMO. If we consider HOMO of benzene to be E±1 and LUMO to be E±2 states, the hν = E2 - E1. In atomic units, electron mass me = = 1, and c = 137.036. Using these we have,

 $E 2 − E 1 = ℏ 2 2 2 2I − ℏ 2 1 2 2I = ℏ 2 2I ( 4−1 )$ E1
So we have,
 $ΔE= 3 2 1 ( 1.4 Α ˙ ) 2 ( 1 a 0 0.529 Α ˙ ) 2 =0.214164 E h$ E2
At the same time,
 $ΔE= hc λ$ E3
Then,

 $λ= hc ΔE = 2π( 137.036 ) 0.214164 E h =4020 a 0 =213nm$ E4

The observed wave length is 255 nm for benzene. In ball park, not too bad for a back-of-envelope.

### Particle on a sphere

Now we're expanding the system from rotation on a plane to rotation on a sphere. However, again, we limit ouselves to have a constant radius, or rigid rotor approximation. Since r is constant, its derivatives are all zero in Eqn. 58. Furthermore, we consider free rotor, which means V = 0 on the sphere. Therefore, Hamiltonian for the system becomes
 $H ^ =− ℏ 2 2m 2 r 2 sin⁡θ [ ∂ ∂θ ( sin⁡θ ∂ ∂θ )+ 1 sin⁡θ ∂ 2 ∂ ϕ 2 ]$ 71

### Separation of variables

In Eqn. 71, we have two variables, θ and φ. We need to separate variables such that we have Schrödinger equation for each variable. We form the total wave function, denoted as Y, in the product form, as we have seen few times when two or more variables are involved
 $Y( θ,ϕ )=Θ( θ )Φ( ϕ )$ 72
Using the product form, Θ has only dependency on θ and Φ depends only on φ. By substituting Eqns. 71 and 72 into Schrödinger equation, we have
 $− ℏ 2 2m 2 r 2 sin⁡θ [ ∂ ∂θ ( sin⁡θ ∂ ∂θ )+ 1 sin⁡θ ∂ 2 ∂ ϕ 2 ]Y( θ,ϕ )=EY( θ,ϕ )$ 73 $− ℏ 2 2m 2 r 2 sin⁡θ [ Φ( ϕ ) ∂ ∂θ ( sin⁡θ ∂Θ( θ ) ∂θ )+ 1 sin⁡θ Θ( θ ) ∂ 2 Φ( ϕ ) ∂ ϕ 2 ]=EΘ( θ )Φ( ϕ )$ 74
Multiplying both sides by sin2θ/Y,
 $sin⁡θ Θ( θ ) ∂ ∂θ ( sin⁡θ ∂Θ( θ ) ∂θ )+ 2IE ℏ 2 sin 2 θ=− 1 Φ( ϕ ) ∂ 2 Φ( ϕ ) ∂ ϕ 2$ 75
The left-hand side of Eqn. 75 depends only on θ, while the right-hand side has φ dependence. In order for the left-hand side and the right-hand side of Eqn. 75 to equal at all times, the both sides must be equal to constant. This is because if, say, only θ is changing, but the right-hand side has no θ dependence. So, the right-hand side must be constant. It can be said the same about when only φ is changing. Then, we already know the answer for the solution to the right-hand side, and they are Eqns. 63 and 70.

The left-hand side is much harder! We set the left-hand side a constant.

 $sin⁡θ Θ( θ ) ∂ ∂θ ( sin⁡θ ∂Θ( θ ) ∂θ )+ 2IE ℏ 2 sin 2 θ= m 2$ 76
I might put the derivation later. It is lengthy, to say the least. The resulting Θ equation is
 $Θ l,m ( θ )= ( 2l+1 2 ( l−| m | )! ( l+| m | )! ) 1/2 P l | m | ( cos⁡θ )$ 77
Here the terms in the parenthesis on the right-hand side is the normalization constant, P(cosθ) is called associated Legendre polynomial. Since the polynomial has cosθ dependence, the independent variable for Legendre polynomial is confined to -1 to +1.

+ Legendre polynomial
By rearranging Eqn. 76 by moving m2 to the left-hand side, replacing 2IE/ℏ2 with l(l + 1), and multiply both sides by sin2(θ)/Θ(θ), we have
 $1 sin⁡θ d dθ ( sin⁡θ dΘ( θ ) dθ )+l( l+1 )Θ( θ )− m 2 Θ( θ ) sin⁡ 2 θ =0$ C1
As we saw in Eqn. 77 Θ function is a polynomial. This polynomial, the associated Legendre polynomial is derived from Legendre polynomial, which is a solution to the second-order differential equation of the form,
 $d dx ( x 2 −1 ) d dx P l ( x )=l( l+1 ) P l ( x )$ C2
such that the the polynomial is normalized with P(1) = 1. In Eqn. C2, we limit ourselves to have m = 0 so that the last term on the left-hand side of Eqn. C1 is zero. Then, the solution is given by
 $P l ( x )= 1 2 l l! d l d x l ( x 2 −1 ) l$ C3
Legendre polynomial is derivative of polynomial of order l+1, hence Legendre polynomial is polynomial of order l, and below lists several of Legendre polynomial.

l Pl(x)
0P0(x) = 1
1P1(x) = x
2P2(x) = (3x2 - 1)/2
3P3(x) = (5x3 - 3x)/2
4P4(x) = (35x4 - 30x2 + 3)/8
5P5(x) = (63x5 - 70x3 + 15x)/8

When m ≠ 0, then we need to seek full solution to Eqn. C1. The solution to Eqn. C1 then is called associated Legendre polynomial,

 $P l | m | ( x )= ( 1− x 2 ) | m |/2 d | m | d x | m | P l ( x )$ C4
The m dependence is given as the derivatives of Legendre polynomials. The values of |m| ≤ l can be understood that Pl is a polynomial of degree l, and the dervatives of Pl is confined to |m| ≤ l. First few associated Legendre polynomial are listed below.
l mPl(x) Pl(cosθ)
00P0(x) = 1 = 1
10${P}_{1}^{0}\left(x\right)$ = x $=\mathrm{cos}\theta$
11${P}_{1}^{1}\left(x\right)$ = (1 - x3)1/2 $=\mathrm{sin}\theta$
20${P}_{2}^{0}\left(x\right)$ = (3x2 - 1)/2 $=\frac{1}{2}\left(3{\mathrm{cos}}^{2}\theta -1\right)$
21${P}_{2}^{1}\left(x\right)$ = 3x(1 - x2)1/2 $=3\mathrm{sin}\theta \mathrm{cos}\theta$
22${P}_{2}^{2}\left(x\right)$ = 3(1 - x2) $=3{\mathrm{sin}}^{2}\theta$
30${P}_{3}^{0}\left(x\right)$ = (5x3 - 3x)/2 $=\frac{1}{2}\left(5{\mathrm{cos}}^{3}\theta -3\mathrm{cos}\theta \right)$
31${P}_{3}^{1}\left(x\right)$ = 3(5x2 - 1)(1 - x2)1/2/2 $=\frac{3}{2}\mathrm{sin}\theta \left(5{\mathrm{cos}}^{2}\theta -1\right)$
32${P}_{3}^{2}\left(x\right)$ = 15x(1 - x2) $=15{\mathrm{sin}}^{2}\theta \mathrm{cos}\theta$
33${P}_{3}^{3}\left(x\right)$ = 15(1 - x2)3/2 $=15\mathrm{sin}\theta \left(1-{\mathrm{cos}}^{2}\theta \right)$

### Put together the whole thing, 3D rotation

Now, let's look at them as a whole. First, the wave function is Θ·Φ called spherical harmonics, Y(θ,φ), and has a form,
 $Y=ΘΦ= ( 1 2π ) 1/2 ( 2l+1 2 ( l−| m | )! ( l+| m | )! ) 1/2 P l | m | e imϕ$ 78

## Hydrogen Atom

Hydrogen atom is an extension 3D rotation, except that we need to to let go of restriction of constant r, and we have attractive potential Read the section of hydrogen atom in Lowe and Peterson! Sorry, I was not able to finish this section up!!