Chapter 6. The Physical Interpretation of the Wave Function

QM 6. Physical interpretation of wave function, June 2, 2005 1 

Chapter 6. The Physical Interpretation of the Wave 

Function 

x1. So far we have focused on the physical interpretation of eigenvalues and 

the transitions between them. But what about eigenfunctions What kind 

of physical information do they carry This chapter answers this question. 

By far the most important eigenvalue equation in Quantum Mechanics is 

the SchrÄodinger equation 

^HÃ n (r(1);:::;r(N)) = E n Ã n (r(1);:::;r(N)); n =0; 1; 2;::: (6.1) 

Here ^H is the Hamiltonian operator, and Ã n and E n are its eigenfunctions 

and eigenvalues. Several names are frequently used for the eigenfunctions 

Ã n (r(1);:::;r(N)) of ^H: Ãn is the wave function of the N-particle system 

having the energy E n ; or the energy eigenfunction corresponding to the energy 

E n ; or we say that the system has a state Ã n with the energy E n . 

It is a postulate of Quantum Mechanics that, if the system is in the state 

Ã n (r(1);:::;r(N)), then the quantity 

p n (r(1);:::;r(N))dr(1) ¢¢¢dr(N) =jÃ n (r(1);:::;r(N))j 2 dr(1) ¢¢¢dr(N) 

(6.2) 

is the probability that particle 1 is located in a cube of volume dr(1) =


dx(1)dy(1)dz(1) centered around r(1), and particle 2 is located in a cube of 

volume dr(2) = dx(2)dy(2)dz = (2) centered around r(2), and . . . . In what 

follows, jÃj 2 ´ Ã ¤ Ã where Ã ¤ is the complex conjugate of Ã. 

This information is valid only if we are sure that the system is in the 

state Ã n (r(1);:::;r(N)). To be sure of that, we must perform an experiment 

that places the system in that state. For example, we can force a molecule 

that is initially in the ground state E 0 to absorb a photon and be excited to 

the state Ã 2 (r(1);:::;r(N)). The probability that the atoms in the excited 

molecule have certain positions r(1);:::;r(N) is given by Eq. 6.2 with n =2. 

I will con¯ne the discussion that follows to the case of a particle moving 

along a straight line. I do this to simplify some of the equations. What you 

learn from these examples can be easily extended to many particles moving 

in three-dimensional space. 

If the particle moves along a straight line (one-dimensional motion), its 

position is speci¯ed by one coordinate x. If the particle is in a state whose 

wave function is Ã(x), then the probability that the particle is located between 

x and x + dx is 

p(x)dx = jÃ(x)j 2 dx (6.3) 

The probability that the particle is located in the region between a and


b is then 

Z b 

a 

p(x)dx = 

Z b 

a 

jÃ(x)j 2 dx (6.4) 

We add up the probabilities that the particle is at points between a and b. 

The probability that the particle is somewhere along the line is 

Z +1 

¡1 

jÃ(x)j 2 dx =1 (6.5) 

This integral is equal to 1 because we are certain that the particle is somewhere 

on the line; the probability of an event that is certain is equal to 1, by 

de¯nition. 

When a wave function satis¯es Eq. 6.5, we say that it is normalized. 

Eq. 6.5 is called the normalization condition. We can generalize this condition 

to the case of N particles moving in three-dimensional space: 

Z +1 

¡1 

jÃ(r(1); r(2);:::;r(N))j 2 dr(1)dr(2) ¢¢¢dr(N) =1 

The probability that the N particles are somewhere in the space must be 

equal to 1. 

It is this interpretation of the wave function that prompted us to require 

that all eigenstates that have a physical meaning must be normalized. 

x2. Application to a vibrating diatomic molecule. To better understand the 

probabilistic interpretation of the wave function and the manner in which it


A 

B 

Figure 6.1: A model for a vibrating diatomic molecule: two atoms connected 

by a spring 

is used, I will examine the example of a vibrating molecule. The simplest 

model for the vibrational motion of a diatomic molecule assumes that the 

interaction between the two atoms is accurately simulated by a spring (see 

Fig. 6.1). The potential energy of such a spring is 

V (x) = 1 2 k (r ¡ r 0) 2 ´ kx2 

2 


where r is the distance between the centers of the atoms and k is called the 

force constant of the bond (the spring). The distance r 0 corresponds to the 

lowest potential energy, and it is the bond length. It is not di±cult to show 

that the potential energy V (x) in Eq. 6.6 leads, in classical mechanics, to an 

oscillatory motion of the distance x between the atoms, with a frequency 

s 

k 

! = 

¹


where 

¹ = m 1m 2 

m 1 + m 2 

is the e®ective mass (m 1 and m 2 are the masses of the two atoms). 

Since we know the potential energy, we can write the SchrÄodinger equation 

for this system and solve it. For the ground state (the lowest energy 

eigenstate), the wave function is 

Ã 0 (x) = 

µ # 

¹! 

1=4 

exp 

"¡ ¹!x2 

¼¹h 

2¹h 


and the ground state energy is 

E 0 = ¹h! 2 


Thewavefunctionofthe¯rstexcitedstateis 

with the energy 

Ã 

4 

Ã 1 (x) = 

¼ 

µ ! 

¹! 

3 1=4 ∙ 

x exp ¡ ¹! 

x2¸ 

¹h 

2¹h 

E 1 = 3¹h! 

2 



We will study harmonic oscillators in more detail in Chapter 16. The 

information given above is all we need here. 

x3. Data for HCl. To perform calculations with the wave functions given 

above, we need to work with a speci¯c molecule. I choose here HCl, for which 

I have the following data.


The reduced mass is ¹ = m 1 m 2 =(m 1 + m 2 ). For m 1 =1:0078 amu = 

1:0078 £ 1:6605 £ 10 ¡24 gram and m 2 =34:968 amu = 34:968 £ 1:6605 £ 

10 ¡24 gram (I use the conversion factor 1 amu = 1:6605£10 ¡24 g; see Appendix 

1), I have ¹ =0:979568 £ 1:6605 £ 10 ¡24 gram (see WorkBookQM.6.1). 

Since the force constant 1 is k =4:9 £ 10 5 dyne/cm, we calculate that the 

frequency ! has the value 

s s 

k 

! = 

¹ = 4:9 £ 10 5 erg 

0:979568 £ 1:6605 £ 10 ¡24 gr =5:4886 £ 1014 sec ¡1 

Using ¹h =1:0546 £ 10 ¡27 erg sec (see Appendix 1) and the values of ¹ 

and ! just calculated, we can write 

Ã 0 (x) = 

Ã 

0:979 £ 1:66 £ 10 ¡24 £ 5:489 £ 10 14 

£ exp 

" 

¼ £ 1:055 £ 10 ¡27 

! 1=4 

g 

sec erg sec 

¡ 0:979 £ 1:66 £ 10¡24 £ 5:489 £ 10 14 gcm 2 

2 £ 1:055 £ 10 ¡27 sec erg sec 

µ # 

x 

2 

cm 


Let us look at units. The unit of the wave function is (g/erg sec 2 ) 1=4 , 

where g stands for grams. Since erg = g cm 2 sec ¡2 ,wehave 

g 

sec 2 erg = 

g 

= 1 

sec 2 g cm2 cm ; 2 sec 2 

1 D. A. McQuarrie, Statistical Mechanics, Harper Collins, New York, 1976, p. 95


and the wave function has units of 

" 

gr 

# 1=4 

= 1 

sec 2 erg cm 1=2 

But cm is a cumbersome unit for length in molecular physics, because it is 

much too large. I will use ºA instead. This means that 

" 

# 1=4 

g 

= 1 

sec 2 erg cm = 1 

1=2 

10 4 ºA 1=2 

and the factor in front of the exponential in Eq. 6.11 is 

Ã 

0:979 £ 1:66 £ 10 ¡24 £ 5:489 £ 10 14 ! 1=4 

1 2:27837 

= (6.12) 

¼ £ 1:055 £ 10 ¡27 10 4 1=2 

ºA ºA 1=2 

The units of the exponent in Eq. 6.11 are 

gcm 2 

sec 2 erg =1 

This means that the exponent is dimensionless, if I use for x units of cm. 

But I want to use ºA. Therefore the exponent is 

¡ 0:979 £ 1:66 £ 10¡24 £ 5:489 £ 10 14 µ x 2 µ x 2 

= ¡42:327 (6.13) 

2 £ 1:055 £ 10 ¡27 10 ºA 8 ºA 

In this expression I must use x in ºA. 

Putting it all together gives 

Ã 0 (x) = 2:27837 

" µ # 

x 

2 

exp ¡42:327 

ºA 

ºA 1=2 

(6.14)


The probability that the di®erence x = r ¡ r 0 , between the interatomic 

distance r and the equilibrium bond length r 0 ,takesavaluebetweenx and 

x + dx, when the vibrating molecule is in the ground state (see x1), is 

p 0 (x)dx = 

Ã " µ #! 

2:27837 

x 

2 2 

exp ¡42:327 dx (6.15) 

ºA 1=2 ºA 

Because the wave function has units of ºA ¡1=2 , dx in this formula must have 

units of ºA andp(x)dx is dimensionless (as it should be). 

A plot of the probability distribution function p 0 (x) versusx is shown 

in Fig. 6.2 as a solid line. 

As you can see, when the molecule is in the 

ground state, it is more likely that the distance between the atoms is equal 

to the equilibrium distance r 0 . However, the probability that the bond length 

deviates from this value (i.e. that x = r ¡ r 0 6= 0) is not negligible. 

The ¯gure also shows (dotted line) the probability distribution function 

p 1 (x) =Ã 1 (x) ¤ Ã 1 (x) ´jÃ 1 (x)j 2 

when the molecule is excited in the ¯rst vibrational state. Here Ã 1 is given 

by Eq. 6.9. p 1 (x)dx is the probability that r ¡ r 0 takesvaluesbetweenx and 

x + dx, when we know that the oscillator is excited to the state Ã 1 . 

x4. Interpretation. Here we go again: probabilities instead of certainty. The 

laws of classical mechanics place no restrictions, in principle, on position


5 

|ψ 0 | 2 

4 

3 

2 

|ψ 1 | 2 

1 

-0.4 -0.2 0.2 0.4 

x=r 12 -r 0 Angstrom 

Figure 6.2: Probability distributions for the bond length x when the oscillator 

is in the ground state (solid line) and the ¯rst excited state (dashed line) 

measurements. 

Only technical di±culties prevent us from measuring the 

position with arbitrary accuracy. 

Quantum Mechanics tells us that if we 

were to measure the interatomic distance r, for a single oscillator, we would 

get di®erent results in di®erent measurements, performed under identical 

conditions. 

As in the case of the measurements discussed in the previous chapter, 

knowing the probability with which something takes place is useful when we 

make a large number N of identical measurements. If N is large enough, we


can be sure that for 

Np 0 (x)dx 

oscillators in the ground state Ã 0 (x), x ´ r ¡ r 0 takesvaluesbetweenx and 

x + dx. 

x5. Average values. The probability is also useful for calculating various 

averages. For example, the average distance between the atoms, when the 

diatomic molecule is in the ground state, is 

hri = 

Z 1 

¡1 

= r 0 

Z 1 

= r 0 

rp 0 (r)dr = 

¡1 

p 0 (x)dx + 

Z 1 

¡1 

Z 1 

¡1 

(r 0 + x) p 0 (x)dx 

xp 0 (x)dx 

In this calculation we have used the fact that x = r ¡ r 0 and 

p 0 (r)dr = p 0 (x)dx 

Here p 0 (r)dr is the probability that the distance between the atoms takes 

values between r and r + dr when the molecule is in the ground state. We 

have also used the facts that 

and 

Z 1 

¡1 

Z 1 

¡1 

p 0 (r)dr =1 

xp 0 (x)dx =0


You can calculate these integrals fairly easily by hand, or by Mathematica 

or Mathcad. 

Exercise 6.1 You have a sample of 10 12 HCl molecules. The force constant 

is k =4:9 £ 10 5 dyne/cm and the equilibrium bond length is r 0 =1:27460 

ºA. Calculate how many molecules that are in the ground state have a bond 

length between 1.24 ºA and1.29ºA. 

Exercise 6.2 Do the same type of calculation for I 2 . 

For this molecule, 

r 0 =2:666 ºA and! =214:57 cm ¡1 . (Data from G. Herzberg, Molecular 

Spectra and Molecular Structure. I. Spectra of Diatomic Molecules, VanNostrand, 

New York, 1950, p. 540.) Calculate how many molecules, out of 10 12 , 

have an interatomic distance between 2.53 ºA and2.66ºA. 

Note: ! =214:57 cm ¡1 is a strange unit. 

In Appendix 2, there is a table 

that tells you that 1 cm ¡1 = 1:9862 £ 10 ¡16 erg. This means that 

¹h! = 1:9862 £ 10 ¡16 £ 214:57 erg. 

From this information, I can calculate 

the frequency: since ¹h =1:0546 £ 10 ¡27 erg sec, I have ! =¹h!=¹h = 

(1:9862 £ 10 ¡16 £ 214:57) = (1:0546 £ 10 ¡27 )=4:041 £ 10 13 sec ¡1 . 

x6. The e®ect of position uncertainty on a di®raction experiment. Let us look 

at a system consisting of diatomic molecules adsorbed on a surface. Under


(a) 

(b) 

Figure 6.3: Upper ¯gure shows the surface atoms undisturbed by uncertainty 

in positions. Lower one shows the real instantaneous positions. 

certain conditions, these molecules position themselves to form a periodic 

array (see Fig. 6.3). If all the molecules have the same bond length, they form 

a perfect \grating" on the surface (Fig. 6.3a). If we scatter from the surface 

a beam of X-rays (or electrons) having a certain wavelength, the grating will 

di®ract the beam and produce a di®raction pattern. From this pattern, we 

can, by using di®raction theory, determine the positions of atoms. 

However, we know that the molecules will not have the same bond length. 

AccordingtoQuantumMechanics,thebondlengthineachmoleculehasa 

given value with a given probability. 

The instantaneous positions of the


Figure 6.4: The dotted lines show the locations the atom can reach due to 

uncertainty in position 

atoms are shown, with some exaggeration, in Fig. 6.3b. The grating formed 

by the molecules is not perfect. 

We can calculate the di®raction pattern 

produced by the molecules in Fig. 6.3a and by those in Fig. 6.3b. When calculating 

the di®raction pattern produced by Fig. 6.3b, we use the probability 

p 0 (x)dx. The experiments ¯nd that the observed pattern is that calculated 

for the imperfect grating. This is an indirect con¯rmation of the quantum mechanical 

postulate that di®erent distances between atoms are observed with 

di®erent probabilities; moreover, the agreement with experiment is quantitative, 

con¯rming the formula used for computing the probability that the 

molecule has a given the bond length. 

x7. The e®ect of position uncertainty in an ESDAID experiment. Consider


another experiment, performed by John Yates and Ted Madey, when they 

were working at NIST. They started with a perfect metal surface whose 

atoms are arranged in a nearly perfect periodic array (see Fig. 6.4). Then 

they adsorbed atoms or molecules on the surface. In Fig. 6.4, I show one such 

atom, adsorbed on top of a surface atom. The most likely position of the 

adsorbed atom is shown by heavy lines in Fig. 6.4. However, the positions 

shown with dotted lines | and those between them| also occur with some 

probability. 

This surface is bombarded with high-energy electrons, which break the 

metal-atom chemical bond. When the bond is broken the adsorbed atoms 

come o® the surface. The atoms whose bond is perpendicular to the surface 

leave it, when the bond is broken, in a direction perpendicular to the surface 

(see the solid line and arrow in the ¯gure). The ones that are slanted come 

o®, when the bond is broken, at an angle with the surface. The experiment 

detects the point where the atoms leaving the surface hit a position-sensitive 

detector (represented by a slab at the top of the ¯gure). An atom that leaves 

the surface on a trajectory perpendicular to the surface hits the center of the 

detector. One that leaves the surface on a slanted trajectory hits the detector 

o®-center. From the distribution of the points of impact on the detector, one


can calculate the probability of the initial position of the atom. The result 

agrees with Quantum Mechanical calculations. 

I should caution you that the real experiment is close in spirit to the 

description just given, but is more complicated. 

x8. Why did we think that we could measure position accurately You have 

just learned that we cannot know with certainty the positions of the particles 

in a system. Why isn't this true for the large objects we encounter in everyday 

life Quantum mechanics has a simple answer: the larger a particle, the less 

uncertain its position. 

Let us see how we reach this conclusion, in the case of the bond length 

of a vibrating diatomic molecule. If the oscillator is in the ground state, the 

probability that x = r ¡ r 0 has a value between x and x + dx is 

I can write Eq. 6.16 as 

jÃ 0 (x)j 2 dx = 

jÃ 0 (x)j 2 dx = 

µ # 

¹! 

1=2 

exp 

"¡ ¹!x2 dx (6.16) 

¼¹h 

¹h 

µ 1 1=2 

" µ # 

x 

2 

exp ¡ dx (6.17) 

¼¾ 2 ¾ 

where 

¾ 2 ´ ¹h 

¹! 


is a quantity with units of length squared.


From Eq. 6.18 you can see that if x À ¾, thenjÃ 0 (x)j 2 is practically zero. 

Therefore I am sure that the bond length r is never much longer than r 0 + ¾, 

nor is it much shorter than r 0 ¡ ¾. Thesmaller¾ is, the more accurate is my 

knowledge of the bond length. Thus ¾ is a good descriptor of the accuracy 

with which I can know the bond length. 

Let us calculate ¾ for a diatomic whose reduced mass is a gram and whose 

frequency is ! =10 14 sec ¡1 .Inthiscase 

¾ 2 = ¹h 

¹! = 1:0546 £ 10¡27 erg sec 

1gr£ 10 14 gr/sec 

I¯ndthat¾ 2 » = 10 ¡41 cm 2 and ¾ » = p 10 £ 10 ¡20 cm. 

This uncertainty in position is beyond the accuracy of any conceivable instrument. 

Because of this, in the classical world (large mass), we are entitled 

to claim that it is possible to know the position of a particle with \unlimited" 

accuracy. We know that this is not quite right, but no one will be able to 

prove us wrong in our lifetime. 

The statement that position is precisely known when the mass is large is a 

dumb statement, even though it is made frequently. The word `large' makes 

sense only if we compare one mass to another. When I say that a friend is 

heavy, I am implicitly compare him to the average person, or compare his 

weight to what it was twenty years ago. So what do I mean when I say that


for a heavy oscillator, the position is known accurately Heavy compared to 

what 

This is easy to answer for the case of the oscillator. Let us say that the 

highest accuracy of the best position measurement is ¾ 0 .If 

s 

¹h 

¹! ¿ ¾ 0 (6.19) 

then the quantum uncertainty in the position of the oscillator is not detectable. 

The inequality (6.19) is satis¯ed if the mass satis¯es 

s 

¹h 

¹ À 

!¾ 2 0 

If ¾ 0 =10 ¡8 cm, then 

s 

1:0546 £ 10 

¡27 

¹> 

g= p 10 

10 14 £ 10 ¡25 g= p 10 £ 10 ¡12 g 

¡16 

For a diatomic with reduced mass exceeding 3 £ 10 ¡12 gram, the di®erence 

in the bond length between di®erent copies of the same molecule is not detectable. 

Because classical physics deals with objects heavier than 10 ¡12 

gram, we can claim that position is accurately determined by the laws of 

mechanics. 

Exercise 6.3 One way of describing the uncertainty in the position is to use 

the mean-square error 

Z +1 

`2 ´ 

¡1 

x 2 jÃ 0 (x)j 2 dx


Calculate this quantity and show that ` is of order ¾ = q ¹h=¹!.

Chapter 6. The Physical Interpretation of the Wave Function

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