A. La Rosa Lecture Notes

Q U A N T U M M E C H A N I C S

C O N T E N T

CHAPTER-8

FROM THE HAMILTONIAN EQUATIONS TO THE SCHRODINGER EQUATION

REPRESENTATION OF THE WAVEFUNCTION IN THE SPATIAL AND MOMENTUM SPACE

8.1Hamiltonian for an electron propagating in a crystal lattice

8.1.A Defining the BaseStates and the Hamiltonian Matrix

8.1.B Stationary States

Energy bands

8.1.C Time-dependent States

Electron wavepacket and group velocity

Effective mass

Electrical conductivity

8.2Hamiltonian equations in the limit when the lattice space tends to zero

From a general reference of the states I,

to adetailed account of the amplitude-probability on positionx

8.3 The Schrodinger Equation

8.3.A Non-relativistic free particle motion

8.3.BParticle in a potential V=V(x,t)

8.3.CBorn’s interpretation of the wavefunction

Statistical prediction of quantummechanics

Deterministic evolution of the wavefunction

8.3.DNormalization condition for the wavefunction

8.4Representation of the Wavefunction in the spatial and momentum space

8.4.AWavefunction in the Spatial Space

8.4.BWavefunction in the momentum Space

8.5Expectation Values

CHAPTER-8

REPRESENTATION OF THE WAVEFUNCTION IN THE SPATIAL AND MOMENTUM SPACE

8.1Hamiltonian for an electron propagating in a crystal lattice[1]

In our study of a two-state system, we learned that there is an amplitude probability for the system to jump back and forth to the other energy state.

Similarly in a crystal,

One can imagine an electron in a ‘well’ at one particular atom and with some particular energy.

Suppose there is an amplitude probability that the electron move into another ‘well’’ at one of the nearby atom.

From its new position it can further move to another atom or return to its initial ‘well’.

This study will allows to understand a ubiquitous phenomenon in nature that if the lattice is perfect, the electrons are able to travel through the crystal smoothly and easily, almost as if they were in vacuum.

8.1.A Defining the base states and the Hamiltonian matrix

We would like to analyze quantum mechanically the dynamics of an extra electron put in a lattice (as if to produce one slightly bound negative ion).

Some considerations first:

a) An atom has many electrons. However many of them are quite bounded that to affect their state of motion a lot of energy (in excess of 10 eV) is needed. We will assume that the motion of an extra electron (weakly bounded to the atom (the depth of the ‘well’ is much smaller than 10 eV.) It is the dynamics of this weekly bounded electron that we are interest in.

b) What would be a reasonable set of base states?

We could try:

Any statet of the one-dimensional crystal can then be expressed as,

t=nAnt=nnt (2)

where An(t)=n(t)is the amplitude probability, at the time t,to find the state is in the base state n.

c)How would the Hamiltonian look like?

(3)

Similar to the case on the ammonia molecules treated in the previous chapter, since all the atoms sites in the crystal are energetically equivalent, we can assume

Hnn = Eo , n= 1,2, 3, (4)

On the other hand, an electron can move from one atom to any other atom, thus transferring the negative ion to another place. We will assume however that the electron will jump only from one atom to the neighbor of either side.That is, for a given n, Hnj=0 unless j=n±1.

Hn(n+1)≡ -W , Hn(n-1) ≡ -W (5)

The equation of motion (3) takes the form

…

-WAn-2 +EoAn-1 - WAn

-WAn-1 +EoAn -WAn+1 (6)

-WAn +EoAn+1 -WAn+2

…

8.1.B Stationary States

Let’s look for a solution in which all the amplitudes An vary with time at the same frequency; that is, a state of single energy E (the latter to be determined.)

t=nAnt=nan

Replacing (7) in (6)

Ean = -Wan-1 +Eoan -Wan+1 (8)

If we label the atoms by their positions,

xn=nb

where b is the spacing between contiguous atoms in the crystal

equation (8) takes the form,

Ea(xn) = -Wa(x n-1) +Eo a(xn) -Wa(x n+1) (9)

Trial solution

a(xn) = (10)

The equation becomes,

E = - W +Eo - W

E = - W +Eo - W

E = Eo- W[ + ]

E(k) = Eo- 2W Cos (k b)(11)

For any choice of k there is a solutiona(xn) =

There is an infinite number of solutions (one for each k.)

For a given k,

  • We get a stationary state of energy E(k) =Eo-2WCos (k b) where the amplitude probabilities vary with time as,

Ank(t) =

  • Spatially, the electron is equally likely to be found at every atom; all the Ank(t)2 are equal. Only the phase is different for different atom location, changing in the amount (kb) from atom to atom.

Fig 8.1 Real part of the amplitude probability An for finding the atom location of the extra electron in the crystal, at a given time.

The energy Eas a function of k is shown in the figure below.

  • Only energies in the range from E -2W toE +2W are allowed. If an electron in a crystal is in a stationary state, it can have no energy other than values in this band.

(This allows understanding the energy bands formation in a crystal, which explains the working principle of electrical conductivity in metals and semiconductors.)

  • The smallest energies E~ (Eo -2W) are obtained for the smallest values of k 0. As k increases, the energy increases as welluntil it reaches a maximum at k=/b

For k  larger than /b the energy would start to decrease, but there is no need to consider such values, because they do not give new states.

Fig 8.2Energy of the stationary states as a function of k, where k is the wavevector of the state k=nAnk= n

In effect, while a wavevectork2 = k1+2/b has a profile of higher spatial and thus is indeed different than that the profile corresponding to the wavevectork1, it turns out that both profiles have the same value at each atom location (see Fig. 8.3 below). Hence, as far as the evaluation of the probability-amplitude at the atoms-site is concerned, both wave-vectors describe the same quantum state.

Fig 8.2Real part of A(xn) = for two different values of k, with k1and k2 separated by 2/b). Although both profiles are in general different, both have the same value at the atom-sites. Thus, both describe the same quantum state.

Formally, this is a consequence of the fact that.

since xn =nb

=

= (11)

Accordingly, only the values within the range –/b < k/b are considered in the description of the quantum states.

In this range, the higherkthe higher the energy of the stationary state.

For small values ofk, the following approximation holds,E(k) = Eo- 2W Cos (k b) = Eo- 2W [1 - (½) (k b)2] ;

E(k) = (Eo- 2W) + W b2k2 (12)

8.1.C Time-dependent States

In a stationary state, an electron is equally likely to be found in any atom in the crystal.

In contrast, how to represent a situation in which an electron of a certain energy is located in a certain region? That is, a situation in which an electron is more likely to be found in one region than at some other place?

Electron wavepacket

An alternative is to make a wavepacket, out of a proper linear combination of the stationary states found in the previous section; that is, using components of the type

nF(k)

with the proper weighting factorsF(k).

For example, we can make a wavepacket with a predominant wavenumber ko, and the other wavenumbers within a range k.

Fig. 8.3 Profile of the real part of the wavepacket at a given time t.

Using

Group velocity

We learned in Chapter-4 that , events though the components travel at different phase velocities still we can define a velocity for the wavepacket called the group velocity

If ko is in the low-energy range, then we can use expression (12) for the energy, which gives

(13)

Velocity of a low-energy wavepacket having a predominant wavenumberko,

Effective mass

Since E(k)=(Eo- 2W) + W b2k2 , we realize that low-energy electrons move with energy proportional to the square of its velocity, like a classical particle. Indeed, this resemblance with a classical particle can be made more transparent through the definition of an effective mass. For that purpose, firstlet’s move our zero energy reference to (Eo- 2W), as to obtain E= W b2k2=W b2=

(14)

This expression can be written as

(15)

as to resemble the motion of the quantum packet with a classical description of the motion of an electron. Waves of amplitude-probability behave like a particle.

The constant is called the “effective mass”.

(16)

has nothing to do with the electron’s mass. It has been defined as to resemble the motion of a classical particle. However, in a real crystal it turns out to have a magnitude of about 2 to 20 time the free-space mass of an electron.

Notice also,

(17)

We understand now how an electron can ride through the crystal, flowing perfectly free (with v=group) despite having all the atom to hit against. It does by having its amplitude probability flying from one atom to the next, with W providing the quantum probability to move from one atom to the next. That is how a solid conduct electricity.

8.2Hamiltonian equations in the limit when the lattice space tends to zero

We will consider now a more elaborate procedure which will allow us to describe the wavefunction in more detail, given its value at a particular space coordinate.

In the previous section we realized that

waves of amplitude-probability (wavepackets) in a crystal

behave like a particle.

One might expect that

the general quantum mechanical description of a particle

would show the same kind of behavior observed for the lattice.

In that sense, let’s make the analysis of wavefuntions in a discrete lattice to become our initial starting point from which deducea more general quantum mechanical description.

Strategy

  • Let’s take a linear lattice, and imagine that the lattice spacing “b” were to be taken smaller and smaller. In the limit, we would have the case in which the electron could be anywhere along the line.
  • In other words, assume the possibility oflabeling the space with infinity of points (in a continuous fashion.)

If we can work out the equations that relate the amplitude probabilities at one point to the amplitude-probability at neighboring points, then we would have the quantum mechanical description of an electron’s motion in the continuum space.

xn  x (18)

We know that xn=[A (xn)]

(Theexpression on the left constitutes a better notation, emphasizingthat there is no need to include the character A.)

This expression gives the amplitude probability of the general stateto be found atxn. This suggest to use the notation

xn=[A (xn)]≡(xn) x= A (x) ≡(x).

amplitude probability to find the stateatx.

(19)

Implementation

For a lattice we have thehamiltonian equation.

-WA(xn-1) +Eo A(xn) -WA(xn+1) (20)

which determines the amplitude-probabilities An ≡A(xn)ofthe general solution

This equation has stationary solutions k (one for each value of k within the range/bk/b)

k= xn [Ak (xn ,t)] (22)

where Ak (xn)=

and E(k) = (Eo- 2W) + W b2k2

For convenience, let’s move the origin of the energy scale to the minimum energy of the band. The value of Eobecomes equal to 2W in the equations above.

Note: When understood that we are referring to the amplitude probabilities of the state k, below we will use simply A(xn,t) instead of Ak (xn ,t) (as to simplify the notation.)

-WA(xn-1) +2W A(xn) -WA(xn+1)

E(k) = W b2k2 (23)

W[A(xn)-A(xn-1)] - W[A(xn+1) -A(xn)]

= Wb - Wb

= Wb - Wb

= -Wb[ - ]

= - Wb2

- Wb2 (24)

If we blindly take b --> 0, then E(k) =0 and , and we get nothing. Recall however that W gives the likelihood the electron jumps to the neighbor atom and , thus, it is plausibly to think that as the spacing “b” decreases the value of W simultaneously increases. We could consider that as “b” decreases, the productWb2 remains constant. In fact, expression (16) gives . So we will consider that as b --> 0, the value of W increases such that the product Wb2remains constant and equal to .

Consequently, as ,

-

We realize the amplitude probabilities depend on both x and t.Accordingly, it is more proper to use the partial derivative notation,

-

More specifically,

(25)

In summary,

The wavefunction of electron in the crystal’s stationary statek is described by,

k= xn [Ak (xn ,t)] k= x[Ak(x,t)dx]

using the notation defined in (19,)Ak(x,t)=k(x,t)

k= x[k(x,t)dx]

where the amplitude probabilities k(x,t) satisfy the differential equation,

8.3The Schrodinger Equation

8.3.A Non-relativistic free particle motion

The correct quantum mechanical equation for the motion of an electron in free space was first discovered by Schrodinger.

Schrodinger equation(26)

Notice that the Schrodinger equation is very similar to the equation (25) above. Indeed, the purpose of the discussion in the previous paragraph was to show that:

the correct fundamental quantum mechanics equation (26),

has the same form one gets for

the limiting case of an electron moving along a line of atoms.

Such limiting case of a linear lattice can help us understand better the Schrodinger equation. We can think of the equation (26) as a differential equation describing thediffusion of a probability amplitudefrom one point to the next along the line.If an electron has a certain amplitude probability to be at one point, it will have some amplitude to be at neighboring points.

In fact the Schrodinger equation looks very similar to the diffusion equation , the latter governing, for example, the time evolution of a gas spreading along a tube. But, there is a substantial difference due to the complex number in front of the time derivative of the Schrodinger equation:

while the diffusion equation leads to exponentially decaying solutions (as tends to a uniform distribution of molecules),

the Schrodinger equation leads to complex propagating waves.

8.3.B Particle in a potential V=V(x,t )

What would be the limiting case of the Hamiltonian equations if the atoms were subjected to an external potential?

For comparison, let’s take a look to the Hamiltonian equations corresponding to the case V(x)=0, which was given in the expression (6) and reproduced below,

…

-WAn-2 +EoAn-1 - WAn

-WAn-1 +EoAn -WAn+1 (27)

-WAn +EoAn+1 -WAn+2

…

What to do in expression (27) if V≠0?

Some clues can be drawn from the two-state system studied in Chapter 7, when the application of an external electric field was considered. In that case, the dynamic equation became,

that is, the additional energy appears in the diagonal terms.

One can foresee that the following equation (modified version of (27))

-WAn-1 + (Eo+Vn)An -WAn+1 (29)

would lead to

Or, equivalently

(30)

The Schrodinger equation is indeed

“This equation marked a historic moment constituting the birth of the quantum mechanical description of matter.

For many years the internal atomic structure of the matter had been a great mystery. No one had been able to understand what held matter together, why there was chemical binding, and specially how it could be that atoms could be stable. Although Bohr had been able to give a description of the internal motion of an electron in a hydrogen atom which seemed to explain the observed spectrum of light emitted by this atom, the reason that electrons moved this way remained a mystery.

Schrodinger’s discovery of the proper equations of motion for electrons on a atomic scale provided a theory from which atomic phenomena could be calculated quantitatively, accurately and in detail.” Feynman’s Lectures,” Vol III, page 16-13

8.3.CBorn’s interpretation of the wavefunction

In 1926 Max Born gave postulated the following interpretation for the wavefunction:

If, at the instant , a measurement is made to locate the particle associated with the wavefunction , then the probability that the particle will be found within the volume element around the position is equal to

= (32)

where * stands for the complex conjugate number.

That is, plays the role of probability density.

Pictorially, the particle must be at some location where the wavefuntion has an appreciable value.

Note:

The predictions of quantum mechanics are statistical

In order to know the state of motion of a particle, we must make a measurement

But a measurement necessarily disturbs the system in a way that can not be completely determined

Notice however that , being the solution of a differential equation (the Schrodinger equation), varies with time in a way that is completely deterministic. That is, if were known at t=0, the Schrodinger equation determines precisely its form at any future time.

But, what happens is that the wavefunction at t=0 can not be completely be uniquely determined. This is because, a set of measurements at t=0 at most may lead to the determination of but not uniquely defines.

8.3.DNormalization condition for the wavefunction

The probabilistic interpretation of the wavefunction implies, therefore, the following requirement:

(33)

because given a particle, the likelihood to find it anywhere should be one. Inherent to this requirement is that,

(34)

Notice that if is a solution of the Schrodinger equation, the function (being a constant) is also a solution. The multiplicative factor therefore has to be chosen such that the function satisfies the condition (34). This process is called normalizing the wavefunction.

In general, there will be solutions to the Schrodinger equation (31) whose integral (33) tend to an infinite value. This means they are non-normalizable and therefore can not represent a particle probability density. Such functions must be rejected on the grounds of Born’s probability interpretation.

Quantum mechanics states are represented by square-integrable functions that satisfy the Schrodinger equation

  • Suppose that is normalized . As the time evolves, will change. How do we know if it will remain normalized?

Here we show that the Schrodinger equation has the remarkable property that it automatically preserves the normalization of the wavefunction:

Proof:

Let’s start with

(36)

On the other hand,

(37)

According to the Schrodinger equation

Taking the complex conjugate, and assuming that the potential is real,

Adding the last two expressions,

(38)

Replacing (38) in (37) and (36),

The expression on the right is zero because

1

[1] The Feynman Lectures, Vol III, Chapter 13