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Lectures 5- Quantum Mechanics is constructed from a set of postulates about the way microscopic particles behave. These postulates have the same logical role in Quantum Mechanics that axioms do in Mathematics, or laws do in Thermodynamics. Each of these
- postulates, axioms, and laws – act as the logical foundation on which the theory is built. They have different origins, and lead to systems with different bases for truth. Axioms are invented, and then the mathematical system inferred from them. The only basis for truth in mathematics is internal consistency based on logical inferences – i.e. whether a theorem based on the axioms is proven to be correct. There is no basis for deciding on the truth or falsehood of an axiom however, just for the conclusions from the axioms. Laws are summaries of experiments, and some of them are used to develop complex theoretical systems, as in the case of Thermodynamics. Unlike axioms, the laws themselves can be proven true or false. There are two bases for this – first, refined experiments may directly demonstrate that the laws are untrue. Second, experiments can demonstrate that logically correct predictions based on the laws are untrue, which logically shows that the laws themselves are untrue. Thus scientific systems differ from mathematical systems in that internal consistency is no longer sufficient to determine truth or falsehood – successful comparison to an external reality – experimentation – is also necessary. Postulates are similar to axioms in that they are invented by the theorist. They are different from axioms in that they are invented for the purpose of explaining experimental observations. The postulates themselves are often not directly testable,
however. They are used to develop a theoretical framework – i.e. to make predictions. These predictions are then compared to experiment. The truth or falsehood of the postulates is then based on the success or failure of the predictions. It is important to understand this, because a common trap for students first encountering Quantum Mechanics is to try to understand the direct justification for the postulates. This is not possible, since they are not based on direct experience (are not laws), and most certainly are not based on common sense. Remember that common sense is based on the direct experience of our senses (hence the name), and the realm of quantum mechanics is in dimensions of size, energy and mass that are too small to be detected by our senses. Thus it is pointless to try to understand where the postulates come from. It is only useful to understand their implications, the way that predictions are based on them, and the truth or falsehood of these predictions. The first of the postulates of quantum mechanics is that the wavefunction (x) contains all available information about the system. If the information is not contained in (x), then it is information that quantum mechanics says we cannot obtain. There are two types of wavefunctions. One type of wavefunction is a wavefunction that is a solution to the Schrödinger equation. These wavefunctions are called eigenfunctions, and have special properties that we will discuss later. The other type of wavefunction is a wavefunction that is not a solution to the Schrödinger equation. In either case, the wavefunction provides a complete description of the measurable properties of a system. However, we still need to learn how to extract this information once we’ve determined a wavefunction.
g(x)= Af(x)=^ ^ (^) x^2 2 (2x)= (^) x (2)= 0
Example 2) A = ^ (^) x , f(x, y)= sin ( xy ).^2 What is g(x)?
g(x)= Af(x)= ^ (^) ^ x ( sin ( xy ))= y^2 2 cos (x y ).^2 Quantum mechanics uses only a class of operators called linear, Hermitian
operators. An operator A is linear if
A c f ^1 (^) 1 ( ) x c f 2 (^) 2 ( ) x c Af 1 ^ (^) 1 ( ) x c Af 2 ^ (^) 2 ( ) x
In words, this means that an operator is linear, if the operator operating on a linear combination of operands results in the same linear combination of the operator operating
on each individual operand. For example the operator (^) dxd is linear. To see this we plug
our operator into the definition and get d dx c f (x)+ c f^ (x) = c
df dx + c
df (^1 1 2 2 1 1 2) dx^2 ,
which satisfies the condition for a linear operator. It is also useful to note that a linear
combination of linear operators is also a linear operator. For example, since dxd^22
and x are both linear operators, (^5) dx^ d^22 ix , a linear combination of the operators, is also a
linear operator. Prove this as an exercise.
An example of an operator that is not linear is the operator A = square, i.e. the operator that squares a variable. To see this we plug our operator into our definition of a linear equation and get
( c f 1 (^) 1 c f 2 (^) 2 )^2 = c 1^2 f (^) 12 + c c f f +c 21 2 1 2 2^2 f (^) 22 c f 1 (^) 12 c f 2 22. An operator is Hermitian if for two functions 1 and 2
^ 1 ^ (^ A^ ˆ^ ^^2 )^ d ^^ ^2 (^ A ˆ^^1 ) d
The reason that quantum mechanical operators must be Hermitian is that the eigenvalues (solutions) obtained by solving quantum mechanical equations are always real when the operators are Hermitian. Since we will be using the solutions of our equations to describe physical observables the requirement that the operators are Hermitian ensures that the results of the equations will be physically meaningful.
As an example of how this equation works, let A ˆ be the operator P ˆ x ih dx^ d ,
the operator for the linear momentum in the x-direction, and let
1 ( ) x (^1) 1/ 4 e x^2 / 2 x
and
2 ( ) x 2 1/ 21/ 4 xe x^2 / 2 x
Therefore,
A ˆ^ (^) 2 ( ) x i d^2 1/ 21/ 4 xex^2 / 2
dx
^
i^2 1/ 21/ 4^ [ e x^2^^ / 2^ x e^2 x^2 / 2]
^
and
1 * ( ) x A ˆ 2 ( ) x dx i^2 1/ 2 ( e x^2^^ x e^2 x^2 ) dx
(^)
^ ^
by a constant. We call a function which satisfies an eigenvalue equation an
eigenfunction of the operator A , and we call the constant a an eigenvalue of A . In
other words if I have some function f(x) and an operator A , and I operate on my function
with A , and get back my function times a constant, then my function is an eigenfunction and my constant is an eigenvalue. Why do we care about eigenfunctions and eigenvalues? This is where our third postulate comes in. It says that once we know the identity of an operator that corresponds to an observable, the only values of that observable which can be measured are the eigenvalues ai of that operator. As a first example, the one dimensional Hamiltonian operator , H ^ 2 m^2 ^ x^2 2 V x ( ) , is the operator which corresponds to the total energy of a particle
moving along the x-axis. Its eigenvalues are the only values of the energy that can be measured. This is exactly the same as saying that these are the only values of the energy that the particle can have. The Hamiltonian operator in quantum mechanics corresponds to a function of classical mechanics called the Hamiltonian that represents the total energy of a conservative system. The classical Hamiltonian is the sum of the kinetic energy T, and the potential energy, V(x), i.e.,
H T V x ( ) pmx V x ( ) 2 2 By comparing the classical Hamiltonian and the quantum Hamiltonian, we can figure out the operators for several classical observables. First note that the potential energy , V(x) appears in both the classical and quantum Hamiltonians. This must mean that the
operator which corresponds to the potential energy is simply the potential energy itself,
i.e., V = V(x). Since the only other component of the Hamiltonian is the kinetic energy
in the x direction, Tx, the operator for the kinetic energy , T x 2 m^2 ^ x^22. As a final
example, if we want to figure out the operator for the linear momentum in the x
direction, we note that the classical kinetic energy is given by T x pmx 2 2 and the quantum
mechanical kinetic energy is given by T x 2 m^2 ^ x^22. This suggests that p x^2 ^2 ^ x^22
and that therefore p i ^ x.
Comparisons like these lead to the following rules for generating the operators that correspond to various classical observables.
- The operator for a position variable, q , is the position variable itself. Thus
the operator for position in the y direction, is y = y, and the operator for the potential
energy in a conservative system, which is a function of position only, is V = V(x).
- The operator for momentum, p , is (^) i ^ q , where q indicates a position
variable. For example, the operator for the momentum in the z direction is p z i ^ z.
All other operators can be generated as a function of position and momentum operators. Therefore, we need to learn some rules for creating functions of operators. For the sum of two operators we simply have
(A+ B)f(x)= Af(x)+ Bf(x) ^ ^ ^
In addition to the operators I've already shown you, we will determine other operators as we need them. Once we obtain the operator for an observable, another postulate says that if the wavefunction of the system is an eigenfunction of an operator, then the only measured value of the observable will be the eigenvalue corresponding to this eigenfunction. Let me repeat this, because it is very important. If the wavefunction that defines the state of a system is an eigenfunction of some operator, then the only measured value of the observable associated with that operator will be the eigenvalue associated with that eigenfunction. However, if the wavefunction of the system, which we also call the state of the system, is not an eigenfunction of the operator, then each measurement made on the system will still be one of the eigenvalues of the operator. We are just unable to predict which of the eigenvalues it will be. However, later we'll learn how to calculate the probability that a given eigenvalue will be observed.
Lecture 7 Let’s see how what we’ve talked about so far works by applying it to a pair of simple systems. The simplest problems in quantum mechanics are the free particle and the particle in a one-dimensional box. A free particle is one that can move unconstrained through space, with no potential impeding its motion. In other words, for the free particle moving in one dimension, V(x) = 0. Thus the Schrödinger equation for the free particle is
- 2m^ ^2^^ ^2 x (x) 2 E ,
which we can rewrite as 2
x^^ (x) 2 = - 2mE 2 = -k 2 ,
where k = ( 2mE 2 )1/ 2 , and is called the wavevector of the particle. The most general
solution to this equation is
(x)= (A cos kx+ B sin kx)^.
We can see that this satisfies the Schrödinger equation by plugging this value of (^) into
our equation. When we insert our solution into the Schrödinger equation, we get
2 x^2 (^ A^ cos^ kx^ ^ B^ sin^ kx^ )^ k^^^2 (^ A^ cos^ kx^ ^ B^ sin^ kx^ ) k^2 , just as our equation requires. There are two main conclusions we can draw from this result. The first we can
draw by calculating *^ dx , the probability. This shows that the probability of finding a
particle is the same anywhere along the x axis, just as we would expect for a free particle.
V x ( ) in these regions it is impossible for the particle to be there, and therefore the
probability amplitude, ( ) x , must be zero everywhere in these regions.
In the region 0 x a, V(x) = 0 and the Schrödinger equation is 2
x^ (x) 2 = - 2mE 2 = -k 2
where k = ( 2mE 2 )1/ 2 , just as it was for the free particle. Once again the general solution
to this equation is
(x)= A cos kx+ B sin kx.
Remember that ( ) x must be continuous for all values of x. The region where we have
to pay particular attention to this is at the boundary of the region in which the particle
moves. At x = 0, must be equal to zero, since = 0 for x<0. Similarly at x = a,
must also equal zero, since = 0 for all x > a. We call these constraints on the value of
at the boundary of our potential well boundary values.
These boundary values place constraints on our solution. We see the first of
these by setting x = 0 and setting = 0 in compliance with our first boundary value. This
gives us (0)= 0 = A cos 0+ B sin 0 = A.
Therefore A = 0 and our wavefunction simplifies to (x)= B sin kx.
We see our second constraint by applying our second boundary condition, setting x = a and = 0. This gives us
(a)= 0 = B sin ka
This will only equal zero when ka = , 2, 3,..., n,.... Therefore we can write
k = na^ .
and our wavefunction becomes
(x)= B sin n x a
If we normalize this as before, we get finally,
(x)= (^2 a )^ 1/ 2 sin n x a
Note that the requirement k = na^ leads to the quantization of the energy of the
particle in a box, since
k = ( 2mE 2 )1/ 2 = na^
Solving this equation for E yields
E = 8man h^2 2 2 ,
which is quantized because of the presence of the integers in the equation. Notice that the general solutions to the Schrödinger equation for the free particle and the particle in a box are identical. The only difference between the two problems is the presence of the boundaries that constrain the movement of the particle in the particle in a box. We will find that every time a constraint is placed on the motion of a particle, the energy of that particle will be quantized. In the absence of these constraints the particle can take on any energy. What are some of the physical consequences of these results? First, notice that there is a minimum energy for the particle in a box. The lowest energy is when n= 1 and
Thus we will model the electrons as moving in a 5.78 x 10-10^ m box. According to the solution for the particle in a one dimensional box,
E = n 8man h
2 2 2 ,
where in this case, as we have just shown, a, the length of the box, is equal to 5.78 x 10- m. The Pauli exclusion principle, which we will discuss in detail later, tells us that there can only be two electrons for each quantum number n. Butadiene has four electrons. We place two in the lowest state of our particle in a box, n= 1, and the final two in n = 2. Thus for butadiene, the highest occupied molecular orbital is n = 2, i.e., HOMO = 2. The lowest unoccupied molecular orbital, LUMO, is n = 3. The electronic absorption should occur at the energy necessary to promote an electron from the HOMO to the LUMO, i.e., from n = 2 to n= 3. The energy of this transition is
E(3 2)= (^) 89 mah^2 2 - (^) 84 ma h^2 2 = 9.02x 10 ^19 J
This corresponds to an absorption at 220 nm. Butadiene has an absorption at 217 nm, so we see that despite its simplicity, the particle in a box is a fairly good model for the ultraviolet spectra of simple conjugated models. If we plot the wavefunctions for n = 1, n = 2, etc., we see that they take on the appearance of standing waves. Notice that n=1 contains 1/2 of a wave, n=2 contains a full wave, n = 3 contains 3/2 wave, and in general level n will contain n/2 waves, i.e., each level contains an integral number of half waves. If we look at the probability
density * for each of these levels, we find that for n = 1 the maximum probability is
in the center of the box, at 2^ a. For n=2, there are two maxima, at 4^ a and 34^ a. As n
increases, the probability density spreads out more and more, until at high n, the distribution is completely even, which is what we would expect classically. This is an important result. In general, a quantized system will approach the behavior predicted by classical mechanics (the classical limit) when the quantum numbers become very large. This is called the correspondence principle , due to Neils Bohr. Notice that in our probability distribution there are points, besides the fixed points at x = a and x = 0, where the probability density is zero. These points are called nodes. In many problems of interest to us, including the one and three dimensional harmonic oscillator, the one and three dimensional particle in a box, the rigid rotator and the hydrogen atom, the higher the number of nodes in a state the higher the energy of that state. For the particle in a box, the number of nodes is related to n by
nodes = n - 1.
where ij is called the Kronecker delta function which is defined by ij = 0, i j ij = 1, i = j. Sets of functions that are both orthogonal and normalized are called orthonormal functions. It turns out that all discrete solutions to the Schrödinger equation are complete orthonormal sets of functions. Complete orthonormal sets of functions are important
because any arbitrary function (x) can be expressed as a linear combination of members
of a complete orthonormal set of functions. Thus any arbitrary wavefunction (x) can be expressed as a linear combination of eigenfunctions, i(x). In other words, ( ) x c 1 1 ( ) x c 2 2 ( ) x c 3 3 ( ) ... x
where (x) is any arbitrary wavefunction, the ci ’s are constants, and the i(x) are
eigenfunctions of some operator. This rule, that any arbitrary wavefunction can be expressed as a linear combination of eigenfunctions, is sometimes referred to as the superposition principle.
This explains why we observe only eigenvalues even when our wavefunction (x)
is not itself an eigenfunction. We are observing the eigenvalue of one of the eigenfunctions that make up our wavefunction. The probability of observing one of these eigenvalues is equal to the square of its coefficient, ci^2. This implies that the sum of the squares of the coefficients must be one, i.e.,
i ci^2 ^1.
Sometimes two eigenfunctions will have the same energy. In this case we say that the eigenfunctions are degenerate. One property of degenerate eigenfunctions is that all linear combinations of degenerate eigenfunctions will also be eigenfunctions. This is not true for nondegenerate eigenfunctions. Note that while the solutions we have obtained for the particle in a box are eigenfunctions of the Hamiltonian operator they are not eigenfunctions of the position
operator X . If this was so, X (x)= x (x) would equal a constant times (x) , which is
not the case. According to our postulates, if we have a state ( ) x which is an
eigenfunction of the Hamiltonian, we will obtain eigenvalues of some eigenfunction of
the operator X when we measure x, but we won't be able to predict which one. Repeated measurement of x will yield the probability densities we have calculated. Even though we cannot predict the value of a given measurement when a wavefunction is not an eigenfunction of the operator in question, we can always predict the average value of the measurement. For this we need another postulate. To introduce this postulate, we'll begin with a review of averages. Most of us are
introduced to averages of what are called discrete distributions. The numbers you
obtain by rolling a die are a discrete distribution, with the integers 1-6 the only possible
results, and no intervening numbers possible. Discrete distributions are intrinsically
discontinuous. In this case, the procedure for taking an average is simple. If you roll the
die 5 times, you add the five numbers, divide by 5 and you have the average. For
