1. The Stale of a System Is Completely Specified by lts Wave Function
Classical mechanics deals with quantities called dynamical variables, such as position, momentum, angular momentum, and energy.
A measurable dynamical variable is called an observable.
The classical-mechanical state of a one-body system at any particular time is specified completely by the three position coordinates (x, y, z) and the three momenta or velocities ( x , y ,z ) at that time.

2. The Postulates and General Principles of Quantum Mechanics
The state of a quantum-mechanical system is completely specified by a function (r,t) that depends on the coordinates of the particle and on time.
This function, called the wave function or state function, has the important property that *(r,t)(r,t)dxdydz is the probability that the particle lies in the volume element dxdydz located at r at time t.
If there is more than one particle, say two, then we write

3. Because the square of the wave function has a probabilistic interpretation, it must satisfy certain physical requirements. For example, a wave function must be normalized, so that in the case of one particle, for simplicity, we have

4. We let nx /a be z

5. and thus the above wave functions are normalized.
In addition to being normalized, or at least normalizable,
we require that (r, t) and its first spatial derivative be single-valued, continuous, and finite. We summarize these requirements by saying that (r,t) must be well behaved.

8. In quantum mechanics, we deal only with linear operators
In quantum mechanics, we deal only with linear operators. An operator is said to be linear if
where c1 and c2 are (possibly complex) constants. Clearly, the "differentiate" and "integrate" operators are linear because
and
The "square" operator, SQR, on the other hand, is nonlinear because

9. Consider an eigenvalue problem with a two-fold degeneracy
Postulate 2
To every observable in classical mechanics there corresponds a linear operator in quantum mechanics.
Consider an eigenvalue problem with a two-fold degeneracy

10. Classical-Mechanical Observables and Their Corresponding Quantum-Mechanical Operators

13. Generally, an operator will have a set of eigenfunctions and eigenvalues, and we indicate this by writing

14. As a specific example, consider the measurement of the energy
As a specific example, consider the measurement of the energy. The operator corresponding to the energy is the Hamiltonian operator, and its eigenvalue equation is

16. SOLUTION: The normalization integral of (x) is elementary:

17. In other words, show that the only values of the energy that can be observed are the energy eigenvalues,
SOLUTION: The operator that corresponds to the observable Eis the Hamiltonian
operator, which for a particle in a box is
The average energy is given by

18. Similarly,

19. The Commutator of Two Operators

20. if
For example

25. Quantum-Mechanical Operators Must Be Hermitian Operators
If two operators do not commute, then their corresponding observable quantities do not have simultaneously well-defined values
Quantum-Mechanical Operators Must Be Hermitian Operators
In an equation, we have
multiply from the left by * and integrate to obtain

26. take the complex conjugate of Equation
where the equality a* = a recognizes that a is real. Multiply from the left by  and integrate
Equating the left sides of Equations 4.25 and 4.27 gives
The operator A must satisfy Equation 4.28 to assure that its eigenvalues are real. An operator that satisfies Equation 4.28 for any well-behaved function is called a Hermitian operator. Thus, we can write the definition of a Hermitian operator as an operator that satisfies the relation

27. Eq. 1
Postulate 1’
To every observable in classical mechanics there corresponds a linear; Hermitian operator in quantum mechanics.
Consider the operator A = d / dx. Does A satisfy Equation 4.29? Let's
substitute A = d/ dx into Equation 4.29 and integrate by parts:

28. For a wave function to be normalizable, it must vanish at infinity, and so the first term on the right side here is zero. Therefore, we have
For an arbitrary function f (x), d/dx does not satisfy Equation 4.29 and so is not Hermitian.

30. Thus, Equation 4.29 is satisfied, and the kinetic energy operator is Hermitian.

31. Consider the two eigenvalue equations
We multiply the first of Equations by m*, and integrate
then we take the complex conjugate of the second, multiply by n, and integrate to obtain

32. By subtracting Equations, we obtain
Because A is Hermitian
When n = m, the integral is unity by normalization and so we have
Now if the system is nondegenerate, and

33. A set of eigenfunctions that satisfies the condition in Equation 4
A set of eigenfunctions that satisfies the condition in Equation 4.42 is said to be
orthogonal.
The particle in a box is a nondegenerate system. The wave functions for this system are

34. Because n and m are integers
both integrals start at zero and go over half or complete cycles of the cosine and equal zero if m n.
The wave functions of a particle in a box are orthogonal.

35. When n = m in Equation
cos 0 = 1, The second integral on the right side vanishes and
so we have

36. A set of functions that are both normalized and orthogonal to each other is called an orthonormal set. We can express the condition of orthonormality by writing
The symbol nm occurs frequently and is called the Kronecker delta
The eigenfunctions of a particle constrained to move on a circular ring of radius a are

38. Suppose that two operators A and B have the same set of eigenfunctions, so that we have
Equations 4.50 imply that the quantities corresponding to A and B have simultaneously sharply defined values. According to Equations 4.50, the values that we observe are an and bn.
if two operators have the same set of eigenfunctions, then they necessarily commute.

39. Postulate 5
The wave function or state function of a system evolves in time according to the timedependent Schrodinger equation
H does not contain time explicitly, and in this case we can apply the method of separation of variables and write

40. If H does not contain time explicitly, then the left side of Equation 4.70 is a function of x only and the right side is a function oft only, and so both sides must equal a constant.
If we denote the separation constant by E , then Equation 4.70 gives
time-independent Schrodinger equation