The Basic Postulate of Quantum Mechanics

And now we can determine states for which <DA2> = 0; A is the exact value (precise measurement, for instance of impulse)

Δ A y = 0 (A− A) y = 0

A y = A y

The wavefunction that satisfies an equation p y = py or the wavefunction which satisfies an equation H y = E y is called the eigenfunction and A (p and E here) is the eigenvalue.

Operator · Eigenfunction = Eigenvalue · Eigenfunction A . y = A · y

for instance y = eax; A = d/dx →dy/dx = a a: Eigenvalue y: Eigenfunction

The Basic Postulate of Quantum Mechanics:

The eigenvalues A are identical to measured meanings

In general the equation A y = Ay has its solutions only for definite values of physical quantity A. They produce either discrete series A1, A2,.... or continuous series of values in the corresponding range. The measurement of physical value A (for instance, measurement of energy E) in a state yA1 gives finally the exact value of A1 (for instance, the definite value of energy E1).

The physical quantity measurement in system that is characterized by eigenfunction gives its precise meaning, its eigenvalue. If we repeat our measurements of this value we will find the same eigenvalue. Considered particle is situated in definite state and its eigenfunction is exposed to the corresponding operator.

kinetic energy Dy = −(π/a)2(2/a)½ sin(πx/a) Δ = ∂²/∂x²

<Ekin> = o∫a(2/a)½ sin(πx/a)·(−h²/2m) (−(π/a)2(2/a)½ sin(πx/a))dx =

h²π²/ma³o∫a sin2(πx/a)dx = h²π²/2ma² = h²/8ma² = E1

The total energy E average of distribution is in general:

<E> = -¥ò+∞y* | (−h²/2mΔ + | V(x)) y dx
| ê | |
kinetic energy | ¿ + | èpotential energy

<E> = -¥ò+∞y* H y dx

If we compare all such equations then we will draw a rule according to which one can calculate the quantity A average of distribution in the quantum mechanics:

<A> = -¥ò+∞y* A y dV

dV = dx dy dz : the volume element

A is the operator here, which corresponds to physical quantity A. In the Dirac's style one can obtain:

<A> = <y|A|y>

If we consider the normal energy average of distribution: <E> = -¥ò+∞y*H ydx ; where y is the Schroedinger equation solution H y = Ey, then it's obviously: <E> = -¥ò+∞y*H ydx = E · òy* ydx = E. I.e. the average of distribution is apparently the energy value E! There is no any overlapping and hence uncertainty E since when making energy measurements the energy operator H gives the measured value as its eigenvalue.

Example:

We measure the particle impulse and always obtain the measured value p. It means that now the particle is in the state in which the eigenfunction is the impulse operator. The wavefunction after making impulse measurements is as follows

yp = C eipx/h

since the eigenvalue equation for impulse p yp= p yp:

h/i∂/∂x | (C eipx/h) | = p · | (C eipx/h)
| yp |  | yp

And if now we measure the kinetic energy then this measurement corresponds mathematically to the application of kinetic energy operator, −h²/2m∂²/∂x², on the wavefunction yp:

−h²/2m∂²/∂x² C eipx/h = −h²/2m·(ip/h)2. c eipx/h = p²/2myp

It finally gives us the eigenvalue p²/2m. And it means that the wave keeps like wavefunction yp. The kinetic energy measurement doesn't annihilate the result of first measurement. Hence there the possibility to make two measurements simultaneously with any accuracy.

If y isn't eigenfunction then it's possible to expand y on eigenfunctions φn with coefficients an in the following way:

y = Σn an φn

|y> = Σn an|n> <A> = <y|A|y> = Σm,n am* <m|A|n> an = Σm,n am* an· An <m|n>
<A> = Σn |an|2 An
da <m|n> = δm,n | A φn = An φn <A> = òy*A ydV = ò Sm am*φm*A Σn anφndV = ò Sm,n am*anAnφm*φndV
<A> = Σn|an|2 An
da φm*φn = δm,n
Σn|an|2 = 1

Ok, but what are the transformation coefficients an?

Multiplication by <m| to the left gives:

<m|y> = Σ an <m|n> <m|n> = δm,n
<m|y> = am
| ò fm* y dV = ò fm* Σ an φn dV ò fm* y dV = Σn an òfm* φn dV
ò fm* y dV = am