# 1 Recap Heisenberg uncertainty relations  The product of the uncertainty in momentum (energy) and in position (time) is at least as large as Planck’s.

## Presentation on theme: "1 Recap Heisenberg uncertainty relations  The product of the uncertainty in momentum (energy) and in position (time) is at least as large as Planck’s."— Presentation transcript:

1 Recap Heisenberg uncertainty relations  The product of the uncertainty in momentum (energy) and in position (time) is at least as large as Planck’s constant

2 Conjugate variables (Conjugate observables)  {p x,x}, {E,t} are called conjugate variables  The conjugate variables cannot in principle be measured (or known) to infinite precision simultaneously

3Example  The speed of an electron is measured to have a value of 5.00 x 10 3 m/s to an accuracy of 0.003%. Find the uncertainty in determining the position of this electron  SOLUTION  Given v = 5.00  10 3 m/s; (  v)/v = 0.003%  By definition, p = m e v = 4.56 x 10 -27 Ns;   p = 0.003% x p = 1.37x10 -27 Ns  Hence,  x ≥ h/4  p = 0.38 nm xx 4.56±1.37)  10 -27 Ns p = (4.56±1.37)  10 -27 Ns  x = 0.38 nm 0 x

4 Example  A charged  meson has rest energy of 140 MeV and a lifetime of 26 ns. Find the energy uncertainty of the  meson, expressed in MeV and also as a function of its rest energy  Solution  Given E  m  c 2 = 140 MeV,  = 26 ns.   E ≥h/4 .03  10 -27 J = 1.27  10 -14 MeV; = 1.27  10 -14 MeV;   E/E = 1.27  10 -14 MeV/140 MeV = 9  10 -17 “Now you see it” “Now you DONT” Exist only for  = 26 ns E ±  E

5 Example Example estimating the quantum effect on a macroscopic particle  Estimate the minimum uncertainty velocity of a billard ball (m ~ 100 g) confined to a billard table of dimension 1 m Solution For  x ~ 1 m, we have  p ≥h/4  x = 5.3  10 -35 Ns,  So  v = (  p)/m ≥ 5.3  10 -34 m/s  One can consider  v = 5.3x10 -34 m/s (extremely tiny) is the speed of the billard ball at anytime caused by quantum effects  In quantum theory, no particle is absolutely at rest due to the Uncertainty Principle 1 m long billard table A billard ball of 100 g, size ~ 2 cm  v = 5.3  10 -34 m/s

6 A particle contained within a finite region must has some minimal KE  One of the most dramatic consequence of the uncertainty principle is that a particle confined in a small region of finite width cannot be exactly at rest (as already seen in the previous example)  Why? Because… ...if it were, its momentum would be precisely zero, (meaning  p = 0) which would in turn violate the uncertainty principle

7 What is the K ave of a particle in a box due to Uncertainty Principle?  We can estimate the minimal KE of a particle confined in a box of size a by making use of the UP  Uncertainty principle requires that  p ≥ (h/2  a (we have ignored the factor 2 for some subtle statistical reasons)  Hence, the magnitude of p must be, on average, at least of the same order as  p:  Thus the kinetic energy, whether it has a definite value or not, must on average have the magnitude

8 Zero-point energy This is the zero-point energy, the minimal possible kinetic energy for a quantum particle confined in a region of width a Particle in a box of size a can never be at rest (e.g. has zero K.E) but has a minimal KE K ave (its zero-point energy) We will formally re-derived this result again when solving for the Schrodinger equation of this system (see later). a

9 Recap  Measurement necessarily involves interactions between observer and the observed system  Matter and radiation are the entities available to us for such measurements  The relations p = h/ and E = h are applicable to both matter and to radiation because of the intrinsic nature of wave-particle duality  When combining these relations with the universal waves properties, we obtain the Heisenberg uncertainty relations  In other words, the uncertainty principle is a necessary consequence of particle-wave duality

10 Introductory Quantum mechanics

11 Probabilistic interpretation of matter wave

12 A beam of light if pictured as monochromatic wave (, ) A = 1 unit area Intensity of the light beam is A beam of light pictured in terms of photons A = 1 unit area Intensity of the light beam is I = Nh N = average number of photons per unit time crossing unit area perpendicular to the direction of propagation  h Intensity = energy crossing one unit area per unit time. I is in unit of joule per m 2 per second

13  Consider a beam of light  In wave picture, E = E 0 sin(kx–  t), electric field in radiation  Intensity of radiation in wave picture is I =  On the other hand, in the photon picture, I = Nh  On the other hand, in the photon picture, I = Nh  Correspondence principle: what is explained in the wave picture has to be consistent with what is explained in the photon picture in the limit N  infinity: Probability of observing a photon

14 Statistical interpretation of radiation  The probability of observing a photon at a point in unit time is proportional to N  However, since  the probability of observing a photon must also  This means that the probability of observing a photon at any point in space is proportional to the square of the averaged electric field strength at that point Prob (x) Square of the mean of the square of the wave field amplitude

15 What is the physical interpretation of matter wave?  we will call the mathematical representation of the de Broglie’s wave / matter wave associated with a given particle (or an physical entity) as The wave function,   We wish to answer the following questions:  Where is exactly the particle located within  x? the locality of a particle becomes fuzzy when it’s represented by its matter wave. We can no more tell for sure where it is exactly located.  Recall that in the case of conventional wave physics, |field amplitude   is proportional to the intensity of the wave). Now, what does |   physically mean?

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