Week 8b - Also we have E = h!E = h since!t 1 E = h h!t!E!t...

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Also we have Equations (I) and (II) are basic formulations of the uncertainty principle. e.g. the product ! p x ! x cannot be infinitesimally small but must equal in magnitude at least Planck's constant, h. Likewise for ! E ! t. E = h ! " ! E = h ! ! since ! t ! ! # 1 " ! E = h ! ! # h ! t " ! E ! t # h (II) In text books sometimes see: ! p ! x " h 2 = h 4 # ! E ! t " h 2 This comes about from a more precise determination.
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These fundamental limitations show that if we use a particle description for electromagnetic (e.m.) radiation or a particle to explain a phenomenon then the wave aspect is suppressed and vice versa. ! p x " h ! x = # For ! t = 0 then: ! E ! t " h => ! E = and since ! E = h ! ! ! ! -> ! ! and therefore the wave aspect is suppressed since all the quantities ! , E, " and p ( )) > # ) are completely uncertain. For example, if the particle nature of say an electron is to be displayed, both ! x and ! t must be zero: i.e ! x = 0 then: ! p x ! x " h Also we have ! p x = h " 2 !" so !" = # .
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Alternatively, if the wave characteristics of a particle or e.m. radiation are to be defined perfectly then ! ! = 0 and ! " = 0 (so that ! E = 0 and ! p = 0) and the particle characteristics of precise location in space and time are completely uncertain. Therefore, as first pointed out by Bohr , the wave-particle duality is complementary rather than contradictory. As different as they are, both wave and particle aspects are required and they complement each other to fully describe matter or electromagnetic radiation. The uncertainty principle is a direct consequence of the wave character of moving bodies (particles) and has nothing to do with experimental uncertainties or how good the measuring equipment is!
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Examples 1. electron v = 300 m/sec ! v = 0.010% i.e. velocity accurate to 0.010% p = mv = (9.11 x 10 -31 kg) x 300 m/sec = 2.7 x 10 -28 kg m/sec & ! uncertainty in momentum ! P is also 0.010% since ! p = m ! v ! ! p = 0.010 100 x 2.7 x 10 -28 kg.m sec = 2.7 x 10 -32 kg m/sec Hence the minimum uncertainty in position is _______________________________ ! x = h ! p = 6.6 x 10 -34 J.S 2.7 x 10 -32 kg m/s = 2.4 cm. 2. Bullet v = 300 m/sec m = 50 gm ! v= 0.010% p = mv = 15 kg m/sec. ! p = 0.010 100 x 15 kg.m sec = 1.5 x 10 -3 kg.m/sec hence ! x = 6.6 x 10 -34 J.S. 1.5 x 10 -3 kg.m/sec = 4.4 x 10 -31 m.
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This is beyond possibility of measurement! This is beyond possibility of measurement! So if the bullets velocity is known to within 0.010% then we can determine exactly where the bullet is! - h 2 2 m " 2 " x 2 # $ % & ( ) ( x , y , z ) + V ( x ) ) ( x ) = E ) ( x ) Schrödinger Wave equation 1925 In order to account for the wave nature of particles, a new mechanics was developed (wave mechanics) to give the following equation of motion: This is called the time independent Schrödinger Equation and is the equation of motion for systems with quantum characteristics. In wave mechanics it replaces Newton’s Laws (i.e. ). The solutions of the time- independent equation gives stationary states i.e. energy states that do not depend upon time or have a definite energy for a long time m d 2 x dt 2 = " F
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To demonstrate how the Schrödinger equation can be used to give energies, lets look at the problem of an electron trapped in an infinite potential well. (Similar to the particle in the box.) We wish to find the $ ’s which satisfy this system. We look for solutions to the time independent S.E.
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