finte potential well 2.ppt - \u201cAll of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics.It

# finte potential well 2.ppt - u201cAll of modern physics is...

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“All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics...It has survived all tests and there is no reason to believe that there is any flaw in it….We all know how to use it and how to apply it to problems; and so we have learned to live with the fact that nobody can understand it.” --Murray Gell-Mann
Lecture 11: Particles in Finite Potential Wells n=1 n=2 n=3 n=4 0 L U 0 I II III U(x) (x) AlGaAs GaAs AlGaAs U(x) x
Act 1 Act 1 (a) (b) 2. Which of the following wave functions corresponds to a particle more likely to be found on the left side? (x) 0 x (x) 0 x (c) (x) 0 x
Solution Solution (a) (b) 2. Which of the following wave functions corresponds to a particle more likely to be found on the left side? (x) 0 x (x) 0 x (c) (x) 0 x None of them! (a) is clearly symmetrical. (b) might seem to be “higher” on the left than on the right, but only the absolute square determines the probability.  0 x
Probabilities n=1  0 x L n=2  0 x L  0 x L n=3 U= U= 0 x L 0 x L 0 x L Often what we measure in an experiment is the probability density, | (x)| 2 . 1 ( ) sin n n x B x L Wavefunction = Probability amplitude 2 2 2 1 ( ) sin n n x B x L Probability per unit length (in 1-dimension)
“Again an idea of Einstein’s gave me the lead. He had tried to make the duality of particles – light quanta or photons - and waves comprehensible by interpreting the square of the optical wave amplitudes as probability density for the occurrence of photons. This concept could at once be carried over to the -function: | | 2 ought to represent the probability density for electrons (or other particles). It was easy to assert this, but how could it be proved?” M. Born, Nobel Lecture (1954).
Probability and Normalization 1 ( ) sin n n x B x L We now know that . How can we determine B 1 ? We need another constraint. It is the requirement that total probability equals 1 . The probability density at x is | ( x )| 2 : Therefore, the total probability is the integral: In our square well problem, the integral is simpler, because = 0 for x < 0 and x > L: Requiring that P tot = 1 gives us:  0 x L n=3 |B 1 | 2 Integral under the curve = 1 1 2 B L 2 tot P x dx   2 2 1 0 2 1 sin 2 L tot n P B x dx L L B
Probability Density 2 2 sin n P x N x L In the infinite well: . (Units are m -1 , in 1D) Notation: The constant is typically written as “N”, and is called the “ normalization constant ”. For the square well:

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